Straight from the boook by Nguyễn Quang Huy

Straight from the Book

Titu Andreescu Gabriel Dospinescu

Straight from the Book

XYZ~ Press

Titu Andreescu University of Texas at Dallas

Gabriel Dospinescu Ecole Normale Superieure, Lyon

The only way to learn mathematics is to do mathematics. , Library of Congress Control Number: 2012951362

-Paul Halmos ISBN-13: 978-0-9799269-3-8 ISBN-IO: 0-9799269-3-9

ÂŠ

2012 XYZ Press, LLC

All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (XYZ Press, LLC, 3425 Neiman Rd., Plano, TX 75025, USA) and the authors except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of tradenames, trademarks, service marks and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights.

9 8 7 6 5 4 3 2 1 www.awesomemath.org Cover design by Iury Ulzutuev

Foreword

This book is a follow-on from the authors' earlier book, 'Problems from the Book'. However, it can certainly be read as a stand-alone book: it is not vital to have read the earlier book. The previous book was based around a collection of problems. In contrast, this book is based around a collection of solutions. These are solutions to some of the (often extremely challenging) problems from the earlier book. The topics chosen reflect those from the first twelve chapters of the previous book: so we have Cauchy-Schwarz, Algebraic Number Theory, Formal Series, Lagrange Interpolation, to name but a few. The book is one of the most remarkable mathematical texts I have ever seen. First of all, there is the richness of the problems, and the huge variety of solutions. The authors try to give several solutions to each problem, and moreover give insight about why each proof is the way it is, in what way the solutions differ from each other, and so on. The amount of work that has been put in, to compile and interrelate these solutions, is simply staggering. There is enough here to keep any devotee of problems going for years and years. Secondly, the book is far more than a collection of solutions. The solutions are used as motivation for the introduction of some very clear expositions of mathematics. And this is modern, current, up-to-the-minute mathematics. For example, a discussion of Extremal Graph Theory leads to the celebrated Szemenยงdi-Trotter theorem on crossing numbers, and to the amazing applications of this by Szekely and then on to the very recent sum-product estimates of Elekes, Bourgain, Katz, Tao and others. This is absolutely state-of-the-art material. It is presented very clearly: in fact, it is probably the best exposition of this that I have seen in print.

viii

As another example, the Cauchy-Schwarz section leads on to the developements of sieve theory, like the Large Sieve of Linnik and the Tunin-Kubilius inequality. And again, everything is incredibly clearly presented. The same applies to the very large sections on Algebraic Number Theory, on p-adic Analysis, and many others. It is quite remarkable that the authors even know so much current mathematics I do not think any of my colleagues would be so well-informed over so wide an area. It is also remarkable that, at least in the areas in which I am competent to judge, their explanations of these topics are polished and exceptionally well thought-out: they give just the right words to help someone understand what is going on. Overall, this seems to me like an 'instant classic'. There is so much material, of such a high quality, wherever one turns. Indeed, if one opens the book at random (as I have done several times), one is pulled in immediately by the lovely exposition. Everyone who loves mathematics and mathematical thinking should acquire this book.

Imre Leader

Professor of Pure Mathematics University of Cambridge

Preface This book is a compilation of many suggestions, much advice, an even more hard work. Its main objective is to provide solutions to the problems which were originally proposed in the first 12 chapters of Problems from the Book. We were not able to provide full solutions in our first volume due to the lack of space. In addition, the statements of the proposed problems contained' typos and some elementary mistakes which needed further editing. Finally, the problems were also considered to be quite difficult to tackle. With these points in mind, we came up with a two-part plan: to correct the identified errors and to publish comprehensive solutions to the problems. The first task, editing the statements of the proposed problems, was simple and has already been completed in the second edition of Problems from the Book. Although we focused on changing several problems, we also introduced many new ones. The second task, providing full solutions, however, proved to be more challenging than expected, so we asked for help. We created a forum on www.mathlinks.ro (a familiar site for problem-solving enthusiasts) where solutions to the proposed problems were gathered. It was a great pleasure to witness the passion with which some of the best problem-solvers on mathlinks attacked these tough-nuts. This new book is the result of their common effort, and we thank them. Providing solutions to every problem within the limited space of one volume turned out to be an optimistic plan. Only the solutions to problems from the first 12 chapters of the second edition of Problems from the Book are presented here. Furthermore, many of the problems are difficult and require a rather extensive mathematical background. We decided, therefore, to complement the problems and solutions with a series of addenda, using various problems as starting points for excursions into "real mathematics". Although

we never underestimated the role of problem solving, we strongly believe that the reader will benefit more from embarking on a mathematical journey rather than navigating a huge list of scattered problems. This book tries to reconcile problem-solving with "professional mathematics" . Let us now delve into the structure of the book which consists of 12 chapters and presents the problems and solutions proposed in the corresponding chapters of Problems from the Book, second edition. Many of these problems are fairly difficult and different approaches are presented for a majority of them. At the end of each chapter, we acknowledge those who provided solutions. Some chapters are followed by one or two addenda, which present topics of more advanced mathematics stemming from the elementary topics discussed in the problems of that chapter. The first two chapters focus on elementary algebraic inequalities (a notable caveat is that some of the problems are quite challenging) and there is not much to comment regarding these chapters except for the fact that algebraic inequalities have proven fairly popular at mathematics competitions. To provide relief from this rather dry landscape (the reader will notice that most of the problems in these two chapters start with "Let a, b, c be positive real numbers"), we included an addendum presenting deep applications of the Cauchy-Schwarz inequality in analytic number theory. For instance, we discuss Gallagher's sieve, Linnik's large sieve and its version due to Montgomery. We apply these results to the distribution of prime numbers (for instance Brun's famous theorem stating that the sum of the inverses of the twin primes converges). The note-worthy Tunin-Kubilius inequality and its classical applications (the Hardy-Ramanujan theorem on the distribution of prime factors of n, Erdos' multiplication table problem, Wirsing's generalization of the prime number theorem) are also discussed. The reader will be exposed to the power of the Cauchy-Schwarz inequality in real mathematics and, hopefully, will understand that the gymnastics of three-variables algebraic inequalities is not the Holy Grail. Chapter 3 discusses problems related to the unique factorization of integers and the p-adic valuation maps. Among the topics discussed, we cover the local-global principle (which is extremely helpful in proving divisibilities or arithmetic identities), Legendre's formula giving the p-adic valuation of n!,

a beneficial elementary result called lifting the exponent lemma, as well as more advanced techniques from p-adic analysis. One of the most beautiful results presented in this chapter in the celebrated Skolem-Mahler-Lech theorem concerning the zeros of a linearly recursive sequence. Readers will perhaps appreciate the applications of p-adic analysis, which is covered extensively in a long addendum. This addendum discusses, from a foundational level, the arithmetic of p-adic numbers, a subject that plays a central role in modern number theory. Once the basic groundwork has been laid, we discuss the p-adic analogues of classical functions (exponential, logarithm, Gamma function) and their applications to difficult congruences (for instance, Kazandzidis' famous supercongruence). This serves as a good opportunity to explore the arithmetic of Bernoulli numbers, Volkenborn's theory of p-adic integration, Mahler and Amice's classical theorems characterizing continuous and locally analytic functions on the ring of p-adic integers or Morita's construction of the p-adic Gamma function. A second addendum to this chapter discusses various classical estimates on prime numbers, which are used throughout the book. Chapter 4 discusses problems and elementary topics related to prime numbers of the form 4k + 1 and 4k + 3. The most intriguing problem discussed is, without any doubt, Cohn's renowned theorem characterizing the perfect squares in the Lucas sequence. Chapter 5 is dedicated again to the yoga of algebraic inequalities and is followed by an addendum discussing applications of Holder's inequality. Chapter 6 focuses on extremal graph theory. Most of the problems revolve around Turan's theorem, however the reader will also be exposed to topics such as chromatic numbers, bipartite graphs, etc. This chapter is followed by a relatively advanced addendum, discussing various topics related to the Szemeredi-Trotter theorem, which gives bounds for the number of incidences between a set of points and a set of curves. We discuss the theorem's classical probabilistic proof, its generalization to multi-graphs due to Szekely and its application to the sum-product problem due to Elekes, as well as more recent developments due to Bourgain, Katz, and Tao. These results are then applied to natural and nontrivial geometric questions (for instance: what is the least number of distinct distances determined by n points in the plane? what is the maximal number of triangles of the same area?). Finally, another addendum

Xll

completes this chapter and is dedicated to the powerful probabilistic method. After a short discussion on finite probability spaces, we provide many examples of combinatorial applications. Chapter 7 involves combinatorial and number theoretic applications of finite Fourier analysis. The central principles of this chapter include the roots of unity and the fact that congruences between integers can be expressed in terms of sums of powers of roots of unity. To provide the reader with a broader view, we beefed-up this chapter with an addendum discussing Fourier analysis on finite abelian groups and applying it to Gauss sums of Dirichlet characters, additive problems, combinatorial or analytic number theory. For instance, the reader will find a discussion of the P6lya-Vinogradov inequality and Vinagradov's beautiful use of this inequality to deduce a rather strong bound on the least quadratic non-residue mod p. At the same time, we explore Dirichlet's L-functions, culminating in a proof of Dirichlet's theorem on arithmetic progressions. This section is structured to first present the usual analytic proof (since it is really a masterpiece from all points of view), up to some important simplification due to Monsky. At the same time, we also discuss how to turn this into an elementary proof that avoids complex analysis and dramatically uses Abel's summation formula. Chapter 8 focuses on diverse applications of generating functions. This is an absolutely crucial tool in combinatorics, be it additive or enumerative. In this section, the reader will have the opportunity to explore enumerative problems (Catalan's problem, counting the number of solutions of linear diophantine equations or the number of irreducible polynomials mod p, of fixed degree). We also discuss exotic combinatorial identities or recursive sequences, which can be solved elegantly using generating functions, but also rather challenging congruences that appear so often in number theory. The chapter is followed by an addendum presenting a very classical topic in enumerative combinatorics, Lagrange's inversion formula. Among the applications, let us mention Abel's identity, other derived combinatorial identities, Cayley's theorem on labeled trees, and various related problems. Chapter 9 is rather extensive, due to the vast nature of the topic covered, algebraic number theory. While we are able only to scratch the surface, nevertheless the reader will find a variety of intriguing techniques and ideas. For

xiii

instance, we discuss arithmetic properties of cyclotomic polynomials (including Mann's beautiful theorem on linear equations in roots of unity), rationality problems, and various applications of the theorem of symmetric polynomials. In addition, we present techniques rooted in the theory of ideals in number fields, finite fields and p-adic methods. We also give an overview of the elementary algebraic number theory in the addendum following this chapter. After a brief review on ideals, field extensions, and algebraic numbers, we proceed with a discussion of the primitive element theorem and embeddings of number fields into C. We also briefly survey Galois theory, and the fundamental theorem on the prime factorization of ideals in number fields, due to Kummer and Dedekind. Once the foundation has been set, we discuss the prime factorization in quadratic and cyclotomic fields and apply these techniques to basic problems that explore the aforementioned theories. Finally, we discuss various applications of Bauer's theorem and of Chebotarev's theorem. The next addendum is concerned with the fascinating topic of counting the number of solutions modulo p of systems of polynomial equations. We use this as an opportunity to state and prove the basic structural results on finite fields and introduce the Gauss and Jacobi sums. We go ahead and count the number of points over a finite field of a diagonal hypersurface and to compute its zeta function. This is a beautiful theorem of Weil, the very tip of a massive iceberg. Chapter 10 focuses on the arithmetic of polynomials with integer coefficients. An essential aspect of the discussion concentrates on Mahler expansions, the theory of finite differences, and their applications. The techniques used in this chapter are rather diverse. Although the problems can be considered basic, they are challenging and require advanced problem-solving skills. Chapter 11 provides respite from the difficult tasks mentioned above. It discusses Lagrange's interpolation formula, allowing a unified presentation of various estimates on polynomials. The longest and certainly most challenging chapter is the last one. It explores several algebraic techniques in combinatorics. The methods are standard but powerful. The last part of the chapter deals with applications to geometry, presenting some of Dehn's wonderful ideas. The last problem presented in the book is the famous Freiling, Laczkovich, Rinne, Szekeres theorem,

xiv

a stunningly beautiful application of algebraic combinatorics. We would like to thank, again, the members of the mathlinks site for their invaluable contribution in providing solutions to many of the problems in this book. Special thanks are due to Richard Stong, who did a remarkable job by pointing out many inaccuracies and suggesting numerous alternative solutions. We would also like to thank Joshua Nichols-Barrer, Kathy Cordeiro and Radu Sorici, who gave the manuscript a readable form and corrected several infelicities. Many of the problems and results in this book were used by the authors in courses at the AwesomeMath Summer Program, and students' reactions guided us in the process of simplifying or adding more details to the discussed problems. We wish to thank them all, for their courage in taking and sticking with these courses, as well as for their valuable suggestions.

Titu Andreescu titu.andreescu@utdallas.edu

Contents 1

Some Useful Substitutions 1.1 The relation a 2 + b2 + e2 = abc + 4 . . . . . . . . . . . . . . . 1.2 The relations abc = a + b + e + 2 and ab + be + ea + 2abe = 1 1.3 The relation a 2 + b2 + e2 + 2abe = 1 1.4 Notes . . . . . . . . . . .

21 27

Always Cauchy-Schwarz. . . 2.1 Notes . . . . . . . . . . . . . . . . . . . . . . . Addendum 2.A Cauchy-Schwarz in Number Theory

29 61 62

Look at the Exponent 3.1 Introduction..... 3.2 Local-global principle 3.3 Legendre's formula . . 3.4 Problems with combinatorial and valuation-theoretic aspects 3.5 Lifting exponent lemma 3.6 p-adic techniques . . . . 3.7 Miscellaneous problems 3.8 Notes . . . . . . . . . . Addendum 3.A Classical Estimates on Prime Numbers Addendum 3.B An Introduction to p-adic Numbers

91 91 92 96 104 110 116 125 134 135 141

Primes and Squares 4.1 Notes . . . . . . .

189 203

Gabriel Dospinescu gdospi2002@yahoo.com

1 2 9

Contents

xvi

T 2 's Lemma

5.1 Notes . . . . . . . . . . . . . . . . . . . Addendum 5.A Holder's Inequality in Action.

205 226 . 227

Some Classical Problems in Extremal Graph Theory 6.1 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . Addendum 6.A Some Pearls of Extremal Graph Theory. Addendum 6.B Probabilities in Combinatorics . . . . . .

235 251 252 265

Complex Combinatorics 7.1 Tiling and coloring problems 7.2 Counting problems . . . 7.3 Miscellaneous problems .. . 7.4 Notes . . . . . . . . . . . . . Addendum 7.A Finite Fourier Analysis

281 281 285 299 309 310

Formal Series Revisited 8.1 Counting problems . . 8.2 Proving identities using generating functions 8.3 Recurrence relations .. 8.4 Additive properties . . . 8.5 Miscellaneous problems 8.6 Notes . . . . . . . . . . Addendum 8.A Lagrange's Inversion Theorem

337 339 349 354 361 371 375 377

A Little Introduction to Algebraic Number Theory 9.1 Tools from linear algebra 9.2 Cyclotomy . . . . . . . . . . . . . . . . . 9.3 The gcd trick . . . . . . . . . . . . . . . 9.4 The theorem of symmetric polynomials. 9.5 Ideal theory and local methods 9.6 Miscellaneous problems . . . . . . . . . 9.7 Notes . . . . . . . . . . . . . . . . . . . Addendum 9.A Equations over Finite Fields Addendum 9.B A Glimpse of Algebraic Number Theory

395

396 402 408 410 420 426 433 434 456

Contents

xvii

10 Arithmetic Properties of Polynomials 10.1 The a - blf(a) - f(b) trick . . . . . . 10.2 Derivatives and p-adic Taylor expansions. . 10.3 Hilbert polynomials and Mahler expansions 10.4 p-adic estimates . . . . . 10.5 Miscellaneous problems 10.6 Notes . . . . . . . . . .

485 485 494 497 506 513 520

11 Lagrange Interpolation Formula 11.1 Notes . . . . . . . . . . . . . .

521 537

12 Higher Algebra in Combinatorics 12.1 The determinant trick . . . . . 12.2 Matrices over lF2 . . . . . . . . 12.3 Applications of bilinear algebra 12.4 Matrix equations . . . . . . . 12.5 The linear independence trick 12.6 Applications to geometry 12.7 Notes . . . . . . . . . . .

539

Bibliography

585

541 546 552 561 568 576 583

Chapter 1

Some Useful Substitutions Let us first recall the classical substitutions that will be used in the following problems. All of these are discussed in detail in [3], chapter 1 and the reader is invited to take a closer look there. Consider three positive real numbers a, b, c. If abc = 1, a classical substitution is x z a= - , b= ~ , c= -. y z x A less classical one is

a=_x_, y+z

b=-Yz+x'

c=_z_ x+y

(for some positive real numbers x, y, z) whenever ab + bc + ca its equivalent version y+z x+y b=Z+x a=-Y , c=x ' z

+ 2abc =

1, or

when abc = a + b + c + 2 (the equivalence between the two relations follows from the substitution (a, b, c) --t (~, ~)). Two other very useful substitutions concern the relations a 2 + b2 + c2 = abc + 4 and a2 + b2 + c 2 + 2abc = 1. In the first case, with the extra assumption max(a, b, c) 2: 2" one can find positive numbers x, y, Z such that xyz = 1 and

1 1 1 a=x+-, b=y+-, c=z+-. x y z

Chapter 1. Some Useful Substitutions

In the second case one can find an acute-angled triangle ABC such that

= cos(A),

b = cos(B),

Thus (b 2 - 4)(e2 - 4)

= cos(C).

The relation a2 + b2 + c2

= (be ± 2a)2,

abc + 4

(a - 2)(b - 2) + (b - 2)(c - 2)

abc, then

2)(a - 2) 2

x2 + y2 + z2 = xy z + 4 { xy -+- yz + zx = 2(x + y + z) Proof. By the second equation we have max(x, y, z) 2 2 and so the first equation yields the existence of positive real numbers a, b, c such that

1 a

= a+-,

and abc = 1.

Proof. If max(a, b, c) < 2, then everything is clear, so assume that max(a, b, c) 2 2. -

abc = 4, so there exist positive numbers x, y, z such that 1 a = x+-,

But then a, b, e

b = y +-, y

Since abc = 1, we have

= z +-. z

22 and we are done again.

so the second equation can be written

Proof. The most natural idea is to consider the hypothesis as a quadratic equation in a, for instance. It becomes a 2 ± abc + b2 + c2 - 4 = 0, and solving the equation yields a =

=Fbe ± J(b 2 - 4)(c2 - 4) 2

The left-hand side is also equal to .

The following exercise is trickier and one needs some algebraic skills in order to solve it. We present two solutions, neither of which is really easy.

Titu Andreescu, Gazeta Matematidi

Then a2 + b2 + c2 xyz = 1 and

2. Find all triples x, y, z of positive real numbers such that

41 =

+ (c -

(be ± 2a)2 > (b+2)(c+2) -

Writing similar expressions for the other two variables, we are done.

We start with an easy exercise, based on the resolution of a quadratic equation. 1. Prove that if a, b, e 2 0 satisfy la 2 + b2 + c2

which can also be written as

(b - 2)(e - 2) =

Of course, in practice one often needs to use a mixture of these substitutions and to be rather familiar with classical identities and inequalities. But experience comes with practice, so let us delve into some exercises and problems to see how things really work.

1.1

1.1. The relation a2 + b2 + e2 = abc + 4

1 c

z=c+ -

Chapter 1. Some Useful Substitutions

because

2 ~ ~ _ a + b _ (2 b+a ab - ea

+ b2) . Since 2::: a 2': 3 and 2::: ab

We deduce that CL a-I) (2::: ab - 1) = 4. the AM-GM inequality and the fact that abc a = b = e = 1 and thus when x = y = Z = 2.

2': 3 (by

= 1), this can only happen if 0

Proof. If x + y = 2, the second equation yields xy = 4, so that (x - y)2 = -12 which is a contradiction. Thus x + y =f 2 and similarly y + z =f 2, z + x =f 2. The second equation yields

1.1.

The relation a 2 + b2 + e2

= abc + 4

Proof. The hypothesis yields the existence of positive real numbers x, y, Z such that x y x z a = - +-, b = Y + .:, e = - +-. y x z y z x The miracle is that both sides of the inequality have very nice factorizations. For the right-hand side, this is easy to observe, since 1 1 1) a+b+e+3=(xy+yz+zx) ( -+-+xy yz zx and

4- xy z=2+---x +y - 2' and a rather brutal insertion of this expression in the first equation gives

(X_y)2+( 4-X y )2 x+y-2

For the left-hand side, things are more subtle, but one finally reaches the identity

(4 - xy)2 2-x-y

Unless x = y = 2, this implies the inequality 2 > x + y. If two of the numbers x, y, z are equal to 2, then trivially so is the third one. If not, the previous argument shows that 2 > x + y, 2> y + z and 2 > x + z. But then the second equation yields

x+y+z> LX(Y;z) =xy+yz+zx=2(x+y+z), eye

a contradiction. Thus, the only solution is x

= Y = z = 2.

The following problem hides under a clever algebraic manipulation a very simple AM-GM argument. The inequality is quite strong, as the reader can easily see by trying a brute-force approach. 3. Prove that if a, b, e 2': 2 satisfy a 2 + b2 + e2

The desired inequality becomes simply the AM-GM inequality for two numbers! 0 The following problems are rather tricky mixtures of algebraic manipulations and elementary number theory. 4. Find all triplets of positive integers (k, l, m) with sum 2002 and for which the system x Y -+-=k Y x

= abc + 4, then

a + b + e + ab + ae + be 2': 2J(a + b + e + 3)(a2 + b2 + e2

-+ -

3).

Marian Tetiva

has real solutions. Titu Andreescu, proposed for IMO 2002

Chapter 1.

Some Useful Substitutions

1.1.

The relation a 2 + b2 + c2

~. ~

Proof. The system has solutions if and only if k 2 + l2

= lkm

+ 2)(l + 2)(m + 2) =

Jv [(no + ;vn)

n _ ( aD -

;vn) nJ.

Part of the following problem can be dealt in a classical way, but we do not know how to solve it entirely without using the trick of substitutions. 6. The sequence (a n )n2:0 is defined by ao =

+ l + m + 2)2.

As k + l + m = 2002, we deduce that any solution of the problem satisfies k + l + m = 2002 and (k

abc + 4

+ 4.

An easy computation shows that this relation is equivalent to (k

+ l + m + 2)2 =

an+l = anan-l

= 97 and

+ J(a~ - l)(a~_l

for all n 2: 1. Prove that 2 + y2

+ 2an

- 1)

is a perfect square for all n.

20042 = 24 . 32 .1672.

A simple case analysis shows that the only solutions are k and its permutations. The result follows.

Titu Andreescu

= l = 1000, m = 2

Proof. Writing

(an+! - a n a n _d 2 = (a; - l)(a;_l - 1)

and simplifying this expression yields

5. Solve in positive integers the equation

2 an-l

(x+2)(y+2)(z+2) = (x+y+z+2)2.

2 + an2 + an+l

2anan-lan+l

= 1,

thus Titu Andreescu Proof. A simple algebraic manipulation shows that the equation is equivalent to x2+y2+z2 = xyz+4 and, seeing this as a quadratic equation in z, we obtain the equivalent form (x2-4)(y2-4) = (xy_2z)2. If x2 < 4, then y2 < 4 as well and so x = y = 1, yielding z = 2. If x = 2, then y = z. In all other cases, we can find a positive square-free integer D (which is easily seen to be different from 1) and positive integers u, v such that x2 - 4 = Du 2 and y2 - 4 = Dv 2. Thus, solving the problem comes down to solving the generalized Pell equation a 2 - Db 2 = 4, which is a classical topic: this equation always has nontrivial integer solutions and if (ao, bo) is the smallest solution with ao, bo > 0, then all solutions are given by

(2an_l)2

+ (2a n )2 + (2a n+d 2 -

(2a n-l)(2a n )(2an+d

= 4.

Since we clearly have an > 2 for all n, this implies the existence of a sequence x n > 1 such that 2an = Xn + x;:;-l and such that Xn+l = XnXn-l. Thus logxn satisfies a Fibonacci-type recursive relation and so we can immediately find out the general term of the sequence (an)n. Namely, a small computation shows that if we define a = 2 + yI3, then Xn = a 4Fn , where Fn is the nth Fibonacci number. Thus an

="2

a 4Fn + a 4Fn

and so 2 + y2

+ 2a n = 2 + ( a 2Fn +

The result follows, since an

a 21Fn )

+ a- n E Z for

a~n) 2

all n, by the binomial formula.

Chapter 1. Some Useful Substitutions

1.2.

Remark 1.1. Here is an alternative proof of the fact that all terms of the sequence are integers, without the use of substitutions. The method that we will use for this problem appears in many other problems. As we saw in the previous solution, the sequence satisfies the recursive relation

1.2

Writing the same relation for n + 1 instead of n and subtracting the two yields the identity a~+2 - a~-I = 2anan+l(an+2 - an-I). Note that (an)n is an increasing sequence (this follows trivially by induction from the recursive relation), so that we can divide by a n+2 - an-I i- 0 in the previous relation and get an+2 = 2a n a n+l - an-I. The last relation clearly implies that all terms of the sequence are integers (since one can immediately check that this is the case with the first three terms of the sequence). Note however that it does not seem to follow easily that 2 + V2 + 2an is a perfect square using this method.

Remark 1.2. There are a lot of examples of very complicated recurrence relations that rather unexpectedly yield integers. For instance, the reader can try to prove the following result concerning Somos-5 sequences: let al = a2 = ... = a5 = 1 and let

The relations abc = a + b + e + 2 and'ab + be + ea + 2abe = 1

The relations abc = a + b + e + 2 and ab + be + ea + 2abe = 1

The first inequality in the following problem is very useful in practice and we will meet it very often in the following problems. 7. Prove that if x, y, z

> 0 and xyz = x + y + z + 2, then

+ yz + zx 2: 2(x + y + z)

and

3 Vx + vY + vZ ::; '2vxyz.

Proof. With the usual substitutions b+e

e+a y- -b-'

x=-' a

a+b e '

z=--

the first inequality comes down (after clearing denominators and canceling out similar terms) to Schur's inequality

a(a - b)(a - c)

+ b(b -

a)(b - c)

+ e(e -

a)(e - b) 2: 0,

while the second one follows. by adding up the inequalities

o for n 2: O. Then an is an integer for all n. Similarly one defines Somos-6, Somos-7, etc sequences by the formulas

ao ao

= al = ... = a5 = 1,

= al = ... = a6 =

etc. One can prove (though this is not easy) that all terms of Somos-6 and Somos-7 sequences are integers. Surprisingly, this fails for Somos-S sequences (in which case al7 is no longer an integer!).

The reader may find a bit strange the first method of proof of the following problem, but it is actually a quite powerful one. We will use again this kind of argument, see problem 11 for instance. Also, the third solution uses a very useful technique. S. Let x, y, z

> 0 be such that xy + yz + zx = 2(x + y + z). Prove that xyz ::; x

+ y + z + 2. Gabriel Dospinescu, Mircea Lascu

Chapter 1. Some Useful Substitutions

Proof. We will argue by contradiction, assuming that xyz > x + y + z + 2. We claim that we can find 0 < r < 1 such that X = rx, Y = ry, Z = r z satisfy XY Z = X + Y + Z + 2. Indeed, this comes down to the vanishing of f (r) = r3 xy z - r (x

+ y + z)

between 0 and 1, and this is clear, since f(O) condition xy + yz + zx = 2(x + y + z) yields XY

- 2

< 0 and f(1) > O. Next, the

+ YZ + ZX = 2r(X + Y + Z) < 2(X + Y + Z).

This contradicts the first inequality of problem 7.

1.2.

The relations abc

= a + b + e + 2 and,ab + be + ea + 2abe = 1

and defining X,y,z to be the three roots of this polynomial (in some order). Increasing 173, i.e. lowering the constant term, corresponds geometrically to lowering the graph. As we lower the graph, the smallest root increases, thus we maintain three positive real roots until the smallest root becomes a double root. If the double root is at t = a and the larger root at t = b, then we have 2 2 171 = 2a + b, 172 = a + 2ab and 173 = a b. If we fix 171 and 172 with 172 = 2171 as hypothesized , then we find b = ~ 2(a-1 J and because 0 < a ::; b, we see that 1 < a ::; 2. By the discussion above xyz = 173 ::; a2 b, so it suffices to show that

Proof. The condition can also be rewritten in the form (x - 1)(y - 1)

+ (y -

1)(z - 1) + (z - 1)(x - 1) = 3

or in the form

xyz - x - y - z - 1 = (x - 1)(y - 1)(z - 1). We will discuss several cases. If x, y, z ::::: 1, then by the A~\1-GM inequality and the first identity, we get

which yields, thanks to the second identity, the desired estimate. If x, y, z ::; 1 or if only one of the numbers x, y, z is smaller than or equal to 1, then (x - 1)(y - 1)(z - 1) ::; 0 and so xyz ::; x + y + z + 1 in this case. Finally, if two of the numbers are smaller than 1, say x, y ::; 1, the desired inequality can be written in the form 0 ::; x + y + z(1 - xy) + 2, which is obvious. 0

Proof. For three positive real numbers x, y, z consider fixing the first two elementary symmetric polynomials 171 = x + y + z and 172 = xy + yz + zx and letting 173 = xyz vary. This amounts to varying only the constant term in the polynomial

for 1

< a::; 2. But this rearranges to (a-2)2(a 2 -1) ::::: 0 and we are done. 0

The technique used in the first solution of the next problem is rather versatile and the reader is invited to read the addendum 5.A for more examples. 9. Let x, y, z

> 0 be such that xy + yz + zx + xyz = 4. Prove that

Gabriel Dospinescu

Proof. Using the usual substitution 2a x = b+e'

2b y=e+a'

Z=--

a+b'

the problem reduces to proving the inequality 2

3 (

Iff! -

> 16

b+e

)

(a+b+e)3

(a + b)(b + e)(e + a)'

Chapter 1. Some Useful Substitutions

This is a quite strong inequality and it is easy to convince oneself that most applications of classical techniques fail. However, the following smart application of Holder's inequality does the job:

1.2.

The relations abc

= a +b+e +2

and'ab + be + ea + 2abe

the inequality becomes v'3r+R> 2rs R - R2' or (R+r)Rv'3 2': 2rs. This splits trivially into R+r 2': 3r (Euler's inequality) and s ::; R, the last one being well-known and easy. 0

Here is yet another easy application of problem 7. so it is enough to prove that 10. Let u, v, w

> 0 be positive real numbers such that u + v + w + ';uvw = 4.

This reduces after expanding to

2: a(b -

e)2 2': 0, which is clear.

Proof. First, we get rid of those nasty square roots, via the substitution xy

2 = 4e,

Then

2be

x=~,

2 = 4a,

2ea y=b'

2ab

z=~

3(a + b + e)2 2': 16(a + be)(b + ea)(e + ab). The hypothesis becomes a 2 + b2 + e2 + 2abe = 1, so that there exists an acuteangled triangle ABC such that a = cosA, b = cosB, e = cosC. Next, observe that

= cosC + cos A . cosB = - cos(A + B) + cos A . cosB = sin A . sinB.

Using this (and similar identities obtained by permuting the variables), the desired inequality becomes

v'3 L

cos A 2': 4

II sin A.

Using the well-known identities s

= 4R II cos ~ ,

= 4R II sin ~ ,

VfUV --;;; + VfvW --;; + VfWU --;; 2': u + v + w.

= 4b2 â&#x20AC;˘

and replacing these values in the inequality yields the equivalent form

e + ab

Prove that

China TST 2007 Proof. To get rid of those nasty square roots, let us perform the substitution

~=e, ~=a,

(f-=b.

Then u = be, v = ea and w = ab, so that the inequality becomes a + b + e 2': ab + be + ea and the hypothesis is ab + be + ea + abc = 4. Another substitution a

=~, x

b- ~ - y'

= ~z

yields xyz = x+y+z+2. We need to prove that xy+yz+zx 2': 2(x+y+z), which is the first component of problem 7. 0 The following problem has some common points with problem 7 and one can actually deduce it from that result. But the proof is not formal. 11. Prove that if a, b, e

> 0 and x = a + -b , y = b + ~, z = e + ~, then 1

xy+yz+zx 2': 2(x+y+z).

L cos A = 1 + ~,

Vasile Cartoaje

Chapter 1. Some Useful Substitutions

Proof. The method is the same as the one used in problem 8. The starting point is the observation that xyz 2: 2 + x + y + z. Indeed, xyz = abc +

1 abc

1 a

+ a + b+ e+ - +

1 -b

1 e

+ YZ + ZX < 2r(X + Y + Z)

2(X

and tlb + be + ea

+ 2abe = 1

12. Prove that for all a, b, e > 0,

Japan 1997 Proof. With the usual substitution

b+e

= -a-'

a+e

a+b e

z=--,

= -b-'

the problem asks to prove that for any positive numbers x, y, z such that xy z = x + y + z + 2 we have

+ Y + Z).

Proof. As in the previous proof, we obtain that xyz 2: x + y + z + 2. Since we obviously have min(xy, yz, zx) > 1, this implies that z 2: 2;:yX_+l and similar inequalities obtained by permuting the variables. Next, we have

+ b+ e+ 2

This contradicts the first inequality in problem 7.

The relations abc = a

+ - 2: 2 + x + y + z.

Let us assume for sake of contradiction that xy + yz + zx < 2(x + y + z) and let us choose r E (0,1] such that X = rx, Y = ry, Z = rz satisfy XY Z = 2 + X + Y + Z. This equality is equivalent to r3 xy z = 2 + r (x + y + z) and such r exists by continuity of the function f(r) = r 3 xyz-2-r(x+y+z) and by the fact that f(I) 2: 0 and f(O) < O. The hypothesis xy + yz + zx < 2(x + y + z) can also be written as XY

1.2.

(x-I)2

-'-x-,-----"---2+ 1 +

(y y2

(z-I)2 z2

> -. +1 - 5

Applying the Cauchy-Schwarz inequality (or what is also called Titu's lemma), we obtain the bound

+ y + z = a + -a + b +b - + ec +- > 6. -

(x 1)2 x2+ 1

(Y-I?

+1 +

(z 1)2 (x+y+z 3)2 z2 + 1 2: x 2 + y2 + z2 + 3 .

In particular, there are two numbers, say x, y, such that x + y 2: 4. The inequality to be proved is equivalent to z 2: 2(~~~~2XY, so we are done if we can prove that 2+ x +Y 2(x + y) - xy ------~ > . xy-I x+y-2

It remains to prove that the last quantity is at least ~. Let S = x P = xy + yz + zx, so that the inequality becomes

But this is equivalent, after an easy computation (in which it is convenient to denote S = x + y and P = XV), to (xy - y - x)2 2: (x - 2)(y - 2). If (x - 2)(y - 2) ~ 0, this is clear; otherwise x, y 2: 2 as x + y 2: 4. Let u = x - 2,v = y - 2, then the inequality becomes (uv + u + v? 2: uv, with u, v 2: 0 and it is obvious. 0

Notice that as %+ ~ 2: 2 and cyclic permutations of this, we have S 2: 6. Also, by problem 7 we have P 2: 2S. Therefore,

The form of the following inequality strongly suggests the use of the Cauchy-Schwarz inequality. It turns out that this approach works, but in a rather indirect and mysterious way, which makes the problem rather hard.

+ y + z and

(S - 3)2 > (S 3)2 = S - 3 = 1 _ _ 2_ > ~. S2 - 2P + 3 - S2 - 4S + 3 S - 1 S - 1 - 5

Remark 1.3. There are other solutions for this problem: some of them are shorter, but not easy to find. Here is a particularly elegant one, based on

Chapter 1. Some Useful Substitutions

the linearization technique: the inequality is homogeneous, so we may assume that a + b + e = 1. We need to prove that

1.2.

The relations abc = a

+ b+e + 2

Letting e -+ 0, we deduce that any k as in the statement must satisfy

k(k+1)22:

The point is to bound from below (I-a (1-~t)2 2 by an affine function of a, suitably +a chosen. The best choice is the following

and ab + be + ea + 2abe = 1

which is equivalent to 4k2 the problem.

+ 2k 2:

(k+~)3

1. We claim that any such k is a solution of

Pick such k and perform the usual substitution

a X=-b+e'

b y=e+a'

Z=--

a+b

to reduce the problem to Of course, the question is how one came up with something like this. Well, it is actually very easy: we need constants A, B such that

2 (x

+ y + z) + k(xy + yz + zx) + xyz 2:

+~) 3

Now, we know that xy + yz + zx + 2xyz = 1 and it is by now a classical fact (use problem 7 for l/x,l/y,l/z) that x+y+z 2: 2(xy+yz+zx). Thus, it is enough to ensure that for all a E [0, 1], with equality for a = ~. Impos~ng also th~ vanisfi~g of the derivative of the difference between the left and nght- hand SIde at 3" YIelds the desired constants A, B. Once we have the previous estimates, it is very easy to conclude by adding them and taking into account that a + b + e = 1. 13. Find all real numbers k with the following property: for all positive numbers a, b, e the following inequality holds

+ (2k2 + k)(xy + yz + zx) + 1 -

(xy

~ yz + zx) 2:

+~) 3

This can be rewritten as

(2k2

+k -

(Xy

+ yz + zx -

i) 2: o.

But the last inequality follows from the fact that 2k2

xy + yz

+k -

2 -

and

+ zx 2: 4'

which, by returning to the substitution Vietnam TST 2009

Proof. Take first of all a

= b=

a X=-b+e'

1 and arbitrary e to get that is equivalent to

L, a(b - e)2 2:

b y=--, e+a

e z = a + b'

In condusion, the solutions of the problem are the real numbers k such that 4k2 + 2k 2: 1. 0

Chapter 1. Some Useful Substitutions

We end this section with a very challenging problem, which answers the following natural question: what would be the analogue of the classical substitution x = b!e, y = eta, Z = a~b when we have five variables? We warn the reader that the first solution is really not natural. ..

1.2.

The relations abc

= a + b + e + 2 and ab + be + ea + 2abe =

aI, a2, ... , a5

The crucial claim is that under the previous hypothesis, the system has solutions whose unknowns Xi are positive. Probably the easiest way to prove this is to exhibit such a solution. Well, simply take Xl = 1, X5 =

14. Let

be positive real numbers such that

1 + al + a2 + a3 + ala3 -----=---=---=-:::. 1 + a5 + a3a5 + a2 a 5

and the define

What is the least possible value of

a\ + :2 + ... + a5

Gabriel Dospinescu, Mathematical Reflections

Proof. Consider the linear system

1 + X5 = ---,

II i=l

= 2+L

+ X5

X3 X4 --=--~ a2

The question is, of course, how on earth did we choose the value of X5? Well, simply by solving the system, as indicated in the beginning of the solution. Note that we clearly have Xi > 0 and an easy computation shows that these are solutions of the system (we only have to check two equations, since three are satisfied by construction). The conclusion is that if the ai's satisfy the condition

ai =

i=l

then we can find

+ ala4 + a4 a 2 + a2a5 + a5a3 + a3 a l,

i=l Xi

> 0 such that a2 =

+ X4 , ... ,a5 =

Xl X2 -"----=-

So, the problem reduces to finding the minimal value of

We can try to solve this system by expressing, for example, all variables in terms of Xl, X5. Namely, we can use the fifth, first and second equation to express X2, X3, X4 in terms of Xl, X5. Replacing the obtained values in the other two equations and eliminating Xl, x5 between them, we obtain that the system has a nontrivial solution if and only if

+ ala4 + a4 a 2 + a2 a 5 + a5 a 3 + a3 a lÂˇ

Xl X2 X5 ---+ + ... +--X2 + X3 X3 + X4 Xl + X2

i=l

All the previous (painful!) computations are left to the reader, since they are far from having anything conceptual. Note that the same result can be obtained by computing the determinant of the associated matrix. Now, the previous relation is almost the one given in the statement, up to a permutation of the variables. So, since the conclusion is symmetric in the five variables, we will assume from now on that the previous relation is satisfied instead of the one given in the statement.

We claim that the previous expression is always at least 5/2. By Titu's lemma, this reduces to proving that (Xl

+ X2 + X3 + X4 + X5)2 2: ~ L i<j

But since

XiXj.

Chapter 1. Some Useful Substitutions

1.3.

this reduces to (~Xi)2 ::; 5 ~ x;, which is Cauchy-Schwarz. Putting everything together, we obtain that the minimal value of

The relation

+ b 2 + c 2 + 2abc = 1

Using this lemma for the inverses of the

21 ai's

and denoting

111 S=-+-+ ... +_, al

we deduce that is 5/2, attained when all game with 7 variables?

ai's

are equal to 2. Who wants to play now the same 0

Proof. Here is another proof, also far from being evident. " We will use the following nontrivial Lemma 1.4. For all nonnegative real numbers (Xl

Xl, X2, ... ,X5

On the other hand, by Mac-Laurin's and AM-GM inequalities we also have

we have

+ X2 + ... + X5)3 :2: 25(XIX2X3 + X2X3X4 + ... + X5XIX2).

Taking into account these inequalities and the hypothesis, we deduce that

Proof. This is not easy at all, but here is a very elegant (but unnatural) proof: consider the identity XIX2X3

+ X2X3X4 + X3X4X5 + X4X5XI + X5 X I X 2 = X5(XI + X3)(X2 + X4) + X2X3(XI + X4

which immediately implies that S :2: ~.

X5).

We may assume that X5 = min Xi (since the inequality is cyclic), so that Xl + X4 - X5 :2: O. Denoting Xl + X2 + X3 + X4 = 4t and using the AM-GM inequality, we deduce that

1.3

The relation a2 + b2 + c2 + 2abc = 1

The following problem is a rather tricky application of the AM-GM inequality. 15. Prove that in all acute-angled triangles the following inequality holds

Thus, it remains to prove that

COS (

2 X5

(4t-

X 5)3

::;

cosB

(cos cosC

(cos C) 2 cosA

+ 8cosAcosBcosC:2: 4.

(X5+ 4t )3

By homogeneity, we may assume that X5 = 1 and then expanding everything the inequality becomes, with the substitution 4t 1 = 3x, (x-1)2(8x+7):2: 0, which is clear. 0

Titu Andreescu, MOSP 2000

Proof. Since 2

cos A

+ cos 2 B + cos 2 C + 2 cos A cos B cos C

Chapter 1. Some Useful Substitutions

The relation a 2 + b2

+ c 2 + 2abc =

+ (COSB)2 -C + (cosC)2 -A 2: 4(cos 2 A+cos 2. B+cos 2 C). cos cos

b+c

X=--,

Let x

= cos 2 A,

= cos 2 B,

= cos 2 C,

L x

-+-+

z 2: 4 (x+y+z). x

-¢==}

a+b c

Z=--.

+ L --:S -2 -¢==} zyXY

LX + L VxY :S

3 2

LX+ LVxY:S -xyz

~(x + y + z + 2)

-¢==}

vxf

:S 2 (L x +

x+y+z

-+-+->--y z x ijxyz for any positive real numbers x, y, z, as follows by adding up the AM-GM inequalities 2x y

+ '!{ 2: 3 3 (;2.

VYz Thus, it is enough to prove that xyz :S t4'

Proof. Since the triangle is acute, there are positive numbers x, y, z such that a2

= y + z,

= Z + x,

Then

easy to prove.

cos A But this is well-known and 0

= X + y.

= ---,.;===::;:=;=====c=

L 16. Prove that in every acute-angled triangle ABC,

J(x

+ y)(x + z)

and similar identities for cos B, cos C. The desired inequality can be written as

The following problem is a preparation for a hard problem to come, but has also an independent interest.

(cos A

+ y)(x + z) + L

yz 3 (y + z)J(x + y)(x + z) :S 2'

which (after clearing denominators) is equivalent to

+ cosB)2 + (cosB + cosC)2 + (cosC + cosA)2 :S 3.

Proof. Letting

(x 2 (y

+ z) + yzJ(x + z)(x + y)) :S ~(x + y)(y + z)(z + x)

and then to cos A =

YYz

cosB

3) .

This follows from Cauchy-Schwarz

The point is that we actually have

c+a

y= -b-'

The desired inequality can be written

so we need to prove that

for some positive real x, y, z, we obtain the relation xyz = x + y + z + 2 from the classical identity L cos 2 A + 2 cos A = 1. Thus we can find positive real numbers a, b, c such that

the inequality can be written

COSA)2 (cosB

1.3.

0, Y-;;

cosC =

0 Y-;;Y

L eye

Jyz(x

+ y) . yz(x + z) :S 3xyz + 2 LYz(y + z). eye

Chapter 1. Some Useful Substitutions

+ y)

. yz(x

+ z) :S

xyz

1 "2Yz(y

The relation a 2

+ b2 + c2 + 2abc = 1

+ z) a=

Even though the following problem seems classical, it is actually rather hard. Fortunately, we did the difficult job in the previous problem. We also present an independent and very elegant approach due to Oaz Nair and Richard Stong. 17. Prove that if a, b, c 2 0 satisfy a 2 + b2 + c2 + abc = 4, then

o :S ab + bc + ca -

abc=c(a+b)+ab(l-c)20.

The hard point is proving that ab + bc + ca - abc :S 2. Taking into account the hypothesis, this can also be written as

b) 2 + (b; c) 2+ ( c) 2 :S a;

4)(c 2

Thus, it is enough to prove that J(b 2 - 4)(c2 - 4) :S 4 - bc. Note that bc :S 4, since b2 :S 4 and c 2 :S 4. Squaring the previous inequality thus yields an equivalent one which is obvious.

18. Prove that in all acute-angled triangles the following inequality holds a 2b2 c

Proof. The inequality on the left is easy: the hypothesis and the AM-GM inequality imply that abc :S 1, so that min(a, b, c) :S 1. We may assume that c = min(a, b, c), so that

( a;

We end this chapter with another challenging problem, which reduces after some tricky algebraic manipulations to the infamous "Iran 1996 inequality."

abc :S 2.

Titu Andreescu, USAMO 2001

ab+bc+ca

-bc + J(b 2

------~----~----~

and two similar inequalities.

Thus, it is enough to prove that a + bc :S 2. But the given condition and the fact that a is nonnegative imply that

However, this follows immediately from the AM-GM inequality: Jyz(x

1.3.

If we denote a = 2x,b = 2y,c = 2z then x 2 + y2 + z2 + 2xyz = 1 and so there exists a triangle ABC such that x = cos(A), y = cos(B), z = cos(C). Thus, the problem reduces to the previous one (since a, b, c are nonnegative, the triangle ABC is acute angled). D Proof. Here is a very elegant proof for the hard part of the inequality. Among the three numbers a-I, b - 1, c - 1, two have the same sign, say b - 1 and c - 1. Thus (1 b)(1 - c) 2 0, implying that b + c :S bc + 1 and ab + ac - abc :S a.

-2-+

a 2c 2

b2c 2 29R2. a

+-2

Nguyen Son Ha Proof. If A, B, C are the angles of the triangle, the sine law and the identity sin 2 x + cos 2 X = 1 yield the equivalent form of the inequality

Write cos 2 A Then xy + yz such that

yz,

cos 2 B

zx,

cos 2 C

xy.

2xyz = 1 and so there exist positive numbers X, Y, Z

X x = Y

+ Z'

Y y= X+Z'

Z z= X+Y'

We want to prove that ' " --'--(1_ _ y z-,-)-c-(1_-_z_x-,-) > ~. L 1 xy - 4

Chapter 1. Some Useful Substitutions

On the other hand, we have

1.4. Notes Using the identities

X(X + Y + Z) 1 - yz = (X + Y)(X + Z)'

ab + be + ea 4ab this becomes

so that the inequality is equivalent to

(ZX

1.4

+ (e + a)2 + (a + b)2 2':

9 4'

Proof. We may assume that a 2': b 2': e. First, we show that

-----=-+ (a+b)2

(a+e)2

+ (b+e)2 >-+ - 4ab

(a+e)(b+e)

This can be rewritten

(a +1 b +1)2 2': 4ab((a a +b)2b)2 , e-

or equivalently 4ab(a+b)2 2': (a+e)2(b+e)2. This is clear, as (a+b)2 2': (a+e)2 and 4ab 2': (b + e)2. Thus, it remains to prove that

9 [ 1 + (a + e)(b2] + e) 2': 4'

(ab + be + ea) 4ab

e(a + b) 4ab - (a

2e2 (a+e)(b+e)'

= 2 - - -_ __

2e2 . + e)(b + e)

+ b) (a + e) (b + e) 2': 8abe and we 0

Notes

We would like to thank the following people for their solutions: Vo Quoc Ba Can (problems 13, 18), Xiangyi Huang (problems 4, 5), Logeswaran Lajanugen (problem 1, 9), Oaz Nair (problem 17), Dusan Sobot (problems 2, 9, 16), Richard Stong (problems 8, 17), Gjergji Zaimi (problems 2, 3, 6, 7, 8, 11, 12, 15, 16, 17).

Lemma 1.5. For all positive numbers a, b, e we have 1

2(ab + be + ea) (a+e)(b+e)

---,'-_----,---.,--_c-'-

This completes the solution to the problem.

+ ZY)2 2': 4

and it is a consequence of the following famous inequality

(ab + be + ea) ( (b + e)2

= 4+

e( a + b) 4ab '

But this is simply the standard inequality (a are done.

This can also be written in the form

+ Y + Z)Âˇ L

~----'->

~ XY (X + Y + Z) > ~ ~ Z(X + y)2 - 4'

XYZ(X

Chapter 2

Always Cauchy-Schwarz ... As the title suggests, all problems in this chapter can be solved using the Cauchy-Schwarz inequality, even though sometimes this will require quite a lot of work. Let us recall the statement of the Cauchy-Schwarz inequality: if aI, a2,···, an and bl , b2, ... , bn are real numbers, then

This follows easily from the fact that L~=l (aix + bi )2 :2: 0 for all real numbers x. Indeed, the left-hand side is a quadratic function of x with nonnegative values, so its discriminant is negative or O. But this is precisely the content of the Cauchy-Schwarz inequality. Another proof is based on Lagrange's identity

L I

(aibj - a j bi )2.

:Si<j:Sn

This useful identity will be used several times in this chapter. As the best way to get familiar with this inequality is via a lot of examples of all levels of difficulty, we will not insist on any theoretical aspects and go directly to battle. We start with two easy examples, destined to give the reader some confidence. He will surely need it for the more difficult problems to come ...

Chapter 2.

Always Cauchy-Schwarz . ..

Proof. The inequality being symmetric in b, c, but not in a, it is natural to deal first with v1J=1 + JC=l. This is easy to bound using Cauchy-Schwarz:

1. Let a, b, c be nonnegative real numbers. Prove that

v1J=1 + JC=l :::; J(b - 1 + 1)(1 + c - 1) = V;;;;. is enough to prove that VbC + va=I :::; J a( bc + 1). But this

for all nonnegative real numbers x. Titu Andreescu, Gazeta Matematica Proof. This is just a matter of re-arranging terms and applying CauchySchwarz:

(ax 2 + bx + c) (cx 2 + bx + a)

So, it more the Cauchy-Schwarz inequality.

Another easy, but a bit exotic application of the Cauchy-Schwarz inequality is the following Chinese olympiad problem.

(ax 2 + bx + c)(a + bx + cx 2)

2 (ax + bx + cx)

4. Let n be a positive integer. Find the number of ordered n-tuples of integers (a I, a2, ... , an) such that

= (a+b+c)2x 2.

+ a2 + ... + an

2 n 2 and aI

+ a§ + ... + a; :::; n 3 + 1.

D 2. Let p be a polynomial with positive real coefficients.

(t)

p(~) =

proof By the Cauchy-Schwarz inequality we have

is true for x = 1, then it is true for all x > O.

+ alX + ... + anxn

:::; vn(n3

+ 1) < n 2 + 1. n 2, we must have

+ a2 + ... + an 2

l an) xn (1) =(aO+aIX+···+anx)n (aoa+-;:-+···+

2 (ao

+ a2 + ... + an

Since ai are integers and al

and observe that

+ a2 + ... + an =

n 2.

But then (again by Cauchy-Schwarz) we have + a§ + ... + a; 2 n 3 , forcing + a§ + ... + a; E {n 3, n 3 + I}. If + a~ + ... + a; = n 3, then we must have equality in Cauchy-Schwarz, implying that all ai's are equal to n. Assume now that + a§ + ... + a; = n 3 + 1 and let bi = ai - n. Then

p(x)p ;:

The result follows.

China 2002

Prove that if

Titu Andreescu, Revista Matematica Timi§oara Proof. Write p(X)

is once D

+ al + ... + an )2

p(I)2. D

+ b§ + ... + b~ =

+1-

2n . n 2 + n 3 = 1,

The following exercise is already a bit trickier, due to the lack of symmetry.

forcing all but Olle bi vanish. This is however impossible, as bl +b2+ . -+bn = O. Therefore, this second case will not occur and the only solution is

3. Prove that for all real numbers a, b, c 2 1 the following inequality holds:

D We continue the series of easy exercises:

Chapter 2. Always Cauchy-Schwarz . ..

5. Let

Xl, X2, ... , XlO

be real numbers between 0 and

. 2 Xl sm

such that

33 The equation

. 2 x2 + ... + sm . 2 xlO =. 1 + sm

Prove that 3(sin Xl

+ sin X2 + ... + sin XlO)

:=:; cos Xl

+ cos X2 + ... + cos XlO.

can also be rewritten as

Saint Petersburg 2001 Proof. The Cauchy-Schwarz inequality and the hypothesis yield

cos Xl =

This simplifies to

V'sm2 X2 + sm . 2 X3 + ... + sm . 2 XlO 1

2': "3(sinx2 + sinx3 + '"

+ sinxlO)

49(Z2

+ 4y2 + 9x2) = 36(x + y + z)2.

Now, by Cauchy-Schwarz we have

and similarly for the other variables. The result follows by adding up these inequalities. D The following exercise combines an easy application of the CauchySchwarz inequality with some classical formulae from geometry. Recall that r is the inradius and that s is the semi-perimeter of a triangle.

yielding 49(Z2 + 4y2 + 9x2) 2': 36(x + y

+ z)2.

Thus, we are in the equality case of the Cauchy-Schwarz inequality, which is precisely the case when there is a k > 0 such that X = 4k, z = 36k, y = 9k. Then the sides of the triangle are 13k, 40k, 45k. Thus the triangle is similar to the triangle with sidelengths 13,40,45, and the problem is solved. D

6. The triangle ABC satisfies

Show that ABC is similar to a triangle whose sides are integers, and find the smallest set of such integers. Titu Andreescu, USAMO 2002 Proof. Let the incircle be tangent to the sides of the triangle at points which split the sides into segments of length X and y, X and z, and y and z. Thus, the sides of the triangle have lengths X + y, X + z, y + z.

The statement of the following problem looks rather classical. There are however some technical problems which make the problem more difficult than expected. 7. Let n 2': 2 be an even integer. We consider all polynomials of the form n 1 xn + an_Ix - + ... + alX + 1, with real coefficients and having at least one real zero. Determine the least possible value of ai + a~ + ... + a~_l' Czech-Polish-Slovak Competition 2002

Chapter 2.

Always Cauchy-Schwarz . ..

Proof. Suppose that the corresponding polynomial has a real zero x. Using Cauchy-Schwarz, we obtain

The denominators in the following inequality look awful, but a clever application of the Cauchy-Schwarz inequality can make all of them equal. The method of proof is worth remembering, since it appears quite often.

(xn

+ a2 x2 + ... + a n_1 Xn - 1)2 ::; (ai + a~ + ... + a;_d(x 2 + x4 + ... + x 2(n-1)).

+ 1)2 =

(a1 x

Thus (xn+1)2 2 4 2( -1) = f(x) x +x +···+x n and so we need to find first the minimal value of f. Now, looking at the zeros of the derivative of f suggests that the minimal value might be taken at x = 1, so that it equals n~l' This is not easy to prove using derivatives, as the computations are a bit nasty. Instead, we will prove in an elementary way that 2

+ a2 + ... + an -1

8. Prove that for any positive real numbers x, y, z such that xyz ~ 1 the following inequality holds

Than Le, KaMaL magazine Proof. Using Cauchy-Schwarz, we obtain

In order to prove this, note that we have the inequalities xn

+ 1 ~ x 2 + x n - 2,

+ 1 ~ x4 + x n- 4, ... ,

+ 1 ~ x n - 2 + x 2.

Thus x x3

Adding them shows that

n-2

-4-(xn

+ 1) ~ x 2 + x4 + ... + xn-2.

+ y2 + z

<x(~+l+z) -

+ y + z) 2

Writing down similar inequalities for the second and third terms of the lefthand side, we reduce the problem to proving that

On the other hand, multiplying the last inequality by xn and adding the result to the previous inequality gives the inequality n

~ 2 (xn + 1)2 ~ x2 + ... + xn-2 + xn+2 + ... + x2(n-1).

Thus it remains to prove that (xn + 1)2 ~ 4x n , which is clear. The previous paragraph shows that

This is immediate from LXY ~ 3\!(xyz)2 ~ 3

and '"'

+ a2 + ... + an-l

4 --1 n-

if xn + a n _1X n - 1 + ... + a1x + 1 has at least one real zero. On the other hand, choosing a1 = a2 = ... = a n-1 = - n=-l shows that n~l is optimal. D

x ~

(x+y+z)2 3 ~ ijxyz( x

+ y + z)

~ x

+ y + z.

Here is a rather nice-looking inequality taken from a Romanian Team Selection Test. There are plenty of ways to prove it, but what follows is particularly elegant and naturaL

Chapter 2.

9. Let n that

2: 2 and let

aI, a2,·... , an and

Always Cauchy-Schwarz . ..

bl, b2, ... , bn be real numbers such

+ a§ + ... + a~ = bi + b§ + ... + b~= 1 + a2b2 + " . + anb n = O. Prove that

Proof. This is a simple application of the Cauchy-Schwarz inequality. Namely, we have

and alb l

so that Cezar and Tudorel Lupu, Romanian TST 2007 Proof. The proof mimics the proof of Cauchy-Schwarz: take any real number x and apply the Cauchy-Schwarz inequality to obtain

i)a~ xb t )2 2: (L~-l a~

+: L~l b t

J a 2 + b2 + c 2 + d2 + e2 2: (a + b)~. On the other hand,

Va + Vb::=;

J2(a

+ b),

Vc + vfi + y'e ::=;

J3(c

+ d + e) = J3(a + b),

therefore

i=l

By hypothesis, the left-hand side equals 1 + x2. Therefore

r *2 s

+ 1)

for any real number x. The difference between the right-hand side and the left-hand side being a quadratic polynomial which takes nonnegative values on the whole real line, its discriminant has to be negative or zero, thus

Combining these two inequalities yields the estimate

J a2 + b2 + c2 + d2 + e 2 >- 6( V2V30 (Va + Vb + Vc + Vd + y'e)2 + /3)2 . To see that this is optimal, it suffices to keep track of the equalities in the previous inequalities. For instance, we can take a = b = 3 and c = d = e = 2. Thus the answer is T = 6( f~ 0 3+ 2)2' The next problem requires a whole series of applications of CauchySchwarz.

where A = L~=l ai and B = L~l bi . It is immediate to check that this is 0 equivalent to the desired inequality. The following problem is easy, but the lack of symmetry might make it appear more difficult than it really is. 10. Find the largest real number T with the following property: if a, b, c, d, e are nonnegative real numbers such that a + b = c + d + e, then

11. Let Xl, X2, ... ,Xn be positive real numbers such that

1 1 1 --+--+"'+--=1 1 + Xl 1 + X2 1 + Xn . Prove the inequality

v'x1+JX2+"'+Fn2: (n-1) ( -1+ -1+ ... + 1- ) . v'x1 JX2 VX;; Vojtech Jarnik Competition 2002

Chapter 2. Always Cauchy-Schwarz ...

38 I Proof. Let ai = l+Xi' so that ~ D ai Schwarz we can write:

' " Ja2

~vIxi=~

= 1 and Xi =

+ a3 + ... + an al

~jiaj ai'

Then, using Cauchy-

.fii2 + y'a3 + ... + Fn

J(n - 1)al

max(al,a2) ::; 4min(al,a2).

Rearranging terms in the previous sum gives

1 .'" y'al (_1_ + _1_ + ... + _1_) ~ .fii2 y'a3 Fn

A bit more difficult, but with a similar flavor is the following problem. 13. Let n

and using again Cauchy-Schwarz we can bound this from below by '"

> 2 and let

(n-1)2y'al

(Xl

vn=I' ~.fii2 + y'a3 + ... + Fn' 2 ( xl

2 let aI, a2, ... an be positive real numbers such that

(al

+ X2 + ... + xn) (~ + ~ + ... + ~) = n 2 + l.

2 + X22 + ... + Xn)

The idea of the following example is worth keeping in mind, since it turns out to be useful in a wide range of problems. ~

X2, . .. , Xn be positive real numbers such that

Prove that

J(n - 1)(a2 + a3 + ... + an)

yields the desired inequality.

12. For n

Xl,

Using once more Cauchy-Schwarz for the denominators of each fraction

.fii2 + y'a3 + ... + Fn ::;

+ a2 + ... + an)

(1xi + -1+ ... + -1) > -

Proof. The idea is to fix two of the variables, say aI, a2 and apply the CauchySchwarz inequality to get rid of the remaining variables. Explicitly, this can be written in the form

+ ~) 2 2

~ (al + a2 + ... + an) (~ + ~ + ... + ~) al a2 an

;, ( (a +a )(> :,)+n-2) 1

+4+

n(n - 1)

Proof. The crucial idea is to write the hypothesis in a different way: expanding the product and rearranging terms shows that the hypothesis can also be written

(~+ ~ + ... +~) ::; (n + ~)2 al a2 an 2 Titu Andreescu, USAMO 2009

Gabriel Dospinescu

Prove that max( aI, a2, ... ,an) ::; 4 min( aI, a2, ... , an).

Since everything is symmetric in aI, a2,'" ,an, the result follows.

vn=I

where the first inequality is simply the hypothesis, while the second one is Cauchy-Schwarz applied to n - 1 terms, grouping the terms indexed 1 and 2 in each multiplicand. We deduce that (alala2 +az)Z < 25 , which immediately 4 implies that

Let us write ti,j

Now, expanding again gives us

(If; - [~J

Chapter 2.

Since

({!; +

fir

Always Cal1chy-Schwarz . ..

Thus, we need to prove that for any al,a2,' .. ,an

0 we have

= 4 + ti,j,

the inequality to be proved becomes 2

L (4 + ti,j)ti,j > 4 + n(n -1)' l:5,i<j:5,n

Define Si = a I + a2 + ... + ai, with the convention So = O. This is an increasing sequence and so

which is equivalent (taking into account the hypothesis that the sum of all ti,j is 1) to 2

--L tÂˇ . >n(n-1) l:5,i<j:5,n ~,J

This follows from Cauchy-Schwarz inequality, but there is still a detail to be explained: we have to show that we cannot have equality. But if we had equality, all ti,j would be equal to n(n2_1) and so all numbers ~; + ~~ would be equal. Since n > 2, this would force an equality Xj = Xk for some j =I=- k. But then tj,k = 0, a contradiction. 0

Proof. First note that for any nonnegative real numbers c and A we have

We.continue with an inequality which combines a rather direct application of the Cauchy-Schwarz inequality with a nice telescopic identity. We also present a beautiful alternate solution, due to Richard Stong. 14. Prove that for any real numbers Xl, X2,Âˇ .. ,X n the following inequality holds: X2 Xn ~ -Xl- + 2 2 + ... + 2 2 < V n. 1 + xi 1 + Xl + x 2 1 + Xl + ... + xn

Bogdan Enescu, IMO Shortlist 2001

Here the first inequality is just 1 + A2 + x2 ~ 1 + A2 with equality if and only if X = 0; the second is Cauchy-Schwarz with equality if and only if xVc = \11 + A2. Thus we cannot have equality in both cases and the final inequality is actually strict. Applying this inequality repeatedly, it is easy to see by downwards induction on k that Xn -Xl- + ... + 1 + xi 1 + xi + ... + x~

-----,,-----~

Proof. The solution is very short, but far from obvious. The idea is to use the Cauchy-Schwarz inequality to reduce the problem to

Xl Xk <--+ ... + + 1 + xi 1 + xi + ... + X~

The case k

= 0 is the desired result.

n-k 1 + Xl2

+ ... + X2' k

Chapter 2. Always Cauchy-Schwarz . ..

The trick of making the denominators equal thanks to a smart application of Cauchy-Schwarz or AM-GM inequality is also used in the following problem. 15. Prove that if a, b, c, d are positive real numbers, then

Proof. Let us write the condition in the form

xi + (Xl + X2)2 + (X2 + X3)2 + ... + (Xn-l + xn)2 + X; = 2. If we fix 1 :::; k :::; n and apply the Cauchy-Schwarz inequality twice, we obtain

a b c d 4 --=---O--~ + 2 2 2+ 2 2 2+ 2 2 2 > . 2 2 2 a +c +d a +b +d a +b +c b +c +d a + b+c + d

xi + (Xl + X2)2 + ... + (Xk-l + Xk)2 > (Xl - (Xl + X2) + (X2 + X3) - ... + (-I)k-I(Xk_1 + Xk))2

P.K. Hung Proof. Using the AM-GM inequality in the form (x a b2 + c 2 + d2 2': (a2

+ y)2 2': 4xy,

we obtain

k (Xk

+ xk+d 2+ ... + (Xn-l + xn)2 + x; > ((Xk + xk+d - (Xk+l + Xk+2) + ... -

IXk I <-

By Cauchy-Schwarz, 3

One needs rather good gymnastics with Cauchy-Schwarz to deal with the following problem. Xl,

X2, ... ,Xn be real numbers satisfying

xi + x~ + ... + x; + XIX2 + X2X3 + ... + Xn-IXn =

2k(n + 1 - k) . n+ 1

By studying the equality case in the previous inequalities, one immediately sees that this value is attainable for each fixed k. Specifically, if we define

Inserting this in the previous inequality yields the desired result (note that the inequality is strict, since otherwise the equality in the Cauchy-Schwarz inequality would yield a = b = c = d, for which the inequality is strict). Note that even though the inequalities we used were very rough, the constant 4 is optimal: simply take a = band c, d close to O. 0

16. Let n 2': 2 be an integer and let

xn)2

n-k+l Taking into account the hypothesis and these two inequalities, we deduce that 2 2': k(n~i~k) x~ for all k, so that

Adding these inequalities yields

L a 2': (~a:)

and

4a 3

+ b2 + c2 + d2)2Âˇ

For a fixed 1 :::; k :::; n, find the maximum value that IXkl can take. China 1998

= V2k(:~~-k)

and take Xo

Xn+l

= 0 by convention,

then equality occurs

j Xj

interpolates linearly and evenly between these endpoints and if (-I)kXk = akÂˇ The explicit formula is Xj

{

(-I)k-j~

= (-I)j-k (n+l-j)ak (n+l-k)

j:::; k .>k J-

Another ingenious application of the Cauchy-Schwarz inequality can be found in the following problem. 17. Let a, b, c be positive real numbers. Prove that

1 1 1 -+-+-+ a2 b2 c 2

1 7 (1 1 1 1)2 + b + c)2 >- -25 -+-+-+--abc a +b+ c Iran 2010

Chapter 2.

Always Cauchy-Schwarz . ..

Proof. We see that we have equality when a = b = c, so we have to apply the Cauchy-Schwarz inequality in a smart way for the left-hand side of the inequality. Namely, start with

45 Another application of Cauchy-Schwarz thus yields x sin A

+ y sin B + z sin C

::;

V(1 + y2)(x2 + z2 + 2xz cos B + sin2 B).

Thus, it remains to prove that x

+ z2 + 2xz cos B + 1 -

cos 2 B ::; 1 + x 2 + z2

which is equivalent to (xz - cos B? 2':

-+ +->--a b c-a+b+c D

and we are done.

The form of the following problem strongly suggests using CauchySchwarz. However, it is rather easy to check that many attempts fail. .. 18. Let x, y, z be real numbers and let A, B, C be the angles of a triangle. Prove that xsinA + ysinB Proof. Using that C =

x sin A

7f -

+ zsinC::;

J(1

+ x 2)(1 + y2)(1 + z2).

+ ysinB + z sinC = (x + zcos B) sin A + zsinB cos A + ysinB.

Using Cauchy-Schwarz and the fact that sin2 A (x

+ z cos B) sin A + z sin B cos A

::;

+ cos2 A =

V+ (x

= Jx2

z cos B) 2

1, we obtain

+ z2 sin2 B

+ z2 + 2xzcosB.

+ by + cz,

= ay + bz + cx,

+ bx + cy.

Assuming that min(IA - BI, IB CI, IC - AI) 2': 1, find the smallest possible value of (a 2 + b2 + c2)(x 2 + y2 + z2). Adrian Zahariuc, Mathematical Reflections ProofÂˇ Note that by Lagrange's identity and the Cauchy-Schwarz inequality we can write

(a 2 + b2 + c2)(x2 + y2 =

+ z2) (ax + by + cz)2 + (ay -

>A2

B, we obtain the identity

19. Let a, b, c, x, y, z be real numbers and let

Fortunately, this is equivalent to the classical inequality

We now enter the zone of challenging problems, with an unusual one. The first solution seems to come from nowhere (well, it actually came from the author's imagination ... ), but we also present a beautiful geometric proof of Richard Stong, which makes things rather clear.

This reduces the problem to showing that

1 1 1 1 14 (1 1 1 1) +-+-+ 3(a+b+c) >-+-+-+ abc - 15 a b c a+b+c

+ x 2z2 ,

bx)2

+ (bz -

cy)2

+ (cx -

az)2

(ay-bx+bz-cy+cx-az)2 3

=A2+IB-CI2 3

Since everything is symmetric, we obtain 2 (a + b2 + c2)(x 2 + y2 + z2)

2': max (A2

IB - CI

+ IC -

AI2 C 2 + IA B12) 3' 3'

Chapter 2. Always Cauchy-Schwarz . ..

Using the hypothesis, it is immediate to check that the last quantity is at least ~. To see that the answer of the problem is ~, it remains to find a 6-tuple (a, b, c, x, y, z) satisfying the conditions of the problem and for which

A + B + C = 0. Recall that cos a + cos(a + 21r /3) + cos(a + 41r /3) = for any a. If we now take w to be the angle between u and v, then we compute

Taking A

= 1,B = O,C = -1 max (

(this is a triple which minimizes

2 IB - CI B2 3 '

b = 0,

= -3"'

=y=

A2 + B2 + C 2 = IIuI1 2 '1IvI1 2(cos 2 W + cos 2(w + 21r/3) + cos 2(w + 41r/3))

1 211uII2 '1IvI1 2(3

~llul12 ·llvI1 2.

+ cos2w + cos2(w + 21r/3) + cos2(w + 41r/3))

IC - AI2 C 2 IA - B12) 3 ' + 3

under the restrictions of the problem), we must ensure that we have equality in the Cauchy-Schwarz inequality. Solving the corresponding system yields, after some tedious but easy work, suitable values for a, b, e, x, y, z, namely a = -

z = -2.

Proof. Let u be the column vector with entries (a, b, c), v the column vector with entries (x, y, z) and let R be the linear map

The conditions A + B + C = and min(IA - BI, IB - CI, IC - AI) 2 1 imply that A 2 + B2 + c 2 2 2 (without loss of generality, assume that A and B are nonnegative. As IA - BI 2 1, we have max(A, B) 2 1, so A2 + B2 + C 2 = 2A2 + 2B2 + 2AB 2 2 max(A, B)2 2 2) and so

It is easy to see from the proof above that equality is attained. Simply choose any (x,y,z) with x + y + z = and choose (a,b,e) with a + b + e = 0, orthogonal to (x, y, z) so that A = 0, and scaled so that B = 1. For example, taking (x,y,z) = (1, -2, 1) we see that (a,b,e) = (-1,0,1)/3 suffices. 0

We present three short solutions for the following problem. However, none of them is really easy or natural. 20. Let a, b, e, d be real numbers such that Then A = uTv, B = uTRv, C = uT R 2 v and we want to minimize IIul1 2 ·llvI1 2. Note that R acts on ]R3 by fixing the line x = y = z and rotating 1200 in the orthogonal plane x + y + z = 0. Write u = UI + U2, where UI lies on the line x = y = z and U2 lies in the plane x + y + z = and similarly for v. Then

(a 2 + 1)(b2 + 1)(e2 + 1)(d2 + 1)

= 16.

Prove that

-3::; ab + be + cd + da + ae + bd - abed::; 5. Titu Andreescu, Gabriel Dospinescu

Thus the contributions of UI and VI to A, B, C are all the same (hence cancel in IA - BI, etc.). Thus clearly the minimum occurs when UI = VI = 0. In this case, R acts as a 1200 rotation on v, hence V + Rv + R 2 v = and thus

Proof. Write the inequality in the form

(ab+be+ed+da+ae+bd-abed-1)2::; 16.

Chapter 2. Always Cauchy-Schwarz . ..

Next, apply Cauchy-Schwarz in the form

from where the conclusion follows, as the desired inequality can be written as

(a(b + c + d - bed) + be + bd + cd - 1)2 ::; (a 2 + 1) [(b + c + d - bcd)2 + (be + bd + cd - 1)2] . The miracle is that we actually have

Of course, this can be checked by brute force, but perhaps nicer is determining it through two applications of Lagrange's identity. 0 Proof. We will use Lagrange's identity and Cauchy-Schwarz in the form

II(a 2 + 1) =

[(a + b)2 + (1 - ab)2] [(c + d)2

+ (1 -

11 -

L:: ab + abedl ::; 4. o

Straight from the Book, isn't it?

We continue with a really nice, but rather technical problem. It requires some delicate algebraic computations and a rather exotic application of Cauchy-Schwarz. We also present a more conceptual proof, due to Richard Stong, which uses more advanced tools, but proves much more. 21. Let a, b, c, d, e be nonnegative real numbers such that a 2+b2+e2 = d2+e 2 and a 4 + b4 + c4 = d4 + e4 . Prove that a 3 + b3 + e3 ::; d3 + e3 . IMC 2006

cd)2]

2: [(a + b)(c + d) - (1 - ab)(1 - cd)f .

Proof. Since we only deal with even exponents in the hypothesis, let us square the desired inequality and write it in the form

Note that

(a + b) (c + d) - (1 - ab) (1 - cd)

= -1 + ab + be + cd + da + ac + bd - abed. Next, the identity

A moment of thought shows that what we have just proved is exactly the desired inequality. 0 Proof. Here is a rather unnatural, but very elegant proof. Consider the polynomial P(X) = (X - a)(X - b)(X - c)(X - d) and observe that the hypothesis becomes P(i)P( -i) written as iP(iW = 16. On the other hand, we have

P(i) We deduce that

= 16. This can also be

combined with the hypothesis easily yield a

+ b6 + c6 =

3a 2 b2 c2

+ d 6 + e6 .

Thus, we need to prove that

L:: ab + abed + i (L:: a - L:: abc) . With the obvious substitutions, this is equivalent to the inequality

Chapter 2. Always Cauchy-Schwarz . ..

Using Cauchy-Schwarz

(22:: x3

+ 3xyz)2 =

(2:: x(2x2 + yz) f : ; (2:: x2) (~)2x2 + yz)2) ,

it remains to prove that

which follows immediately from x 2y2 equivalent to I:(xy - xz)2 2': O.

+ y2 z2 + z 2x 2 2':

xyz(x

+ y + z),

itself 0

Proof. Consider the polynomial p(t) = t 3 - alt 2 + a2 t - a3 with al and a2 fixed and a3 allowed to vary. Regard the roots x, y, z of p( t) as functions of a3Âˇ Suppose J : ffi. ---t ffi. is any smooth function (at least three times differentiable for the discussion below). Define a function

g(a3)

J(x)

Apply this to x = a2, y = b2, Z = c2, and J(u) = u 3/ 2. This amounts to choosing x, y, z to be the roots of p(t) = t 3 - (d 2 + e2)t 2 + d 2e2t - a3. Since fill (() < 0 for all <:; > 0, we see that g is a decreasing function of a3 for a3 2': O. Thus g(O) 2': g(a3) which, unwinding definitions, is the desired inequality. This argument shows that the same inequality holds for any exponent strictly between 2 and 4 and gives similar (sometimes reversed) inequalities 0 for other exponents. The reader will probably appreciate the beauty of the following inequality. It is more difficult than it appears at first sight, as the obvious application of Cauchy-Schwarz fails rather badly. 22. Prove that for any real numbers Xl, X2, ... ,Xn the following inequality holds

+ J(y) + J(z).

Then g is a differentiable function of a3 and

,a _ J'(x) g( 3)- (x-y)(x-z)

J'(y) + (y-z)(y-x) (z

for some <:; with min(x, y, z) < <:; one merely checks that the curve

J'(z) = ~ J"'(() , x)(z-y) 2

< max(x, y, z). To prove the first equality

IMO 2003

Proof. The first step is to order the xi's, say Xl ::; X2 ::; ... ::; x n . Then n

2:: IXi i,j=l

cs ()

s s s) ( X+ (x-y)(x-z) ,y+ (y-z)(y-x) ,z+ (z-x)(z-y)

preserves aI, has bdu

8 8=0

0, and has

8=0

xjl = 22::(xj - Xi) = 2( -(n - 1)Xl - (n - 3)X2 + ... + (n - 1)xn). i<j

Thus, applying Cauchy-Schwarz yields

1. For the second equality

note that

J'(t)-

J'(x)(t - y)(t - z) J'(y)(t - z)(t - x) J'(z)(t - x)(t - y) (x-y)(x-z) (y-z)(y-x) (z-x)(z-y)

vanishes at t = x, y, z. Hence by two applications of Rolle's Theorem, its second derivative vanishes at some <:; in (min(x, y, z), max(x, y, z)) and this is the required <:;.

It is easy to compute the last expression and the final estimate that we obtain is "

(

~ ~ 00

) 1

< 4n(n -

2 -

2 ~~. i=l

Chapter 2. Always Cauchy-Schwarz . ..

On the other hand, Legendre's identity shows that

i~l (Xi -x;)2 ~ 2n tx1- 2 (tXi)

the last relation being obvious by definition of the ni's. Finally, and most importantly we have 2

Well, unfortunately, when we combine all this we see that we are not done, because of the bad term 2 CL:7=1 Xi)2. Fortunately, it is easy to repair the argument: indeed, we can always add the same number to all xi'swithout changing the hypothesis or the conclusion of the problem. Thus, we may assume that Xl + X2 + ... + Xn = O. But then the previous inequalities allow us to conclude the proof. 0 The following problem is a very tricky application of the Cauchy-Schwarz inequality. The technique used in the proof is worth remembering, since it is quite useful. It is also a standard tool in analysis and probability (it is actually likely that the following problem is inspired by probability theory). 23. Let aI, a2, ... ,an be positive real numbers which add up to 1. Let the number of integers k such that 2 1- i 2: ak > 2- i . Prove that

Putting these inequalities together, we deduce that

Taking N = log2(n), we obtain an even stronger (and strict!) inequality, in which 4 is replaced by 1. 0 We end the series of moderately difficult problems with a very nice-looking improvement of an IMO 2004 problem.

> 2 be an integer. Find the greatest real number k with the following property: if the positive real numbers Xl, X2, ... , Xn satisfy

24. Let n

k> L. Leindler, Miklos Schweitzer Competition

(Xl

+ X2 + ... + xn)

1 ( -Xl

+ -1 + ... + -1 ) , X2

then any three of them are sides of a triangle. Adapted after IMO 2004

Proof. Choose a positive integer N and split the sum in two parts: the one for i ::; N and the one for i > N. Apply the Cauchy-Schwarz inequality for each of them to obtain

Proof. The main point is to solve this problem for n real numbers, let us look at the possible values of

f(x)

and

i>N

= (x + b + c) ( -1 + -1 + -1) X

when X 2: b + c. It is not difficult to check (either directly or by computing the derivative) that f is increasing in this domain of x, so that

On the other hand, we have """' ~ 2i -- 2N

= 3. If b, c are positive

and

"""' ~ n'z_~ < """' n'1, i>N

i2:1

f(x) 2: f(b

+ c) =

1 2(b + c) ( -b

+ -1 + - 1 ) c

b+c

= 2+2

(b+c)2 2: 10. bc

Chapter 2. Always Cauchy-Schwarz . ..

We deduce that if x, b, c satisfy f(x) < 10, then x, b, c are the sides of a triangle (since everything is symmetric in x, b, c). Thus, for n = 3 the answer is 10. To reduce the general problem to the case n = 3, we uSG Cauchy-Schwarz, which allows us to obtain information about Xl, X2, X3 knowing that

We give two proofs for the next difficult problem. The first is based on a very tricky application of Cauchy-Schwarz combined with a mixing-variables argument, while the second one is a pure, but very technical, mixing-variables argument.

(Xl

+ X2 + ... + xn) ( -1 + -1 + ... + -1 ) . Xl X2 Xn

25.

If a, b, c, d, e are real numbers such that a + b +

c + d + e = 0, then

Indeed, using Cauchy-Schwarz and the inequality (X4

+ ... + x n )Âˇ

1 ( -X4

+ ... + -1

)

2:: (n - 3) 2 ,

Vasile Cartoaje Proof. First of all, we may assume that among a, b, c, d, e at least three of the numbers are nonnegative, say a, b, c. This follows immediately from the pigeonhole principle, possibly after changing the sign of all numbers. The key step is the following very tricky application of Cauchy-Schwarz:

we obtain

9(a 4 + b4 + c4+d4 + e 4 ) = [9(a 4 + b4 + c4 ) + 2(d4 + e 4 )] + (7d 4) + (7e 4 )

so that

> -'-2-.:V_2_1--,-( ( 9---,-(a_4_+_b4_+_c4-,-)_+_2-'..(d_4_+_e_4.;. . :))_+_21--.:(_d2_+_e2-,-)~)2 84 + 63 + 63

Using this and the previous case (n = 3), we deduce that if

A small computation yields the equivalent form of this inequality r;7,2 >(Xl+X2+"'+Xn) (1 1) (n-3+v10) -+ 1 +"'+-,

then any three among the numbers Xl, X2,"" Xn are the sides of a triangle. It remains to check that this is indeed optimal. To see this, choose X4 = X5 = ... = Xn = 1,

Then

Xl,

ViO

= X2 = -4-'

ViO

X3--- 2 .

This reduces the problem to showing that

X2, X3 are not the sides of a triangle and

(Xl

+ X2 + ... + xn) ( -1 + -1 + ... + -1) Xl X2 Xn

Thus, the answer is kn = (n - 3 + y'IO)2.

= (n -

3 + vr;7,2 10) .

To exploit the relationship between a, b, c, d, e, we use the fact that D

Chapter 2. Always Cauchy-Schwarz . ..

Thus, it remains to prove that

Proof. We will use a mixing variables argument. Let

for all nonnegative numbers a, b, c. This inequality does not seem to follow easily from well-known results, so we will employ the powerful technique of mixing variables to prove it. Let

We want to prove that for a + b + c + d The basic formula we need is that

f(a, b, c) = 36(a4 + = 15(a4 +

b4 + c4) + (a + b + c)4 - 21(a 2 + b2 + c2)2 b4 + c4) - 42(a 2b2 + b2c2 + c2a 2) + (a + b + c)4.

cb + c) .

f(a,b,c)~f ( a'-2-'-2-

0 we have F( a, b, c, d, e)

d+e -2d+e) F(a, b, c, d, e) - F ( a, b, c, -2-' = (d - e)2(21d2 + 34de + 21e 2 - 7(a 2 + b2 + c2)),

7(a 2 + b2 + c2) ::; 7(a

+ b + c)2 = 7(d + e)2 ::; 17(d + e)2 + 4(d2 + e2) = 21d2 + 34de + 21e 2.

For this we compute

f(a,b,c) - f (a, b;C, b;C) = 15 (b 4 +c4 _ -42a2 (b2+c2_ =

which can be checked by tedious computation. By the pigeonhole principle three of a, b, c, d, e must have the same sign (counting zero as having either sign). By symmetry, we may assume a, b, c ~ O. Then

We will first show that for a = min(a, b, c) we have

(b~C)2)

(b~C)4)

Therefore by the formula above F( a, b, c, d, e) ~ F (a, b, c, d!e, d!e). Thus we may assume three of a, b, c, d, e are nonnegative and the other two are equal. To keep the same basic formula, we invoke symmetry and switch our assumption to c, d, e ~ 0 and a = b = -(c + d + e)/2. Now suppose c::; e. Then we have

C -42 (b 2c2 _ (b;6 )4)

S(b - c)2 [42b 2 + 92bc + 42c2 - 56a2] ~ 0,

where the final inequality follows since 2Sb + 2Sc - 56a 2 ~ O. Thus we reduce to the case where a ::; b = c. In this case we compute 2

15b2a2 + 2ba 3 = 4(b - a)2(b2 + 10ab + 4a 2)

f(a, b, b) = 4(b4 + Sb3 ~

+ 4a4)

The result follows. Note that we have equality for a

= b = c = 2 and d = e = -3.

7 7(a 2 + b2 + c2) = -(c + d + e)2 2

+ 7c2

7 7 ::; 2(d + 2e)2 + 7e 2 = 2d 2 + 14de + 21e 2 ::; 21d2 + 34de + 21e 2. Therefore we again have F(a,b,c,d,e) ~ F (a,b,c, d!e, d!e). We conclude that we can repeatedly average the largest of c, d, e with either of the other two and we will only lower F. By continuity of F, we therefore reduce to the case where a = band c = d = e. In this case it is easy to check that F(a, b, c, d, e) = 0 and we are done. 0

Chapter 2. Always Cauchy-Schwarz . ..

We end this chapter with a challenging inequality, which combines some clever uses of Cauchy-Schwarz with a tricky homogeneity argument. This result also appears in [35] and it is a generalization of a problem discussed in [3], chapter 2, example 11. 26. Prove that for any positive real numbers aI, a2, ... ,an, such that

XI,

and a second application of Cauchy-Schwarz yields

X2,···, Xn

(a2+

n : 1

.~1.+J2.

g:;

Ll<i<j<n XiXj

C2+~l + aJ

(~) ~----------------

XiXj.

lS;i<jS;n

i=l

So, it remains to prove that

(L a2 + .~l. + an) ~ n :

the following inequality holds

(a2

+ .~l. + an) 2

A very elegant approach for this inequality was proposed by Darij Grinberg in [35], where a more general result is proved. We may assume that Vasile Cartoaje, Gabriel Dospinescu

Proof. First of all, it is enough to prove that for any Xl, X2, ... , Xn any aI, 0;2, ... , an > 0 we have ""' ____a__I ____ (X2

~a2+···+an

+ ... + Xn)

~ n

We need to prove that

LI<i<j<nXiXj

Since this is homogeneous, it is enough to prove it when In this case, it is equivalent to

Using T2's lemma (a form of Cauchy-Schwarz), we reduce this to proving that

+ X2 + ... + Xn = 1.

Ll<i<j<n XiXj

G) But Cauchy-Schwarz shows that

> 0 and

Note that

L aiaj i<j

= 1-

a; =

~ L ai(1 -

Thus, if we let bi = ai(1 - ad, it remains to prove that

ai).

Chapter 2. Always Cauchy-Schwarz . ..

This follows directly from Cauchy-Schwarz and the identity

L bib

= (b l

+ b2 + ... + bn )2 -

(bi

+ b~ + ... + b~).

i<j

Remark 2.1. We can also prove the inequality ~

a'a' tJ

i<j

by mixing variables: consider the map

and set x = al !a 2 . We claim that F(al, a2,"" an) 2: F(x, x, a3,'" ,an), A small computation shows that this is equivalent to

Another computation shows that

al a2 2x (al - a2)2 1-al +1-a2 -1-x=2(1-al)(1-a2)(1-x) and

(al - a2)2. 1 - 2x 2 . (1 -x)2(1 - al)(l -a2)' Thus, the inequality F(al, a2,' .. , an) 2: F(x, x, a3,'" ,an) is equivalent to

2L 1 - ai 2: i2:3

But this is easy, since the left-hand side is at least 2 2.:i>3 ai = 2(1 - 2x) and 2(1 - 2x) 2: li'=-~x is equivalent to (1 - 2x)2 2: 0.- Thus, we have F(al' a2,··· ,an) 2: F(x, x, a3,· .. , an). Continuing to mix variables in this way and using the continuity of F implies that F(al' a2, ... ,an) is at least F(m, m, ... , m), where m is the arithmetic mean of the ai's, namely lin. The result follows.

> _n_

~ (1 - a·) (1 - a .) - 2n - 2

2.1. Notes

1- 2x I-x'

2.1

Notes

The following people provided solutions to the problems discussed in this chapter: Vo Quoc Ba Can (problem 20), Ta Minh Hoang (problem 17), Mitchell Lee (problem 6), Dung Tran Nam (problem 25), Dusan Sobot (problems 1, 2, 5, 7), Richard Stong (problems 14, 19, 21, 25), Gjergji Zaimi (problems 3, 4, 10, 13).

2.A. Cauchy-Schwarz in Number Theory

Addendum 2.A

Cauchy-Schwarz in Number Theory

This addendum shows some very beautiful applications of the CauchySchwarz inequality to number theory problems. We present Gallagher's sieve and a be.autiful arithmetic application; we discuss the large sieve, an amazing tool invented by Linnik and extensively developed by a series of brilliant mathematicians; finally we discuss another famous result, the Tunin-Kubilius inequality. The Cauchy-Schwarz inequality plays an important role in the proofs of all these theorems, which are elementary but quite powerful: this ought to show the reader how Cauchy-Schwarz appears in "real mathematics" and not only in olympiad-type problems. Analytic number theory has the reputation of being rather technical and this addendum is no exception. We hope that the results presented here (especially their applications) will compensate for this nonetheless. The following two sections try to give satisfactory answers to the following natural problem: suppose that A is a set of integers such that A

(mod p) = {x

(mod p)lx E A}

Theorem 2.A.1. (Gallagher's larger sieve) Let 8 be a finite nonempty set of integers and let P be a finite set of prime powers. Assume that for each pEP we can find a real number u(p) 2': I{s (mod p) Is E 8} I such that " A(p) ~ -;;-() pEP p where X

> 2 log X,

= maxsES Is I. Then

181 ~

LpEP

A(p) - log 2X

A(p) LpEP u(p) -

log 2X

Proof. Let pEP and let s(r,p) be the number of elements of 8 that are congruent to r modulo p. Then by Cauchy-Schwarz and the fact that u(p) 2': I{s

(mod p)ls E

8}1

we have

thus

is relatively small for all primes p in a finite set P. Are there nontrivial bounds on the size of A? Multiplying this by A(p) and summing over all p yields

2.A.l

Gallagher's sieve and a nice application

Recall that the von Mangoldt function A is defined by A(pn) = logp if p is a prime and n 2': 1 and A(x) = 0 for any other integer x. The crucial property of A is that

LA(d)

181

for all n. The following theorem is a pretty tricky application of the CauchySchwarz inequality, but the application given below reveals its usefulness and power.

L ~~? ~ 181 L A(p) + L L pEP

pEP

S1 :FS2

A(p).

pisl -S2

A(p) ~ log(lsl - S21) ~ log2X,

plsl-s2

logn

din

we deduce that

181 2 L ~((p)) ~ 181 L pEP

A(p)

+ (181 2 -181) log2X,

pEP

from which the result follows immediately.

Chapter 2. Always Cauchy-Schwarz . ..

The promised application (taken from [19]) requires some preliminaries. The following result is very classical, being related to the following natural question: given two positive integers, what is the probability that they are relatively prime?

Proposition 2.A.2. As x -+

00,

2.A. Cauchy-Schwarz in Number Theory

This follows from Euler's celebrated formula L::n2':l computation 1

7& =

~2 and the following

we have 3

L 'P(k) =

+ O(xlnx).

7r 2x2

Finally, as

k~x

Proof. The key point is the equality

'P(k) _ " J-L(d) k -~ d '

we obtain the desired result by combining the previous observations.

dik

Theorem 2.A.3. Let a, b > 1 be integers such that for any prime power p there exists k 2: 1 (depending on p) such that b == a k (mod p). Then b is an integral power of a.

which follows easily from the classical formula

'P(k) = kÂˇ

II (1 - ~) . pik

This relation yields

L'P(k) = "~k. "J-L(d) ~-dk~x

k~x

dik

LJ-L~d) d~x

= L

. Ljd J~J

J-L(d).

[~]

(1: [~])

d~x

x; .L J-L~~) + 0 d~x

Note that L::d~x ~

(L d~x

= O(xlnx). Next, we claim that "J-L(d) _ ~ ~

d2':l

7r 2 Âˇ

~) .

Proof. We may assume that a 2: 3 (as we may work with a 2 and b2 instead of a and b). The most difficult step is to establish that In a and In b are linearly dependent over Q. Let us assume that this is not true and consider a large number x. Let Sx be the set of numbers smaller than x and of the form ai . Oi, with i, j nonnegative integers. As In a and In b are linearly independent over Q, the set Sx has the same number of elements as the number of pairs (i, j) for which i In a + j In b < In x. So, there is an absolute constant c > 0 depending only on a and b such that ISxl > c(1nx)2. We will bound ISxl from above using Gallagher's sieve. For any positive integer y let Py be the set of prime powers dividing at least one of the numbers a-I, a 2 - 1, ... ,aY - 1. For each p E Py , let u(p) be the order of a mod p. Then u(p) ::; y for all p E Py , and, since b is a power of a modulo p, we have ISx (mod p)1 ::; u(p) for all p E P y . To continue, we need a technical lemma. 1 Which

uses the absolute convergence of the double series, as well as the fact that > 1 and 1 for k = 1.

2:dlk J-L(d) equals 0 for all k

Chapter 2. Always Cauchy-Schwarz . ..

Lemma 2.A.4. There exist constants such that for all y 2: 1 we have cly2

Cl, C2

> 0, depending only on a 2: 3,

A(p) ~ c2y2.

pEPy

Proof. One estimate is very easy, since

A(p)

pEPy

A(d)

j=l dlai-l

Lln(a -1) < Y

(y+1) 2

Âˇlna.

j=l

The other estimate is more delicate and crucially uses properties of cyclotomic polynomials2 and proposition 2.A.2. Let t/Jn be the nth cyclotomic polynomial. Remark 9.5 implies that

A(p) 2: L

L u(p)=d

A(p) - LA(p) = lnt/Jd(a) -Ind.

pl<pd(a)

A(p) = L

d~y

Since LIn d = O(y In y), it suffices to use proposition 2.A.2 to conclude.

d~y

The previous lemma shows that by choosing c correctly and y about clnx, we can ensure that

L pEPy

A(p) 2: ~ L A(p) > 2In(2x) u(p) y pEPy

and so by Gallagher's sieve

ISxl

~ In(~x) (L A(p)

-In(2X))

pEPy

2For more details the reader is referred to section 9.2.

c3 lnx

for an absolute constant C3. This contradicts the first paragraph for large x. Hence In a and In b are linearly dependent over Q and so there exists an integer c > 1 and positive integers i, j, relatively prime, such that a = ci and b = d. If p is a prime power divisor of ci - 1, there is kp such that p divides d-ikp - 1 and so p divides d - 1. We deduce that ci - 1 divides d - 1 and so 0 i divides j. The result follows.

Remark 2.A.5. The result does not hold if we consider only primes instead of prime powers. For instance, using properties of quadratic residues (not more than the multiplicativity of Legendre's symbol) one can easily prove that 16 is an 8th power modulo any prime. Of course, 16 is not an eighth power of an integer. In the beautiful papers [4] and [32],3 the following general result is proved fully using techniques of algebraic number theory:

2.A.2

A(p) 2: (L<P(d)) .In(a-1) - LInd.

d~y u(p)=d

Theorem 2.A.6. Let n > 1 be an integer and let a be an integer such that a is an nth power modulo any sufficiently large4 prime. Then either a is the nth power of an integer or 81n and a = 2~ bn for some integer b.

pld

By the very definition of t/Jd we have lnt/Jd(a) 2: <p(d) ÂˇIn(a - 1). Hence

pEPy

2.A. Cauchy-Schwarz in Number Theory

The large sieve

This rather long section presents a very deep result in analytic number theory, known as the large sieve. Introduced by Y. Linnik and extensively developed by Renyi, Bombieri, Davenport, Montgomery, Vaughan, Gallagher (see [8], [9], [14], [25], [37], [44], [46], [57], [55], [56] to cite only a few references), this became a basic tool in modern analytic number theory, with rather spectacular results. We start by presenting a vast generalization of a famous inequality of Hilbert, due to Montgomery and Vaughan, which, combined with some very clever tricks, yields the analytic form of the large sieve inequality. Combined with standard results in finite Fourier analysis (for which the reader is invited to read addendum 7.A) this yields arithmetic forms of the large sieve. This has some amazing applications to the distribution of prime numbers, twin 3 Also

see theorem 9.B.60. the proofs show that it is enough to assume that this holds for a set of primes of Dirichlet density 1. 4 Actually,

Chapter 2.

Always Cauchy-Schwarz . ..

primes, least quadratic residue, prime numbers in arithmetic progressions, etc. We follow rather closely the amazingly well-written article [56].

The analytic form of the large sieve inequality We will spend some time proving the following deep theorem, known as the analytic form of the large sieve inequality. Let Ilxll = minnEz Ix - nl be the distance from the real number x to the discrete set Z.

Theorem 2.A.7. Let X1,X2,'" ,Xn be real numbers such that

for all i =1= j. Let T(x)

Zk'

2i1rkx

2.A. Cauchy-Schwarz in Number Theory

The reader can easily check that A = (aij) E Mn (C) is hermitian if and only if aij = aji for all i, j. A fundamental result of linear algebra is that for any such matrix A we can find real numbers >'1, A2, ... , An and an orthonormal basis VI, V2, . .. , Vn of en (i.e. (Vi, Vj) = 0 if i =1= j and 1 otherwise) such that AVi = Ai' Vi for all i. This can also be stated as: all eigenvalues of a hermitian matrix are real numbers and there exists an orthonormal basis consisting of eigenvectors. 5 The following result will be very useful in the next sections.

Proposition 2.A.9. Let A be a hermitian matrix and let C > O. If the inequality /(Av, v)/ ~ C/v/ 2 holds for any eigenvector. V of A, then it holds for any V E en. Proof. Let VI, V2, .. . , Vn be an orthonormal basis of eigenvectors, with corresponding eigenvalues AI, A2, ... , An. Consider any V E en and write V = 2:::7=1 XiVi for some Xi E C Then

M<kSM+N

be a trigonometric polynomial, where M, N E Nand ZM+1,"" ZM+N EC

(Av, v) = LXiXjAi' (Vi,Vj) = LAi i,j i=1

Then

·IXi/2.

By hypothesis we have /(AVi' Vi)/ ~ C/Vi/ 2 for all i, and since AVi must have /Ail ~ C. Hence Theorem 2.A.7 has a long history and many mathematicians contributed to it: Davenport-Halberstam, Gallagher (who gave a very simple proof of the inequality with 1rN + 1. instead of N + 1.), Montgomery and Vaughan, Selberg. E E l • d One can prove (this is due to Selberg) that the factor N + € can be Improve to N -1 + 1. and that this is sharp. The proof of theorem 2.A.7 will span over E

the next sections.

Ai' Vi, we

I(Av,v)1 ~ CL i=1

IXil2 = Clvl2.

We end this section with a very useful technique. Though elementary, it will playa key role in the proof of theorem 2.A.7.

Proposition 2.A.I0. (Duality principle) Let (aijhSiSm,ISjSn be complex numbers and let C > O. The following are equivalent:

Tools from linear algebra

1) For all Zi

In this section we recall a few standard facts about inequalities concerning matrix norms and we prove a duality principle. Recall that the standard hermitian product on en is defined by (x, y) = 2:::7=1 XiYi, where Xi (respectively 2 Yi) are the coordinates of x (respectively y). If V E en, we denote IvI = (v, v).

Definition 2.A.8. A matrix A = (aij) E Mn(C) is called hermitian if for all x, y E en we have (Ax, y) = (x, Ay).

e we have n

L LaijZi j=1 i=1 Av

CLl zi/ 2 • i=l

5Recall that if A = (aij) E Mn(C) is any matrix and if oX E e and v E en {O} satisfy = oX· v, then we say that v is an eigenvector of A associated to the eigenvalue oX.

Chapter 2.

Always Cauchy-Schwarz . ..

2) For all Yi E C we have m

2.A. Cauchy-Schwarz in Number Theory

Since 2

L LaijYj i=l j=l

'"' (_I)klkl x +k

LLaijZiYj i=l j=l

(the last equality is immediate, since 2~~-=-l~;n -t 0 as n -t 00), we can also write

LaijZi i

= lim '"'

_1f_ sin 1fX

N -+00

L...

Ikl~N

(1 _0]) (_I)k + N

Thus, it is enough to prove that for all N 2:: n we have

::; L IYjI2. L j=l j

L... x2 - k 2

k=l

Ikl~N

::; CLIYiI . i=l

= ~ 2x( _I)kk = o(N)

L...

Proof. Assume for instance that 1) holds. Then for all Zi, Yi E C we have, by Cauchy-Schwarz m

2 ::; CL IZiI . L i

IYjI2.

By choosing for Zi the complex conjugate of Lj aijYj, we obtain the inequality in 2). The converse is proved in exactly the same way. 0

(_I)k

Ikl)

'"' ZiZj L... '"' ( 1 - -N L... i#j

Ikl~N

+ X· 2

X· J

1f n 2 ::; -E '"' L... IZil . i=l

Now, since there are N -Ikl solutions of the equation jl - j2 = k with jl, j2 E

{I, 2, ... , N}, it is not difficult to see that the inequality is equivalent to Montgomery and Vaughan's theorem The key technical ingredient in the proof of theorem 2.A.7 is the following delicate result [57]. Theorem 2.A.l1. (Montgomery and Vaughan) Let Xl,X2"",Xn be real numbers such that IIXi - Xjll 2:: E > 0 for all i::l j. Then for all Zl, Z2, ... , Zn E C

This follows from the following vast generalization of a famous inequality due to Hilbert, applied to the family of real numbers (Xi + j)l~i~n,l~j~N and to the family of complex numbers (( -I)j zih~i~n,l~j~N' Theorem 2.A.12. (Montgomery- Vaughan) Let E > 0 and let Xl, X2, ... , Xn E ~ be such that IXi - Xj I > E for all i ::I j. Then for any complex numbers Zl,Z2"",Zn

The proof of theorem 2.A.ll is a very nice mixture of elementary, analytic and algebraic arguments. A first crucial ingredient is Euler's famous identity (that we will take for granted) 1f

-- =

sin 1fX

(_I)k

N-+oo

'"' - - . L... x + k

Ikl~N

'"' Zi' Zj

L... x ·- x · #j 2 J

Proof. By homogeneity and symmetry we may assume that E = 1 and that Xl < X2 < ... < Xn · Then the hypothesis implies that Xj -Xi 2:: j - i for i < j. Consider the matrix A, where aij = Ii#)' Xi~Xj' Note that A is antisymmetric,

thus i· A is a hermitian matrix, to which proposition 2.A.9 can be applied with

Chapter 2. Always Cauchy-Schwarz ...

C = 7r. Thus, we may assume that Z = (Zl, ... , zn) is an eigenvector of iA, so also of A, say Az = i . AZ for some A E lR. By Cauchy-Schwarz, the square of the left-hand side of the desired inequality is bounded by inequality

2.A. Cauchy-Schwarz in Number Theory

Now, using the AM-GM inequality 12zjZkl ::; terms, we obtain

LaijZj

i #i

IZjI2. L

LHS ::; 3 L

L:i IZiI2. L:i 1L:#i a ijzjl2, so it is enough to prove the

Ii - jl, we have for

IZjl2 + IZkl2

and rearranging

aTj'

As IXi - Xjl ::::

any fixed j that 6

::;71"2Ll ziI 2. i=l

"a 2. < " (i -1 ~ tJ - ~ "...j." trJ

"...j." trJ

71"2

< 2" ~ -2 3'

')2 -

>1 n_

Expanding brutally the left-hand side and using the crucial observation that Combining the last two inequalities yields the desired result.

aijaik = ajk(aij - aik) yields

Proof of theorem 2.A.7

LHS = L ZjZk' L aijaik j,k ii-j,k = L j

IZjl2 . L

aTj

ZjZkajk

It is now time to use the fact that

(L.:

aij - L

tOPJ

aik

+ 2ajk )

topk

Let akj = e2i1rkxj for 1 ::; j ::; nand M < k ::; M that for all (Zk)M<k"'5.M+N we have

+ N.

We need to prove

is an eigenvector: and by the duality principle (proposition 2.A.l0) it is enough to prove that for all Y1, ... ,Yn we have n

L LakjYj M<k"'5.M+N j=l

Doing the same with the other sum, we finally obtain that the term

Expanding the left-hand side, we obtain the equivalent inequality

L YjlYj2 . L e2i1rk(xil-xh) ::; L jdj2 M<k"'5.M+N _C j 6We use here Euler's famous identity

cancels with the other similar term, so that LHS = L j

IZjI2. LaTj + L2zj Zka ]k' ih

IYjI2.

Chapter 2. Always Cauchy-Schwarz . ..

Using the formula for the sum of a geometric series, a small computation shows that for all u

""'

L..t

M<k"5.M+N

e2i7rku =

(e i7rU (2M+2N+l) _ ei7r~(2M+1))

., 1

2zsm(7ru)

By the triangle inequality, it is thus enough to prove that for any s E {2M

+ 2N + 1, 2M + I}

and for any Yl, ... ,Yn we have

2.A. Cauchy-Schwarz in Number Theory

Note that the only difference between this theorem and theorem 2.A.7 is the factor 7rN, which is N in theorem 2.A.7.

Proof. Assume that f is a continuously differentiable complex-valued map on R Integrating by parts, it is easy to establish the equality E' f(Xj) =

Xj+~ x-S f(t)dt

l +

Xj+~

lXj ( E) x_s t - Xj +"2 f'(t)dt J

(t - Xj -"2E) !,(t)dt.

Using the triangle inequality and the fact that 2 - 2 I t-xo+~I<~ J

But this follows from theorem 2.A.1l with Zj = Yj . ei7rsoxj.

for t E [Xj - ~,Xj] (and a similar inequality for t E [Xj,Xj

+ ~]),

we obtain

A quick proof of a weaker form of the sieve inequality In this section we present Gallagher's short and beautiful proof of the following weaker form of the large sieve inequality. It is much simpler than the proof presented in the previous sections and the result it establishes is good enough in most applications of the large sieve.

Take f(x) = T(x)2 and add the corresponding inequalities. The hypothesis "Xi - xjll 2: E ensures that the intervals (Xj - ~,Xj +~) do not overlap mod 1, so using the I-periodicity of f we obtain

Theorem 2.A.13. Let Xl,X2,'" ,Xn be real numbers such that

Using Parseval's equality and the Cauchy-Schwarz inequality, we obtain

for all i =1= j. Let

2i7rkx Zk' e

T(x) = M<k"".5.M+N

be a trigonometric polynomial, where M, N E Nand ZM+l,"" ZM+N E C. Then

M+N

k=M+l

M+N IZ kI 2 .

k=M+l

127rkzk12

Chapter 2.

Always Cauchy-Schwarz ...

If.

So we are done if we can ensure that maxM<k:SM+N Ikl :::; Of course, for arbitrary M and N it is unreasonable to hope for such an inequality, but a moment of thought shows that M plays no role in the theorem we are trying to prove, so we can simply choose M = - [Ntl] to finish the proof. 0

2.A. Cauchy-Schwarz in Number Theory

Now, if the an were uniformly distributed, one would expect that Lk=h (mod q) ak behaves as Eq = ~ Lk ak· Actually, a small computation reveals that 2

L L ak-Eq h=l k=h (mod q)

Arithmetic forms of the large sieve inequality We will apply theorem 2.A.7 for a special family of well-spaced numbers pick an integer Q > 1 and consider all numbers of the form ~, with gcd(a, q) = 1 and 1 :::; a :::; q :::; Q. It is clear that the difference between Applying the large sieve two such numbers is (in absolute value) at least inequality to this collection, we deduce the following

=L L h=l k=h

(mod

ak q)

which combined with the previous formula yields

Xi:

Unfortunately,

Theorem 2.A.14. Let T(x) =

IT (~)12 : :;

q=l

l:Sa:Sq

gcd(a,q)=l

L~:~ IT (~) 12 is usually larger than

l<a<q,~(a,qH IT (~) I'

L ake2i1rkx M<k:SM+N

be a trigonometric polynomial. Then for all Q

ak -Eq k=h(mod q)

> 1 we have

(N +Q2)

2 L lakl . M<k:SM+N

but if q is a prime number, they are actually equal! Using these remarks and the previous theorem, we obtain the following strong inequality:

Theorem 2.A.15. Let (ak)M<k:SM+N be a sequence of complex numbers. Then for all integers Q > 1 we have 1 LP· L L ak - - Lak p:SQ h=l k=h (mod p) p k p

Next, let us observe that

L... ake (-a)q ""' k =

2i7rka q

L... h=l

(

""'

L... ak ) e 2i7rah q •

:::; (N +Q2) L/akI 2, k

the sum being taken over all primes p :::; Q.

By specializing ak = 1kEA for a subset A C [M the following equidistribution result:

k=h (mod q)

Therefore, using the techniques of addendum 7.A, more precisely Plancherel's formula (theorem 7.A.5, but this can also be done by expanding everything), we obtain

Corollary 2.A.16. Let A all integers Q > 1 we have

+ 1, M + N]

+ 1, M + N],

be a set of integers. Then for

L k=h (mod

ak q)

p L LP· p:SQ h=l

1 /2 :::; IA n (h + pZ)I- -IAI

we obtain

(N + Q2)IAI.

Chapter 2.

Always Cauchy-Schwarz . ..

and

This finally yields a "sieve inequality."

Theorem 2.A.17. Let A c [M + 1, M + N] be a set of integers and suppose that for each prime p ::::; Q at least w(p) residue classes mod p contain no element of A. Then N+Q2 IAI ::::; w(p)' Lp::;Qp Proof. Use the previous corollary and the observation that for each p we have 1 /2 L A n (h + p71) - -IAI p

2.A. Cauchy-Schwarz in Number Theory

IAI2 2: w(P)-2 '

as at least w(p) terms in the sum are equal to ~. p

The previous theorem helps understanding the name "large sieve." Indeed, in typical applications A (mod p) will miss a good part of the residue classes modulo p (i.e. the integers w(p) will be quite large) and the sieve inequality will yield nontrivial bounds on IAI. The next result, an amazing theorem of Linnik, is at the very origin of the large sieve and one of its best applications. It concerns the distribution of the least quadratic non residue modulo p. Vinogradov conjectured that it is smaller than c( c )pE: for any c > and if one assumes the Generalized Riemann Hypothesis, then one can actually prove that it is smaller than c(log p? The following deep theorem due to Linnik [47] is a first step in the direction of Vinogradov's conjecture (which is still largely open):

Â°

Theorem 2.A.18. (Linnik) Let n(p) be the least positive integer which is not a quadratic residue mod p. Then for all c > there is a constant c(c) such that for all N we have

Â°

= {n

::::; NI

(~)

= 1 for all

p E PN } .

We will prove the following technical result:

Lemma 2.A.19. There exist positive constants c(c) and Cl(c) such that for all N > Cl(c) we have IANI > c(c)N. Let us assume for a moment that this holds. Since AN (mod p) misses all quadratic non-residues mod p, we can choose w(p) = 2: ~ in the previous theorem and obtain IANI ::::; I~~I' Hence for all N > Cl(c) we have IPNI < ~, which is enough to conclude the theorem. Let us prove the lemma. Note that AN contains all numbers n ::::; N all of whose prime factors are smaller than NE:, as Legendre's symbol is multiplicative. We will consider all numbers n = kPl'" Pd smaller than N,

zs-!

,,2

with Pi E [NE:-T, NE:] in nondecreasing order and some integer k. Note that ~

k ::::; N"2E:, as n ::::; N, so k is relatively prime to any prime greater than NE:-T. This easily implies that all such numbers are different and they clearly belong to AN. Since there are more than PIP2"'Pd N - 1 possible values for k, it follows that we have at least

such numbers. Taking into account the fact 7 that 7r(x) = o(x) and "

110gb

~ - = log-l-+0(1) p oga

pE[a,b]

Proof. Fix c > 0, which for simplicity (but without loss of generality) we take of the form ~ for some integer d greater than 2. For a positive integer N, let

as a, b -7 00, the last quantity is easily seen to be greater than c(c)N for some positive constant c( c) and all sufficiently large (depending on c) N. The desired result follows. 0 7Both these results are proved in addendum 3.A.

Chapter 2. Always Cauchy-Schwarz ...

Though theorem 2.A.17 is already a very strong result, in most applications one needs more refined estimates, in which one takes into account all q :S Q, not only prime numbers. In order to do this, we need another nice application of finite Fourier analysis and Cauchy-Schwarz, but before doing that recall that f.t is Mobius's function, defined by f.t( n) = (_l)k if n is a product of k ~ 0 distinct prime numbers and f.t( n) = 0 otherwise. Theorem 2.A.20. Let A c [M + 1, M that for each prime p :S Q we have

I{a

+ N]

2.A. Cauchy-Schwarz in Number Theory fact that the theorem holds for q (with T ( f, for q', we have

be a set of integers and suppose

instead of T (x)) and then

a b)12 L IT ( -+q q'

b) 12 I ( -q'

g(q) T

~ g(q)g(q')IT(0)1 2 = g(qq')IT(0)12.

(mod p)la E A}I :S p - w(p). =

+ x)

bE(Zjq'Z)* aE(ZjqZ)*

bE(Zjq'Z)*

Then for all trigonometric polynomials T(x) integers q

I:kEA ake2i7rkx and all positive

o Combining the special case T(x) = I:kEA e2i7rkx of the previous theorem with the large sieve inequality (theorem 2.A.14), we obtain the following strong form of theorem 2.A.17:

Proof. Let g(q) = f.t(q)

II p _w(p) w(p) ,

Theorem 2.A.21. (Montgomery) Let A c [M + 1, M and suppose that for each prime p :S Q we have

+ N]

be a set of integers

plq

a multiplicative function. We will prove the theorem in two steps: first we will prove it when q is prime and then we will show that if the theorem holds for two relatively prime numbers q, q', then it also holds for qq'. First, assume that q is prime. Let Xh

kEA,k-=h

I{a Then

IAI :S

(mod p)la E A}I :S p - w(p).

NtQ2 , where

akÂˇ (mod q)

Then at least w(q) of the numbers Xh are zero (by hypothesis), so by CauchySchwarz and Plancherel's identity we get

Some applications We present here two classical applications of the large sieve: a uniform upper bound on 11"( m + n) - 11"( n) and an upper bound for the number of twin primes smaller than n.

from where the result follows easily. Next, assume that the result holds for q and q', with gcd(q,q') = 1. Using the Chinese Remainder Theorem and the

Theorem 2.A.22. If m ~

m+ 1 and m+n.

viii,

then there are at most l~n primes between

Chapter 2. Always Cauchy-Schwarz ...

Proof. Let A be the set of prime numbers between m + 1 and m + n. No element of A is divisible by some prime smaller than or equal to [Vn], so by

2.A. Cauchy-Schwarz in Number Theory

where d( k) is the number of divisors of k. Combining the previous estimates, we obtain

theorem 2.A.21 we obtain

cln

IAI:::; n~n,

L = ' " J-L(q)2 ~

q5.vn

Using the fact that L ~

P~l

= Lj2: 1

'" pi1 ... rJ.s 5.vn

f(n) - f([Vn]) :::; (10gn)2

II _1_. p-l plq

for some constant the result follows.

(~r, we obtain

1 1 . = ' " -: > log vin = - log n. v,l"'m ~ J 2 .

j5.vn

Theorem 2.A.23. There exists an absolute constant c such that for all n there are at most~) primes p < n such that p + 2 is also prime. (logn -

Proof. Let f(n) be the number of twin primes smaller than n and let A be the set of twin primes in ([Vn], n]. If p E (2, [Vn]] is a prime number, then clearly 0, -2 rt. {a (mod p)la E A}, so we can take w(p) = 2 for such primes in theorem 2.A.21 (and w(2) = 0), deducing that

IAI :::; L'

p:2 = Lj2:1 (~r, we obtain (h

k5.vn gcd(2,k)=1

(

L k5. ifii gcd(2,k)=1

> O. Using the trivial bound f([Vn]) :::; vin = 0 Co~n)' 0

Theorem 2.A.24. (V.Brun) The sum of the inverses of all twin prime numbers converges.

Proof. By the previous theorem, there are at most twin primes between 2j and 2j +1 , for some absolute constant c. The sum of the inverses of these twin primes is smaller than 2~ . ~ = :&. As LJ'>l A converges, the result J J - J follows. 0 These are far from being the most spectacular applications of the large sieve (though they are already deep theorems!). We refer the reader to [9], [19], [37], [44J, for many other applications.

L =

It remains to obtain a lower bound on L. Since

d(k) > k

Using the previous theorem, we can easily prove the following famous result, which is probably the origin of all modern sieve theories:

Inserting this result in the previous inequality yields the desired result.

f(n) - f([vin]) =

+ 1)··· (is + 1)

2: c(logn)2,

2.A.3

The Turan-Kubilius inequality

In this section, we use again the Cauchy-Schwarz inequality to prove a famous inequality in analytic number theory, the Tunin-Kubilius inequality. We then apply it to establish a well-known result of Hardy and Ramanujan (which was actually the origin of this inequality) and then a very amusing consequence due to Erdos. We also present a dual form of the Tunin-Kubilius inequality, due to Elliott, which has a similar form to the inequality established in theorem 2.A.15 and which turned out to have some pretty far-reaching consequences (for instance, a proof of the prime number theorem!).

Chapter 2. Always Cauchy-Schwarz . ..

2.A. Cauchy-Schwarz in Number Theory

A function f : {I, 2, ... } -+ C is called additive if f(ab) = f(a) + f(b) for all relatively prime positive integers a, b. It is called strongly additive if moreover f(pk) = f(p) for all primes p and all positive integers k. Thus

1) For all strongly additive maps f : {I, 2, ... } -+ C and all n 1 n ;, L If(k) - Er(n)1 2 :::; CÂˇ Bja(n). k=l

f(n) = L f(pvp(n)) , respectively f(n) = L f(p) pin pin

2) For all additive maps f : {I, 2, ... } -+ C and all n 1 n ;, L If(k) - Ef(n)12 :::; CÂˇ Bf(n). k=l

when f is additive, respectively strongly additive. In particular, if f is additive we have

Proof We will prove only 1), as 2) is proved in exactly the same way. Suppose

while if f is strongly additive, then

first that f takes nonnegative values. Denote E observe that

= Eja(n),

vBja(n) and

If(k) - EI2

k=l

L f(k)2 - 2E L f(k) k:5.n k:5.n

is strongly additive, we have

Thus

[~]

L f(k)2 = L L f(p)f(q) = L k:5.n k:5.n p,qlk p:5.n p

is a good approximation of the average value of f(l), ... , f(n) if f is additive, respectively strongly additive. Our fundamental result will allow us to see

f(p)2

[~]

f(p) :::; -2nE 2 + 2E L f(p). p:5.n

All in all, we obtain

Theorem 2.A.25. (Turrin-Kubilius inequality) There exists an absolute constant C > 0 with the following properly:

[!!.-]

L f(p)f(q) pq:5.n pq p#-q

and the last quantity does not exceed nB2 + nE2, as f takes nonnegative values and [xl:::; x for all x. On the other hand, since [xl> x-I for all x we can write ' -2E L f(k) = -2E L k:5.n p:5.n

as an upper bound for the variance of f as a random variable on {I, 2, ... , n} (with the uniform distribution). Without further ado, we follow [24], which contains a very easy proof of this theorem.

+ nE2.

S:::; nB2

+ 2EL f(p). p:5.n

Next, two applications of Cauchy-Schwarz show that

Chapter 2. Always Cauchy-Schwarz. ..

Using the classical estimates8

Proof. Use the Tunin-Kubilius inequality with f Note that if f = w, then

L P-1 = O(1og log n),

2.A. Cauchy-Schwarz in Number Theory

= w,

respectively f

p:::;n

we deduce that lOglOgn)) logn and the result follows. Suppose now that tions gl, g2 by

is real-valued and define two strongly additive func-

as follows from Mertens' theorem 3.A.5. Similar arguments apply for f = O.

The following theorem is an immediate consequence of the previous result. Its original proof was much more intricate (see [39]).

Theorem 2.A.27. (Hardy-Ramanujan) The functions wand 0 have normal order log log n, i. e. if f E {w, O} , then for all c > 0 we have

Using Cauchy-Schwarz, we get (with E = Ej(n), Ej = Egj(n))

L If(k) k:::;n

E12::; 2

LL Igj(k) -

lim {n ::; xll x-too x

Ejl2

c< 1ogf(n)log x < + c} I 1

= l.

Proof. Let f E {w,O} and let

j=l k=l

and the result follows easily by applying the result of the previous paragraph to gl, g2Âˇ Finally, if f takes on complex values, set gl = Re(f) and g2 = Im(f) and use the same argument. 0

Since for any n E Ax we have

(f(n) -loglogx)2 2: c 2 . (loglogx)2,

Corollary 2.A.26. (Turan) If w(n), O(n) is the number of prime factors of n without (respectively with) multiplicity, then we deduce that

L(w(k) -loglogn)2 = O(nloglogn) k:::;n

c 2 . (loglogx)2 'IAxl::; L

(f(n) -loglogx)2 ::; L(f(n) -loglogx)2.

nEAx

and L(O(k) -loglogn)2

= O(n log logn).

k:::;n

8Which follow easily from the results of the addendum 3.A.

Using the previous corollary, the result follows.

n:::;x

Here is a very beautiful consequence of the Hardy-Ramanujan theorem, which is surprisingly difficult to prove by other means:

Chapter 2. Always Cauchy-Schwarz ...

Theorem 2.A.2B. (Erdos) We have

2.A. Cauchy-Schwarz in Number Theory

2) For all x and all complex numbers an 2

lim l{aÂˇbI1::;a,b::;n}1 =0. n-+oo n2

Proof. By the previous theorem, there are n 2 + o(n 2) pairs (a, b) with 1 < a,b::; n and such that O(ab) is about 2loglogn. The same theorem shows that n 2 + o(n 2) numbers k E {I, 2, ... , n 2} have O(k) about 10glogn2

= log log n + log 2 ÂŤ2loglogn,

so the number of numbers of the form ab (with 1 ::; a, b ::; n) must be o(n 2).

Before looking at the proof, the reader is strongly advised to compare this theorem and theorem 2.A.15.

Proof. We will prove only the first part, as the same argument yields the second part. Observe that Tunin-Kubilius's inequality can also be written in the form

The following difficult theorem (see [28]) refines considerably HardyRamanujan's result. It is also known as the fundamental theorem of probabilistic number theory.

Theorem 2.A.29. (Erdos-Kac) Let f : {I, 2, ... } -+ ffi. be a strongly additive function such that the sequence (f (p))p is bounded (p runs over the prime numbers) and such that lim n -+ oo Bf(n) = 00. Then for all real numbers a 1 lim -I{n::; xlf(n) - Ef(x) x

x-+oo

< a~ Bf(x)}1

1ja

=!7C

27r

e- U 2 /2du.

-00

Let us end this section and this addendum with another beautiful application of the Tunin-Kubilius inequality, due to Elliott. We let vp(n) denote the exponent of the prime number p in the factorization of n.

This suggests defining for j, k E [1, n] the quantity

if j = pU for some u and some prime p and ajk = 0 otherwise. Given any sequence Yl, ... ,Yn of complex numbers, we can certainly define an additive map f such that f(pU) = #Ypu for all U,p such that pU ::; n. Then the previous inequality can be written n

Theorem 2.A.30. There exists an absolute constant C > 0 such that:

LakjYj

k=l j=l

1) For all x and all an E C 2

Applying the duality principle 2.A.1O and unwinding definitions, we obtain the desired inequality. 0 The previous theorem plays a key role in Hildebrand's "elementary" proof (see [40]) of an extremely difficult theorem of Wirsing [85], [86], that confirmed a long-standing conjecture of Erdos and Wintner.

Chapter 2. Always Cauchy-Schwarz . ..

Theorem 2.A.31. (Wirsing) Let f : {I, 2, ... } -+ [-1,1] be a multiplicative function (i.e. f(ab) = f(a)f(b) whenever gcd(a, b) = 1). Then

M(J) = lim

.!. L

x-too X

exists. Moreover, M(J) =

f(n)

n::;'x

Chapter 3

1 - f(p)

00.

To see how subtle this theorem is, take Mobius' function for f: it is fairly elementary that L:p 1 = 00, so the previous theorem yields L:n::;,x J.l(n) = o(x). But it is elementar; to prove that this is equivalent to the prime number theorem! Even though much easier than Wirsing's original proof, Hildebrand's proof is still rather technical and we refer the reader to his article [40] for details.

Look at the Exponent 3.1

Introduction

If p is a prime, we define the p-adic valuation map vp : Z -+ N U {oo} by: vp(o) = 00 and for n "# we have vp(n) = k if pk divides n, but pHI does not divide n. More concretely, for n "# 0, vp(n) is the exponent of the prime number p in the prime factorization of n. The unique factorization of integers into powers of prime numbers easily yields

with equality if vp(a) "# vp(b). The first property allows us to extend this map vp to Ql by vp (~) = vp(a) - vp(b) for all nonzero integers a, b. It is an easy exercise to check that this is well-defined (i.e. if ~ = then we get the same value for vp (~) and Vp (a)).

We will frequently use the following easy properties of the p-adic valuation:

Vp(gcd(Xl' X2,· .. , xn)) = min Vp(Xi), l::;'t::;'n

vp(lcm(xl,x2"" ,xn )) = max Vp(Xi). l::;'t::;'n

Chapter 3. Look at the Exponent

3.2

Local-global principle

3.2. Local-global principle

Remark 3.1. Another natural proof uses the identity

A very useful idea when dealing with divisibilities (or even equalities concerning arithmetic objects) is the local-global principle: if a and b are nonzero integers, then a divides b if and only if for all primes p we have vp(a) ::::: vp(b). This "local-global principle" is the simplest of a series of such statements, concerning the way in which the arithmetic of integers is governed by the "behavior at each prime." Some of these statements are very deep and most of them are an active area of research. Here are some other such examples: a positive integer is a perfect square if and only if it is a perfect square mod p for all primes p (this is already a quite serious result, using the quadratic reciprocity law). Another famous example is Hasse-Minkowski's local-global pdnciple: if ai are nonzero rational numbers, the equation alxI + a2x~ + ... + anx; = 0 has nontrivial rational solutions if and only if it has nontrivial real solutions and nontrivial p-adic solutions for all primes p. This is a deep theorem and we refer the reader to the addendum concerning p-adic numbers for more details. We start with some easy applications of the local-global principle.

lcm(x, y) = _-,x_y_ gcd(x,y) and its analogue

_ 1cm (x,y,z ) -

xyz gcd(x, y, z) gcd(x,y)gcd(y,z)gcd(z,x)·

-~-7-~+~~!...,-;----:

Plugging in these values immediately yields the result. 2. Let all a2, ... ,ak and bl, b2, ... ,bk be positive integers such that ai and bi are relatively prime for all i E {I, 2, ... , k}. Prove that

1. Prove the identity

lcm(a, b, c)2 lcm(a, b) . lcm(b, c) . lcm(c, a)

IMO 1974 Shortlist

gcd(a, b, c)2 gcd(a, b) . gcd(b, c) . gcd(c, a)

Proof. Fix a prime p and let Xi = vp(ai) and Yi = vp(bi ). By hypothesis, we have min(xi' Yi) = 0 for all i and we need to prove that if

for all positive integers a, b, c. USAMO 1972

Proof. Fix any prime p and let x = vp(a), y = vp(b) and z = vp(c). The p-adic valuation of the left-hand side is

then

2max(x,y,z) - max(x,y) - max(y,z) - max(z,x), while that pf the right-hand side is 2 min(x, y, z) - min(x, y) - min(y, z) - min(z, x). We claim that these two quantities are equal. Since the tWQ expressions are symmetric in x, y, z, we may very well assume that x ~ y ~ z. Then we need to prove that 2x - x - y - x = 2z y - z - z, which is clear. 0

min(xl - Yl

+ Z, X2 -

+ z, ... , Xk -

+ z) =

min(xl, X2, ... , Xk).

Note that Xi - Yi + Z ~ Xi for all i, so certainly the left-hand side of the previous inequality is at least the right-hand side. On the other hand, if z = 0, then this forces all Yi to vanish and in this case the equality is clear. So, we may assume that there is some j such that Yj = z > O. Then Xj = 0 and X· - y' + z = 0 · t he equa1·lty clear again. J J rnak mg 0,

Chapter 3. Look at the Exponent

The next problem is quite challenging and requires some preliminaries. A very common situation in analytic number theory is the following one: we have two maps f and g and we assume that g(n) = L:dln f(d) for all n. Mobius found a very nice inversion formula, which expresses f in terms of g. Define the Mobius function J.L by J.L(1) = 1, J.L(n) = 0 if n is not squarefree and J.L(PIP2路路路 Pk) = (_l)k if PI,P2, ... ,Pk are distinct prime numbers. Then the Mobius' inversion formula reads f(n) = L:dln J.L (:;1) g(d). The main ingredient in the proof is the fact that L:dln J.L( d) = 0 for all n > 1 and it equals 1 for n = 1. This is immediate, since if PI,P2, ... ,Pk are the different primes dividing n, then for n > 1 k

L J.L(d) = 1 - L J.L(Pi) i=l din = 1-

J.L(PiPj)

+ ...

3.2. Local-global principle

for all positive integers m, n. Prove that there exists a unique sequence of positive integers (bn)n?l such that an = I1 bd for all n. din Marcel Tena, Romanian TST Proof. In the light of the previous discussion, we are forced to define

bn =

din Of course, the hard point is to prove that bn is an integer. In order to prove this, we will first transform the expression defining bn , using the Mersenne property I of the sequence (an)n. Namely, let n = p~lp~2 ... p~k, then

I1 a--1:L

i<j

b an PiPj n - I1alL . I1a_n _

(~) + (~) - ...

gcd(alL)iEI = a_n_

by the binomial formula. Using this observation, we can write

(~) g(d) =

LJ.L(d)g din

(~)

= L J.L(d)路 L din

f(dI)

dIdln

dl <fi

diin

DiEI Pi

for all subsets I of {I, 2, ... , k}. Therefore we can write

= L f(d l ) L J.L(d) = f(n). The next problem requires the multiplicative version of Mobius's formula, namely if g(n) = I1dlnf(d), then f(n) = I1dlng(d)J.L(%). The proof being exactly the same as above, we leave it to the reader to fill in the details. 3. Let (an)n>l be a sequence of positive integers such that

gcd(a m , an)

PiPjPk

The key remark is that the Mersenne property yields the equality

= (1- l)k = 0,

LJ.L din

II a~(d).

= agcd(m,n)

b n-

I1gcd(alL,alL)

I1 alL . I1 gcd( alL , a~ , a2!.. ) Pi

Finally, the inclusion-exclusion principle coupled with the local-global principle easily yields the formula

I1 Xi . I1 gcd(xi' Xj, Xk) .... _ ( -lcmxI,x2, ... ,Xn) I1i<j gcd( Xi, Xj ) I1 gcd( Xi, Xj, Xk, Xl ) for all nonzero integers

Xl,

X2, ... ,Xn . Using this observation, we deduce that

[ an lcm a~,a~, ... ,a2!..]' PI

ISee chapter 10, the introduction to problem 10.

Chapter 3. Look at the Exponent

which is clearly an integer (by the Mersenne property, if m divides n then am divides an). The result follows. 0

Proof. Let p be a prime. Note that if a sequence (an)n2:1 has the hypothesized property, then so does a'm = pvp(a m ). Further, if the desired conclusion ho~ds for all such a' , then it clearly holds for am. Thus we may assume am = p m and we see that the sequence (U n )n2:1 consists of nonnegative integers and satisfies min( Um, un) = Ugcd(m,n)· Let 0 ::; rl < r2 < ... be the values actually taken on by the sequence (U n )n2:1. For any two elements m and m' of the set Sk =:= {m: Um = rk} we have U d( ') = min(u m, um') = rk and hence gcd(m, m) E Sk· Therefore Sk gc m,m 1. 1 f has a unique minimal element mk and all elements of Sk are mu tIP es 0 mk· Note that if j > k, then min(umj' u mk ) = min(rj, rk) = rk, so gcd(mj, ~k) = mk or mklmj. Thus in the sequence 1 = ml,m2,m3,··· each term dIvIdes the later terms. Now suppose mklm and mk+l t m. Then gcd(mk' m) mk so min(u mk , um) = rk or Um 2': rk. However gcd(mk+l' m) < mk+1 so < rk+l and hence Um < rk+l· Since the ri are all the values m in(u mk+l' u) m taken on by (un), we conclude that U m = rk· . To recap, the sequences rl < r2 < ... and ml, m2,··· determme Um uniquely. If k is the largest index for which mklm, then Uk = rk· Define b = prk-rk- 1 for k > - 2 , and bm = 1 if m is not one of the mk· b1 -- prl ,mk Then we immediately see that if k is the largest index for which mk 1m, then k

II bd II bmi =

dim

prl

i=1

IIpri-r

i- 1

prk

pUm

am·

i=2

3.3. Legendre's formula

such that vp(x) = j. The second equality is a direct computation left to the reader. The purpose of this section is to present some nice applications of this formula. We present two solutions for the next problem: one uses the local-global principle combined with Legendre's formula, while the second one is more exotic, but very powerful. 4. Show that if n is a positive integer and a and b are integers, then 1

,a(a + b)(a + 2b) ... (a n.

+ (n -

1)b)bn - 1 E Z. IMO 1985 Shortlist

Proof. We will prove that for any prime p ::; n we have vp(n!) ::; vp(a(a + b)··· (a

+ (n -

1)b)bn- 1 ),

which is enough to solve the problem. If p divides b, things are rather clear, as v p(n!) ::; n -1 while v p(b n;-l) 2': n -1. So assume that p does not divide b. But then there are at least [nip] multiples of p among a, a+b, ... ,a+ (n-1)b (since among a, a + b, . .. ,a + (p - 1)b there is at least one multiple of p, the same for a + pb, ... , a + (2p - 1)b, ... ), at least [nlp2] multiples of p2 and so on. So, the p-adic valuation of a(a + b) ... (a + (n -1)b) is at least [nip] + [nlp2] + ... and this is exactly the p-adic valuation of n!, by Legendre's formula. 0

= {ai,j h::;i,j::;n where ai,j = (a+(~-I)b). We

Proof. Consider the matrix A claim that we have

n(n-l)

3.3

Legendre's formula

det(A)

Legendre discovered a very beautiful and useful formula for vp(n!):

vp(n!) =

L>1 [2] = J_

?~n) ,

where sp(n) is the sum of digits of n when written in base p. The proof of the first equality is easy, since there are [;] -

[pJ~l]

positive integers x ::; n

b-2-

·a(a+b)···(a+(n-1)b)

To prove this, note that one can evaluate any determinant of the form

Chapter 3. Look at the Exponent

where Pi are polynomials of degree at most n - 1, by multiplying the matrix of the coefficients of the Pi's by the Vandermonde matrix (xrlki and then taking the determinant. Using this observation, it is easy to prove the above formula. Now, since det(A) is an integer (as it is a polynomial expression with integer coefficients in the entries of A), it follows that for any p relatively prime to b, the p-adic valuation of a( a + b) ... (a + (n - 1)b) is at least that of n!. As for primes dividing b the problem is easy (since vp(n!) ~ n - 1), the result follows. 0 5. Prove that for all integers a, b with b i=- 0 there exists a positive integer n such that v2(n!) == a (mod b).

3.3. Legendre's formula

Proof. Let cb(n) be the number of digits of n in base b and consider the sequence defined by

It is trivial to check that d divides ai for all i and that

for any distinct numbers it, i 2 , ... ,il. Since the m-tuples

KaMaL Proof. If s2(n) is the sum of digits of n when written in base 2, we need to find n such that n - s2(n) == a (mod b). Choose n = 2Xl + 2X2 + ... + 2Xk with Xl < X2 < ... < XkÂˇ Then s2(n) = k and we need to ensure that

take only finitely many values, the pigeonhole principle yields the existence of an m-tuple S that repeats infinitely often, say S = Sio = Sil = ... for an infinite increasing sequence io, il, .... Let

Ck =

Write b = 2r C with odd C and choose r < Xl < X2 < ... < Xk such that == 1 (mod <p(c)). Then 2Xi - 1 == 1 (mod c) and so 2Xl

1 + ... + 2Xk

1 == k

aidk

+ aidk+l + ... + aidk+d_l'

We will prove that for all k we have diCk - Sbj (Ck) for all j, which will end the proof. Since clearly dick, it remains to check that d divides Sbj(Ck), which follows from

(mod c). and

Also, we have 2Xl

1 + ... + 2Xk

1 == -k

(mod 2r).

Thus, it is enough to choose k such that k == a (mod c) and k which is possible by the Chinese Remainder Theorem.

Sbj(aidk)

== -a (mod

2r ),

Remark 3.2. This problem admits a vast generalization, that appeared on the IMO Shortlist 2007. Namely, we can prove that for any positive integer d and positive integers bl'~, ... , bm there exist infinitely many positive integers n such that din - sbi(n) for all i. Here sb(n) is the sum of digits of n when written in base b.

== sbj(ai dk + l ) == ... == Sbj(aidk+d_l) (mod d).

A trickier combination of the local-global principle and Legendre's formula can be found in the following problem, which implies the classical inequality Icm(1, 2, ... , n) 2: 2n-l. This nice observation as well as the problem appeared in [31]. Amusingly, the same result appeared much before [31] in the same journal, in the form of a proposed problem! There is also an elementary, but more difficult proof of the fact that lcm( 1, 2, ... , n) ~ 3 n for all n, see for instance [38]. Using the prime number theorem, one can prove that Icm(1, 2, ... , n) behaves like en.

100

Chapter 3. Look at the Exponent

6. Prove the identity

3.3. Legendre's formula

101

Putting these remarks together yields the inequality

(~), (~), ... , (~) )

(n + 1) km (

= km (1, 2, ... , n

L Xr ::; k -

+ 1)

Vp (i

+ 1),

r?:1

for any positive integer n.

which, combined with (3.1), yields the estimate Peter L. Montgomery, AMM E 2686

Proof. We will prove that for any prime p the expressions in the two sides have the same p-adic valuation. Note that

((i + 1) (~ :

:) ) ::; k,

establishing the opposite inequality.

We continue with a rather tricky diophantine equation. Let k denote the number of times p divides the right-hand side of the equality from the problem statement, so pk ::; n + 1 < pk+l. Taking i = pk - 1, we obtain that

which shows that the p-adic valuation of the left-hand side is at least k. The opposite inequality is more delicate. Fix 0 ::; i ::; n and note that Legendre's formula gives

(( n. ++ 11)) = '""" Xn Z

~ r?:1

Xr =

[~] r p

[n p-r i] .

[~] r p

(3.1 )

7. Solve in positive integers x 2007 - y2007 = x! _ yL

Romania TST 2007

Proof. We claim that there are only trivial solutions x = y. Suppose that x > y is a solution of the problem. We will distinguish two cases. In the first case, assume that y ::; 2007. If y = 1, then x 2007 = xl and trivially x = 1 (otherwise x-I would divide x 2007 , so x-I = 1, which is clearly not possible). Thus y > 1 and we may choose a prime ply. Then p divides y2007,x!,y!, so that pix. But then 2007::; vp(x2007 - y2007)

= vp(yl(xl/yl - 1)) = vp(y!) < y,

a contradiction.

Now, since

[a + b]- [a]- [b] = [{a}

+ {b}]

E {O, I}

for any real numbers a, b, we have Xr E {a, I} for all r. Moreover, we have Xr = 0 if r > k, since in this case pr > n + 1. The crucial point is that for all r::; vp(i + 1) we also have Xr = O. Indeed, writing i + 1 = pru for some integer u, we have

xr =

[n; 1] _u_[n; 1_u]

Now, assume that y > 2007. Then x-y is a multiple of any prime p smaller than 2007 such that (2007, p - 1) = 1. Indeed, if p is such a prime then p d"d 2007 2007 . . ' IVI es x - y = x! - y!. If p dIvIdes x, then clearly it also divides y and ~o it divides x y. If not, since p divides x p- I _ yp-I and gcd(2007,p-1) = 1, It follows that p divides x - y. We deduce that x > y + 2007 in this case. But then

xl - yl = y!(x!/y! - 1) > 2007! . x(x - 1) ... (x - 2006) > x 2007 , again a contradiction.

Chapter 3. Look at the Exponent

102

3.3. Legendre's formula

So, for k

The next problem is also tricky.

2 we have

[1 + ~(n!)] >

8. Prove that for all positive integers n different from 3 and 5, n! is divisible by the number of its positive divisors. Paul Erdos, Miklos Schweitzer Competition

Proof. Since the number of positive divisors of n! is precisely ITp::;n (1 +vp(n!)), it suffices to prove the following Lemma 3.3. Let Pn be the set of prime numbers less than or equal to n. For all n =1= 3,5,7 there exists an injective map f : Pn -+ {I, 2, ... ,n} such that 1 + vp(n!) divides f(p) for all p E Pn-

Indeed, if this is true, then IT p::;n(1 + vp(n!)) is a divisor of ITPEPn f(p), which divides n!, because f is injective. Thus the problem is solved for n =1= 3,5,7. For n = 7, we can check by hand that the result holds. It remains to construct f. For p ::; Vn simply choose f(p) = 1 + vp(n!). Note that f(p) ::; n follows from Legendre's formula and for p < q ::; Vn we have [;] ~

~ [;] for all

j, the inequality being strict for j = 1 (because

- ~ ~ ;q > 1). For the other primes, we will define f by induction. Assume

we defined f for all primes q < p, where p E Pn,p >

Vn is given.

There are

multiples of 1 + vp(n!) (we used the fact that p >

Vn to obtain vp(n!) = [~]).

only for p = 3, 5, 7. Let us evaluate

[l+[~1J.

Wdte n = kp

For the next problem, we will need some basic estimates about prime numbers. We refer the reader to the addendum 3.A for proofs and more details. 9. Let n, k be positive integers such that n > 9k . Prove that (~) has at least k distinct prime factors. Paul Erdos, Miklos Schweitzer Competition

Proof. We will prove that for n large enough, (~) is a multiple of a product of k numbers that are pairwise relatively prime and greater than 1. This will clearly imply that it has at least k prime factors. Define Lk = lcm(l, 2, ... , k). The key point of the proof is the following

o ::; r < p, so that

n 1 [kP+ r] [1 + [~] = k + 1

+ Lk, (~)

is a multiple of IT~==-~ gCd(~=tLk)'

dt- L)'

If this happens, we are done, because the numbers gc n-l,i k gcdt-~ n-), k are pairwise relatively prime and greater than 1. To prove the lemma, we will compare p-adic valuations for all primes p. Note that the lemma is equivalent to n

k!l for some

7r(p)-1

and we are done. For k = 1 it remains to see if we can ensure that [~] > But this trivially holds for n > p. So, the only obstruction is n = p and 7r(p) = ~. As we observed, this only happens for n = 3,5,7, which is excluded by the hypothesis. 0

Lemma 3.4. For n ~ k

If we manage to prove that this number is greater than 7r(p) - 1, which is the number of occupied values for f(p), we are done, since we can take for f(p) any of the remaining values. However, note that 7r(p) ::; ~, with equality

103

gcd(i, Lk).

i=n-k+l Thus, we need to prove that for all primes p we have

r-p p [r p] > p -1 + - > P -1- - - . l+kl+k = P+ 1+ k

vp(k!)::;

L i=n-k+l

vp(gcd(i, Lk))'

(3.2)

Chapter 3.

104

Look at the Exponent

Let r = Vp(Lk) = [logp k]. For all i :::; r we can find at least

[~]

multiples of

pi among the k consecutive numbers n - k + 1, n - k + 2, ... , n. Also, if u is a multiple of pi with i :::; r, then so is gcd(Lk' u). We conclude that (3.2) follows

3.4. Problems with combinatorial and valuation-theoretic aspects

Assume that among al, a2,.' ., an there are bi numbers not divisible by Pi. Since al, a2,·· ., an are relatively prime, we have bi 2 1. Consider N

from Legendre's formula, as the only nonzero terms on the left-hand side are of the form [~] with i :::; r. Finally, it remains to prove that k Lk

II p9

p[logp k]

+ Lk < gk. p[logp

Note that

k] :::; 4k .

p'5.v'k

( Ilk),

p'5.v'k

II 'P (p~Pi (bi)+l) .

+ a~ + ... + a~ == bi

(mod p~Pi(bi)+I).

Hence

< k v'k . 4k < gk,

the last inequality being easy to prove.

3.4

· k > V (b.) VPi(bi)+11 a k whenever Pilaj. Using this and Smce j Pi 2 + 1, we have Pi Euler's theorem, we infer that

where here we used theorem 3.A.3. Thus Lk

i=l

II p. II p>v'k

105

Problems with combinatorial and .valuation-theoretic aspects

for all i. Since all prime factors of a1 we deduce that

+ a~ + ... + a~

are among PI, P2, ... ,pN ,

Now, at least one of the ai's is greater than 1, thus

The problems in this section are fairly elementary, but none of them is easy.

10. Let n 2: 2 and let alo a2, ... ,an be positive integers, not all of them equal. Prove that there are infinitely many prime numbers P for which there exists a positive integer k such that

Iranian Olympiad 2004 Proof. By dividing all ai's by their greatest common divisor, we may assume

that they are relatively prime. Suppose that there are only finitely many primes PloP2,'" ,PN such that all prime factors of a1 + a~ + .. , + a~ (where k varies over the positive integers) are among PI,P2,··· ,PN·

+ a~ + ... + a~ 2:

> k > IIp~Pi(bi). i=l

The two relations are clearly contradictory and the problem is solved.

A classical problem of Erdos is the following: if aI, a2,' .. ,an+l are different positive integers not exceeding 2n, then one can find i =I- j such that ai divides aj. The idea is very simple and beautiful: the largest odd divisors of the ai's form a sequence of n + 1 odd numbers between 1 and 2n - 1, so there must be two equal terms in this sequence. But then the corresponding ai and aj have a quotient which is a power of 2. The next problem is a variation on this classical gem.

Chapter 3. Look at the Exponent

106

11. Let f (n) be the maximum size of a subset of {I, 2, ... , n} which does not contain two distinct elements i,j such that il2j. Prove that there exists a constant C > 0 such that for all n we have

If(n) - 4;

I: ; C In n.

Proof. We will actually exhibit the optimal set with the property that it does not contain two distinct elements i,j with il2j. Defining

n/3

< a ::; n,

v2(a)

it is easy to see that An satisfies the conditions of the problem. Indeed, if

Next, we prove that this set is optimal, in the sense that it has the maximal number of elements among all sets satisfying the conditions of the problem. Take any such set A and fix k such that 3 does not divide k and V2 (k) is even. Look at all the numbers k, 3k, 9k, ... and 2k, 6k, 18k, . ... There is at most one element of A among the union of these numbers and by definition there is exactly one element of An among them. On the other hand, if one varies k, the numbers k, 3k, ... , 2k, 6k, ... form a partition of the positive integers. This clearly implies that A has at most as many elements as An. Finally, we have to estimate the size of An. The elements of An are exactly the numbers of the form 4j b with b odd, j 2: 0 such that 3~J < b ::; ~. There are approximately ~ odd numbers b such that 3~j < b ::; ~, with an error ~~ . . h at most 2 if 4j < n and error at most 3~J for 4J > n. The total error IS t us logarithmic in n and since

4n 9

IAnl = -

finishing the proof.

107

12. Find all positive integers n with the following property: there exist natural numbers bI , b2, ... , bn , not all equal and such that the number (bi + k)(b 2 + k)··· (b n + k) is a power of an integer for each natural number k. Here, a power means a number of the form x Y with x, y > 1.

Proof. There are some obvious solutions: for instance, if n is composite, say n = ab with a, b > 1, then we can choose bi = b2 = ... = ba = 2 and all the other bi'S equal to 1. Then for any k we have

== 0 (mod 2)},

i,j E An and il2j, we must have ilj, since we cannot have v2(i) = V2(j) + 1 as both V2( i) and V2(j) are even. So j /i is either 1 or 2. It cannot be 2, because in this case one of v2(i), V2(j) would be odd, so it has to be 1 and i = j.

we have

Problems with combinatorial and valuation-theoretic aspects

Russia 2008

Paul Erdos, AMM E 3403

An = {a E

3.4.

+ O(logn), D

which is certainly a power. So, the real question is to decide whether numbers bI , b2, ... ,bn can exist when n is a prime. It turns out that the answer is negative. Assume that such numbers existed and let CI, C2, .. " CN be the set of distinct numbers among bI, b2, ... , bn , with multiplicities mI, m2, .. " mN. By assumption, we have N > 1 and clearly n = mi + m2 + .. , + mN. Moreover, we know that for any k, the number (C! + k)ml(C2 + k)m2 ... (CN + k)mN is a perfect power. The key point is to choose numbers k for which we can find distinct primes PI, P2, ... ,pN such that vPi (Cj + k) = 1 if i = j and 0 otherwise. In this case, if (CI + krl(C2 + k)m2 ... (eN + k r N = x Y for some x, y > 1, we have YVpi (x) = mi, so that y divides all mi. But then Y divides their sum, which is n and since n is a prime, it follows that n = y. Thus n = Y will divide all mi and this obviously contradicts the fact that N > 1 and ml + m2 + ... + mN = n. Thus, we are done if we can find distinct primes P!'P2,." ,PN and k such that vpi(Cj + k) = 1 if i = j and 0 otherwise. This is very simple: first, we choose some distinct prime numbers PI,P2,." ,PN sufficiently large, say not dividing any of the numbers Ci - Cj with i =f- j, and then choose k such that k + Ci == Pi (mod p;) for all i. Such k exists by the Chinese Remainder Theorem. Of course, vpi(k + cd = 1 and for j =f- i we cannot have PilCj + k, since otherwise Pi would divide Ci - Cj, contradicting the choice of Pi.Thus,

108

Chapter 3.

Look at the Exponent

3·4· Problems with combinatorial and valuation-theoretic aspects

such k satisfies all desired conditions and the answer to the problem is that n must be composite. o

a) the graph G(A) is complete? b) the graph G(A) is connected with all vertices of degree at most 2?

A nice mixture of valuation-theoretic arguments and pigeonhole principle can be found in the following problem. 13. Let a be a positive integer. Prove that the set of prime divisors of 2 2n +a for n = 1,2, ... is infinite. Iranian TST 2009 Proof. Assuming the contrary, let PI,P2, ... ,PN be such that all prime factors 2n of 2 + a are among PI,P2, ... ,PN for all n. Pick a large number r such that n 2N 1 2 r > a + + a and no such that 22 O + a > (PIP2' .. PN Then for all n 2:: no

r·

we have

so that we can find 1 ::; i ::; N with vpi (2 2n + a) > r. This Pi depends on the choice of n, but among the indices i associated to the numbers n = no + 1, no + 2, ... , no + N + 1 there will two identical ones. Thus we can write Tn pr 122n + a and pr I22n + + a for some n 2:: no, some 1 ::; m ::; N + 1 and some m Z 1Z ::; i ::; N. But then 22 n == -a (mod pD, so that 22Tn+n == a 2 ( mo d Pir) andpila2Tn + a. In particular,

109

Miklos Schweitzer, Competition 2009 Proof. The answer to the first question is negative. Let P be the smallest prime different from Pl,P2,··. ,Pk and suppose that G(A) is complete for some finite set A with IAI > p. Then there exist a, b E A different such that P divides a - b, so that a and b arenot connected.

On the other hand, the answer to the second question is positive! We will construct a set A with m elements aI, a2, ... , am such that G(A) is a simple path. It is enough to ensure that for all 1 ::; n ::; m, a n+ I - an E Sand a n+2 - an, a n +3 - an,··. are not in S. But we can choose

Then clearly an+l - an E S. On the other hand, for any i 2:: 2 we have (an+i - an) = n + 1 for all 1 ::; j ::; k. Since an+i - an > (PIP2" . Pk)n+l, it follows that an+i - an is not in S and the result follows. 0 VPj

15. Let m and n be positive integers such that m + 1, m + 2"" ,m + n are composite numbers and m > nn-l. Prove that we can find pairwise distinct prime numbers PI,P2, ... ,Pn such that Pi divides m + i for all 1 ::; i ::; n. Thymaada Olympiad 2004

contradicting the choice of r. The result follows.

We continue with two more unusual problems. 14. Let Pl,P2, ... ,Pk be distinct prime numbers and let S be the set of positive integers all of whose prime factors are among PI,P2, .. ' ,Pk. If A is a finite set of integers, let G(A) be the graph whose set of vertices is A, two vertices a, bE A being connected if a b E S. Is it true that for all m 2:: 3 we can find A with m elements such that

Proof. First, we will deal with those i such that m + i has at most n - 1 prime factors. For such i choose a prime Pi for which p~Pi(m+i) is maximal (note that Pi is unique with this property). Since m + i > nn-l and since m + i has at f t h Vp (m+i) · mos t n - 1 prIme ac ors, we ave Pi ' > n. We claim that if i :I j and m + i, m + j have at most n - 1 prime factors, then Pi :I Pj. Indeed, assume that Pi = Pj = p. Then min(pvp(m+i),pvp(m+j)) divides m + i and m + j and moreover is greater than n. But any common divisor of m + i and m + j divides j - i and so it is smaller than n. This shows that i I---t Pi is injective.

Chapter 3.

110

Look at the Exponent

It is now easy to conclude: make a list with those numbers m + i having at most n - 1 prime factors. For each of them, the previous paragraph yields a prime factor Pi and the associated Pi'S are distinct. If all m + i are in the list, we are done, otherwise successively pick remaining numbers and choose one of their prime factors which is not among the Pi'S or among primes previously selected. This is possible at every step, since any m + i not in the list has at least n prime factors. D

3.5

3.5.

111

Lifting exponent lemma

The reader might wonder what happens for p = 2. There is of course a version of the theorem for p = 2, but the formula is more complicated. The proof is however much easier.

Theorem 3.6. Let x, y be odd integers and let n be an even positive integer. Then

Proof. Write n = 2k a for some odd number a. Then

Lifting exponent lemma

xn _ yn = (xa _ ya)(xa

+ ya)(x2a + y2a) ... (X2k-1a + y2k- 1a).

This section is devoted to some applications of the following useful result:

Theorem 3.5. (Lifting exponent lemma) Let P be an odd prime and let a and b be integers relatively prime to p, such that pia - b. Then for all positive integers n vp(a n - bn ) = vp(n) + vp(a - b). Proof. Consider first the case vp(n) = O. We need to prove that p does not divide which is clear, as by hypothesis

a:=r '

~-~ _ _ _ = an -

+ a n - 2b + .. , + bn- I == nan - I

(mod p)

a-b

and p does not divide nan-I. Next, we prove the result for n = Pi we need to check that p divides exactly once into aP- 1 + ... + bP- 1 . Write b = a + pk i for some integer k. Then by the binomial formula we have bi == a + iai-1pk (mod p2), so that p-l a bP ' " ap-l-ibi - - - = L...t P

a- b

= paP-1

i=O

p-l "'( L...t a p-l

+ ~p . k a p-2)

+ v 2 == 2

Now observe that if u, v are odd numbers, then u 2 V2(X n - yn)

V2(X 2a _ y2a)

(mod 4). Thus

- 1.

Finally, since a, x, yare odd, it is easy to see that X2;_y~a is odd. The result x -y follows. D An easy consequence of these results is the following estimate. It is much weaker than the previous theorems and can be proved directly by induction, too.

Corollary 3.7. If a, b are integers and p is any prime dividing a - b, then for all n we have vp(a n - bn ) 2: vp(a - b) + vp(n). The next problem is an immediate consequence of this corollary. 16. Let a, b, c be positive integers such that cla c

bC â&#x20AC;˘ Prove that

cl a:=r .

1. Niven, AMM E 564

i=O

+ p2kP -2 1 aP-2 = paP-1 an / p

bn / p

Proof. First of all, note that a:=r is an integer, so the statement makes sense.

(mod p2).

Let us apply the case n = p to and (note that they still satisfy the p p n n hypotheses of the problem) to get vp(a - b ) = 1 +vp(an / - bn / ). The result follows now by an immediate induction on vp(n). D

Let p be a prime dividing c. We will prove that vp(c) ~ vp (a:=r). If p does not divide a b, this follows from the hypothesis that c divides aC - bC , so we may assume that p divides a-b. But then everything follows from corollary 3.7. D

Chapter 3.

112

Look at the Exponent

The next problem appeared as a Chinese TST 2009 problem, but the result is much older (see [61]). 17. Let n be a positive integer and let a > b > 1 be integers such that b is odd and bnlan - 1. Prove that a b >

3:.

M.B. Nathanson Proof. Take any prime factor p of b. Since b is odd, we have p

> 2. Note that

3.5.

Lifting exponent lemma

113

the facts that n is odd and gcd(n, q - 1) = 1 imply that p - 1 == -1 (mod q), or p = q. By the lifting exponent lemma (using that n is odd) we have

(p - I)vp(n) = vp(nP -

) :::;

vp((p - I)n + 1) = 1 + vp(n).

Thus (p - 2)vp(n) :::; 1. In particular, p = 3 and vp(n) = 1. Write n = 3a with gcd(a, 3) = 1 and observe that a 2 divides sa + 1. We claim that a = 1. Otherwise, let r be the smallest prime factor of a, so that

Sa == -1 and Sr-l == 1

(mod r),

whence S == -1 (mod r) as gcd(a, r - 1) = 1. But then r impossible. It follows that a = 1 and n = 3.

We deduce that

and the result follows.

Remark 3.S. Using a generalization of the deep Thue-Siegel theorem, Mahler proved in [4S] the following result: if a, b, u, v are nonzero integers with u > v > 1, then there are only finitely many positive integers n such that

3, which is 0

Remark 3.9. Another interesting problem that can be solved using the same ideas is the following one: find all positive integers a and b such that a b divides ba - 1.

19. Find all positive integers a, b, c such that

(2 a

I)(3 b - 1) = d.

Gabriel Dospinescu, Mathematical Reflections It is quite rare to see two very similar problems at the IMO; nevertheless the following problem appeared in weaker forms at the IMO in 1990 and 1999.

IS. Find all primes p and all positive integers n such that n P (p _1)n + 1.

divides

After IMO 1990 and 1999 Proof. Let p and n be as in the statement. Note that if p = 2, then n = 1 or n = 2. From now on, we assume that p > 2. If n is even, then 4 divides n P- 1 , but it does not divide (p 1)n + 1, a contradiction. So, n is odd. Let q be the smallest prime factor of n. Since

(p - I)n == -1 and (p - l)q-l == 1

(mod q),

Proof. We will only give the key ideas, since the computations are a bit tedious. Take a solution (a, b, c) with c > 3 and note that a is even, since 3 divides a 2 -1. Also, as 4 divides d, 4 must divide 3b -1 and so b is even. Then, using the lifting exponent lemma, we deduce that

c - S3(C)

= v3(d) = v3(2 a - 1) = v3(4 a

/ 2 -

1) = 1 + v3(a).

Similarly, by writing b = 2kr with r odd, we have

v2(3

1) = V2

(~(9r

1))

+ v2(b)

v2(4(9 r- 1 + 9r- 2 + ... + 1)) + v2(b) = 2 + v2(b). =

Chapter 3.

114

3.5.

Look at the Exponent

115

Lifting exponent lemma

Taking logarithms, we deduce that n is bounded in terms of a. Since c, b - c < nl, the conclusion follows.

Thus

We continue with a very beautiful, but hard problem.

We deduce that

and b> 2V2 (b) > 2c-l-log2c-2 -

21. Let x, y be relatively prime integers greater than 1. Prove that for infinitely many primes p, the exponent of pin x p- 1 - yp-l is odd.

2c - 3 c

= --.

Barry Powell, AMM E 2948

From here it follows that 3c/2-2 2c- 3 c! ;:: (2-c- - 1)(3 c - 1). It is not difficult (even if it is quite tedious) to check that this does not h~ld for any c ;:: 12. We deduce that in any solution we have c :S 11. yve can easIly exclude the cases c E {8, 9,10, 11}, since in this case we obtam v2~b) ;:: 5, forcing b ;:: 32, which is too large. For c E {4, 5, 6, 7}, we have c! ;:: ~ - 1 f~r only one value of b with v2(b) = C-S2(C) -2, so we simpl.y check t~ see If there IS a compatible value of a. Combining these with the ObVIOUS solutIOns for c :S 3, 0 we end up with (a, b, c) E {(1, 1, 2), (2, 1,3), (2,2,4), (4,2,5), (6,4, 7)}. b

20. Let a be a fixed positive integer. Prove that the equation n! = a - a has only finitely many solutions (n, b, c) in positive integers. Chinese TST 2004

Proof. Choose any integer k > 2. It is a well-known result of Fermat that the equations a 4 + b4 = c2 and a 4 + b4 = 2c2 have only trivial solutions. 1 k 1 Therefore X 2k - + y2 - is neither a perfect square nor twice a perfect square, 1 k 1 whence there is some odd prime p such that Vp(X 2k - + y2 - ) is odd. Since gcd(x, y) = 1, P cannot divide xy. Since p divides X2k - y2k and does not 1 k 1 divide X2k - - y2 - , the order of x/y mod p is 2k and 2k divides p - 1. Hence the lifting exponent lemma gives

The result follows by taking successively k = 3,4, ... and observing that the prime p associated to k in the previous discussion satisfies p ;:: 1 + 2k. 0 Proof. We will argue by contradiction: assume that there exists N > Ix such that for all primes p > N we have

Proof. Let p be an odd prime not dividing a. Then by the lifting exponent

lemma we have Taking n

= b - c and noting that vp(n!) >

vp(b - c) ;:: vp(n!) - vp(aP-

- 1, we conclude that 1

- 1) ;::

for some constant K, independent of n. Letting that b - c ;:: Epn/p for all n. Thus nn

> n!

= a - a

> ab-c

;:: a

p- K

p-K

Epn/ p

> 0, we conclude

We claim that for all p

x:=r.

> N the number x:=~p is a perfect square. Take a

prime factor q of Since q does not divide x (otherwise q divides y, contradicting the fact that gcd(x, y) = 1) and since q divides x P - yP, the order of x/y mod q is 1 or p. If q divides x - y, then q divides px p- 1 and so q = p, contradicting the fact that p > N > Ix - YI. SO the order of x/y mod q is p and so p divides q - 1. Then the lifting exponent lemma implies that vq(x q- 1 - yq-l) = vq(xP - yP) is an even number, finishing the proof of the claim.

Chapter 3.

116

Look at the Exponent

Next, take a prime r == 3 (mod 4) dividing 4xy - 1 and (using Dirichlet's theorem) take a prime p == -1 (mod r - 1) larger than N. Thus there exists z such that

z2 =

x p - yp x-y

~ - 1

x-y

p-1 L k2 = (p-1)p(2p-1)

-b,

~s a multiple of p .. Thus th: numerator of the fraction L~:i when written lowest terms, IS a multIple of p. The next problem is a variation on this idea combined with some ideas from p-adic analysis.

III

~ 22. Let p

For more details on the results and techniques used in this section, the reader is (strongly) advised to read the addendum of this chapter, which is a modest introduction to p-adic numbers. Before passing to the next problem, let us recall the notion of congruence for rational numbers. Let p be a prime. The localization of Z with respect to the ideal pZ is the set Z(p) = {b E Qlgcd(a,b) = 1,

k=l

p-adic techniques

117

since the map x --+ x- 1 is a bijection of (ZjpZ)* and

== - - y == - - == -4 (mod r),

i.e. -1 is a quadratic residue mod r, a contradiction.

3.6

3.6. p-adic techniques

gcd(b,p) = 1}.

It is easy to check that Z(p) is a subring of Q. If a, b E Z(p), we write a == b (mod pJ) if a - bE pJ . Z(p). This is equivalent to the fact that the numerator of the fraction a - b, when written in lowest terms, is a multiple of pJ. It is easy to see that this is an equivalence relation, satisfying the usual properties of congruences over Z. Moreover, we have a natural morphism f of rings Z(p) --+ ZjpZ, sending E to a· 1)-\ (1)-1 being the inverse of b mod p). Then for x E Z(p) we have x == (mod p) if and only if f(x) = 0. This observation is very useful when trying to prove congruences for rational numbers. For instance, let p be a prime greater than 3 and consider the element

> 5 be a prime. Prove that p4 divides the numerator of the fraction p-1 1 p-1 1 2· Ly;;+p· L k 2 k=l

k=l

when written in lowest terms. Gabriel Dospinescu Proof. The first step is to note that p-1

1 (1 1) p-1

2t;y;;=t;

y;;+ p-k

p-1

=t;k(P-k)"

Thus, it is enough to prove that

p-1 1 X=

L k2 k=l

of Z(p). We have p-1 p-1 1 k2 f(x) = L - = L k 2 = 0, k=l

k=l

Now, the crucial remark is that in the field of p-adic numbers we have the convergent expansion

1 _ 1 1t

k (p - k) - - k 2 1 -

1( + y;; +

= - k2

p2 k2

)

+ . .. .

By truncating at level p3 we obtain the congruence 1 1 P p2 k(p - k) = - k2 - k3 - k4

(mod p3).

(3.3)

Chapter 3. Look at the Exponent

118

Of course, one does not need p-adic numbers to find or check the previous congruence, since obtaining the appropriate polynomials in p and k and validating that they work are each formal algebraic matters. The p-adic approach simply gives a straightforward route to both, here. Using (3.3), it remains to prove that

1 k3

p-l

+PL

k=l

1 k4

== 0

1 k3

'at 23.

Let p be a prime and n, s positive integers with n pd divides

d h were

(mod p2).

n-s-l]

Gabriel Dospinescu, Mathematical Reflections

p-l

== 0

(mod p2),

k=l

> s + 1. Prove that

= [ --p::r- .

k=l

We will actually prove that p-l

119

3.6. p-adic techniques

Proof. Fix a primitive root of unity z E C of order p. The field K = Q(z) is an extension of degree p - 1 of Q, as the minimal polynomial of z is

== 0

(mod p).

x p - 1 + X p - 2 + ... + X + 1.

k=l

The same argument as in the preliminary discussion yields p-l

p-l

k4 == Lk4 == 0

L k=l

(modp),

k=l

the last congruence being established either by using the existence of primitive roots mod p (which makes the corresponding sum the sum of a geometric progression with ratio g4, where g is a primitive root mod p) or simply by using explicit formulae for this kind of sum. In order to prove the other congruence, note that

Choosing a prime (ideal) j3 above p in the ring of algebraic integers OK of K and completing j3-adically, we obtain a valuation vp on K (seen inside its j3-adic completion), which extends the usual p-adic valuation on Q and such that vp(l - z) = p~l' To prove the last equality, note that t~.~ is a unit in OK for all i relatively prime to p, so vp(l- z) = vp(l- z~). Since we also have p-l

IT (1- zi)

the result follows. Note that so p-l

2L k=l

The result follows.

p-l

~ Lzkj

1 p-l 1 k 3 == -3p L k4 == 0 . (mod p2).

k=l

The following problem is a generalization of a classical congruence due to Fleck. For the reader not familiar with p-adic numbers, it is worth reading the addendum 3.B before attacking the proof.

= p,

i=l

= 0

j=O

if k is not a multiple of p and 1 otherwise. We deduce that

120

Chapter 3. Look at the Exponent

Now, let n - s - 1 = d(p - 1)

for some 0 :::; r

< p - 1. We will prove that

3.6. p-adic techniques

121

and so

n-s

r+1

p-1

~--=d+-->d.

Thus, the claim is proved and the result follows.

for all 0 :::; j :::; p - 1. This will imply that

(of,;n (_l)k . k

(~))

>d-

We end this chapter with a wonderful result due to Skolem, Mahler and Lech. It truly shows what a versatile tool p-adic analysis can be. Even though the result still holds for p = 2, we will assume that p is odd in order to avoid some technical problems.

plk and since this p-adic valuation is an integer, the result will follow. Now, to prove the claim, we will use the following: Lemma 3.10. The polynomial L~=o k S(~)Xk is a multiple of (1 + x)n-s for all s < n.

te(~)Xk = (l+x)n-sf(X)

Coming back to the proof, write

for some f E Z[X] (note that we necessarily have f E Z[X], as (1 + x)n-s and L~=o k SG)Xk have integer coefficients and (1 + x)n-s is monic). Then for 2 1 :::; j < p we have

2Note that by taking X = -1 in the previous relation we obtain L~=o(-l)kkS(~) = 0,

.k;- "!J ~

A~=o

tpk(~)ak = 0

k=O

for all n ~ 0, where d ~ 1 and Xl, X2, ... ,Xd are integers. Prove that there exists a finite set S and integers CI, C2, ... , CN, dl , d 2 , ... , dN such that

Skolem-Mahler-Lech theorem

Proof. a) Consider the following function f(x) = Lpkak (~). k?O

= (1- zj)n-sf(-zj)

k=O

w, only havo to deal with j ~ 1.

a) Let p be an odd prime and let ao, al,'" be integers such that

b) A sequence (an)n of integers satisfies

k=O then we get the inductive step by differentiating and multiplying both sides byX. 0

V24.

for infinitely many positive integers n. Prove that for all n we have that an = O.

Proof. This is very easy: for s = 0 it is clear, and if

t(-zj)kS(~)

'( X)= X

Of course, such a series cannot converge as a series of real numbers, but it does converge as a series of p-adic numbers, since vp (pk ak (%)) ~ k for all x E Zp

122

Chapter 3. Look at the Exponent

(note that (~) E Zp because the map x ---+ (~) is continuous and takes values in Zp when x is in the dense subset Z of Zp). Thus f defines a map f : Zp ---+ Zp. The crucial property of f is that it interpolates the sequence L~=o pk G) ak and that it converges to its Taylor series (has good analytic behavior). The first claim is obvious, since by definition we have

3.6. p-adic techniques

123

deduce that there exists a E Zp and an infinite sequence of integers nj such that f(nj) = 0 and nj converges p-adically to a. Now, for all x E Zp we can write f(x) =

Cj((x - a)

+ a)j

j?O

The second claim is more delicate. Let us write x) = (k

~k!~ ~bkxj J, j=O

for some integers bj,k' Then we can write

Again, the series defining dk and we also have

= Lj?k Cj (Da j - k converges because v p(Cj) ---+

p-2 vp(d k ) 2: inf vp(Cj) 2: k - - . J?k P- 1

Recall that f(nj) = 0 for all j. On the other hand

'" where Cj -- 6k?j

pkakbj,k

. defi mng . Cj converges, SInce . . N ote t h at t h e senes

(2:

dk(nj - a)k) 2: vp(nj - a) ---+

00,

k?l

pkbj,k) p-2 v ( - - >k·-p k! p- 1

tends to

as k ---+

00.

The same estimate shows that p-2 p-2 vp(Cj) 2: infk· --1 = j--1' k?J Pp-

(this is why we assumed that p > 2). The conclusion is that f converges to its Taylor series and so behaves as a holomorphic function, i.e. has good analytic properties. Now, by hypothesis we know that f(n) = 0 for infinitely many integers n. We want to prove that f = O. This will imply that f(n) = 0 for all nand it is easy to see that this forces an = 0 for all n. Since Zp is compact, we

so that limj-+oo f (nj) - do = O. We deduce that do = O. Dividing the equality f(nj) = 0 by a - nj and repeating the argument yields d 1 = 0, then d 2 = 0 and so on. We deduce that all dj's are zero and so f = O. The result follows. b) We may assume that Xd =I- O. Consider the matrix M defined by

for i < n and whose last row is xd,xd-l, ... ,Xl. This is the companion matrix associated to the characteristic polynomial X d - XIX d- 1 - ... - Xd of the recursive relation. Let Vn be the column vector whose coordinates are an, an+l,·· ., an+d-l· Then the recursive relation becomes Vn+1 = MVn , thus Vn = MnVo. If el is the column vector whose coordinates are 1,0,0, ... ,0 and if (,) is the standard inner product in .lRd , we deduce that an = (Mnvo, el). It

Chapter 3. Look at the Exponent

124

is easy to check that det M equals Xd up to a sign. Pick a prime p > 2 + IXdl, so M is invertible mod p. Using either Lagrange's theorem or the pigeonhole principle, we can find r ;:::: 1 such that MT == Id (mod p). Thus we can write MT = Id + pN for some matrix N with integral coefficients. It is then enough to prove that for any 0 :S j :S r - 1, the set Aj = {n ;:::: Oland+j = O} is either finite or contains all nonnegative integers. In order to prov~ this, the point is to write

where bk = (N k MjVo, el) is a sequence of integers. If Aj is infinite, then part a) of the problem shows that Aj contains all nonnegative integers and the

result follows.

Remark 3.11. Under the hypothesis p > 2, the map f is analytic on the whole Zp. If p = 2, one can still prove (using for instance Amice's theorem 3 ..B.30) that f is locally analytic on Zp. The proof is then exactly the same as m the case p > 2.

Remark 3.12. The result holds for sequences with values in any field of characteristic 0, as Lech proved. The key point is that we have a p-adic version of the Lefschetz principle (the proof is not easy, but elementary, see [13) or [45)): if S is a finite subset of a field K which is finitely ge~erated over .Q, then for infinitely many primes p there is an embedding of K mto Qp sendmg all elements of S to Zp. Applied to the roots of the characteristic polynomial of the recurrence relation, this reduces the proof to the p-adic case, which has already been discussed.

Remark 3.13. The result does not hold for fields of positive characteristic. n For instance, the sequence an = (1 + t)n - 1 - t is linearly recursive with values in IF (( but the reader can easily check that it vanishes precisely at {pnln ;:::: O}~ which is not the union of a finite set and finitely many arithmetic

tÂť,

progressions.

3.1. Miscellaneous problems

3.7

125

Miscellaneous problems

Before attacking the following problem, let us recall aclassical property 2n of ~he Fermat number Fn = 2 + 1, namely that any prime factor p of Fn satIsfies p == 1 (mod 2n+1). Indeed, if p divides Fn , then p divides 22n +1 - 1 2n but does not divide 2 - 1. Thus the order of 2 mod pis 2n +1 and since thi~ order divides p - 1, the result follows. The next problem is a generalization of some very classical problems, such as the following one: find all integers n such th~t the congruence xy == 1 (mod n) implies the congruence x == y (mod n). It IS easy to see that this is equivalent to nlx 2 - 1 for any x relatively prime to n and that this happens if and only if n divides 24. The next problem is a bit trickier. 25. Let m be an integer greater than 1. Suppose that a positive integer n satisfies nla m - 1 for all integers a relatively prime to n. Prove that n :S 4m(2m - 1) and find all cases of equality. Gabriel Dospinescu, Marian Andronache, Romanian TST 2004

Proof. Write n = 21f ,. with r odd. By the Chinese Remainder Theorem there is an a with a == 3 (mod 2k) and a == 2 (mod r). Such an a is clearly rel~tivelY prime to n, so nla m - 1. Hence 2m == am == 1 (mod r) and 3m == am == 1 (mod 2k). Thus rl2m - 1 and 2kl3 m - 1. One easily checks that v2(3 m 1) = 2 + v2(m), hence 2k14m. We conclude that nl4m(2m - 1) and in particular n:S 4m(2m - 1). Now suppose equality holds. The above argument actually shows that n :S 4Âˇ 2v2 (m) . (2m 1) so for equality to hold we must have m = 2v2 (m), that is, m must be a power of 2. Suppose m = 28 â&#x20AC;˘ Then n

28 +2 (228 -1)

28 +2 .3.5 ... (2 28 -

+ 1),

that is, n is a power of 2 times some number of consecutive Fermat numbers. The cases s = 1 and s = 2 give two equality cases (m, n) = (2,24) and (m, n) = 1 (4,240). Suppose now that s ;:::: 3, and let p be a prime divisor of 228 - + 1. The preliminary discussion shows that p == 1 (mod 28 ), in particular p == 1 (mod 8), so that by quadratic reciprocity, 2 is a square mod p. The Chinese

Chapter 3. Look at the Exponent

126

Remainder Theorem gives an a relatively prime to n with a 2 == 2 (mod p). 28 1 28 Then nlam - 1, so 2 - == a = am == 1 (mod p). But this contradicts the 28 1 fact that 2 - == -1 (mod p). Thus there are no solutions with s 2: 3. 0 Here is a beautiful problem, for which we present two elementary proofs, both ingenious, neither quite natural. In the addendum to this chapter, we will present a more natural proof which uses p-adic analysis. 26. Let Xn 2 of - + 1

be the exponent of 2 in the prime factorization of the numerator 22 2n - + ... + - , when written in lowest terms. Prove that 2 n lim Xn

n--+oo

and that

X2n

2n - n

3.7. Miscellaneous problems

of terms in the sum

2:::%:0

(

(n) -1

+ 1.

+ ... +

(

n) n

-1

2N - N + 1, ending

the proof of the second part of the problem. Now, let us prove the lemma. We will present two completely different approaches, the first one very down-to-earth, the second one more exotic, but more powerful also. Denote the left-hand side by an and the right-hand side by bn . Clearly, ao = bo = 1. Note that

n + 1 n (2 =-- - + 2n 2n 1

22 2n) - + ... + + 1 2 n

n+ 1 =- bn - 1 + 2n

n+1 _ n+1~i!(n-1-i)! 2n an-1 + 1 - 1 + 2n ~ (n _ 1)!

Lemma 3.14. For any positive integer n we have +

x2N

On the other hand,

Proof. The first proof is based on the following nice identity known as Staver's identity, which also appeared as a problem in the USA TST 2000.

-1

(2 N:-1) ' it follows that

Adapted from a K vant problem

127

(2 22 2n+1 _+_+ ... + _ _) . 2M1 1 2 n+ 1

= n+ 1

= 1

i=O

((i + 1) + (n - i)) . i!(n - 1 - i)! n . (n - 1)!

~ ~ (i + 1)!(n -

+2 6

i=O

1 - i)! + i!(n

, n.

i)!

Let us accept this lemma for a moment and see how we can conclude. Using the extension of V2 to all rational numbers, we can write

(~ 2:) ~ n + 1 - v, (n + 1) - o~~~n

V2 ( ( ; ) ) .

Since

and since v2(n+1) ::; log2(n+1), this trivially implies that Xn ~ 00. Moreover, since each of the numbers eNk- 1) is odd and since there are an even number

Taking into account these two relations, it follows by induction that an = bn for all n and the lemma is proved. Now, let us turn to the second approach. We will use here the classical formula 1 a b alb! t (1 - t) dt = ( o a+ b + 1)'. '

which can be proved separately for each value of a + b and then by induction

Chapter 3. Look at the Exponent

128

3.7. Miscellaneous problems

129

is of the form xn (ao + a 1 X + ... +an -1 xn-1 ) for some integers ai and moreover Fn(O) = 0, we deduce that we can write

on say a and integration by parts. We deduce that

1 n 1 1 n L n- = n' . L(n + I)! io tk(1 - tt-:-kdt k=O (k) . k=O 0 1 tn+1 - (1 - t)n+1 = (n + 1) dt. o 2t-l

Fn(X)

= Xn+1 (~ + ~X + .. , + an-l x n+ 1

Therefore, since Ln(2)

n+ 2

n- 1) .

= -!Fn(2), we can write

Making the 'substitution 2t - 1 = s yields 1 tn+1 - (1 - t)n+1 -----'----'----dt o 2t - 1

1 -n -2 2 +

+ s)n+1 -

-1

(1 - s)n+1

and noting that the integrand is an even function of s, the last expression is also equal to 1 2n+l

((1

+ s)n+1 -1 + 1- (1- s)n+1) ds s s

1 =

r ((1 +

i=O

1 = 2n +1

t; n

s)i

+ i +1

from where the result follows trivially.

Proof. This second elementary solution actually mimics the p-adic analytic proof, without mentioning p-adic numbers. Define the sequence of polynomials X

Ln(X) and Fn(X) = L n (2X - X2)

2 ( -1

+ -222 + ... + -2n) 2: min (n + k n O::;k:'Sn-1

V2 (n

+ k + 1)).

As v2(n + k + 1) ::; log2(n + k + 1) ::; log2(2n), the fact that Xn --+ 00 is now 0 clear. The other inequality is also trivial using the previous estimates,

+ (1- s)i) ds

(2 + - 1 i +1

n-1 2n+k "\"'" ak L n +k+l' k=O

immediately implying the estimate V2

2n+1 . L

-+-+ ... +-= 1 2 n

We continue with a rather technical problem. The solution we present is long and complicated, but nevertheless rather natural. This problem also appeared on the IMO 2007 Shortlist. 27. Find the exponent of 2 in the prime factorization of the number

= 1 + 2 + ... + -;;; 2Ln(X), Since

J. Desmong, W.R. Hastings, AMM E 2640

F~(X) = 2(1 - X)L~(2X - X2) - 2L~(X)

2xn(1- (2 - x)n) I-X = -2Xn(1 + (2 - X) + ...

+ (2

xt- 1)

Proof. Note that 1·3· ... (2 n +1 - 1) (1 . 3· ... (2n - 1))2

(2n

+ 1)(2n + 3)· .. (2n + 2n 1·3· .... (2n - 1)

Chapter 3. Look at the Exponent

130

3. 7. Miscellaneous problems

131

so that we need to compute

n n n n V2(( 2n) ((2 +1)(2 +3)"'(2 +2 -1) -1)) 2n- 1 1·3· .... (2n - 1) = 1 + v2((2 n + 1)(2n + 3)··· (2n + 2n - 1) - 1 ·3· .... (2n - 1)).

2n -

XiXj = (LXif - LX;

== -

(mod 2n+1).

i=O

O~i<j~2n-l_l

Since ~

Looking at small values of n, we conjecture that for n

2 we obtain 2n-l_l

To prove this, we start by expanding the product

XiYi

i=O

as if it were a polynomial in 2n. If we work mod 23n , the only terms that count are the first three: if a = 1 ·3· .... (2n - 1), then

XiXj

2n-l_l

== -2

2n-l_l

L xI - 2 L xr n

i=O

2n-l_l

(2n + 1)(2n + 3)··· (2n + 2n - 1) == a + a· 2n

L O~i<j~2n-l_l

(mod 2n+l).

i=O

Finally, we clearly have 2n-l_l

i=O

L xr == 0

(mod 2n+l)

i=O

and

For simplicity, let Xi = 2i~1 and Yi = 2n-1i-l' Then

which combined with the previous observation yields

= 2n - 1 + V2

2 (

L t=O

XiYi

because the remainders mod 2n of the numbers 2i~1 are the same as the remainders of the numbers 2i + 1. Putting everything together, the result follows. 0 )

L O~i<j~2n-l_l

We will assume that n> 2. Then

>n-1 ,

XiXj'

Remark 3.15. There are incredibly many congruences concerning binomial coefficients. A very good exercise for the reader is to prove the following theorem of Ljunggren from 1952: for any p ~ 5 and any integers a and b, the number (~~) - (~) is a multiple of p3. A more difficult result, due to Zieve

(1999) is the fact that (~;:=~) (~) is a multiple of p3n for any n ~ 1 and any integers a, b, if p is a prime greater than 3.

Chapter 3. Look at the Exponent

132

The last problem of the chapter is quite technical, but also very beautiful. A much weaker result was proposed as problem 3 at the IMO 2008. This kind of result is actually very old and classical. 28. Prove that for any c > 0 there are infinitely many n such that the largest prime divisor of n 2 + 1 is greater than cn. Chebyshev, Nagell Proof. Let f(n) be the largest prime factor of Pn = (12+1)(22+1)'" (n 2 +1). We will prove that lim f(n) = 00. This is enough to deduce the desired result: n-+oo

indeed, choose any C > 0, then for sufficiently large n we have f(n) > cn. By definition there exists k n ::; n such that f (n) Ik~ + 1. Then the largest prime factor of k~ + 1 is greater than ckn and we are done (note that k n -+ 00 as n -+ 00, since f(n) divides k~ + 1). To prove the result on f(n), we will bound the p-adic valuation of prime factors of Pn . Let An be the set of prime numbers smaller than or equal to n and such that p == 1 (mod 4). Any prime factor of Pn is in the set A n 2+1' The following lemma is the first crucial ingredient: Lemma 3.16.

3.7.

Miscellaneous problems

133

as p does not divide j, k and p > 2. This being said, if jl is the smallest index with piljr + 1, then all k such that pilk2 + 1 are of the form jl + spi for some o ::; s ::; n;!l or of the form - jl + tpi for some I t ::; . We easily

;/1 ::;

deduce that ni ::; 1 + ~r; when p is odd and the result follows by adding up these estimates. 0 Using the previous observations, we can write (CI > 0 is an absolute constant and 7r(n) is the number of prime numbers not exceeding n) 2logn! < logPn

I.:

vp(Pn)

logp + V2(Pn ) log 2

pSf(n) PEAn2+1

::; I.:

2 (lOgp(n + 1) + p 2::, 1) logp + CIn

pSf(n) PEAn2+1

a) For any odd prime p we have Vp(Pn) ::;

2 2n logp(n + 1) + - - .

p-1

Now, corollary 3.A.2 shows that 7r(n) ::; C21o~n for all n and some absolute constant C2 > O. The crucial fact is that "

logp

~ p-1

= log f(n) 2

0(1)

41p-1

Proof. The second part follows from the fact that the factors of Pn alternate 2 (mod 8) and 1 (mod 4). To prove a), let ni be the number of j E {I, 2, ... ,n} such that pi jj2 + 1. Clearly [1~gp(n2+1)1

vp(Pn)

I.: vp (k k=1

+ 1) =

I.:

niÂˇ

i=1

Note that if p > 2 and pi divides both j2 + 1 and k 2 + 1, then pi divides j - k or j + k. Indeed, pi divides (j k) (j + k) and p cannot divide both j - k, j + k

pSf(n)

a nontrivial result that follows from our proof of Dirichlet's theorem (see addendum 7.A for more details). Taking this into account and dividing the previous inequality by n log n yields log n! 3c2f(n) 2--< n log n - n log f (n)

1 ) + loglogf(n) +0 ( -. n log n

Combining this with the fact that limn-+ oo ~~~;~ want here.

1 easily yields what we

Chapter 3. Look at the Exponent

134

Remark 3.17. Nagell [60] actually proved the following general result: if f E Z[X] is not a product of linear factors with integer coefficients and if P(x) is the largest prime factor of f(1)f(2)ÂˇÂˇÂˇ f(x), then there exists e > 0 (depending on f) such that P(x) > ex log x. The proof consists in a refinement of the previous method, combined with rather deep estimates in analytic and algebraic number theory (more precisely, the prime ideal theorem). Erdos [27] proved that there exists e > 0 depending on f such that P( x) > x(log x ) clog log log x. It is a very deep theorem of Hooley [41] that there exists e > 0 such that for all x > 2 the 11 largest prime factor of TIn<x(n2 + 1) is greater than exIO. The proofs of these theorems are very involved.

3.8

Notes

We would like to thank the following people for providing solutions to some of the problems in this chapter: Xiangyi Huang (problems 8,26), Ofir Nachum (problem 10), Fedja Nazarov (problem 8), Richard Stong (problems 3, 11, 20, 25, 27), Victor Wang (problems 8, 21), Gjergji Zaimi (problems 3, 4,25), Alex Zhu (problem 27).

Addendum 3.A

Classical Estimates on Prime Numbers

Since we are using quite a lot of estimates about prime numbers in various places of this book, gathering these results in one addendum seemed more than appropriate. All results here are absolutely classical and go back to the beautiful ideas of Chebyshev, who was probably the first person to put some order in the chaotic world of prime numbers. These ideas were revisited by Erdos and their exposition is heavily influenced by this "update." Basically, everything follows from some very smart applications of Legendre's formula to middle binomial coefficients. For the reader's convenience, let us recall some standard notation. We let 7r(n) be the number of primes less than or equal to n. If g is a map taking positive values when the argument is large enough and if f is any complexvalued map, we say that f = O(g) (respectively f = o(g)) if ~~:j is bounded (respectively tends to 0) when x -+ 00. The crucial estimate that we will use when dealing with the behavior of 7r( n) is the following easy consequence of Legendre's formula.

Theorem 3.A.1. For any n 2': 2, ([~]) divides TIp::;np[logpn j and is a multiple

of TI[n!l]<p::;nP. Proof. The second part follows immediately from the identity (note that n [~]

+ [ntl])

and the fact that TI[n!l]<p::;nP divides the right-hand side and is relatively prime to [~]!. For the first part, Legendre's theorem yields

136

Chapter 3. Look at the Exponent

As for all a, b E 1Ft we have [a + b] - [a] - [b] E {O, I}, all terms in the sum are equal to 0 or 1 and all terms for j and the result follows.

> logp n are zero. Thus vp (([~])) ::; [logp n]

Proof. Using theorem 3.A.1 for N =

(2:),

we obtain

L vp(N) logp::; L [logp(2n)]logp ::; 1f(2n) log(2n). p5.2n

e;)

n Next, N is the largest among the (2;) and I:k = 4 , hence N ~ 2~:1' We obtain logN 2nlog2-log(2n+1) 1f (2n > . ) > - log(2n) log(2n)

Using this inequality as well as 1f(2n+ 1) ~ 1f(2n), a small computation yields the lower bound for n ~ 6 as in the original statement; it is easy to check directly the cases 2 ::; n ::; 5. Using theorem 3.A.1 again yields

n 7r (2n)-7r(n) <

Combined with the obvious inequality 1f(2k+l) ::; 2k, this implies the inequality

II n<p5.2n

which easily implies that 1f(2n)

3.~n. Finally, we have

Before tackling the next result, we will prove a very elementary, yet powerful inequality due to Erdos:

2 we have

n log2 n 6 log 2 - - > 1f(n) > - - - . logn 2 logn

log N

137

The famous (and deep) prime number theorem asserts that 7r(n)~ogn converges to 1 as n --+ 00. The following result gives a uniform lower bound estimate. Of course, it is weaker than the prime number theorem, but it is rather amazing that with so few tools it already gives the "correct" lower bound. Note that log 2 is approximately 0.69.

Corollary 3.A.2. For all n

3.A. Classical Estimates on Prime Numbers

p::;

c:) : ;

4n ,

Theorem 3.A.3. For any n ~ 2 we have I1p:Sn p < 4n-l. Proof. The proof is by strong induction, the inequality being clear for n = 2. Suppose that it holds for all numbers smaller than or equal to n and let us prove that I1p5.n+l p < 4n. If n + 1 is even, this is clear, so suppose that n = 2k. Note that I1 k+25.p5.2k+l p divides (2k:l) , so it is certainly at most equal to (2k:l). Combining this with the induction hypothesis for k gives

the last inequality being a consequence of

2. 4k

= (1 + 1)2k+l = ... + (2k + 1) + (2k + 1) + ... > 2 (2k + 1) k

k+1

A slightly trickier argument yields the following beautiful theorem of Mertens.

Theorem 3.A.4. We have which applied to n = 2k gives

'"' logp L p5.n p

= logn + 0(1).

Chapter 3. Look at the Exponent

138

Proof. We will use the prime factorization of nL Legendre's formula yields logp

t; (1

n-l

7r( n) . log n < log n! - n "~ plogp _ 1<

p5.n

= 0(1),

so it remains to prove that

n-l (

L k=2

p5.n

Using Erdos' inequality IT <n P < 4n , the previous estimates on 7r(n), and the inequalities n log n > log nf> n(log n - 1) (the first one is obvious, the second one follows easily by induction using the inequality log (1 + ~) < ~) yields 8log2>

log k - log(k + 1)

rkÂˇ

Multiplying this by logp and summing over p ::; n yields

II p -

139

The triangle inequality yields

+ lOgn) < vp(n!) < ~1.'

_n_ _ (1 p -1

- log

3.A. Classical Estimates on Prime Numbers

Note that for k 2: 2 logk 1 1 - log(k + 1) = 10g(k + 1)

logp

- - -logn > -l. L p-l p5.n

1 k ) 1 - 1 (kg 1) = loglogn + 0(1). og +

TÂˇ

Hence k

The theorem follows from this estimate and the fact that the series

Ep p~;~)

converges (since p~;~Jb < p~ if p is large enough).

tlogt

The next result is a simple application of Abel's summation and of the previous theorem. Theorem 3.A.5. We have

+ -dt- - ( 1

log k ) log(k + 1)

+1 (log(k + 1) -logt) dt

tlogtlog(k + 1) 1

- k 2(log 2)2' where we have used

L - = log log n + 0(1).

log(k + 1) -logt::; 10g(k + 1) -log(k) = log(1

p5.n p

Proof. Define an = lo~n if n is a prime and 0 otherwise. By the previous theorem there is a bounded sequence r n such that Sn

= a2 + ... + an = log n + r n.

~ _ n Sk - Sk-l = ~ logk

logn

+ n-l Sk.

L k=2

( _1_ _ logk log(k + 1)

::; 11k.

Since the sum of the upper bounds converges, we have n-l ( 1 k ) in dt 1 - 1 (kg 1) = -1k=2 og + 2 t og t

+ 0(1)

log log n

+ 0(1).

Remark 3.A.6. One can actually say much more and the true content of Mertens' theorems is the following: there is a constant c such that

Thus

L p- L p5.n k=2

+ 11k)

p5.n p

log log n

+c+0

(_1_) logn

Chapter 3. Look at the Exponent

140 and if 1= limn--+=

(1 + ~ + ... +

logn), then

The first estimate is not difficult and uses again Abel's summation f~rn:ula combined with basic integral calculus. The second estimate is much trICkIer.

Addendum 3.B

An Introduction to p-adic Numbers

This rather long addendum is an introduction to the wonderful theory of p-adic numbers and their applications. This is a vast subject and the literature concerning it is huge, so that we cannot even properly scratch its surface. However, even a glimpse into the subject reveals amazing things ... For a variety of reasons, reduction mod p is awfully insufficient. The best way to reduce modulo arbitrary powers of a prime p while still working in a reasonable algebraic (and especially analytic) context is using p-adic numbers. Very roughly, a p-adic number is a kind of "analytic function of p" with coefficients taken mod p. So, p-adic numbers will be infinite expansions LkÂť-= akpk, where ak are integers between 0 and p - 1. This is a mixture of the classical idea of decimal expansion and the more analytic Taylor expansion of a nicely behaved complex-valued function around O. The idea of seeing integers as analytic functions of primes is incredibly powerful and appears all the time in modern number theory. Though the best way to define the field of p-adic numbers is by completing

Q with respect to the p-adic absolute value, we will introduce p-adic numbers algebraically and then develop their analytic properties. We think that this is a bit easier to digest. Next, we briefly discuss what happens when one takes a finite extension of the field of p-adic numbers, which gives us the opportunity to discuss valuations and absolute values from a more abstract (and useful) point of view. This discussion reveals a huge complete extension of Qp, denoted <C p and called the field of complex p-adic numbers. Its mere existence has some amazing consequences, for instance a beautiful geometric result of Monsky, concerning tiling of squares by triangles of the same area. After discussing classical analogues of the exponential and logarithm maps, we focus on the p-adic analogue of the complex Gamma function. This is the most technical part of the addendum, but also the most rewarding. It requires a preliminary discussion of Mahler expansions and p-adic integration, which are rather complicated, but once the machine is sufficiently developed one can prove deep congruences in a quite straightforward way. We highly

142

Chapter B.

Look at the Exponent

143

B.B. An Introduction to p-adic Numbers

Theorem 3.B.2. Any nonzero p-adic integer x can be uniquely written = pku for some nonnegative integer k and some unit u.

recommend the wonderful books [13J, [49J and [67J for a much more thorough treatment and many good examples.

3.B.!

Proof. Before going on to the proof, let us characterize the units of 7l,p. This will also play an important role in the proof of the theorem:

Arithmetic of p-adic integers and p-adic numbers

A nice way (though maybe not the most intuitive) to see a p-adic integer is to understand it as a compatible system of residue classes mod pn, for all n. That is, following Hensel, a p-adic integer is a sequence (Xn)n?:l, where each xn is a residue class mod pn, say the class of an integer an and where an+l == an (mod pn) for all n. This simply says that Xn+1 maps to xn under the natural map 7l,jpn+l7l, -+ 7l,jpn7l,. With this description, it is fairly clear that the set of p-adic integers becomes a ring, if we define the addition and multiplication component-wise (i.e. the sum of the sequences (xn)n and (Yn)n is declared to be the sequence (xn + Yn)n, similarly for multiplication). To avoid useless repetitions, let us give a name to such sequences:

Definition 3.B.1. A sequence (Xn)n>l, xn E 7l,jpn7l, is called compatible if x n+1 == Xn (mod pn) for all n, where x: E 7l, is any lifting of xn- Let 7l,p be the ring (for the previously defined operations) of all compatible sequences and call it the ring of p-adic integers.

Note that 7l, lives inside 7l,p, as any integer n can be thought of as the compatible sequence (n (mod pk))k. The map sending n to this sequence gives an injective ring morphism from 7l, to 7l,p. We always identify an integer and its image in 7l,p. It turns out that our new ring 7l,p has very nice properties, both algebraically and topologically, making it by far easier to handle than 7l, (you might have not noticed, but 7l, is a very complicated object, actually... ). We discussed quite a lot the p-adic valuation for integers and rational numbers and we will see that it naturally extends to 7l,p, making 7l,p a beautiful place to do analysis. However, we need some preliminaries in order to make this dream reality. The following result is crucial for the arithmetic of 7l,p and shows the big difference between 7l, and 7l,p as far as arithmetic is concerned. Recall that a unit in a ring is an element that has a multiplicative inverse in that ring.

Lemma 3.B.3. A compatible sequence (xn) defines a unit in 7l,p if and only if its first component is nonzero. Proof. One direction being obvious, let us assume that the first component is nonzero. By compatibility, all Xn are relatively prime to p, thus their classes mod pn are invertible. Simply choose Yn to be the inverse of Xn mod pn and check that it forms a comp?-tible sequence, which is the inverse of x by construction. 0

To prove uniqueness of the representation given in the theorem, we need the following easy

Lemma 3.B.4. If x E 7l,p and pmx = 0 then x

= o.

Proof. By induction on m we may assume that m = 1. Next, write x as a compatible sequence (Xn)n and observe that the condition px = 0 simply says that PXn = 0 in 7l,jpn7l,. This means that pn-1l xn for each n 2: 1. But since pn divides Xn+l - Xn , we see necessarily that pnlxn, so in fact all components of x are zero. 0

Assume now that x = pku = plv for some u, v units and some nonnegative integers k,l. If k > l, lemma 3.B.4 yields pk-lu = v. As v is invertible (in other words a unit), we have a contradiction with lemma 3.B.3. Similarly, we cannot have k < l, so k = t. Applying lemma 3.B.4 once more, we get u = v, which proves the uniqueness part of the theorem. To prove the existence, write x as a compatible sequence and let m be the largest integer j such that Xj == 0 (mod pi). Then Yn = Xn-;;,m are integers, since by compatibility Xn+m == Xm == 0 (mod pm). More6ver, since Xn is compatible, so is Yn. Then by construction (and the compatibility of (xn)n), the sequence Yn defines a p-adic integer Y such that pmy = x. We claim that Y is a unit, which will finish the proof of the first part of the theorem. But

Chapter 3. Look at the Exponent

144

the first component of Yn does not vanish, so the result follows from lemma 3.B.3. 0 Here is an important consequence of the theorem:

Corollary 3.B.5. Zp is an integral domain. In other words, if ab = 0 then a = 0 or b = O. Proof. This follows immediately from the previous theorem and the second lemma in its proof. 0 If R is an integral domain, one can define its field of fractions: formally, it is the set of all symbols ~ with a, b E Rand b -I 0, it being understood that we identify ~ and if ad = bc. Addition and multiplication of fractions being done as in elementary school, this yields a field. Applying this to Zp, we obtain the field of p-adic numbers.

Qp =

{~Ia, b E Zp, b -I O}

{;n la E Zp, n E N},

the second equality being a consequence of theorem 3.B.2.

3.B.2

The p-adic valuation revisited

We will give a more analytic flavor to Qp, by endowing it with an absolute value, which plays the same role as the usual absolute value on real numbers.

Definition 3.B.6. Let x E Qp - {O} and write (according to theorem 3.B.2) x = pku for a unique unit u and a unique integer k. Call k = vp(x) the padic valuation of x and Ixlp = p-vp(x) the p-adic absolute value of x. Define 10lp = O. The following is an immediate consequence of the definition:

Proposition 3.B. 7. For all x, Y E Qp we have IxYlp = Ixlp . IYlp and Ix + yip::; max(lxl p, IYlp), with equality if Ixlp on Q c Qp.

-I

IYlpÂˇ Moreover, I . Ip extends the p-adic absolute value

3.B. An Introduction to p-adic Numbers

145

Note that the inequality Ix + yip ::; max(lxlp,lylp) satisfied by the padic absolute value is stronger than the usual triangle inequality for real or complex numbers. This has a whole variety of consequences, which make padic numbers a rather exotic object from a geometric point of view. However, this will not affect us, since we will deal with analytic aspects and for that it is enough to introduce a distance on Qp, a measure of how close numbers are to one another. The distance between x, Y E Qp is defined by d(x, y) = Ix - yip. We can then define analytic objects as in the "real world" (i.e. in the world of real numbers, but we will leave it to you to decide whether that is really the real world ... ). For instance, there are obvious notions of convergent sequences, continuous functions, etc. Basically, any real-analytic object has a p-adic counterpart. Just to see how it works, let us give the definition of a convergent sequence:

Definition 3.B.8. Say a sequence of p-adic numbers Xn converges to a p-adic number a if IX n - alp converges to 0 in the usual sense, that is for all N > 1 there is no such that IX n - alp < ljN for all n > no. Intuitively, the sequence Xn converges to a if the difference Xn - a is more and more divisible by p when n is large, that is if vp(x n - a) goes to infinity as n --+ 00. If you think of a p-adic number as a compatible sequence, this means that for any k, if n is large enough (depending on k) then the first k components of Xn - a are zero. The following result is absolutely fundamental:

Theorem 3.B.9. If Xn E Qp converges to 0 then the series Ln>o Xn converges in Qp. That is, the sequence whose general term is Xo + Xl + ... Xn converges in Qp.

Note that this is NOT true for real numbers (think about the harmonic series!). Also, note the following important consequence: a sequence Xn E Qp converges if and only if Xn - Xn-l tends to 0 in Qp, a fact that will be used a lot in subsequent sections. Proof. Write Sn = Xo + Xl + ... + Xn , so that Sn - Sn-l goes to O. Note that we may assume that all Xn are p-adic integers: indeed, since Xn goes to 0, Xn is a p-adic integer for n large enough. Multiplying all Xn by the same large power of p so that the first terms also become p-adic integers does not affect the

Chapter 3.

146

Look at the Exponent

3.B. An Introduction to p-adic Numbers

147

hypothesis or the conclusion. Next, write Si = (Sil' Si2, ... ) as a compatible sequence. Thinking of these sequences as infinite rows of some infinite matrix, the crucial fact is the following:

Finally, let us give another fundamental property of p-adic integers, which shows that they are basically "formal power series in p" or "infinite base-p expansions."

Lemma 3.B.I0. For any n there exists k n such that Sin = Sjn for all i, j ::::: k n . That is, every column of this infinite matrix eventually becomes constant.

Theorem 3.B.12. For any p-adic integer x there exists a unique sequence an E {O, 1, ... ,p - 1} such that 00

Proof. Indeed, note that for i

> j we have

_""' n xLanp.

n=O

and the last quantity goes to infinity as j -+ 00. Thus for i we have Vp(Si - Sj) > n, which implies that Sin = Sjn.

> j large enough

By definition, this equality means that the sequence whose general term is ao + alP + ... + anpn converges to x. Moreover, if an is the first nonzero term of this sequence, then vp(x) = n.

This lemma gives us a candidate for the limit of the sequence Sn: define the sequence a = (ih, 0,2, ... ), where an is the common value of the elements Sin for i large enough (using the notations of the lemma we have an = Sknn). It is then easy to check that this sequence is compatible and defines a p-adic 0 integer which is the limit of the sequence Sn.

Proof. If x is a p-adic integer, there exists a unique ao E {O, 1, ... ,p - 1} such that x ao E pZp. Indeed, it is clear that ao has to be (the lifting to {O, 1, ... ,p - 1} of) the first term of the compatible sequence x. Using this remark, we deduce by induction that for any n there are unique ao, aI, ... ,an E {O, 1, ... ,p-1} such that x-(ao+alP+" '+anpn) E pn+lzp. But this implies that x = limn-too(ao + alP + ... + anpn). The rest is essentially immediate using lemma 3.BA and theorem 3.B.3. 0

The following basic result will be used frequently below. So any p-adic number x can be uniquely written as a Laurent series Proposition 3.B.l1. Zp is a compact subset of Qp. Proof. Consider any sequence Xn of elements of Zp. Since the first component of Xn (seen as a compatible sequence) takes only finitely many values, there exists a subsequence x<pl(n) and an integer al such that x<pl(n) == al (mod p) for all n. The same argument yields a subsequence x<pl(<P2(nÂť and an integer a2 such that x<pl(<P2(nÂť == a2 (mod p2) for all n, etc. Considering

we obtain a subsequence such that x<p(n) == ak (mod pk) for all n ::::: k. It follows that (ak (mod pk))k is a compatible sequence, defining a p-adic integer a. By construction, we have limn--+oo x<p(n) = a and the result follows. 0

for some N and some ak E {O, 1, ... ,p - 1}. Moreover, we have the following nice criterion to establish when x E Q. The proof is a bit tricky. Proposition 3.B.13. The p-adic number

is a rational number if and only if the sequence (ak)k becomes periodic from a certain point.

148

Chapter 3. Look at the Exponent

Proof. It is immediate to check that if (ak)k is eventually periodic, then x is rational (simply because pa + p2a + ... = 1~~a in Qp for any a > 0). The amusing point is proving the converse. Multiplying x by a power of p, we may assume that x E Zp, say x = "Ek2:0 akpk. Write x = ~ for some relatively prime integers u and v and consider the sequence Xk = "E j 2:k ajpJ-k. Then clearly Xk = ak + PXk+l· As Xo = x is rational, it is clear that all Xk are rational. But much more is true: we claim that we can find Yk E Z such that IYkl :S max(lul, Ivl) (using the ordinary absolute value!) and Xk = Y:;. Indeed, if this holds for Xk, then we can take Yk+ 1 = Yk ~vak (clearly IYk+ 11 :S max(lul, Ivl); to see that Yk+l E Z, note that Xk - ak E pZp, so p must divide Yk - vak). Now, the sequence (Ykh is a bounded sequence of integers, so we can find i < j such that Yi = Yj. Then Xi = Xj and by uniqueness (proved in the previous theorem) we must have ai+l = aj+l, ai+2 = aj+2, .... This finishes the proof. D The following is also absolutely crucial. It basically says that in many cases solving a polynomial equation in p-adic numbers is the same as solving it mod p, since any solution mod p will automatically lift to a compatible sequence of solutions mod pn.

Theorem 3.B.14. (Hensel's lemma) Let f E Zp[X] and let a E Zp be such that If (a) Ip < 1 and If' (a) Ip = 1. Then there exists unique b E Zp such that f(b) = 0 and Ib - alp < 1.

Proof. We will prove by induction that one can find a sequence of p-adic integers an with ao = a, an+! == an (mod pn+l) and vp(f( an)) 2: n + 1. By the previous theorem, the sequence an will converge to a p-adic integer band since vp(f(a n )) 2: n + 1 and f(a n ) converges to f(b), then f(b) = O. To prove the existence of a sequence an, assume we constructed ao, ... , an and search for an+! = an + pn+lbn for some p-adic integer bn . We need to ensure that f(a n + pn+ 1 bn ) == 0 (mod pn+2). Since . f(a n + pn+lbn ) == f(a n ) + pn+lbn f'(a n )

(mod pn+2),

it is enough to take bn such that f(a n )+pn+lbn f'(a n ) == 0 (mod pn+2), which is possible as f' (an) is a unit (because an == ao (mod p) and f' (ao) is a unit). D

3.B. An Introduction to p-adic Numbers

3.B.3

149

Absolute values and their extensions

Definitions and Ostrowski's theorem Let us start with an easy observation: Qp is not algebraically closed. Indeed, the equation x2 = p has no solution in Q , since if x E Q satisfies . of pthe form p-a for an P x 2 -_ p, th en p -1 -_ Ip Ip = Ix 21 p = Ixlp2 and Ixlp IS integer a, a contradiction. Thus, it is meaningful to study finite extensions of Q as . f· d p, one IS 0 ten mtereste in solving polynomial equations over Q . It turns out that all finite extensions of Qp also have natural absolute vah:es that extend the absolute value of Qp, though this is not trivial at all. It is thus better to abstract the situation, using the following

Definition 3.B.15. 1. An absolute value on a field K is a map 1·1 : K --t lR.+ such that Ixl = 0 if and only if x = 0, IxYI = Ixl . IYI and Ix + yl Ixl + lyl· The absolute value is called non-archimedean if Ix + yl max(lxl,lyl)· The absolute value is called trivial if Ixl = 1 whenever x :f: O.

:s :s

2. Two absolute values are called equivalent if there exists c Ixl2 = Ixli·

> 0 such that

3. ~ valuation. on a field K is a map v : K --t lR. U {oo} such that v(x) = 00 If and only If x = 0, v(xy) = v(x) + v(y) and v(x + y) 2: min(v(x), v(y)). It is clear that if I . I is an absolute value, then v(x) = -log Ixl defines a valuation. If 1·1 is a non-archimedean absolute value on K, the ring of integers of K is by definition OK = {x E Kllxl :S 1}. It is easy to check that this is a ring and that mK = {x E Kllxl < 1} is a maximal ideal of OK. Thus kK = OK/mK is a field, called the residue field of K. It is clear that any non-archimedean absolute value is bounded by 1 on Z, but the nice and somewhat tricky fact is that the converse holds. Indeed,

150

Chapter 3. Look at the Exponent

Ixlklyln-k :::; (n

+ 1) max(lxl, IyltÂˇ

k=O

Taking the nth root of this inequality and letting n -t 00 yields Ix + yl :::; max(lxl, Iyl), proving the claim. With these remarks being made, we are ready to prove the following beautiful result: Theorem 3.B.16. (Ostrowski) Any nontrivial norm on Ql is equivalent to the p-adic absolute value for some prime p or to the usual absolute value. Proof. Suppose first that the absolute value 1Âˇ1 is non-archimedean. We know that Imlnl = Iml/lni oil for some nonzero m, n E Z. Without loss of generality, then, let Iml < 1, so that for some prime plm we have Ipi < 1. If q is another prime with Iql < 1, then we may find integers a and b with ap+bq = 1, and then

1 = 111

lap + bql :::; max(lapl, Ibql) :::; max(lpl, Iql)

151

Now, we claim that for any integer x > 1 we have Ixl > 1. Indeed, if Ixl :::; 1, by!writing n j in base x and using the same argument as before, we deduce that

if Inl :::; 1 for all n, then for all x, y and all n we can write

: :; L

3.B. An Introduction to p-adic Numbers

< 1,

a contradiction. We conclude that Inl = Iplvp(n), and so I . I is equivalent to the p-adic absolute value. The difficult case is when I . I is archimedean. We saw that in this case there is an integer n > 1 such that Inl > 1. Pick any such n and write for all x> 1 the number x in base n, say x = Xo + Xln + ... + xknk. Note that k :::; logn x and that if en = maxl:Sj:Sn-l Ijl, then

for some constant A, independent of x. Applying this to x N for N large enough yields Ixl :::; x 10gn Inl.

As Inl > 1, this is certainly not true for j large enough, proving the claim. We have, therefore, that for all x, n > 1 both Ixl :::; x 10gn Inl and Inl < n10gx lxi, so that (e.g.) the first inequality is in fact an equality. This implies d that logn Inl is a constant function of n > 1. Thus, there is d such that Inl = n for all integers n > 1 and the result follows. 0 Extensions of absolute values We prove now the following fundamental and nontrivial theorem. The proof is pretty acrobatic and uses a nice mixture of analytic and algebraic arguments. Theorem 3.B.17. Let K be a finite extension of Qlp. There is a unique extension of the absolute value on Qlp to an absolute value on K. This absolute value is non archimedean and it is given by

if x E K, where n

[K : Qlp] and NK/Qp is the norm.

Proof. We prove uniqueness first. We claim that for any two absolute values I . 11, I . 12 on K that extend I . Ip on Qlp, one can find Cl, C2 > 0 such that cli xiI:::; IX12 :::; c21 x iI for all x. Assume that this happens for a moment. Then cllxl~

cllxnh :::; Ixl2 :::; c2Ixl~,

and taking nth roots and letting n -t 00 we get IxiI = Ix/2, proving the uniqueness part. Now, to prove the claim, it is enough to prove the following: if el, e2, ... ,en is a basis of Kover Qlp and if we define

Chapter 3.

152

Look at the Exponent

then there are Cl,C2 > 0 such that cllxl oo S; Ixl! S; c21x1 00 . But clearly for n

X = LXiei i=l

we have

Ixl1 S ~ IXilp 路Ieil, S

so we can take C2

= L: leil!. S

(~leiI1) Ixl

oo ,

Obtaining Cl is more subtle. Let =

3.B. An Introduction to p-adic Numbers

153

Taking dth roots and letting d ---700, we obtain Ix+yl S; max(lxl, Iyl), finishing the proof. It remains to prove the existence of c. This is similar to the first part of the proof. Namely, let el, ... , en be a basis of Kover Qp and let 1路100 be as above. As the norm of an element of Qp is a polynomial expression of the coordinates of that element in the basis el, ... ,en, it follows that x ---7 Ixl is a continuous map. Since it does not vanish on the compact set {xllxl oo = I}, there are positive numbers a, b such that a S; Ixl S; b whenever Ixl oo = 1. An obvious scaling argument implies that alxl oo S; Ixl S; blxl oo for all x, from where

{x E Kllxl oo = I}.

If we equip K with the product topology induced from its vector-space structure over Qp, then S is easily seen to be a bounded, closed subset of K, whence compact. Moreover the map x ---7 Ixll is continuous on S, as say

Because this map does not vanish on S, there is Cl > 0 such that Ixll 2 Cl for E K can be scaled by an element of Qp to become an element of S, the claim is proved. Existence is harder. Defining Ixl = yflNK/lQp (x) Ip, standard properties

whenever Ixl S; 1. The existence of c is thus proved and we are done.

The uniqueness property in the previous theorem ensures that if K c L are two finite extensions of Qp, then the restriction to K of the unique absolute value on L extending that on Qp is the unique absolute value on K extending that on Qp. This implies the following very useful

xES. As any x

of the norm yield Ixyl = Ixl . IYI, Ixl = 0 {:} x = 0 and Ixl = Ixl p for x E Qp. The difficult point is proving that Ix + yl S; max(lxl, Iyl)路 By multiplicativity, it is enough to prove that Ix + 11 S; max(l, Ixl), which reduces to if Ixl S; 1, then Ix + 11 S; 1. This is however quite subtle. We will actually prove the following result: there exists c > 0 such that Ix + 11 S; c whenever Ixl S; 1. Assume that we proved this for a moment. Applying it to x/y or y/x (according to whether Iyl 2 Ixl or not), we deduce that for all x, y we have Ix + yl S; cmax(lxl, Iyl)路 But then for all d we have

(~)xd-iyil S; c(max(lxl, IYI))d.

Ix + yld = I(x + y)dl S; c max I O::;i::;d z

Corollary 3.B.lS. Fix an algebraic closure Qp of Qp. There is a unique extension of I . Ip to a non archimedean absolute value on Qp. From Qp to

<c p

We have a bad news: after all the hard work in the previous section, we have to tell you that Qp is not a very good object. When dealing with p-adic numbers, analysis is intensively used and finite extensions of Qp are very good places to do analysis because they are complete. This means that all Cauchy sequences converge in such a finite extension (this also happens in lR or <C or in a compact interval, but not in (0,1) for instance: the sequence lin is Cauchy, but does not converge to an element of (0,1)). On the other hand, Qp is not complete (this is not really easy to prove, actually, but those with a good analytic background will observe that it follows immediately from Baire's lemma), so one cannot do reasonable analysis on this field.

154

Chapter B. Look at the Exponent

Let us explain why finite extensions of Q!p are complete, since this is very important. The fact that Q!p is complete was essentially proved while proving theorem 3.B.9. To see that a finite extension K is complete for the unique absolute value I . I extending I . Ip, choose a basis el,···, en of Kover Q!p and define Ixl oo = maxi IXil if x = Li Xiei and Xi E Q!p. This is a norm on K (but not an absolute value) and the same argument as in the first paragraph of the proof theorem 3.B.17 shows that there exist Cl, C2 > 0 such that cllxl ::; Ixl oo ::; c21xl for all x. Thus, the notions of Cauchy sequence and convergent sequence are the same for I . I and I . 100. But it is clear (from the fact that Q!p is complete) that K is complete for 1·100, thus K is complete for 1·1, too. _ Now, we would like to have a field that contains Q!p and which is still complete. It turns out that there exists such a field which is moreover minimal. The situation is very similar to that of Q! endowed with the p-adic absolute value: it is not complete for this absolute value, but adding all possible limits of all Cauchy sequences in Q! one ends up with a much bigger field, Q!p. One can play the same game starting with Q!p and one ends up with a huge field C p, endowed with an absolute value 1·lp extending that on Q!p and having the properties: 1) The field C p is complete with respect to I·\p, that is any Cauchy sequence in C p (with respect to I . Ip) converges in Cpo Just as in theorem 3.B.9, we deduce that if an E C p is a sequence converging to 0, then Ln2':l an converges in C p (the notion of convergence is defined just as for Q!p).

B.B. An Introduction to p-adic Numbers

155

Ixl p = limn-+oo lanl p· One checks that this is well-defined (i.e. independent of the choice of the sequence (an)n) and it is an absolute value that extends the one on Q!p. It is not difficult to prove that C p is complete for this absolute value and essentially by definition Q!p is dense in Cpo It is more difficult to prove that Cp is an algebraically closed field. The previous construction does not use any special property of Q!p except the fact that it has an absolute value. In general, for any field K endowed with an absolute value, one can construct (in exactly the same way as above) a field K that contains K and has an absolute value /·1 extending that of K, such that K is dense in K. This field is called the completion of K. If K is algebraically closed, then K is also algebraically closed, though this is not so easy to prove.

A summary The upshot of this technical section is that finite extensions of Q!p behave as Q!p both algebraically and analytically, that Q!p is a pretty bad field from the analytic point of view and that if one wants to do analysis, then one has to work with its completion Cpo Moreover, in C p a series Ln an converges if and only if an converges to 0 (which means that limn-+oo lanl p = 0) and then one can permute its terms as one wishes and still get the same value of the sum (this is definitely wrong in real or complex analysis!). Finally, one can deal with double sums in a rather leisurely way, since if limmax(m,n)-+oo am,n = 0, then

2) The field C p is algebraically closed, in particular it contains Q!p. Moreover, Q!p is dense in Cpo 3) The residue field of Cp is an algebraic closure of lF p . Here is the way one constructs C p : consider the set C of all Cauchy sequences in Q!p. It is easy to check that this is actually a ring, addition and multiplication being defined component-wise. Next, one checks that the subset m of C consisting of sequences that converge to 0 is a maximal ideal of C. One defines C p = C / m. By definition, this is a field. It has a natural absolute value 1·l p, defined by: if x E C p is the class of a Cauchy sequence (an)n, then

and all series converge. Note that we did not prove the last two assertions, but since they are rather easy, we leave them as good exercises for the reader.

3.B.4

p-adic analogues of classical functions

Recall that for any complex number x, the series Ln>o ~~ converges to a complex number called eX and x·~ eX is a surjective group-morphism C ~ C*. Let us study the p-adic analogue of this construction: the problem is that

156

Chapter 3.

Look at the Exponent

vp(n!) is quite large, so we cannot expect that the previous series converges for all x. Actually, by theorem 3.B.9 the previous series converges for some x E C p if and only if vp (~~) ----t 00. Using Legendre's formula vp(n!) = n-;~in), where sp( n) = O(log n) is the sum of digits of n when written in base p, we deduce that the series converges iff 1 ) lim n vp(x) - --1 n--+oo ( p-

sp(n) + --1 = p-

00,

which happens if and only if vp(x) > P~l' i.e. Ixl < p . Moreover, one can easily check (using the remark on double sums made in the previous section) that if x, y satisfy these conditions, then so does x + y. and eX . eY = eX+Y . It turns out that one can construct an inverse to the exponential map, which is however defined on all Cpo More precisely, we have the following nontrivial Theorem 3.B.19. There exists a unique continuous homomorphism logp C; ----t C p such that logp(p) = 0 and

for

lip

< l.

Proof. (sketch) The proof is pretty long, so we only give the main steps. The crucial point is the following

3.B. An Introduction to p-adic Numbers

157

For uniqueness, it is clear that r = vp(x) is uniquely determined. It is thus enough to check that no root of unity ( of 1 of order prime to p satisfies 11 - (I < 1. If ( has order n > 1, then the norm (from Qlp() to <Qp) of 1 _ ( is a divisor of n, and so cannot be a multiple of p. 0

Now, let us study logp' Let x E and write x = pT . ( . v as in the lemma. Note that if we admit that logp exists, then necessarily N logp () = logp(N) = 0 if (N = 1, so necessarily logp() = o. As logp(p) = 0, we must have V - l)n logp x = logp (v) = ( -1 t-l-'-------'~ n

(

I: n21

This shows that if logp exists, then it is unique. It is harder to prove existence. First, by the previous paragraph we must define V - l)n log x = log (v) = (_I)n-l~-----..:_ p

(

n21

if x = pT. ( . V. Note that the series converges, as

Moreover, since the series converges uniformly, it is easy to see that v ----t logp(v) is continuous for Iv -11 < 1. From here it is not difficult to check that x ----t logp (x) is continuous on It remains to check that it is additive. This easily reduces to

C;.

Lemma 3.B.20. Any x E can be uniquely written x = pT . ( . v for some r E <Q, ( a root of unity of order prime to p and v E C p such that Iv - 11 < l. Proof. Let us prove the existence part. By construction, vp(C;) = Ql, so that given any x E C; there is r E Ql and u E C; such that x = pTÂˇU and vp(u) = O. Consider the image of u in the residue field IFp of Cpo It is a nonzero element of some lF~ for some power q of p. Thus vp(u q - 1 1) > 0 and then easily qn u(q-l)qn ----t 1 as n ----t 00. We see similarly that ( = limn--+oo u converges and then (q-l = 1 with vp(u () > O. So one can take v = u/(.

for lui < 1 and Ivl < 1, which is the tricky point. First, one checks that as formal series in X, Y we have log(1

+ X) + log(1 + Y)

= log(1

+ (X + Y + XY)),

for instance by differentiating both sides in X, respectively Y. Next, the series defining logp(1 + u), logp(1 + v) and logp(1 + u + v + uv) converge absolutely

Chapter 3.

158

Look at the Exponent

and one can permute their terms as one wants, without changing the value of the series. This implies that we can substitute X = u and Y = v in the formal series equality and finishes the proof of the theorem. D With the same arguments as in the proof of the previous theorem we 1 obtain 10gp(e X) = x if Ixl < p-p-l (it is easy to check that lex -lip < 1 for such x) and e10gp (x) = x if x is close enough to 1 so that vp (logp (x)) > p~ 1 . We end this section with another useful p-adic analogue, the binomial functions and power functions. Define, for x E Qp and n 2': 0 x) = x(x - 1)路路路 (x - n (n n!

Proposition 3.B.21.

3.B. An Introduction to p-adic Numbers

Some applications We discuss here some immediate applications of the preceding theoretical results. The reader will probably appreciate better the power of p-adic techniques, since none of the following problems are easy to solve by other means.

If Example

3.B.22. (Kiran Kedlaya, USA TST) Let p > 5 and let p-l

= {; (px: k)2'

Jp(x)

+ 1).

1) (Vandermonde's identity) If x, y E Qp, then

159

Prove that for any integers x, y, p3 divides the numerator of Jp(x) - Jp(y) when written in lowest terms. Proof路 Using the tools previously introduced, this is very simple: working in Qp, we can write p-l

2) If x E Zp, then (~) E Zp for all n.

Jp(x)

'"' 1 ( 1 + PX)-2 ~ k2 k k=l

3) If a E C p satisfies lal p < 1 and x E Zp, define

p-l

Then the series converges and x --t (1 homomorphism from Zp to

c;.

+ a)X

is a continuous additive

(-2)~.J j

kjX

k=l

j?O

p-l

(1 _

== '"' ~2 ~k

2px k

k=l p-l

Proof. 1) If x, yare positive integers, simply compare coefficients in

L k2 L

1 p-l 1 k2 - 2px k3

k=l

The result then follows by density and continility. The same argument works for 2). The convergence of the series in 3) follows immediately from 2) and theorem 3.B.9. The continuity follows from the uniform convergence of the series, while the equality (1 + a)X . (1 + a)Y = (1 + a)x+y follows either by a simple computation using 1) or from the case x, y E {1, 2, ... } by continuity and density. D

k=l

P2x2 ) k2

+ p 2x 2

p-l

1 k4

(mod p3).

k=l

It suffices thus to show that p-l

1 l~k3 '"'

k=l

but these congruences have already been discussed in the solution of problem 22, chapter 3. D

Chapter S.

160

Look at the Exponent

Example 3.B.23. (how not to prove Fermat's last theorem) Let p be a prime and let k, N 2: 1. There exist integers x, y, z, not all of them divisible by p and such that x N + yN == zN (mod pk). N Proof. It is enough to show the existence of x, z E Zp such that x + 1 = zN, since then x (mod pk), 1 and z (mod pk) is a solution. We would like to take z = (1 + xN)I/N. Using the results of the previous section, we are tempted 1

to take z = I:n>O (~) xnN. Unfortunately, N is not necessarily prime to p, so we cannot apply directly those results. However, Vp ((!)xnN) 2: Nnvp(x) - p: 1 -nvp(N)

and this tends to 00 as n ---+ 00 if Nvp(x) > p~l +vp(N). We thus choose such x and define z by the previous series. Then z E Zp (by the previous estimate) N and the usual argument with formal series shows that zN = 1 + x . 0 The next problem has already been discussed in chapter 3, problem 26, where two rather difficult solutions were given. Using 2-adic numbers, it becomes almost obvious.

Example 3.B.24. Write

S.B. An Introduction to p-adic Numbers

3.B.5

161

A geometric application

In this section we reward the reader with a mathematical gem, due to Paul Monsky. This uses the existence of an absolute value on C p extending the one on Qp, a result which was explained in previous sections. It is a nontrivial fact from field theory that C p is isomorphic as a field with Co The choice of an isomorphism allows us to transfer the absolute value on C p to one on 'l.-, rr that still extends the p-adic absolute value on Q. The reader who finds this construction very indirect will probably spend some time trying to construct directly such an absolute value on Co Inevitable failure will probably convince him of the power of the arguments in previous sections. Theorem 3.B.25. (Monsky) One cannot dissect a square into an odd number of triangles of the same area. It is absolutely remarkable that no geometric proof is known for this pretty innocent-looking problem. Monsky's proof (see [53]) is a stunning combination of arithmetic and combinatorics. Proof. Consider the square with vertices (0,0), (1,0), (0, 1), (1, 1). Using the extension of the 2-adic valuation to ]R, color the point (x, y) E ]R2 in red if max(l x I2,lyI2) < 1, in blue if Ixl2 2: max(l, lyl2) and in green if lyl2 > Ixl2 and lyl2 2: 1. We will repeatedly use the easy observation that translation by a red point is color-preserving. Here is the crucial point:

> n - 10g2(n) (this is the

Lemma 3.B.26. If T is a triangle whose vertices have three different colors, then IA(T)12 > 1, where A(T) is the area ofT.

Proof. Let us work in 1:h The series I:n 2: suggests considering 10g2( -1). Indeed, the series defining this is exactly 2:. On the other hand, since 10g2 is additive and since 1)2 = 1 and 10g2(1) = 0, we must have 10g2(-1) = 0, that is in Q2 we have the equality I:n:2:1 2: = 0. But then

Proof. By the remark on translations by red points, we may assume that one of the vertices of T is (0,0). Let b = (b l , b2 ) and c = (CI' C2) be the other vertices, say b is blue and c is green. Then

for relatively prime integers an, bn . Then v2(a n ) ordinary logarithm here!) for n 2: 2.

2:n

IA(T)12 V2

I.: k2k) = V2 (- I.: k2k) n

(

k=l

k>n

2: ~~~ (k -10g2 k) > n -10g2(n).

C = /b1 2 -

o as Ib l 12l IC212 2: 1 and

2C b l/

= 21b l l2 ÂˇI C212 ./1 _ CI . b2 / > 1,

I~Âˇ ~12 < 1.

bl 2

162

Chapter 3. Look at the Exponent

Consider now a dissection of the square into n triangles of the same area, which is necessarily lin. Color only the vertices of the triangles, as above. If we can prove that there is a triangle with vertices of .different colors, we deduce from the previous lemma that Inl2 < 1 and so n is even. The existence of such a triangle is a trivial consequence of Sperner's lemma, but it is perhaps useful to recall how things work in this easy two-dimensional case: consider segments on the boundary of the square whose endpoints are red and blue (i.e. one endpoint is red and the other one blue). All vertices on the edge [0,1] x {o} are either red or blue. All vertices on the edge {o} x [0,1] are either red or green. All vertices on the edges [0,1] x {I} or {I} x [0,1] are either blue or green. Therefore all segments on the sides with one endpoint red and the other blue are on the side [0,1] x {O}. As (0,0) is red and (1,0) is blue, there must be an odd number of such segments. On the other hand, assume that no triangle has vertices of different colors. It is easy to check that all triangles have an even number of sides whose endpoints are red and blue. As the triangles partition the square, we deduce that the number of red-blue segments on the border of the square is even, a contradiction. Thus, there must be a "colorful" triangle and the theorem is proved. D

3.B.6

Mahler expansions

One of the miracles of p-adic analysis is that one has a fairly explicit description of all continuous functions on Zp. Of course, this is far from being true in real or complex analysis, so the following theorem is surprising to say the least. It is however absolutely crucial when dealing with more delicate aspects of p-adic numbers and we will use it constantly in the following sections. Theorem 3.B.27. (Mahler) For any continuous function f : Zp -+ Qp there is a unique sequence (an (J) )n2:0 of p-adic numbers such that limn -+oo an = and for all x E Zp

Â°

Moreover, we have minxEzp vp(J(x))

minn 2:o vp(an(J)).

3.B. An Introduction to p-adic Numbers

163

Proof. Note that if the equality

f(x) =

L an(J) (x)

n2:0

holds for all x E Zp, then for all n

Either by considering the exponential generating function of (J(n))n and (an(J))n or by using the theory of finite differences (see chapter 10, section 10.3), we deduce that

Assume for a moment that we proved that Iimn -+ oo an(J) = 0, which is the difficult point of the theorem. Then, since (~) E Zp for x E Zp, we deduce that g{x) = Ln>o an(J)(~) converges uniformly for x E Zp and so 9 is a continuous function. Moreover, by construction g(n) = fen) for all n ;:::: 1, so by density of {1, 2, ... } in Zp we obtain f = 9 and the first part of the theorem follows. Finally, from the previous relations between the values of f at positive integers and the an (J) we obtain

so another density argument yields minxEzp vp (J (x) ) ';" minn 2:o vp (an (J) ). Note that those minima exist, as vp{an(J)) diverges to 00 and as f is continuous on the compact set Zp. It remains to prove that vp(an(J)) -+ 00. As f is bounded (because it is continuous and Zp is compact), by multiplying f by some power of p we may assume that f(Zp) C Zp. As Zp is compact, f is uniformly continuous on Zp, so there is no such that vp(J(x + pnO) - f(x)) ;:::: 1 for all x E Zp. Let

b.f(x) = f(x

+ 1) -

f(x),

Chapter 3. Look at the Exponent

164

then

t:,.n f(x)

-It- k

(~) f(x + k)

k=O

and an(f) = t:,.n f(O). As p divides (P~o) for all 1 ::; k < pno, it follows that vp(t:,.pno f(x)) 2': 1 for all x E tl p and so vp(t:,.n f(x)) 2': 1 for all n 2': pno and all x. The map g = 1t:,.pn 0 f is continuous and g(71p) C 7l p. Applying the same argument to g, we find nl such that vp(t:,.pnl g(x)) 2': 1 for all x. Then vp(t:,. n f (x)) 2': 2 for all n 2': pno + pnl. Continuing like this, we find integers ni such that vp(t:,.n f(x)) 2': d for all n 2': pno + ... + pn d - 1 and all x E 7l p. Taking x = 0 shows that vp ( an (f)) ---7 00 and finishes the proof. 0

Remark 3.B.2S. The numbers an(f) are called the Mahler coefficients of the function f. We discussed only the case of Qp-valued functions, but the result holds for K-valued functions, where K is any complete subfield of C p, with basically the same proof. Here is a nice application of the previous theorem, proposed at the USA TST 2011 by Josh Nichols-Barrer. "V Example 3.B.29. Let p be a prime .. We say that a sequence of integers {Zn}~=~ is a p-pod if for each e 2': 0, there IS an N 2': 0 such that whenever m 2': N, P divides the sum

165

3.B. An Introduction to p-adic Numbers

continuous functions on tl p is clearly a continuous function, from which the result follows. 0 One also has a characterization of locally analytic functions f : tlp ---7 Qp in terms of their Mahler coefficients, though the proof is much more difficult. Recall that a function f : tl p ---7 Qp is called locally analytic if for any a E 7lp there exists na and p-adic numbers f(n) (a) such that whenever vp ( x - a) 2': na we have

Theorem 3.B.30. (Amice) A continuous function f : tl p ---7 Qp is locally analytic on tl p if and only if its Mahler coefficients an(f) satisf'll

3.B.7

The p-adic Gamma function: preliminaries

Recall that the Gamma function is defined for Re( s)

> 0 by

f)-l)k(~)Zk' k=O

Prove that if both sequences {xn}~=o and {yn}~=o are p-pods, then the sequence {xnyn}~=o is a p-pod. Proof. See the sequence Zn as a map on N in the obvious way. The Mahler coefficients of this map are precisely the numbers

am(z)

f)-l)k(~)Zk' k=O

By hypothesis, Z is a p-pod if and only if am(z) tends to 0 in Qp and by Mahler's theorem this happens if and only if Zn extends to a continuous function on tl p (namely x ---7 Ln2:0 an (z) . (~)). But the pointwise product oftwo

It has the nice property that it interpolates the numbers n!, as r(n) = (n -I)! for any n. One would like to have a p-adic analogue of the Gamma function, as this function plays an amazingly important role in real and complex analysis. Unfortunately, it is not difficult to see that there is no continuous function f : tl p ---7 Qp such that f (n) = (n - I)! for all positive integers n. On the other hand, we have the following beautiful result, due to Morita. To simplify the exposition, we will assume from now on that p > 2. There are natural extensions of all the following results to the case when p = 2, but that would force us discuss two cases in both the statement and proof of the following results. 3This means that there is r

> 0 such that

(an (f))

rn for all sufficiently large n.

Chapter 3. Look at the Exponent

166

Theorem 3.B.31. (Morita) There is a unique continuous map r p : Zp -+ Qp such that for all n 2': 2 we have

n-1 rp(n)

II j=l

= (-It

gcd(p,j)=l Proof.

Defi~ng

n-1 g(n)

II j=l

= (-It

gcd(p,j)=l for n 2': 2, let us prove the following

Lemma 3.B.32. g(n + pk) == g(n) (mod pk) for all n and all k 2': 1. We have

167

3.B. An Introduction to p-adic Numbers

However, as p > 2, this appears precisely when 9 = 1 or 9 = -1. Thus, the product of all g's equals -1 and we are done. The previous lemma easily implies that vp(g(m) - g(n)) 2': vp(m - n) for all distinct positive integers m and n. Choose any p-adic integer a and any sequence Xn of positive integers such that limn---+= Xn = a in Zp. Since Vp(g(Xi)-g(Xj)) 2': Vp(Xi-Xj), it follows that the sequence (g(xn))n is a Cauchy sequence and so it converges to some p-adic integer g( a). If Yn is another sequence that converges to a, then applying the result we have just obtained to the sequence X1,Yl,X2,Y2, ... , we deduce that g(Yn) converges to g(a), i.e. g(a) is independent of the choice ofthe sequence (xn)n. Thus, we obtain a map rp : Zp -+ Zp which clearly extends g. Passing to the limit in the inequality vp(g(m) - g(n)) 2': vp(m - n), we deduce that vp(r p(x) - r p(Y)) 2': vp(x - y) for all x, Y E Zp, showing that r p is continuous. This proves the existence of r p' The uniqueness part is a trivial consequence of the density of N in Zp. 0 The following proposition summarizes the basic properties of r po We will prove much deeper results in later sections, once we have developed enough tools.

Proposition 3.B.33.

1) For all positive integers n we have

( ) -_ ( -1 )n+1 rpn+l

so it is enough to check that

n+pk-1 pkll

j=n gcd(j,p)=l

n+pk-1 (

j=n gcd(j,p)=l

) j

[~]! .p

[n]' p

r p(Zp) c Z;.

3) IfTp(x) = -x for x E Z; and Tp(X) = -1 for x E pZp, then

But if 7r : Z -+ ZjpkZ is the natural reduction map, it is clear that

II g,

rp(x + 1) = Tp(X)rp(x). 4) If x E Zp and r(x) E {I, 2, ... ,p} is the unique integer such that

gEG

where G = (ZjpkZ) *. The elements 9 in the previous product come in pairs (g,g-l), but one has to pay attention to the fact that one might have g2 = 1.

x - r(x) E pZp, then

168

Chapter 3.

Look at the Exponent

Proof. 1) This follows immediately by definition of the p-adic Gamm~uncÂ tion. 2) By construction, vp(rp(n)) = 0 for integers n 2': 2. As these integers form a dense subset of Zp and as vp 0 r p is continuous, 2) follows. 3) This follows immediately from the definition if x is a positive integer. The general case follows by density and continuity. 4) By density and continuity, it suffices to prove that rp(-n)rp(n+ 1) = (-It+1-[n/ p]

3.B. An Introduction to p-adic Numbers

Proof. This is very easy using Mahler's theorem 3.B.27. Namely, look for

and observe that Sf (x

1) -

Sf (x) =

II Tp(-J) j=l II(-l) II

plj

gcd(p,j)=l p = (-1)[n/ ](_1)n+1r p(n+ 1)

and the result follows.

3.B.8

Mahler expansions and discrete antiderivatives

Exploiting Mahler's theorem, we develop basic p-adic calculus in this section. This will then be applied to the p-adic Gamma function and then to establish some fairly deep congruences. We start by proving that any continuous p-adic function has a continuous (discrete) antiderivative. Theorem 3.B.34. Let f : Zp ---+ C p be a continuous function. There exists a unique continuous function Sf: Zp ---+ C p such that S f(O) = 0 and Sf(x

+ 1) -

Sf(x)

f(x)

for all x E Zp . Moreover an (S J) = a n -1 (J) for n Mahler's coefficients of g.

n20

Taking into account the condition that S f(O) = 0, we have bo = O. Also, Sf satisfies the desired equation if and only if bn+1 = an(J) for all n 2': O. Thus any solution must satisfy an(SJ) = an-l(J) for n 2': 1 and ao(SJ) = 0, yielding uniqueness. But this also gives the existence, since if an (J) ---+ 0, then a n -1 (J) ---+ 0 and so Sf (x) = Ln20 bn (~) is a continuous function if f is. D

= Tp(-j)r p( -j)

from 3) yields

r (_ ) =

L bn (n ~ 1) = L bn+ (~)

n21

for positive integers n. But multiplying the relations

rp(l - j)

169

Next, we define the notion of integrable and C 1 class functions. Unfortunately, there isn't really a very good notion of p-adic integration and each such construction has its limitations. For instance, the one we have chosen in this book has the drawback that not all continuous functions are integrable. But since we will deal only with sufficiently nice functions, this will be enough for our purposes. Definition 3.B.35. 1) A function f : Zp ---+ Cp is called Volkenborn integrable (or simply integrable) if pn_1

f(x)dx = lim

n-*oo

pn ~

f(j)

j=O

exists in Cpo 2) Say a continuous function f : Zp ---+ C p is of class C 1 if

1, where an (g) are

Chapter 3. Look at the Exponent

170

These definitions might seem a bit strange at first. However, the average whose limit defines the integral of f should be seen as a Riemann sum. So Volkenborn integrable functions are precisely those for which Riemann sums converge. The definition of C 1 class functions is a bit more subtle, but one can prove (although this is nontrivial) that it agrees with the intuitive definition. We will prove part of this assertion when proving the next theorem. The following remark will playa very important role in the proof. Remark 3.B.36. If x

i- y E Zp,

3.B. An Introduction to p-adic Numbers

171

1) f is continuously differentiable, i.e. f'(x) Cp

for all x and x

= limu-tx f(u)-f(x) exists in u-x

f'(x) is continuous.

2) We have S(f')(x) =

(f(x

+ u) -

f(uÂťdu.

Proof. The observation that

then

Indeed, Vandermonde's identity (proposition 3.B.21) yields

1 L --n

p k2:1

ak-1 (f) (pn) k

~ ak-1(f) (pn

~ k=l

k-l'

yields the estimate which immediately yields the result (using the fact that (~) E Zp if x E Zp). Note that Ilcm(l, 2, ... , n)lp -_ p -[logp n] > _.!., n so we obtain the very useful estimate

The following result is fairly technical, so the reader might want to skip the proof at a first reading.

Theorem 3.B.37. A function f of class C 1 is integrable and

where x

n,k

= lak-I(f)lp I(pn

Ikl p

-1) _

k - 1

. p

Thus, to establish the first formula of the theorem, it is enough to check that sUPk xn,k tends to as n -+ 00. Clearly, Ikl p :::; k, so Xn,k :::; klak-1 (f)lp.

Â°

Moreover, since (_I)k-1 = (k~ll)' remark 3.B.36 yields Xn,k :::; lak-1(f)lp;~, Putting everything together, we deduce that Xn,k :::; klak-1(f)lp' min

Moreover, in this case we also have:

(_I)k-l/

(1,;) .

Combined with klak(f)lp -+ 0, this easily proves that sUPk xn,k -+ 0, establishing the first formula.

172

Chapter B.

Look at the Exponent

1) By remark 3.B.36, there are polynomials Pn E Q[X, Y] such that

(~) - (~) =

B.B. An Introduction to p-adic Numbers

173

It is apparent from this formula that G is continuous. Finally, if x 2:: 1 is an integer, we can write by definition

Pn(u, x)

u-x

G(x)

for all u,x and IPn(u,x)lp:; n. Since nlan(f)lp -t 0, it follows that the series

= n--+oo lim ~ pn

pn_1

L (f(x j=O

1 x-I L(f(pn j=O

= n--+oo lim pn

+ j) -

f(j))

+ j) -

f(j))

x-I

converges uniformly and its limit as u -t x is ~n>O an(f)Pn(x, x), which is continuous (as the series converges uniformly). 2) First, we compute the Mahler coefficients of gx(u) = f(x + u) - f(u) by using Vandermonde's identity 3.B.21: an (f) (x: u) - L

gx (u) = L n20

an (f)

(~)

n20

L an (f) tt=1 (~) (n ~ i)

n2I

L 1'(j) = S1'(x). j=O

The conclusion follows now by continuity of G and S l' together with the density of the set of positive integers in Zp. D

3.B.9

Application to the p-adic r-function

We are now ready to prove the following deep theorem, that will allow us to prove very difficult congruences: Theorem 3.B.38. logp or p is an analytic function on pZp and has a powerseries expansion

where

G(x)

r }Zp

(f(x

+ u) -

f(u))du

in terms of Mahler coefficients of gx, using the first formula of the theorem:

r }z*

x- 2n dx

= lim ~ k--+oo

pk_1

L.t t=1 gcd(i,p)=1

zÂˇ-2n .

Proof In this proof we will use the simpler notation Ixl for Ixlp. Using proposition 3.B.33, we deduce that if f(x) = logp(rp(x)), then f(x

+ 1) -

f(x)

= ll x l=llogp(x).

174

Chapter 3. Look at the Exponent

This functional equation combined with the fact that f (0) = 0 (this follows from proposition 3.B.33) shpws that f = SA, where A(x) = llxl=llogp(x). But it is easy to find an antiderivative of A: reasoning from the original power series, one chooses B(x) = 1Ixl =1(X .logp(x) - x) and checks that B' = A. Note that thanks to the hypothesis x E pZp, we have lul p = 1 if and only if lu + xl p = 1, for all u E Zp. Therefore, theorems 3.B.30 and 3.B.37 yield the integral representation of f:

Using the fact that logp(x

+ u) =

logp(l

logp u

+ x/u) =

+ logp(l + x/u)

can write

( 1)n-l xn .L..t n un

all manipulations being easily justified by the fact that fn,x takes small values (and so its Mahler coefficients are small). We finally deduce that

'"' -

= 1 yields

+ u) logp(x + u) = (x

x - ulogp u

+ u) logp u + (x + u) . '"' L..t

= x logp u - x + '"' L..t n2:1

(_l)n-l n

( l)n-l xn - n . -un - x - u logp u Xn+l

. -un- + X + '"' L..t

(_1)n-l n

. -u n- 1

n2:2

and an easy calculation (based on the change of variable n - 1 = j in the last sum) yields 11ul=1 ((x+u) logp(x+u) - (x+u) -ulogp u+u)

= xÂˇ1 Iu l=llogp u+ L fn,x(u), n2:1

where

175

and expanding

n2:1

for lui

3.B. An Introduction to p-adic Numbers

It remains to simplify this a little bit thanks to the following

Lemma 3.B.39. If 9 is of class C 1 and odd, then

r jzp

g(u)du =

g'(O) -2-'

Proof. We leave this as an easy exercise to the reader. We give a hint, however: write g(pn _ j) = _g(j _ pn) = _(g(j) _ png'(j) + o(pn))

and observe that L~:Ol g'(j) = (Sg')(pk).

Applying this to the functions fn,x with n odd, we obtain that (_1)n-lxn+l -n fnx(u) = ( ) 1lul=lU . , n n+ 1

Of course, we integrate this, but the difficult point is to check that the integral commutes with the infinite sum. Again using theorems 3.B.30 and 3.B.37, we

jzp

fn,x(u)du = 0

for n odd. Putting everything together, the result follows.

Chapter 3. Look at the Exponent

176

We would like to have more information about the numbers An that appear in the previous theorem. A first step in doing this is the following lemma: Lemma 3.B.40. We have An - (1 - p2n-l)

r Jz

x 2n dx E Zp.

Proof. Using the fact that 9 --t g-l is a permutation of the finite group (ZjpkZ)*, we deduce that pk_l

i 2n = i=l gcd(i,p)=l

i=O

i=l gcd(i,p)=l

Dividing by pk and letting k --t

00,

i 2n

i=O

the result follows.

-!,

This allows us to compute Bo = 1, Bl = B2 = ~ and to deduce that all Bn are rational numbers. Actually, these numbers Bn are called Bernoulli numbers and appear all over the place in mathematics. Their rather subtle arithmetic properties will help us prove the following important Theorem 3.B.41. One has PAn E Zp for all n 2': 1. Moreover, if p > 3, then Al E Zp,

Proof. Using the previous lemma and discussion, it suffices to prove that pBn E Zp for all n (the second part follows from the observation that B2 = ~ is prime to p if p > 3, so we will not say more about it). We will prove this by induction on n. For n = 1, it is clear, so assume that it holds for j < n.

pk-l_l

i 2n - p2n

Recall that

x-1

Next, let Bn =

r Jz

177

3.B. An Introduction to p-adic Numbers

xndx.

= '"' Bn Xn

L..t n!

n20

Let k be a positive integer. Multiplying this by e kX yields the equality

By linearity of the Volkenborn integral we can write for n > 0

On the other hand, ekXX ---:-=-- _

-1

X eX - l

+ X(1 + eX + ... + e(k-l)X).

Identifying the coefficients of Xn+l in this equality shows that for all n 2': 1 = lim pnk k-+oo

=0, which immediately yields the exponential generating function of the sequence (Bn)n:

p::;i

Take = p and note that (ntl)Bi = (pBi ) G) :;;~i E for all 0:::; i :::; n - 1. Combining this observation and the previous equality, we deduce that pBn E Zp, finishing the proof of the theorem. 0

/ Chapter 3. Look at the Exponent

178

3.B.I0

Some deep congruences

where

p-adic numbers play a crucial role in establishing difficult congruences, which are almost impossible to get by other means. The following theorem is a very famous such example. We present here Robert's and Zuber's nice proof, following [68]. Theorem 3.B.42. (Kazandzidis) If p we have

(~~) == (~)

where j

3 + vp(nk(n - k))

3.B. An Introduction to p-adic Numbers

> 3, then for all positive integers n, k (mod

an =

+ y)2n+1

_ x2n+l _ y2n+1].

It is thus enough to prove that vp(a n ) 2': vp(xy(x+y)) for all n 2': 1. For n = 1, this follows from Al E Zp, which was proved in theorem 3.B.41. So, suppose that n 2': 2. Since (X + 1)2n+1 - x2n+l - 1 vanishes at 0 and -1, there is f E Z[X] of degree 2n - 2 such that

~),

+ vp((~)).

A n [(x 2n(2n + 1)

179

+ x)2n+l -

X 2n +1

1 = X(l

+ X)f(X).

Since y2n-2 f(x/y) is a homogeneous polynomial of degree 2n - 2 with integer coefficients in x, y E pZp, we deduce that

Proof. Proposition 3.B.33 shows that for all positive integers x we have f (px)

Therefore, if we denote x

= (_l)px. (px)! .

vp((x + y)2n+l - x2n+l - y2n+1) = vp(xy(x + y))

px. x!

= kp and y = (n - k)p, we have

2': vp(xy(x + y))

+ vp (y2n-2 f (~)) + 2n -

It is thus enough to prove that

fp(x+y) fp(x)fp(Y) . The definition of logp(x) as a power series shows that Ix lip = Ilogp(x)lp for all x E 1 + pZp. So, if we let f(x) = logp(fp(x)), the congruence we have to prove is equivalent to

vp(f(x + y) - f(x) - f(y)) 2': vp(xy(x + y)). But theorem 3.B.38 yields

vp(f(x + y) - f(x) - f(y)) An [(x + y)2n+l _ x 2n+1 _ y2n+l]) ~ 2n(2n + 1)

= v ('"'"' p

which follows easily from vp ( An) 2': -1 (theorem 3.B.41) and the inequalities

vp(2n(2n

+ 1)) ::; max(vp(2n) , vp(2n + 1)) ::; 10g5(2n + 1) ::; 2n -

the last one being true for n 2': 2.

Similar techniques apply to prove the following nice congruence, taken from [17]. Theorem 3.B.43. If p > 5 and n 2': 1, then

(np)! == ((p _ l)!)n pn .n!

(mod p3+vp(n 3 -nÂť)

n2:1

and this does not hold modulo p4+vp(n -n) unless p is a Wolstenlholme prime p -l) , z.. e. (2p-l = 1 (mo d p4) .

Chapter 3. Look at the Exponent

180

There are similar sharp congruences for p reader to [17J.

= 2,3,5, for which we refer the

Proof. (sketch) As in the proof of the previous theorem, one starts by noting

that

181

3.B. An Introduction to p-adic Numbers

Theorem 3.B,4S. Let n be a multiple of an odd prime p. Then we have an equality in Qp: n-1 'L.. " .!:.k = _ '" Aj 2j L..2Âˇ n , k=l '>1 J gcd(k,p)=l

where Aj are as in theorem 3.B.38. Proof. (sketch) Note that f(x) = logp(r p(x)) is locally analytic on Zp, as follows from theorem 3.B.38 and from the functional equation

so the congruence reduces to Vp (rp(np) - rp(pt) ~ 3 + vp(n

n).

If f(x) = logp(r p(x)), the equality vp(y -1) = vp(logp y ) (true for y E 1 + pZp) reduces the problem to showing that vp(f(np) - nf(p)) ~ 3 + vp(n

f(x It follows easily that

+ 1) -

f(x)

f is differentiable and that g(x + 1) - g(x)

n).

Thus One uses again theorem 3.B.38, but one also has to analyze the term containing A2' As the details are routine, they are left to the reader. 0

1Ixlp=110gpX.

n-l

f' satisfies

= 1Ix1p =1-' X

g(n) - g(l).

k=l,gcd(k,p)=l

Here is yet another similar congruence, with the same idea. Theorem 3.B.44. (F. Rodrigo- Villegas) Let (an)n be a sequence of integers, finitely many of which are nonzero and such that I:n2:1 nan = O. Define A(n)

II (knWk. k2:1

Then for any prime p

> 3 and any

s ~ 1 we have

Proof. (sketch) The point is to study logp

(A1~~~{Âť),

by expressing it in terms of values of the function x H logp r p(x). For instance, if s = 1, then this equals I:k ak logp r p(kp). One finishes as usual by using the analytic expansion of f(x). 0

Next, by differentiating the Taylor expansion of f in theorem 3.B.38, we obtain that for x E pZp A' g(x) = AO 2~x2j. '>1 J J_

The result follows by combining these two identities (with x = n in the second one) and by observing that g(l) = g(O) = Ao. 0

3.B.ll

More on Bernoulli numbers

In this section we focus on some classical congruences satisfied by Bernoulli numbers. Recall from section 3.B.9 that they are defined by Bn = J:z p xndx. We saw that (Bn)n is a sequence of rational numbers, with exponential generating function X -1'

182

Chapter 3. Look at the Exponent

Lemma 3.B.39 shows that Bn = 0 for all odd numbers n > 1 (this can also be deduced from the generating function, by checking that the function x -+ + ~ is even). We saw during the proof of theorem 3.B.41 that Bernoulli numbers satisfy the crucial identity

3.B. An Introduction to p-adic Numbers

hence it is enough to show that we can find k such that But S2n(pk) _ B k

This identity will play a crucial role in the proof of the following beautiful result, which considerably refines theorem 3.B.41. Theorem 3.B.46. (von Staudt-Clausen theorem) For all n 2: 1 we have

B 2n +

L -P1 E Z.

183

S2n)t) p

B2n E Zp.

_1_ 2~1 (2n + l)Bo k(2n-j) 2n + 1 L . JP j=O J

is certainly in Zp for k large enough, as each of the terms in the sum is in Zp. This proves the claim. Finally, a standard argument 4 shows that Sn (p) == 0 (mod p) unless p - 1 divides n, in which case Sn(P) == -1 (mod p). Combining this with the result of the previous paragraph, we deduce that B2n E Zp unless p - 112n, in which case B 2n + ~ E Zp. All in all, this shows that the rational number B2n + L: q-1 2n belongs to Zp for all primes p and so it is an integer. The result follows. 0 1

p-112n

Proof. Let p be any prime. We need the following elementary result:

Lemma 3.B.47. Ifn is even, then for all k 2: 1 we have Sn(pk+l) (mod pk+1).

== p. Sn(pk)

Proof. This is a simple application of the binomial formula: pk-Ip-I Sn(pk+1) = L L(j

The following classical result is considerably more difficult to prove. The proof is actually rather mysterious, though elementary. It is highly related to the existence of p-adic analogues of the Riemann zeta function, though this may not be apparent at all ... Theorem 3.B.48. (Kummer's congruences) Let m, n be positive integers and let p be a prime such that p - 1 does not divide n, but p - 1 divides m - n. Then Brn - !1.n. E pZ . m n P

+ lÂˇ pk)n

j=O l=O

pk_1 p-I L L(jn

Proof. It suffices to prove that ~k

+ nlpk. jn-I)

== pSn(pk)

l)pk+ISn_l(pk)

(mod phI).

Note that if p is odd, then the hypothesis that n is even is useless. Next, we claim that

S2;(P) -

B 2n E Zp for all primes p and all n. Note

that the previous lemma implies that

S2npk+l (pk+l)

S2npk(pk) E

Z P for all k

(mod p) whenever p - 1 does not

divide k 2: 1. Note that it is not even clear that ~ E Zp, but this will be the case (of course, this is clear if k is odd, as then Bk = 1). Let us fix an integer 1 ~ a ~ p - 1 such that a is a primitive root modulo p. The following lemma is crucial.

j=O l=O

== pSn(pk) + ~(p -

== ~~+;~Il

> _ 1,

4Let 9 be a primitive root mod p. If p - 1 does not divide n, then gn

Sn(P) =_

p-2 '"

L.,.g i=O

(p-l)n

= 9

n 9

1 =0.

1 and

Chapter 3. Look at the Exponent

184

+ ;)a -1

~~~~11

Zp. Also, Fermat's little theorem yields Sj,n == Sj,n+p-l (mod p) for all j 2: 0 and all n 2: 1. Hence Un == Un+p- 1 (mod p) for all n 2: 1. As ak+p-l == a k (mod p), this congruence is equivalent to

1, this is equivalent to

eX -

Lbnyn. n~O

The binomial formula and the fact that gcd( a, p) = 1 yield the existence of 9 E YZp[Y] such that (1 + y)a - 1 = aY(1 + g(Y)). We then have

+ y)a -

_~_~( Y - Y

-1)=L(-lt~9(Yt.

1 1 + g(Y)

n~l

(a k _ 1) Bk k

== (a k _ 1) Bk+p-l

(mod p).

k+p-l

The desired result follows from this congruence and the fact that p does not divide a k - 1. 0

The result follows, as 9 E YZp[Y].

Let use denote Un = (an -1)· ~ and observe that by definition of (Bn)n we have a 1 1 xn xn -ax =-- = - ""'nUn = LUn+1 - , • e - 1 eX - 1 X ~ n! >0 n. n~O

185

for all n 2: o. It is now easy to conclude. Recall that p - 1 does not divide k, so p does not divide a k - 1. The previous relation and the fact that bj , Sj,n E Zp for all j, n show that Uk E Zp and so ~ E Zp. The same argument shows that

J., exist bn E Zp such that (as formal series) L emma 3 .B . 49. 'T'here

Proof. By changing the variable Y =

3.B. An Introduction to p-adic Numbers

Remark 3.B.50. Here is the real meaning of this proof. As we have already said, the crucial fact is that Ja(X) = (1+':)a-l - ::k- E Zp[[X]] , say fa(X) = Ln>o bnxn. This allows us to define a measure J-La on Zp (i.e. continuous linear form on the space of continuous functions 9 : Zp -+ Qlp) by

To fully exploit lemma 3.B.49, let us denote

Sn,k = k! . [Xk](e X

It =

_1)n-

j=O

so that

(~)jk, J

for all continuous functions 9 : Zp -+ Qlp. Here an(g) is the nth Mahler coefficient of 9 and the series converges by Mahler's theorem (as limn-too an (g) = 0 and bn E Zp). Since bn E Zp for all n, Mahler's theorem shows that

Xk sn,kkf·

(ex - It = L k~O

Replacing these relations in lemma 3.B.49 and identifying coefficients yields 5

Un+l = L

bj . Sj,n

j~O

5Note that there are no convergence issues, as

Sn,k

= 0 for n

> k.

for all continuous maps g. Note that Sn,k = an(x k ), so the equality UnH = Lj~o bjsj,n established during the proof can be written as

Chapter 3. Look at the Exponent

186

Since vp(x n - x n+p- 1) 2: 1 for all x E Zp and n 2: 1, the previous paragraph shows that Un == Un+p- 1 (mod p) for all n 2: l. This new interpretation of the proof of Kummer's congruence yields other nontrivial congruences. Assume for instance that m == n (mod pN (p - 1)) and m, n > N. By Euler's theorem we have vp(x m - xn) 2: N + 1 for all x E Zp and we deduce that Um == Un (mod pN+l). This observation is the beginning of the construction of the p-adic zeta function of Kubota and Leopoldt.

3.B. An Introduction to p-adic Numbers

It is thus enough to prove that (k~l)Bn+l-kPkk E p2Zp for all 3 ::::; k ::::; n + l. As pBn+l-k E Zp (by the von Staudt-Clausen theorem) and ( n ) E Z it is k-3 k-l p, enough to check that E Zp. This is easy and left to the reader (here one crucially uses the hypothesis p > 3). 0

r,;-

Applying the previous lemma to n (as n is even), we obtain

p-l 1

k == 0

(mod p2)

and

Theorem 3.B.51. For all primes p > 3 we have

and

k=l

t == -~

B p -3

~Bp-3 (mod p)

:2 2:

== B p - 3 (mod p2). k=l One could apply a similar method for the second congruence, but we prefer to deduce it from the first one. Note that

The following result considerably refines these congruences.

p-l

zp(p2) - 2 and noting that B n -

It remains to use Kummer's congruence to obtain B<p(p2)_2 and finally p-l

== 0 (mod p).

k=l

p-l 'L" k<p(p2)-2 -= pB<P (P2) -2 k=l

We end this addendum with an application of the previous results to harmonic numbers. It is standard that for any prime p > 3 we have

187

(mod p3).

p-l

Proof. Euler's theorem yields

p-l (

t; t = t; t +

p-l

~ k) = p t; k(p ~ k)'

Using this identity and the congruence we have just proved, it is enough to prove that p-l

We will use the following general result: Lemma 3.B.52. If p > 3 and n 2: 1, then

+ 2n + ... + (p -

)n _

= pBn

+ np2 Bn-l 2

n+ 1 ( n

--1 n + k=l

+ 1) k

Bn+1-kP = pBn

np2 B n 2

n+ 1

k=3

p-l

~ k) + :2 ) == 0

(mod p2)

Proof. The left-hand side is equal to 1

(k (p

k2 ( _ k) == 0 (mod p) k=l P p-l 1 Â˘} k 3 == 0 (mod p). k=l Â˘}

n 1

(n)

1 Bn+1-k k

The last congruence is also standard and it is proved using primitive roots mod p. 0

Chapter 4

Primes and Squares This chapter is concerned with arithmetic properties of primes of the form 4k + 1. It is very elementary and most problems use the following two results. Theorem 4.1. (Fermat) Any prime number of the form 4k+ 1 can be written as the sum of squares of two integers.

There are many proofs of this classical result, see for instance the first example in chapter 4 of [3]. For a different proof, using properties of Gauss and Jacobi sums, see the addendum 9.A. Proposition 4.2. Let p be a prime of the form 4k + 3 and let a and b be integers such that p divides a 2 + b2 â&#x20AC;˘ Then p divides a and b.

Proof. If p does not divide a, there exists c such that ca == 1 (mod p). Then (bc)2 == -1 (mod p), thus

(-1)~ == (bc)p-l == 1 (mod p), the last congruence being Fermat's little theorem. But this is clearly impossiE=.! ble, since by hypothesis (-1) 2 = -1. The result follows. 0 An easy consequence of proposition 4.2 is that v p ( a 2 + b2 ) is even for any prime p of the form 4k + 3 and for any integers a, b. This is a very useful tool when studying some diophantine equations, as the following problems show.

190

Chapter

Primes and Squares

191

1. Prove that the number 4mn - m - n cannot be a perfect square if m and n are positive integers.

Proof. We clearly have no solutions for Y < -1, so let us suppose that y+2 > 0. Taking the equation modulo 4, we easily obtain that y == 1 (mod 4). The key point is to rewrite the equation as x 2 + 112 = y7 + 27 or, by factoring the right-hand side, as

Fermat Proof. If 4mn - m - n = x 2 for some integer x, then

(4m - 1)(4n - 1) = (2x)2

+ 1.

But there exists a prime p == 3 (mod 4) such that pl4m (2x)2 + 1, contradicting proposition 4.2. 2. Prove that the equation y2 = x 5

1. Then p divides

Since y == 1 (mod 4), we have y + 2 == 3 (mod 4), thus there exists a prime q such that vq(y + 2) is odd. Note that q does not divide

4 has no integer solutions. Balkan Olympiad 1998

Proof. Write the equation as x 5 + 25 = y2 + 62 . We claim that there is always

a prime p == 3 (mod 4) such that vp (x 5 + 25 ) is odd. Since for any such prime p we have v (y2 +6 2 ) == (mod 2), we will thus get a contradiction. Note that x is odd, ot~erwise x = 2Xl, Y = 2Yl and 8 divides YI + 1, which is impossible. If x == 1 (mod 4), there is a prime p == 3 (mod 4) such that vp(x + 2) == 1 (mod 2). One cannot have plx4 - 2x 3 +4x 2 - 8x+ 16 (otherwise p would divide 5.2 4 ), so v p (x 5 + 25 ) = vp(x + 2) == 1 (mod 2) and we are done. If x == -1 (mod 4) then x4 - 2x3 + ... + 16 == 3 (mod 4) and we can repeat the argument , 3 +···+16)=1 (mo d) bytakingaprimep==3 (mod4)suchthatv p (x 4 -2x 2. The claim being proved, the result follows. 0

Proof. We can work modulo 11: since xll == x (mod 11) for any x, we deduce that x 5 == 0,1, -1 (mod 11) for any x. On the other hand, the quadratic residues mod 11 are trivial to find: those are 0,1,4,9,5,3. One readily checks that the equation has no solution mod 11 using these observations. So, it does not have integral solutions either. 0

3. Solve in integers the equation x 2 = y7 + 7. Titu Andreescu, USA TST 2008

y6 _ 2 y 5 + 4y4 - 8 y 3 + 16y2 - 32y + 64,

as otherwise q would divide 7·64 and x 2+112, a contradiction. Thus v q (y7 +2 7 ) is odd, which is impossible, as it equals vq (x 2 + 112) and q == 3 (mod 4). The result follows. 0 4. Find all pairs (m, n) of positive integers such that

Gabriel Dospinescu Proof. First, assume that n 2

> 2. Then m cannot be odd, since otherwise

8 would divide m - 1, but 3m + (nl - 2)m is odd. So m is even. But then m 2 -1 == -1 (mod 4), so there exists a prime p == -1 (mod 4) dividing m 2 -1. But then p divides 3m + (nl - 2)m and since m is even, this implies that p divides 3 and nl - 2. Thus 3 divides nl - 2, a contradiction. Thus, we must have n = 1 or n = 2. If n = 1 then either m 2 - 113m + 1 and m is even or m 2 - 113 m - 1 and m is odd. In the first case we can use the same argument as before to get a contradiction (choose a prime p == -1 (mod 4) dividing m 2 - 1), while the second case is impossible since 3m - 1 is not a multiple of 8 when m is odd. Thus n = 2 and m 2 - 113 m . Thus there is k :::; m such that (m - 1)(m + 1) = 3k . But then m - 1, m + 1 are powers of 3 which differ by 2. Thus clearly m - 1 = 1 and m = 2. We deduce that m = n = 2 is the only solution of the problem. 0

Chapter

192

4. Primes and Squares 2

5. Find all pairs (x, y) of positive integers such that the number x

+ y2 x-y

a divisor of 1995. Bulgaria 1995

Proof. Note that 1995 = 3·5·7 ·19. The key observation is that 3,7,19 are all primes of the form 4k + 3. If p is any of these primes and if p divides x;~t2 then p divides x 2 + y2 and so it divides x, y. Writing x = PXI, Y = PYI, we

obtain

x-Y .. . x 2+ y 2 . h 2 Domg thIS for every pnme factor P of x-v and notmg t at x

+ Y2 =

x - Y

has no solutions with x, Y ~ 1, we just have to solve the equation X;~t2 = 5. This is easy, since we can write it as (2x - 5)2 + (2y + 5)2 = 50 and since 50 can <,mly be written as sum of two squares in two ways 1 + 49 = 25 + 25. Thus 2y + 5 E {-7, -5, -1, 1,5, 7} and since y is positive we must have y = l. But then x = 2 or x = 3. Putting everything together, we deduce that the solutions are all pairs (2k, k), (3k, k) with k E {1, 3,7,19,21,57,133, 399}. 0 6. Find all n-tuples (al,a2, ... ,an) of positive integers such that

193

ones are equal to 2. The equation becomes 5m - 16 = k 2 • As k is odd, a consideration mod 8 shows that m is even. But then (5 m / 2 _k)(5 m / 2 +k) = 16, which easily implies that m = 2. Thus the sequence (aI, a2, .. . ,an) consists of two numbers equal to 3 and the remaining equal to 2. Clearly, all such sequences are solutions of the problem. 0 7. Prove that there are infinitely many pairs of consecutive numbers, no two of which have any prime factor of the form 4k + 3.

Proof. Since no number of the form n 2 + 1 has a prime factor of the form 4k + 3, it is clear that the pairs (( n 2 + 1)2, (n 2 + 1)2 + 1) yield solutions of the problem. 0 The following problem is trickier and its solution uses the theory of Pell equations. 8. Let P be an odd prime. Prove that P == 1 (mod 4) if and only if there are integers x, y such that x 2 - py2 = -1.

Proof. One direction follows directly from proposition 4.2, so let us assume that P == 1 (mod 4). The key point is to consider the positive Pell equation x 2 - py2 = 1. By the general theory of Pell equations, this has a smallest nontrivial solution (xo, YO) (so Yo > 0 is minimal among all solutions with y =I- 0). Note that Xo is odd, since otherwise 4 would divide + 1. If Xo = 2a + 1, we deduce that Yo is even and a 2 + a = pb2 for b = ~. If p divides a, then a = pc2 and a + 1 = d? for some integers c, d (because a and a + 1 are relatively prime), so that d2 - pc2 = 1. But obviously c < Yo, contradicting the minimality of (xo, yo). Thus p divides a+ 1 and we can write a = c2, a + 1 = pd2 for some integers c, d. But then c2 - pd 2 = -1 and we are done. 0

is a perfect square. Gabriel Dospinescu

Proof. Suppose that

First, we claim that ai E {2,3} for all i. It is clear that ai =I- 1 for all i, so assume that ai > 3 for some i. Then ai! - 1 == 3 (mod 4), thus there is a 2 prime p == 3 (mod 4) such that p divides ail - 1. Then p divides k 2 + 4 , a contradiction. Say m among the numbers ai are equal to 3 and the remaining

We continue with another beauty, which became classical in mathematical contests. It has the nice feature of having a completely elementary solution that uses quite a lot of different ideas.

194

Chapter

Primes and Squares

9. Find all positive integers n such that the number 2n of the form m 2 + 9.

1 has a multiple

IMO 1999 Shortlist Proof. Assume that 2n - 1 divides m 2 + 9 for some integer m and that n 2: 2. Then the only prime factor of 2n - 1 which is 3 modulo 4 is 3. Indeed, if p is such a prime, then p divides m 2 + 32 and so p divides 3. Now, if n has a nontrivial odd divisor d, then 2d - 1 == -1 (mod 4) and 3 does not divide 2d _ 1. Thus 2d - 1 has a prime factor p == 3 (mod 4) different from 3. Since 2d _ 1 divides 2n - 1, Choosing m = 3a, it is enough to find a such that (22 + 1)(24 + 1) ... (2 2k - 1 + 1) divides a 2 + 1. this is impossible, by the previous remark. Thus, n must be a power of 2. Conversely, assume that n = 2k and observe that 2k 1 2n _ 1 = 3路 (22 + 1)(24 + 1)路路路 (2 - + 1). 2k 1

Choosing m = 3a, it is enough to find a such that (2 2 +1)(2 4 +1) ... (2 - +1) divides a 2 + 1. The crucial point is that the Fermat numbers 22' + 1 are pairwise relatively prime, so by the Chinese Remainder Theorem it is sufficient to prove that for any i there is a such that a2 + 1 == 0 (mod 22i + 1). But this is clear, since a = 22i - 1 works. Thus, the answer to the problem is: all powers of 2. 0 The next problems are concerned with properties of sums of two squares. A crucial fact is that the set of numbers which can be written as a sum of two squares of integers is closed under multiplication, by Lagrange's formula

Actually, combining theorem 4.1 and proposition 4.2, it is easy to completely characterize the elements of this set: they are precisely the nonnegative integers n such that vp(n) is even for all primes p = 3 (mod 4). 10. Prove that a positive integer can be written as the sum of two perfect squares if and only if it can be written as the sum of the squares of two rational numbers. Fermat

195

Proof. One direction being clear, let us prove that if n is the sum of the squares of two rational numbers, then it is also the sum of the squares of two integers. Thus, we know that na 2 = b2 + c2 for some integers a, b, c with a =J. O. For any prime p, we deduce that

so that vp(n) has the same parity as v p(b 2 + c2). Since for any prime p == 3 (mod 4) we have vp (b 2 + c2 ) == 0 (mod 2), it follows that for any such prime p we have vp(n) == 0 (mod 2). The result now follows from the preliminary 0 discussion. Remark 4.3. Actually, the following result holds: if n 2: 2 is an integer and if an integer x can be written as a sum of n squares of rational numbers, then it can also be written as a sum of n squares of integers. For n = 3, this follows from Davenport-Cassels' lemma ([3], chapter 13, example 12), while for n 2: 4, this follows from the famous theorem of Lagrange, according to which any nonnegative integer is the sum of four squares ([3], chapter 13, example 5). 11. Prove that each prime p of the form 4k + 1 can be represented in exactly one way as the sum of the squares of two integers, up to the order and signs of the terms. Fermat Proof. The fact that any such prime p is a sum of two squares is the content of theorem 4.1. Let us focus on the uniqueness part. Assume that we have p = x 2 + y2 = z2 + t 2. Then (x - z)(x + z) = (t - y)(t + y). We will need the following very useful Lemma 4.4. If a, b, c, d are nonzero integers such that ab = cd, then there are integers m, n, p, q such that a = mn, b = pq, c = mp, d = nq.

Proof. Let ~ = ~ = ~ be the representation of the fraction ~ in lowest terms. Since ap = nc, we have pic, so we can write c = mp for some m. Then also a = mn. Doing the same with dp = nb yields the conclusion. 0

196

Chapter

Primes and Squares

Coming back to the problem, assume that Ixl =1= Izl and It I =1= by the lemma we can find nonzero integers m 1, n I, PI ,Q1 such that

Iyl.

Then

197 Proof. We will show that 32k = m 2 + n 2 + 1 has infinitely many solutions. Indeed, start with the observation that 3 2 .1 = 22 + 22 + 1. On the other hand, if (k,m,n) is a solution with m 2': n, then (2k,3 k m - n,3 k n + m) is also a solution. Indeed, this follows from

But then In the following two problems we will use the fact that the density of the set of positive integers all of whose prime factors are of the form 4k + 1 (or 4k + 3) is zero. This is a nontrivial result, for a proof of which we refer to [3], chapter 4, example 10.

so that using Lagrange's identity we obtain

mI qr,

nI PI

We may assume that divides + so that + is equal to 1,2 or 4. As ml, n1, PI, q1 are nonzero, we obtain a contradiction unless nl = PI = ±1. But in this case we get x - z = ±(t - y) and x + z = ±'(t + y), thus

{lxi, Izl} = {Itl, Iyl} D

and we are done again. We continue with an easy exercise in Lagrange's formula. 12. Prove that the equation 3k in positive integers.

+ n 2 + 1 has infinitely many solutions Saint-Petersburg Olympiad

Proof. Guided by the formula

13. It is a long standing conjecture of Erdos that the equation

-=-+-+n x y z

has solutions in positive integers for all positive integers n. Prove that the set of those n for which this statement is true has density 1.

Proof. We look for solutions with y = z and x = na for some positive integer a. The equation becomes 4xy = n(y + 2x) or, equivalently, y(4a - 1) = 2na. Thus, if we can find a prime factor p of n of the form p = 4a - 1, then we can take y = 2~a and we have a solution in positive integers. Thus, it is enough to prove that the set of integers having at least one prime factor of the form 4k - 1 has density 1, which has already been discussed. D 14. Let T be the set of positive integers n for which the equation n 2 = a 2 +b2 has solutions in positive integers. Prove that T has density 1. Moshe Laub, AMM 6583

we will choose k = 2a • Since all factors in the product are sums of two squares and since the set of numbers which are sums of two squares is stable 2a by multiplication, it follows that 3 - 1 is always a sum of two squares. Since it is trivially not a perfect square (it is of the form 3k + 2), the conclusion follows. D

Proof. We will prove that nET if and only if n has at least one prime factor of the form 4k + 1. Suppose first that nET and choose positive integers a b . ' such that n 2 = a 2 + b2 . If all pnme factors p of n are of the form 4k - 1, then 2 2 for any such p we have pla + b , so that p divides a and p divides b. Dividing the previous relation by p2 and repeating the argument, we deduce that pvp(n)

198

4. Primes and Squares

Chapter

divides a and b. Since this happens for all pin, it follows that n divides a and b, which is clearly impossible. Conversely, if n has a prime factor p == 1 (mod 4), by Fermat's theorem we can find integers c, d such that p = c2 + d 2 . We may assume that c, d are positive (they are nonzero since primes are not perfect squares). But then

n (n( c d 2

)) 2

;

(2:Cd)

199 Note that

and since

We continue with two very nice problems concerning primes of the form 4k + 1. The method used in the solution of the following problem is standard.

L lv'JPJ =

j=l

16. Find all functions 2k

L[v'JPl =

j=l

j=l i250jp

1= L

i=l

L 1, k>_J_ ">:2.p

hand, the condition j 2: is equivalent to j 2: 1 + integer. Thus we can also write

f;[v'JPl =

[ k -

.2] ) ~ =2k 2 -

and it remains to prove that

t [i2] i=l

The following functional equation is rather nonstandard.

2k (

i=l

and the result follows.

2k2 - 2k

f : Z+ ---+ Z with the properties:

1. f(a) 2: f(b) whenever a divides b.

2. for all positive integers a and b,

f(ab) + f(a 2 + b2 ) = f(a) + f(b).

since the inequality i ::; .j)p with 1 ::; j ::; k implies that i ::; 2k. On the other

i=l

Proof; Write p = 4k + 1 and note that k

+ 1)

2k 2k LXi = L(p - Xi)

1; .

pk(2k

it remains to prove that the sum of the quadratic residues mod pis pk. But if Xl, X2,ÂˇÂˇÂˇ, x~ are the nonzero quadratic residues mod p (any residue mod p is implicitly taken between 0 and p - 1), then p - XI,P - X2, ... ,p - X2k are a permutation of the Xi'S (since -1 is a quadratic residue mod p, which follows from p == 1 (mod 4)). Thus

15. Let p be a prime number of the form 4k + 1. Prove that p-l -4-

2 =

i=l

and so nET. Now, the density of those numbers which are not divisible by any prime of the form 4k + 1 is 0 and we are done. 0

L2k i

[%], since %is not an

t; [.2~ ] 2k

Gabriel Dospinescu, Mathlinks Contest Proof. Since 1 divides any integer, it follows that f(1) 2: f(x) for all x. Let k = f(1).

The first step is to prove that n and not their multiplicities, i.e.

f (n) only depends on the prime factors of

f(n) = f

(IIp) . pin

200

Chapter

Primes and Squares 201

Indeed, consider two positive integers a, m and choose b = am. Thus

f(a 2m)

+ f(a 2(m 2 + 1)) = f(a) + f(am) condition f(a) ~ f(a 2(m 2 + 1)) and f(am)

and by the first ~ f(a 2m). Thus both of these inequalities must be equalities and so f(am) = f(a 2m) for any positive integers a, m. This immediately proves the claim. In the second step, we prove that f(n) does not depend on the prime factors of n that are congruent to 1 or 2 modulo 4. Indeed, the proof of the previous claim shows that for all n and x we have f(n) = f(n(x 2 + 1)). In particular, f(n) = f(2n) and so we don't have to care about possible powers of 2 in the prime factorization of n. Also, if p == 1 (mod 4) and x is chosen such that plx 2 + 1, then

f(n) ~ f(np) ~ f(n(x 2 + 1))

f (

17. Prove that the equation x 8 = n! nonnegative integers.

g(81 U 8 2 )

L ka -

Then

II pvp(n!).

PEAn

> nip - 1 (by Legendre's formula), we obtain

lnlnn > In rnT >

p),

L (~-1) lnp.

PEAn

+ g(81 n 8 2 )

The first two relations are obvious. For the third one, note that if the sets of prime factors congruent to 3 mod 4 of a and of b are A, B, then the set of prime factors of the form 4j + 3 of ab is exactly A U B and the set of prime factors of the form 4j + 3 of a2 + b2 is An B. If g( {n}) = k n for some integers k n :::; k, then an easy induction on IAI shows that

g(A) =

+ 1.

(x 2 - 1)(x 2 + 1)(x4 + 1).

rnT ~ x 2 Then, using that vp(n!)

where P3 is the set of prime numbers of the form 4k + 3. Let Pl.P2, ... be the elements of P3 and define g(A) = f(I1aEAPa)' We obtain a function 9 defined on the set of all subsets of N with integral values. Moreover, we claim that g(0) = k, 8 1 C 82 =* g(8 1) ~ g(82) and finally

+ g(82) =

many solutions in

Let An be the set of prime numbers P :::; n of the form 4k + 3. The key point is that no P E An can divide x 2 + 1 or x4 + 1, so that pvp(n!) must divide x 2 _ 1. Thus we obtain

pln,pE P3

g(81)

+ 1 has only finitely

Proof. Suppose that (x, n) is a solution of the equation x 8 = n!

f(n)

and so f(np) = f(n). In conclusion, we have

f(n)

The following challenging problem requires some estimates about prime numbers that follow from Dirichlet's theorem, for which we refer the reader to addendum 7.A.

(IAI- 1)k.

A classical inequality of Erdos (theorem 3.A.3, chapter 3) yields

~ lnp < In

PEAn

(II

p) < nln4.

p~n

We deduce that ' " In p I 4 In n L-< n +PEAn p 4 for any solution (x, n). N.o,:, it remains to prove that there are only finitely many such integers n. ThIS IS a consequence of the proof of Dirichlet's theorem, which establishes, among many other things, that

aEA Conversely, any choice of such k n yields a corresponding 9 and the previous construction yields a solution f of the equation. This ends the solution. 0

1 ' " lnp 1 Inn L -+ 2"

PEAn

for n -+

00.

See addendum 7.A for a proof.

Chapter

202

Primes and Squares

It is really amazing that the following result has a purely elementary proof. We present here the beautiful idea of John H.E.Cohn and we refer the reader to [18] for other similar results (including the fact that 1 and 144 are the only squares in the Fibonacci sequence).

"18. Let Lo = 2, L1 = 1 and Ln+2 = Ln+1 + Ln be Lucas's famous sequence. Then the only n > 1 for which Ln is a perfect square is n = 3. Cohn's theorem Proof. If Xl and X2 are the roots of the polynomial X2 - X - 1, then Ln = + x'2, which combined with X1X2 = -1 yields L 2n = L~ - 2( _1)n. This already shows that if Ln is a perfect square, then n is odd, for y2 Âą 2 is never a perfect square. The case when n is odd is much more subtle. There are two key properties of the Lucas numbers that make everything work. The first is that Lk == 3 (mod 4) whenever k is an even number not a multiple of 6 (use the previous formula for L2n and the fact that Ln is odd whenever n is not a multiple of 3). Call such a number k good. The second key ingredient is the fact that Lk divides L n+2k + Ln for all good numbers k and all nonnegative integers n. This follows from k == 0 (mod 2), the equality X1X2 = -1 and the computation

L n+2k + Ln = xr(xi x~+k(x~ + x~)

+ 1) + x'2(xÂ§k + 1) =

+ x~+k(x~ + x~) =

LkLn+k.

Assume now that n == 1 (mod 4) and n > 1. Then we can write n = 1 + 2 . 3T k for a nonnegative number r and a good number k. Applying the second key point 3T times, we deduce that Ln == -L1 == -1 (mod Lk). As Lk == 3 (mod 4), it has a prime divisor of the form 4j + 3 and so Ln cannot be a perfect square. Assume finally that n == 3 (mod 4) and n > 3. Then we can write n = 3 + 2 . 3T k for a nonnegative number r and a good number k. The same argument shows that Ln == -L3 == -4 (mod L k ) and we reach the same conclusion, as numbers of the form x 2 + 4 have no prime factors of the form 4j + 3. 0

4.1. Notes

4.1

203

Notes

~any of the solutions to the problems in this chapter were provided by the followmg ~eo~le: Alexandru Chirvasitu (problem 14), Daniel Harrer (problems 2, 3), BenJamm Gunby (problems 1, 6, 9), Fedja Nazarov (problems 12 15) ' , Gjergji Zaimi (problems 5, 7, 13, 16).

Chapter 5

T2'S Lemma All of the following problems fall to a rather handy inequality, known as T2 's lemma even though it is a special case of the Cauchy-Schwarz inequality, This result says that for all real numbers aI, a2, ' , , ,an and all positive real numbers XI,X2,,', ,X n the following inequality holds

To get the reader familiar with this trick, we start with a series of more direct applications, 1. Let

Xl, X2, ' , , , X n , Yr, Y2, ' l' ,Yn

be positive real numbers such that

+ X2 + ", + Xn 2: XIYI + X2Y2 + '" + XnYn'

Prove that Xl

+ X2 + ' , , + Xn

Xl ::; -

Xn + -X2 + ' , , + -, Y2

Romeo Ilie, Romania 1999

Chapter 5. T2

206

Lemma

Proof. Using T2'S lemma, we can write

207

Proof. T2'S lemma gives a lower bound

the last inequality being exactly the hypothesis. 2. Let a, b, e be nonzero real numbers such that ab + be + ea 2': that ab 1 -::---::+ be + e2 ea > -a 2 + b2 b2 + e2 + a2 - 2

Since La(b+e+d)

Prove

Titu Andreescu Proof. The trick is to add ~ to each fraction, in order to exploit the identity

it remains to prove that

= (Lar - La 2 :::; 3La2,

a 2 2': 1. But again by Cauchy-Schwarz we have

1 = (ab + be + cd + da)2 :::; (a 2 + b2 + e2 + d 2)2, which proves that

2': 1 and finishes the solution.

4. Prove that if the positive real numbers a, b, e satisfy abc = 1, then abc

- - + e+a+1 + a+b+1->1. b+e+1 Vasile Cartoaje, Gazeta Matematica With this observation, the inequality becomes Proof. The solution using T2'S lemma is straightforward: L b+

and it is then a trivial consequence of T2 's lemma, combined with the hypothesis ab + be + ea 2': O. 0 3. Prove that for any positive real numbers a, b, e, d satisfying ab + be + cd + da =

b+e+d

b3 e+d+a

a a (b

(La)2

+ e + 1) 2': 2 L ab + La'

so that we only need to prove the inequality L a 2 2': L a. This follows from L a 2 2': (E3a)2 and La 2': 3 (the first being a consequence of Cauchy-Schwarz, the second being the AM-GM inequality). 0 5. Let a, b, e be real numbers such that

111 -2 - + - 2- + - -2 > 2 . a +1 b +1 e +1-

the following inequality holds

-:---- +

a e+

e3 d+a+b

d3 1 > -. a+b+e - 3 IMO

1990 Shortlist

Prove that ab + be + ea :::;

2"' Titu Andreescu

208

Chapter 5. T2 's Lemma

Proof. Observe that

209

Proof. Using T 2 's lemma, it is sufficient to prove the inequality a2

On the other hand, we have

Note however that the obvious application of T2'S lemma fails. The previous inequality is equivalent to

,,~> (a+b+c)2 L a2 + 1 - a 2 + b2 + c2 + 3Âˇ

Combining the two inequalities immediately yields the result.

The following problems are a bit less straightforward. However, they do not require any heavy machinery. 6. Prove that for any positive real numbers a, b, c, 1

Unfortunately, the only reasonable way to prove this is to expand it in the form 2 3L" a2b2 + 2"1 abc L ab(a + b ) 2': 2" a.

Titu Andreescu, MOSP 1999

Proof. The solution using T 2 's lemma is a bit tricky: the point is to look at the denominator of the right-hand side, because

(a + b)(b + c)(c + a) = (a + b)c2 +(b + c)a 2 + (c + a)b 2 + 2abc.

It is possible to solve the following problem using T 2 's lemma, but the proof is not really elegant. In the addendum we discuss some applications of Holder's inequality, which makes this problem really easy. 8. Let a, b, c be positive reals such that abc 1

This suggests writing the left-hand side in the following way c2 -::-c--

c2(a + b)

a2 a2(b + c)

b2 b2(a + c)

( rabC)2 2abc

(

b )

c+a

(

a+b

) 2

1. Show that 1

+ b5 (c+2a)2 + c5 (a + 2b)2 -> -3Âˇ

+ --'----'-Titu Andreescu, USA TST 2010

Proof. Start by making the substitution

7. Prove that for any positive real numbers a, b, c the following inequality holds 2

a5( b+2c) 2

A direct application of T2 's lemma finishes the proof.

(b +a)c

Fortunately, this is trivial, since

(a + b + c + rabC)2 --+--+--+--> . a+b b+c c+a 2rabC - (a+b)(b+c)(c+a) 1

+ b2 + c2)2 > 3 a2 + b2 + c2 2 2 2 a (b + c)2 + b (c + a)2 + c (a + b)2 - 4" . ab + bc + ca

L a2 + 1 = 3 - L a2 + 1 ~ 1.

a+ b + c 2': 4" . ab + bc + ca . 3

1 x =~,

Gabriel Dospinescu

The inequality becomes

= b'

1 Z= -.

210

Chapter 5. T2 's Lemma

It is really not easy to prove the following inequality using T2 's lemma, but the trick is worth remembering.

Applying T 2 's lemma, it is enough to prove that

3(x~

+ y~ + z~)2 ~

211

L x 2(z + 2y)2.

The right-hand side is equal to 5 I: x 2y2 + 4 I: x. Thus, we need to prove that

9. Prove that for any positive real numbers a, b, c the following inequality holds

-+--+--> 3a + b 3b + c 3c + a -

1 2a + b + c

+ 2b + c + a +---2c + a + b'

Note that

M.O. Dnimbe

ProofÂˇ Choose three positive real numbers 0:, {3" form the first inequality being Chebyshev's, the second one by the AM-GM inequality and the fact that xyz = 1. Thus, it suffices to prove that

_0:_ + _{3_' 3a + b 3b + c

and use T 2 's lemma in the

+ _,_ > (0: + (3 + ,)2 3c + a - a(30: +,) + b(3{3 + 0:) + c(3, + (3)'

Now, we impose the conditions 13

But 3x 5 + y2 z2 ~ 4x 4 and so it is enough to prove that I: x 4 ~ I: x. This follows from the power-mean inequality and the fact that x + y + z ~ 3. 0

Proof. As in the previous solution, we reduce the problem to proving the following inequality

30: +, = 2,

3{3 + 0: = 1,

3, + {3 = 1.

Solving this linear system yields the solution 0: = ~,{3 = ~" we obtain the inequality

= ~. Therefore,

411121 1 7 3a + b 7 3b + c 7 3c + a - 2a + b + c

-. --+_. - - + - . - - > - - - Proceeding in the same way with the two other terms of the left-hand side and adding up the resulting inequalities yields the desired result. 0

Using the AM-GM inequality, we can write

x3 (2y+z)2

2y + z 2y + z x + 27 + 27 ~ 3"

Proof. We can also use the trick of integrating polynomial inequalities to deduce fractional inequalities. Namely, the inequality

and two similar inequalities. Adding them yields the following estimate ~ ----::+

(z+2y)2

(x+2z)2

x+y+z > ---.:::.--

(y+2x)2 -

and we end up using once more the AM-GM inequality.

can be easily proved using T 2 's lemma, since it can be written in the form

-z + -x + -y

+ y + z.

Chapter 5. T2 's Lemma

212

Using this inequality, we can write

for all 0

213

The proof is a bit tricky. If n is even, the inequality follows trivially from the chain of inequalities

::; 1. Integrating this between 0 and 1 yields the desired inequality.

4(XIX2

+ ... + XnXI)

o Remark 5.1. The technique used in the second proof looks unusual. It is actually quite powerful and we refer the reader to [3], chapter 19 for many more applications. The following two problems are closely related and use a rather useful inequality. 10. Prove that for all n 2:: 4 and all Xl --+ X n +X2

Xl, X2, ... , Xn

X2 XI+X3

+ ... +

> 0, Xn

2:: 2.

Xn-I+XI

Tournament of the Towns 1982

::;

+ X3 + ... + X n -I)(X2 + X4 + ... + xn) (Xl + X2 + ... + Xn)2,

4(XI

the first one being obvious and the second one being the AM-GM inequality. For n odd, things are subtler, but we can reduce the problem to the case when n is even by the following mixing argument: we may assume that Xl 2:: X2, so that we trivially have

Thus, replacing Xl, X2,Âˇ . . ,Xn by Xl, X2 + X3, X4, ... ,Xn , we preserve the sum of the xi's while not decreasing the quantity XIX2 + ... + XnXI. Since n - 1 is 0 even, everything follows from the previous step. 11. Let n 2:: 4 be an integer and let aI, a2, ... , an be positive real numbers such that + aÂ§ + ... + a~ = 1. Prove that

Proof. We start in the usual way by using T 2 's lemma:

Mircea Becheanu and Bogdan Enescu, Romanian TST 2002

Proof. Applying T 2 's lemma we obtain Since '"

ai+l

it remains to prove the following very useful

Lemma 5.2. lfn 2:: 4 and

XI,X2, ... ,X n

are nonnegative real numbers, then

= '"

622

a i (ai+l

Since I: a; = 1, we have I: a;a;+l

::;

(I:2ai2yIai) 2 . + 1) > - I: a i (ai+l + 1)

1/4 by lemma 5.2. The result follows.

We give two proofs for the following beautiful problem. The first one is a standard application of T2 's lemma, the second one uses a very useful technique.

215

Chapter 5. T2 '8 Lemma

214

12. Prove that for any positive real numbers a, b, e, the following inequality holds abe --;:=== + + 2 > 1. 2 va + 8be Vb 2 + 8ea + 8ab -

The problem asks to prove that under this assumption we have x Assume that this is not the case, so x + y + z < 1. Let

Hojoo Lee, IMO 2001 so that X

x+y+z

> y,

+ y + z 2: 1.

x+y+z

+ Y + Z = 1 and

Proof. The most natural application of T2 's lemma turns out to work very smoothly. Indeed,

512

> (_1 X2 _ 1) (~ y2 - 1) (~ Z2 - 1) .

We deduce that 512X 2y 2Z 2

The only problem is to show that

> (X + Y)(Y + Z)(Z + X)(2X + Y + Z)(2Y + X + Z)(2Z + X + Y), which is impossible since X + Y 2: 2VXY, 2X + Y + Z 2: 4{1X2yZ and the similar inequalities obtained by cyclic permutations. 0

This suggests using Cauchy-Schwarz and, indeed, we have (L:a

a 2 + 8be) S VL:a. vL:a 3

+ 24abe.

There are two traps in the following problem, making the problem harder than it appears at first sight.

Thus, the problem is solved if we can prove the inequality

Fortunately, after expansion, this becomes

L: a(b -

e)2 2: 0, which is obvious.

13. Let a, b, e be positive real numbers such that ab + be + ea = 3. Prove that abe --~+ + 2c + a 2 -<1. 2a + b2 2b + c2 T.Q. Anh

Proof. Make the substitution

VT+8bc'

~. Y=V~'

z=V~'

Multiplying the relation ~ - 1 = ~ and the two similar ones yields the relation

Proof. The inequality we have to establish seems to be opposite to the usual applications of T2 's lemma. This is actually not a very serious problem, since we can always change the terms of the sum a bit and change the sign of the inequality. Indeed, note that a

2a + b2 =

1( 2"

1 - 2a + b2

)

217

Chapter 5. T2 '8 Lemma

216 Thus, the inequality is equivalent to 2

2a + b2

2b + e2

b __ -::-+

>1. 2e + a2 -

And now, another trap: one would be tempted to use T2 's lemma in the obvious form, but it is not difficult to check that this does not work. Instead, we take advantage of the hypothesis ab + be + ea = 3 to write

b2 2a + b2

(bVb)2

(L bVb) 2 2ab + b3 2: 6 + L b3 .

Thus, it remains to prove that L(ab)3/2 2: 3, which is immediate by the power-mean inequality and the hypothesis. 0

is close to n - 2. This can be done easily, by taking n - 1 of the variables equal to some x very close to 0, and the last variable equal to xl-no The difficult part is proving that F(al' a2, ... , an) :S n - 2 holds for any sequence al, a2, ... , an as in the statement. Using the basic inequality, this reduces to proving that n 1

L <n-2 i=l (1 + ai)(l + aHI) -

for any positive numbers ai with ala2 ... an = 1. To prove this, write ai = X:~l' with Xn+l = Xl. Subtracting 1 from every fraction, we obtain the equivalent inequality XiXHI + Xi x H2 + x;+1 i=l (Xi + Xi+1)(XHl + XH2) 2: 2.

This is too complicated to try a Cauchy-Schwarz approach, but it simplifies a lot if we observe that

The following problem is also quite tricky.

14. Determine the best constant kn such that for all positive real numbers aI, a2, .. . , an satisfying ala2··· an = 1, the following inequality holds

ala2

-----,~--=--:-~----:-

(ai

+ a2)(a~ + al)

a2 a3

(a~ + a3)(a~ + a2)

+ ... +

anal

(a~

+ al)(al + an

< kn· -

Gabriel Dospinescu, Mircea Lascu

Proof. The basic inequality that will be used is

(x2

+ y)(y2 + x) 2: xy(l + x)(l + y).

XiXHl

(Xi

+ Xi XH2 + X;+l

2 X i +l

+ xHd(XHl + XH2)

+ XHl + (Xi + xi+d(Xi+l + Xi+2)·

Since

""' Xi > ""' L Xi + xHl L Xl it remains to prove the inequality ""'

+ X2 + ... + Xn

2 i l

( Xi

+ + Xi+1 )( XHl + Xi+2) 2: X

T2 's lemma reduces this to the easier inequality This reduces immediately to (x2 - y2)(x - y) 2: 0, which is clear. This already shows that for n = 2 the maximum value is k n = ~, since (1 + x)(l + y) 2: 4 if xy = 1. Let us assume now that n > 2. We will prove that kn = n - 2. To prove that kn 2: n - 2, it is enough to exhibit sequences all a2, ... ,an for which the value of the expression

Expanding the left-hand side makes the previous inequality obvious and fin0 ishes the proof of the fact that k n = n 2 for n 2: 3. The following difficult problem seems to be exactly the opposite of what is usually called a standard application of T2 's lemma. It turns out that we can actually apply T2 's lemma, but in a very nontrivial way.

Chapter 5. T2

218

Lemma

15. Prove that for any positive real numbers a, b, c, (2a+b+c)2 2a2+(b+c)2

(2b+c+a)2

(2c+a+b)2 <8

+ 2b2 +(c+a)2 + 2c2 +(a+b) 2 - ·

Titu Andreescu and Zuming Feng, USAMO 2003

Proof. Writing b+c

x=--, a

a+c b

y=--,

a+b c

+ 2x 1 + 2y 1 + 2z < 5 ( 2 + x)2 + (2 + y)2 + (2 + z? < 8 ¢1= }--+--+---'-2-+-x-'2'2 + y2 2 + z2 x2 + 2 y2 + 2 z2 + 2 - 2 1)2 (y - 1)2 (z - 1)2 > 1 + + -. x2 + 2 y2 + 2 z2 + 2 - 2

(X -

In order to prove the last inequality, we use T2 's lemma in the obvious form and so it suffices thus to show that (x+y+z-3)2 >~. x 2 + y2 + z2 + 6 - 2

Expanding (x

+y +z -

is the obvious equality case in the original inequality). Thus, we should have f (~) = J+v and also f' (~) = u, since ~ would be a minimum of ux+v- f(x) on [0, 1]. A straightforward, but tedious computation shows that u = 4 and v = We need to prove now that this pair (u, v) really works, i.e. that f(x) ::; ux + v holds for any x E [0,1]. Clearing denominators and expanding everything yields (after a tedious computation) the equivalent inequality

36x 3

15x 2 - 2x

Z=--,

we can write the inequality as

¢=}

219

+ 1 2: o.

But the conditions we imposed were made so that the left-hand side is divisible by (3x - 1)2. Doing the euclidean division shows that the left-hand side is (3x _1)2( 4x + 1), which is obviously positive. Finally, we just have to add the inequalities f (x) ::; ux + v for x E {a, b, c} to end the proof. D The following problems are harder than the previous ones. They are still based on T2 's lemma, but applied in more subtle ways and often combined with other tricks. 16. Let a, b, c, d be positive real numbers such that abed

1. Prove that

1 1 1 1 -:------:"""'" + + + > 1. (1 + a)2 (1 + b)2 (1 + c)2 (1 + d)2 -

3)2, we reduce this to proving that Vasile Cartoaje

L x 2 - 12 L x + 4 L xy + 12 2: o. This is not obvious, but the observation that x 2 + 4 2: 4x reduces it to proving that :E xy 2: 2 :E x, which is problem 7 in chapter 1. D

Proof. This solution uses the linearization method. To simplify computations, we may assume (since the inequality is homogeneous) that a+b+c = 1. Define

Proof. The proof using T 2 's lemma is rather mysterious. By performing the substitution t _ z a - JL d= ~ - x' b - y , c= -, t' z we obtain the equivalent inequality

the map f (x) = 2x2

+ (1 _ x) 2 .

The inequality can also be written as f(a)+ f(b)+ f(c) ::; 8. We will try to find u, v such that f(x) ::; ux + v for any x E [0,1], with equality for x = ~ (which

Of course, an immediate application of T2 's lemma fails rather badly, so we need something more clever. Let us apply T2'S lemma for the first two and

Chapter 5. T2

220

Lemma

221

then for the last two terms of the inequality. If we try to prove the stronger inequality (X+y)2 (z+t)2 >1

The following problem is a bit tricky, especially because of the strange hypotheses.

+ (z+t)2+(t+x)2 - , we easily realize I that it is equivalent to (y + z)(t + x) :::; (x + y)(z + t). Unfortunately there is no reason to have (y + z)(t + x) :::; (x + y)(z + t). The

17. Let n 2: 16 be a positive integer and suppose that the positive numbers aI, a2,ÂˇÂˇ., an satisfy al + a2 + ... + an = 1 and al + 2a2 + ... + nan = 2. Prove that

(x+y)2+(y+z)2

miracle is that if this fails, then we can apply T2 's lemma for the first and fourth terms of the initial inequality and then for the middle terms. In this case, we obtain the stronger inequality

(x + t)2 + (y + z)2 > l. + y)2 + (x + t)2 (y + z)2 + (z + t)2 -

Gabriel Dospinescu Proof. The first step is to apply Abel's summation formula to rewrite the inequality as

A similar argument shows that this holds precisely when (x + y)(z + t) :::; (y+z)(x+t), in particular whenever (y+z)(t+x):::; (x+y)(z+t) fails. The result follows. D

v'2) + ... + an-l (vn - vn=I) > an vn. Using the obvious bound al v'2 > al (v'2 - 1) and the formula al v'2 + a2( V3 -

Y'k- Vk=l =

Proof. This solution is not natural, either, but it is very elegant. We claim that for all positive numbers a, b we have 1

+ a)2 +

Y'k+ vik=1'

an application of T2'S lemma shows that the left-hand side is greater than

+ b)2 2: 1 + ab'

+ ... +an-l)2 L~ll ai( VI + vT+I) .

(al +a2

This is rather easy to prove, though it requires some nasty computations. Namely, by clearing denominators and performing the obvious simplifications, we reduce the problem to proving that

ab(a 2 - ab + b2) - 2ab + 1 2: 0,

On the other hand, Cauchy-Schwarz together with the hypothesis give the estimate

which is equivalent to

ab(a - b)2

+ (ab -

1)2 2: O.

and also n-l

L aiJi+1 < V3.

Once we have this inequality, we deduce that

1 (1

+ a) 2 2: 1 + ab + 1 + cd = 1 + ab +

is convenient to denote a

ab 1 + ab = 1

i=l

and the result follows. 1 It

and b = ;t~.

Therefore, taking into account that al + a2 + ... + an-l to prove the inequality

= 1 - an, we still need

222

223

Chapter 5. T2 's Lemma

The hard point is to figure out that such an inequality holds, since proving it is a very easy matter. Note that the right-hand side is clearly positive, so by squaring and canceling similar terms we obtain the equivalent inequality

But since

we have an < n~l and so it is enough to prove the previous inequality with an replaced by n~l' But this is immediate for n ~ 16, which ends the solution.

2+6 y

+ yz

9y2z 2 > 6yz(3y+z) (2y+z)2 2y+z

Â˘=}

4y2(y-z)2 (2y+z)2

~ O.

Using this estimate we conclude: We end this chapter with two challenging inequalities. The first one uses a combination of T2'S lemma and a rather subtle linearization technique. 18. Prove that for all positive numbers a, b, c the following inequality holds

~~~ -+ -+ 8b + c 8c + a

(Va? Ja(8b + c)'

Define x

~ 2:: J

42::xy -

3xyz

2:: _1_ 2y+z

2:: _1_ > __3 _

+ z - x + y + z'

42:: xy -

9xyz :::; x+y+z

2:: x 2 + 2""'~" xy.

It is not difficult to check that the last inequality is actually equivalent to Schur's inequality, which finishes the proof of this hard problem. 0

The next problem requires some preliminaries. We will prove the following beautiful discrete variant of a classical inequality of Wirtinger.

so it remains to prove that

(2:: Va)

:::;

it remains to prove that

Proof. We use T2'S lemma in the form

V8"b+c =

and since

+ z2

-->1. 8a + b -

Vo Quoc Ba Can

r-a-

2::XV8y2

a(8b + c).

= Va, y = Vb, z = y'c;, so that the inequality becomes

Theorem 5.3. (Fan's inequality) If xl, X2," ., Xn are real numbers which add up to zero, then

2 cos -21l' (Xl2 + x22 + ... + Xn) n

This is actually a very strong inequality and it is easy to see that it resists any attempt to prove it using Cauchy-Schwarz (even if its form invites us to use such a technique). We will use a linearization technique, by approximating first J8 y 2 + z2 by a rational function in y, z. The crucial ingredient is the following estimate: J 8y2 + z2 :::; 3y + z - -3yz -. 2y+z

~ XIX2

+ X2X3 + ... + XnXIÂˇ

Proof. We will actually mimic the proof of Wirtinger's inequality by using finite Fourier transforms (for more on this fascinating subject, the reader is invited to read the addendum 7.A). Namely, if Zl, Z2,' .. , Zn is a sequence of complex numbers, define n

AI"",,,

Zj = -

. ~ Zke

k=l

2i1fki n.

224

Chapter 5. T2 's Lemma

Using the fact that

225

follows from the definitions. Thus, using Parseval's identity we obtain

n """'

L...te

2i rrk j

LY; = LIYjl2

k=l

if and only if j is not a multiple of n (in which case it equals n), it is easy to check that we can recover our sequence from the sequence of its finite Fourier transforms using the inversion formula 1

Vn L

11 e~121xj12

2i7l"kj

Zk e

k=l

'We also have the following discrete version of Parseval's identity: n

L k=l

Zkl =

IZkI 2 .

and another application of Parseval's identity yields the desired inequality.

k=l

Indeed,

We are now ready to prove the following version of Shapiro's inequality. The proof is based on a very tricky application of T2 's lemma combined with Fan's inequality.

19. Let

J2 cos1

271"_1 n

Prove that for all

Xl, X2, ... ,X

n E [al an], ,

the

Shapiro inequality holds:

Vasile Cartoaje, Gabriel Dospinescu Now, the proof of Fan's inequality is very easy: write the inequality in the form

Proof. First, we write the inequality in the form

> n - 2 Note that the hypothesis Xl +X2+' . '+X n = 0 can be written in the very simple ~ form xn = O. Now, let 2 Yj = Xj - Xj+1 and observe that ih = (1- e n )Xj, as is the analogue of the derivative of a function when establishing discrete inequalities.

Note that

Chapter 5. T2

226

Lemma

Addendum 5.A

Using T2'S lemma, it suffices to prove that

(1- al~)

(LXif 2:

~ ( L Xi(Xi+1 +Xi+2) - 2~~ L(Xi +Xi+l)2).

A crucial step is to note that we have the identity

Xi(Xi+l + Xi+2) = L(Xi + Xi+I)(Xi+l + Xi+2)

which allows us to perform the substitution Xi problem to proving that

(1- al~)

(Lbi )2 2:

n(2

+ Xi+1

Lbibi+l -

~ L(Xi + Xi+l?, =

2bi and reduce the

(1 + al~)

= bi

which holds for any positive real numbers rather with the following: Theorem 5.A.1. Let

+ b2 + ... + bn

-=---=-----n

q such that ~

+ ~ = 1, but

be positive real numbers. Then

II (ajl + aj2 + ... + ajn) 2: ({;Iau a 2l ... akl + ... + {;Ial n a2n··· akn)k. j=l

Proof· This is a very easy application of the AM-GM inequality. Indeed, just add up the following inequalities

(1 + a~) L c; 2: 2L CiCi+l· Cl + C2 + ... + Cn

aij

ai, bj , p,

A rather tedious, but straightforward computation shows that the previous inequality is equivalent to

Of course, we have

The purpose of this addendum is to present some applications of Holder's inequality, which are quite similar to the problems considered in this chapter. Of course, Holder's inequality is important because of its applications in measure theory, probability theory and analysis and not really for the amusing problems to be discussed here. Actually, we will not even deal with the classical version

Lb;)

for nonnegative real numbers bi . This simplifies drastically if we make another substitution bl

Holder's Inequality in Action

0. Finally, note that

1 2?T 1 + 2 = 2cos-. an n Thus, the result follows from Fan's inequality.

5.1

Notes

We thank the following people for providing solutions: Alexandru Chirv8,situ (problem 16), Xiangyi Huang (problem 14), Michael Rozenberg (problem 15), Dusan Sobot (problems 1, 3), Gjergji Zaimi (problems 1, 10,

11).

o We are now able to obtain a generalization of T2 's lemma that turns out to be very handy in quite a lot of situations when T2 's lemma fails. There are of course much more general results than the following one, but our purpose is not to delve into the greatest generality.

228

Chapter 5. T2 's Lemma

Theorem 5.A.2. Let aI, a2, . .. ,an and Xl, X2, . .. ,Xn be positive real numbers and let q > P be two positive integers. Then ai ag aK l+p-q (al -+-+路路路+->n

:?n -

(Xl

5.A. Holder's Inequality in Action

229

A.3. For any positive real numbers a, b, c the following inequality holds b+c

+ a2 + ... + an)q .

---;=:~= 2

va + bc

+ X2 + ... + xn)p

c+a Vb 2 + ca

a+b

> + ab -

Proof. Note that by the previous theorem we have

Pham Kim Hung Proof. Let S be the left-hand side. Using Holder's inequality, we can write

~ (af+l

so it is enough to prove the stronger inequality

+ a~+l + ... + ai;+l )p+I

and the result follows from the Power Mean inequality.

The first theorem is particularly useful when working with sums of square (or cubic or ... ) roots, which are a nightmare most of the times. Here are a few examples that will probably convince you how useful this result is. Most of the problems are quite hard to solve by other means.

An easy computation shows that this follows from Schur's inequality. A.4. Let a, b, c be positive reals such that abc = 1. Show that 1 1 1 a 5 (b + 2c)2 + b5 (c + 2a)2 + c5 (a + 2b)2

A.2. Prove that for all positive real numbers a, b, c such that abc = 1 we have \!a3 +7+ \!b3 +7+

\!c3 +7:S 2(a+b+c).

~ 3路

Titu Andreescu, USA TST 2010

Vasile Cftrtoaje

Proof. The substitution a

Proof. This is quite a strong inequality, but the previous results are effective in such a context: (\!a 3 +7+\!b3 +7+ \!c3 +7)3

= ~,b = ~,c = ~

reduces the problem to

(2y + z)2 + (2z + x)2 + (2x + y)2 2:: 3.

By theorem 5.A.2,

:S (1 + 1 + l)(a + b + c)((a 2 + 7bc) + (b 2 + 7ca) + (c 2 + 7ab))

X3 ____ +

and the result follows immediately from the inequality ab + bc + ca :S a 2 + b2 + c2.

(2y+z)2

Y (2z+x)2

z > x+y+z (2x+y)2 9

and the result follows from the AM-GM inequality.

Chapter 5. T2 's Lemma

230

A.5. Prove that if Xl, X2, .. . , Xn are positive real numbers with product 1, then

5.A. Holder's Inequality in Action

Proof. This is a pretty tricky application of Holder's inequality: 4(1

2:: ( Xl

+ X2 + ... + Xn + -1 + -1 + ... + -1 Xl

231

+ a2)2(1 + b2)2(1 + e2)2 =

(a 2b2 + a2 + b2 + 1)(b2 + e2 + b2e2 + 1) . (1 + 1 + 1 + 1)(a2 + a2e2 + e2 + 1)

2:: (1 + ab + be + ea)4.

The result follows. Gabriel Dospinescu

Proof. This follows easily by adding the inequalities

n/(x~ + 1)(x2 + 1) ... (x~ + 1) 2:: Xi + XIX2···Xi-IXi+I",Xn V

= Xi

+~ Xi

Proof. Using Holder's inequality, we can write

obtained from Holder's inequality.

abc + ab + e 2:: ab + be + ea,

~a. a+b. a+b+e > a+Vab+ ij(j])C. 3

Kiran Kedlaya

Proof. Using theorem 5.A.1, we obtain

a+b (a+a+a) (a+ -2- +b) (a+b+e) 2::

One concludes by observing that

(

ab(~+b) 2:: Vab.

ab a + b (2)

ij(a3 + 1)(b3 + 1)(e3 + 1) 2:: ab + e. We would like to have

A.6. For any positive real numbers a, b, e the following inequality holds

A.S. Prove that for all positive real numbers a, b, e the following inequality holds abc + ij(a3 + 1)(b3 + 1)(e3 + 1) 2:: ab + be + ca.

which is equivalent to (a - 1)(b - 1) 2:: O. There is no reason for this to hold, but at least one of the inequalities (a - 1)(b - 1) 2:: 0, (b - 1)(e - 1) 2:: 0 and (a - 1)(e - 1) 2:: 0 holds, as two of the numbers a - 1, b - 1, e - 1 must have the same sign. The result follows. 0 A.9. Prove that for all positive real numbers a, b, e the following inequality holds

3 3

+ rabc )

A.7. Prove that for all real numbers a, b, e we have

Titu Andreescu, USAMO 2004

Proof. Of course, we cannot apply theorem 5.A.1 directly in this case, but the form of the inequality is a temptation to find a way to apply theorem 5.A.1. The key point is the inequality Michael Rozenberg

a 5 - a2 + 3 2:: a3 + 2,

Chapter 5. T2 's Lemma

232

which is equivalent to (a 2 - 1) (a 3 - 1) ;::: 0 and thus obvious. Using this, the result follows immediately from theorem 5.A.1, by writing

5.A. Holder's Inequality in Action

233

so that after substituting x = a 2 , y = b2 and z = c2 , it is enough to prove the inequality (x + y + z)6 ;::: 27(xy + yz + zx)2(x 2 + y2 + z2). But this is immediate from the AM-GM inequality and the identity

A.10. Let a, b, c be the sides of a triangle. Prove that

~+ ~+ ~>a+b+c.

A.12. Prove that for all positive real numbers a, b, c, d

V~-

Titu Andreescu, Gabriel Dospinescu

a 3 + 1 b3 + 1 c3 + 1 d 3 + 1 _._ - . 2- - . - >abed - -+-1 2 2 2 a +1 b +1 c +1 d +1 2

Vasile Cartoaje, Gazeta Matematica

Proof. This is a hard inequality. The point is to use Schur's inequality a2(a _ b)(a - c)

+ b2(b -

a)(b - c)

+ c2 (c -

Proof. This is also a very hard problem. Of course, one is again tempted to use the theorem, but one needs a trick. The key point is that for all positive numbers a we have

a)(c - b) ;::: 0,

a++ > 1a +

which can also be written as

1 -

2Âˇ

Indeed, using the inequality

with the theorem 5.A.2. Indeed, observe that we can write

which is just Cauchy-Schwarz, one reduces this to

The result follows.

which, after division by (a A.ll. Prove that for all a, b, c > 0 we have a2

_ +

( b

2 -b c

c2 )

(a-1)4;:::0. 4

;:::

27 . (a

+b +c

Proof. If X is the square root of the left-hand side, then Holder's inequality

yields

+ 1)2 and a small computation is equivalent to

Once we have this key inequality, the rest is an immediate application of theorem 5.A.1. 0

Chapter 6 â&#x20AC;˘ Some Classical Problems In Extremal Graph Theory

This elementary chapter is a variation on a classical topic in extremal graph theory: Turan's theorem on graphs without cliques. Recall that if Gis a graph and k is a positive integer, then a k-clique is a set of k vertices, any two of which are connected. A graph is called k-free if it does not contain a k-clique. We then have the following standard result. Theorem 6.1. (Turtin) The maximal number of edges in a k-free graph with k- 2 n vertices is k _ 1 . 2 + 2 ,where r is the remainder of n when divided

n2- r2 (r)

by k - 1.

In particular, the maximum number of edges in a k-free graph with n vertices is at most ~2 , which is really the estimate that we will constantly use. We start with a series of rather direct applications of these two theorems. The other problems are however different in nature and more difficult.

ti .

1. Let

xl, X2,' .â&#x20AC;˘ ,Xn

be real numbers. Prove that there are at most

pairs (i,j) E {1,2, ... ,n}2 such that i < j and 1 < /Xi - XjJ < 2.

MOSP 2001

Chapter 6. Some Classical Problems in Extremal Graph Theory

236

Proof. Consider the graph whose vertices are 1,2, ... ,n. Connect two vertices IXi - Xj I < 2. We claim that this graph contains no triangle. If we manage to prove this, the result follows from Tunin's theorem. Suppose that the graph contains a triangle, thus we can find distinct a, b, c such that

i, j by an edge if (i, j) satisfies 1 <

IXa -

xbl

< 2,

IXb -

xci < 2,

By symmetry, we may assume that Xa < Xb equalities become Xb - Xa > 1, Xc - Xb > 1, so inequality Ixc - xal < 2. The result follows.

Ixc -

xc'

Then the previous in> 2, contradicting the 0

Xc -

xal

< 2.

2. Prove that if n points lie on a unit circle, then at most 3 segments connecting them have length greater than

J2. Poland 1997

Proof. Consider the graph whose vertices are the n points and connect two vertices if their distance is greater than J2. The point is that this graph contains no 4-clique. This is clear, since a chord of length J2 subtends a central angle of ~. Thus, by Turan's theorem the number of vertices is at most [~2], finishing the proof.

237 know Ail"'" A ik · Thus, if k ·1959 - (k - 1) . 1999 :::: 1, we can always add one more person. If k ~ 48, then there are at least 79 people who know all of Ail>"" Aik and one of them is among AI, A 2 , ... , A 1995 . If k = 49, then there are at least 39 people who know all of Ail' ... ,Aik and (although that person may not be among AI, A 2 , ... , A 1959 ) he completes a set of 50 all of 0 whom know each other. 4. We are given 5n points in a plane and we connect some of them so that 2 10n + 1 segments are drawn. We color these segments in 2 colors. Prove that we can find a monochromatic triangle.

Proof. Note that the corresponding graph (with vertices the 5n points and edges between points connected by a segment) contains a 6-clique. Indeed, otherwise by Turan's theorem it has at most (5~)2 . ~ = 10n 2 edges, which contradicts the hypothesis. Now, consider a 6-clique C/ of our graph G. The edges of C/ are colored in two colors and we claim that there is always a monochromatic triangle in C/. This is standard, but we recall the proof. Pick a vertex VI of C/. By the pigeonhole principle, we may assume that VI V2, VI V3, VI V4 have the same color. If one of the edges V2V3, V2 V 4 or V3 V 4 has the same color as VI V2, we are done. Otherwise, the triangle V2 V 3 V 4 is monochromatic and we win again. 0

3. There are 1999 people participating in an exhibition. Out of any 50 people, at least two do not know each other. Prove that we can find at least 41 people who each know at most 1958 other people.

The following problem does not use Turan's theorem, but it is still a very classical topic, namely Ramsey's numbers.

Taiwan 1999

5. A group of people is called n-balanced if the following two conditions are satisfied:

Proof. This is a special case of the proof of Zarankiewicz's lemma. For the reader's convenience, let us recall the proof. Suppose that the conclusion does not hold, so we can find at least 1959 people, say AI, ... ,A I95 9, each having at least 1959 friends. Start with AI. There is a person among A 2 , .•. ,A 195 9 that knows AI. Assuming that we found persons Ail' ... ,Ak (il = 1) among AI"'" A 1959 , every two knowing each other, let us try to add a new person A·tk+l to the group . But there are at least k ·1959 - (k -1) ·1999 persons that

a) among any three people, there are two who know each other-, b) among any n people, there are at least two not knowing each other. Prove that there are always at most (n-I~(n+2) people in an n-balanced group. Dorel Mihet, Romanian TST 2008

238

Chapter 6. Some Classical Problems in Extremal Graph Theory

Proof. We will prove the result by induction. For n = 2, this is trivial, so assume that it holds for n - 1. Consider an n-balanced group and pick an arbitrary person P. Let A be the set of friends of P and let B be the set of all the other persons in the group. By hypothesis, any two persons in B are friends and B has at most n - 1 elements. By induction, A has at most (n-2)2(n+l) persons (by b) and the fact that P knows all elements of A, it follows that A is an n - I-balanced group). Thus there are at most

239

Applying Cauchy-Schwarz and taking into account that n

L d(Xi) = 2k i=l

yields the desired result. The following problem is very similar to the previous one.

(n-2)(n+l) (1) (n-l)(n+2) 2 +1+ n= 2 persons in the group, which is enough to prove the inductive step.

The following problem and the method of proof are absolute classics. k 6. Prove that a graph with n vertices and k edges has at least - (4k - n 2 ) 3n triangles.

APMO 1989

Proof. Let xl, X2, ... ,Xn be the vertices of the graph G and let d(x) be the degree of x. Observe that for a given edge e = XiXj with endpoints Xi, Xj, there are at least d(Xi) + d(xj) - n triangles containing e. Indeed, there are d(Xi) + d(xj) - 2 edges having as endpoints one of Xi,Xj and another vertex among the n-2 remaining vertices, so there are at least d(Xi)+d(xj)-2-(n-2) triangles containing Xi, Xj as vertices. Summing over all edges and taking into account that we count three times every triangle in this way, shows that the number of triangles is at least (we denote by E(G) the set of edges of G)

7. A graph with n vertices and k edges has no triangles. Prove that we can choose a vertex such that the subgraph obtained by deleting this vertex and all its neighbors has at most k

~~)

edges. USAMO 1995

Proof. Each edge xy gets killed when we remove X and its neighbors, when we remove y and its neighbors, or when we remove any common neighbor of X and y and its neighbors. Thus, this edge is killed a total of d( x) + d(y) times. Summing over all edges and taking into account that d(x) appears as a summand exactly d(x) times, we obtain that the average number of edges killed by removing a vertex and its neighbors is ~ L:x d(x)2. Since

we have

k ;;;1 "L..td(x)2 ~ (2n

Thus, on average we kill at least ~ edges. The result follows.

The previous sum is also equal to 8. A graph G has n vertices and contains no complete subgraph with four vertices. Prove that G contains at most ~; triangles. Ivan Borsenco, Mathematical Reflections

Chapter 6. Some Classical Problems in Extremal Graph Theory

241

Proof. The result is easy for n = 2, 3, 4, so let us assume that n > 4 and that the result holds for all k < n. We may assume that G contains a triangle ABC. Let G I be the subgraph formed by the vertices of G different from A, B, C. Since G I has no 4-clique, it has at most (n-;3)2 edges by Turan's theorem. By

find A E lFp such that V4 = AV2. Then A = 1, as (VI, V4) = (VI, V2) = 1. We conclude that V2 = V4, a contradiction. For the second part, take any prime p and set n = p2. The graph G n in the first part of the problem has n vertices. Let us evaluate the number of edges. If V = (x, y) is a nonzero vector in V, then the equation (w, v) = 1 has p solutions. Thus the number of edges is P(P~-l) and it is immediate to check

240

the inductive hypothesis, G I has at most (n~i)3 triangles. Any other triangle in G consists either of a vertex of G I and an edge of ABC or of an edge of G I and a vertex of ABC. Moreover, any edge of G I forms a triangle with at most one vertex of ABC and any vertex of G I forms a triangle with at most one edge of ABC, because G contains no 4-clique. Thus G contains at most (n-3)3 27

(n-3)2 3

n3 + n - 3 + 1 = 27'

establishing the inductive step. Note that the result is optimal if n is a multiple of 3, since we can consider the tripartite graph K n j3,nj3,nj3' 0 It is a standard result that a graph with n 2: 4 vertices and more than n+n~ edges has a four-cycle (see, for instance [3], example 3, chapter 22). The following nice problem shows that this result is almost optimal. 9.

a) Let p be a prime. Consider the graph whose vertices are the ordered pairs (x, y) with x, y E {O, 1, ... ,p-1} and whose edges join vertices (x, y) and (x', y') if and only if xx' + yy' == 1 (mod p). Prove that this graph does not contain a 4-cycle . b) Prove that for infinitely many n there is a graph G n with n vertices and at least nf - n edges that does not contain a 4-cycle. Hungary-Israel Competition 2001

Proof. One can give a rather down-to-earth proof of a), based on explicit computations, but we prefer the following approach. Consider the lFp = Z/pZvector space V = lF~ and the standard inner product ((x, y), (x', y')) = xx' + yy'. We are asked to prove that we cannot find four distinct vectors VI, V2, V3, V4 E V such that (Vi, Vi+l) = 1 for all 1 :::; i :::; 4 (here V5 = vI). If such vectors existed, V2 and V4 would be both orthogonal to the nonzero vector VI - V3 and so they would be contained in a line of V. Thus, we can

that this is greater than

nf - n = r; - p2.

The result follows.

We continue with two very nice problems concerning graphs with n 2 + 1 edges and 2n vertices. Note that this is the first case when Thran's theorem ensures the existence of a triangle. The following results show that we can do better. 10. Prove that a graph with 2n vertices and n 2 + 1 edges contains two triangles sharing a common edge. Chinese TST 1987

Proof. We will prove the result by induction. For n = 2, this is trivial. Assume now that the result holds for n and consider a graph G with 2n + 2 vertices 2 and n + 2n + 2 edges. By Turan's theorem and the pigeonhole principle, we can find a triangle xyz such that d(x) == d(y) (mod 2). Consider the 2n points different from x, y. If there are at least n 2 + 1 edges among them, we are done by the inductive hypothesis. If not, then there are at least 2n + 2 edges whose endpoints are x or y. But since d(x) == d(y) (mod 2), it follows that there are actually at least 2n + 2 edges whose endpoints are x or y and which are different from the edge xy. Let Al be the set of vertices V =I=- x, y that are connected to x and let A2 be the set of vertices v =I=- x, y that are connected to y. Then the previous result implies that IAII + IA21 2: 2n + 2. Since IAI U A21 :::; 2n, it follows that IAI n A21 2: 2 and so we can find two distinct vertices z =I=- t that are connected to both x, y. Then the triangles xyz and xyt share a common edge and we are done. 0 11. A graph has 2n vertices and n 2 + 1 edges. Prove that it contains at least n triangles.

242

Chapter 6. Some Classical Problems in Extremal Graph Theory

Proof. Again, the proof is by induction. For n = 2 this is trivial, so assume that the result holds for n and consider a graph with 2n + 2 vertices and n 2 + 2n + 2 edges. By T\min's theorem we can find a triangle XIX2X3. Let Ai be the set of vertices v i=- Xl, X2, X3 that are connected to Xi. There are obviously at least

triangles different from XIX2X3, since any element of Ai n Aj forms a triangle with the vertices XiXj' Thus, if we have S 2: n, we are done. Assume that S S n - 1. Then, using the inclusion-exclusion principle, it follows that

2n - 1 2:

IAI

IAII + IA21 + IA31 n + 1, IAII + IA21, IA21 + IA31, IA31 + IAII is smaller

U A2 U A31

thus one of the numbers 2n - 1, say IAII + IA21. But this means that there are at least

+ 2n + 2 -

(2n + 1) = n

than

edges among the vertices A 3 , A 4 , .. . ,A2n +2 . Thus, by induction we find at least n triangles among these vertices and adding the triangle AIA2A3 yields at least n + 1 triangles. The inductive step is thus proved and the problem solved. D The following problems are more difficult. The next one concerns coverings of the edges of a graph by cliques.

"\V

12. There are n aborigines on an island. Any two of them are either friends or enemies. One day they receive an order saying that all citizens should make and wear a necklace with zero or more stones so that i) for any pair of friends there exists a color such that each of the two persons has a stone of that color; .

243

Proof. Let G be the graph whose vertices are the aborigines, two of them being connected if they are friends. Let Bi be the set of aborigines who receive a bead of the i-th color. Then by condition ii) each Bi forms a clique in G and by i) each edge is in one of these cliques. Conversely, if we have a collection of k cliques that cover every edge of G, then we can give a bead of color i to every aborigine in the i-th clique and satisfy the conditions of the order. The previous paragraph shows that we need to find the smallest number f(n) of cliques that cover the edges of any graph on n vertices G. Clearly f(2) = 1 and f(3) = 2. To find an optimal graph, consider a complete bipartite graph on n vertices. The key point is that such a graph has no triangles, so if we want to cover its edges by cliques, we need at least E( G) cliques, where E( G) is the number of edges. Therefore, we have a clear bound on f(n), namely f(n) 2: [r~n. The hard part is to prove the opposite inequality. We will prove this by induction, going from n to n + 2 (which, combined with the first two values of f(n), will be enough to conclude). Consider a graph G with n + 2 vertices. We may assume that at least two vertices are connected, say X, y. Consider the graph G I obtained by deleting X, y and all edges adjacent to these two vertices. By induction, we know that we can cover its edges by f(n) cliques. If v is a vertex of G I , consider the subgraph spanned by X, y, v. By adding either a triangle or an edge, we can cover all edges of this subgraph incident to v. Finally, by adding one more edge to cover the edge xy, we covered all edges of G by using at most f (n) + n + 1 cliques (by the way, all the cliques used were only edges or triangles). Therefore

f(n Since

ii) for any pair of enemies there does not exist such a color. What is the least number of colors of stones required (considering all possible relationships between the inhabitants of the island)? Belarus 2001

it follows that

+ 2) S

f(n)

+ n + 1.

[(n:2)2] = [:2] +n+1, f (n) S [~2].

Therefore the answer is

[~2].

It is not hard to guess the answer of the following problem, but proving that this is the correct answer is quite a pain in the neck.

244

Chapter 6.

Some Classical Problems in Extremal Graph Theory

13. What is the least number of edges in a connected n-vertex graph such that any edge belongs to a triangle? Paul Erdos, AMM E 3255 Proof. Let f(n) be the desired number and let g(n) = [3n;2]. We will prove that f(n) = g(n) for all n. To save words, call good a connected graph in which every edge belongs to a triangle. We can easily establish that f(n) ::; g(n) by explicit constructions: if n = 2k + 1, consider k triangles sharing a common vertex, this being the only common point of any two triangles. This is a good graph with 3k edges, so f(2k + 1) ::; 3k = g(2k + 1). If n = 2k, start from the above good graph with order 2k -1, choose an edge AB and add a vertex X and edges AX, BX. We obtain a good graph with n vertices and g(n) edges. The hard point is proving that any good graph on n vertices has at least g( n) edges. We will prove this by strong induction. For n = 3,4, this is easily checked. The crucial observation that makes the induction work is that a good graph with n vertices and less than g(n) edges must have a vertex of degree 2. This is trivial, since by connectedness all vertices have degree at least 1 and since every edge is in a triangle, every vertex must have degree at least 2. However, the sum of the degrees of all vertices is twice the number of edges, thus smaller than 3n. Suppose now that f(j) = g(j) for all j < n and let us prove that f(n) 2: g(n). Suppose G is a connected graph with f(n) edges such that each edge belongs to a triangle and suppose f(n) < g(n). There is a vertex x of G of degree 2, which by assumption must be in some triangle xyz. Suppose the edge yz is in a second triangle. Then the graph G' obtained from G by removing x and the edges xy and xz is connected, has n - 1 vertices, every edge belongs to a triangle, and f(n) - 2 < g(n) - 2::; g(n - 1) edges. This is a contradiction. Thus yz is not contained in another triangle. Form a new graph Gil by deleting the vertex x and collapsing the vertices y and z into a single new vertex w, where a vertex v of Gil is joined to w if in G it. was joined to either of y or z. Clearly Gil is a connected graph with n - 2 vertices and every edge of Gil is contained in a triangle. In forming Gil, we lost three edges xy, yz, and xz. We would also lose an edge if some vertex v were adjacent to both y and z, but by

245 assumption this does not occur. Thus Gil has f(n) - 3 < g(n) - 3 = g(n - 2) edges. Again this contradicts the induction hypothesis. Thus we must have f(n) 2: g(n). 0 The following result is classical and very nice. We give two proofs, the first one being an explicit inductive construction, the second one using the powerful probabilistic method of Erdos. We refer the reader to the addendum 6.B for more details on this versatile tool. 14. Prove that for every n there is a graph with no triangles and whose chromatic number is at least n. Mycielski's theorem Before starting the proof, let us recall some definitions. A k-proper coloring of a graph G is a coloring of its vertices with at most k colors such that each vertex receives one color and no two adjacent vertices have the same color. The chromatic number of a graph is the smallest number k for which G has a proper k-coloring. Proof· We will prove this by induction on n. For n = 1 everything is clear, so assume that we have a graph G with no triangles and chromatic number at least n and let us construct another graph with no triangles and chromatic number at least n+ 1. Let x( G) be the chromatic number of G. We may assume that X(G) = n (otherwise keep G for the inductive step). Let VI, ... , vk be the vertices of G and consider the following new graph G': the set of vertices consists of VI, •. . , Vk, together with vertices Xl, . .. ,Xk, Y (such that the 2k + 1 vertices thus obtained are distinct). Connect Xi with all neighbors of Vi and connect y with Xl,·· ., Xk· Also, keep the edges in G. We claim that G' is triangle free and X( G') 2: n + 1. It is clear by construction that G' has no triangles. Also, we easily have X(G') ::; X(G) + 1, since any proper n-coloring of G extends to a proper n + I-coloring of G' (assign the same color to Xi as to Vi and assign the color n + 1 to y). The difficult part is proving that this is actually an equality. Suppose thus that G' has chromatic number at most n and take any proper n-coloring of G'. We will construct a proper n - I-coloring of G, which will contradict

246

Chapter 6. Some Classical Problems in Extremal Graph Theory

247

the fact that G has chromatic number n. Assume, without loss of generality, that Y has color n and let A be the set of vertices of G whose color is nand for each Vi E A change its color with that of Xi. We claim that this gives a proper n - I-coloring of G. Note that no two vertices of A are adjacent, as all these vertices have the same color. Now, if Vi E A is adjacent to Vj tt A, then Vj is adjacent to Xi, so that they have different colors. Consequently, we found an - I-proper coloring of G, which is impossible. We deduce that the chromatic number of G' is at least n + 1. D

independent vertices in (~) ways and the probability that a given set of a

Proof. We will actually prove a stronger result, which is a classical theorem of Erdos. We will use the probabilistic method, for which the reader is referred to the addendum 6.B. . Theorem 6.2. For any positive integers k, n there exists a graph with chromatic number greater than k and such that each cycle has length greater than n. Take N sufficiently large and consider random graphs with N vertices GN,p, each edge appearing independently, with probability p = Nc-l for some o < c < ~. If X is the number of cycles of length at most n in GN,p (which cycles we will call short in the sequel), then clearly n

E[X] :::;

Ncn

:L N i . pi < 1 _ N-c

vertices is independent is certainly (1 - p) (~)). An easy computation shows that the last quantity tends to 0 as N -+ 00. Thus for N sufficiently large we have P(a(G) 2: a) < and so there is a graph Gp,N = G with X < ~ and a(G) < a. Now, delete one vertex from each short cycle arbitrarily. The remaining graph G l has at least N /2 vertices, no cycles of length at most n and independence number smaller than a. Thus

for sufficiently large N. The result follows.

We end this chapter with three gems in extremal graph theory, all taken from mathematical contests. \fI5. For a pair A plane, let

(Xl, Yl) and B = (X2' Y2) of points on the coordinate

d(A, B) = IXl -

x21 + IYl -

Y21·

We call a pair (A, B) of (unordered) points harmonic if 1 < d(A, B) :::; 2. Determine the maximum number of harmonic pairs among 100 points in the plane. USA TST 2006

i=3

since there are at most N i cycles of length i, each appearing with probability pi. Since the last quantity is smaller than for sufficiently large N, it follows by Markov's inequality that

Let a( G) be the size of the largest independent set in G (the independence number of the graph) and let X(G) be the chromatic number of G. Then it is easy to see that X( G) 2: Qfci) if G has N vertices. Taking a = [~ln NJ, the probability that a(G) 2: a is at most (~) . (1 - p)m (one can choose a

Proof Consider the graph with vertices on the 100 points and whose edges connect points of harmonic pairs. The key point is that this graph contains no 5-clique. Indeed, if Pi(Xi, Yi) are five vertices, any two of which are connected, then for all i =I- j we have 1 < d(Pi, Pj) :::; 2. Order these points such that Xl :::; X2 :::; ... :::; X5· It is easy to see that among the numbers Yl, Y2,·· . ,Y5 one can find three forming a monotonic sequence. Call these three numbers Yil' Yi2' Yi3' Then d(Pi1 ,Pi2 ) + d(Pi2 ,Pi3 ) = d(Pi 1,Pi3 )· This is however impossible, since d(Pil' Pi2 ),d(Pi2 , Pi3 ), d(Pi3 ,Pi1 ) E (1,2]. This finishes the proof of the fact that our graph contains no K 5 . By Tunin's

248

lOt

theorem, it has at most ~ . = 3750 edges and so there are always at most 3750 harmonic pairs. To finish the proof, it remains to exhibit a configuration having 3750 harmonic pairs. It is enough to distribute the 100 points near the points (0, ±~) and (±~, 0), having 25 points near each of these four points. D 16. For a finite graph G let f(G) be the number of triangles formed by the edges of G and let g( G) be the number of tetrahedra formed by the edges of G. Find the least constant c such that g(G)3 :::; c· f(G)4 for any finite graph G. IMO Shortlist 2004

Proof. Let us begin by considering the case of a complete graph Kn· Then obviously

f(G) Thus

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Chapter 6. Some Classical Problems in Extremal Graph Theory

(~) and g(G) = (~).

> g(Kn)3 _ (~)3

C- f(Kn)4 -

(~)4

and this happens for all n. Taking the limit as n --+ 00, we deduce that c ~ :2' The hard part is to prove that the value 332 actually holds for arbitrary graphs G. To do this, we will first consider the two dimensional version of the problem, which is comparing the number of triangles and edges in a graph. The reason is simple: computing tetrahedra in a graph G comes down to computing triangles in all subgraphs induced by the vertices of G and taking the sum of these numbers of triangles (and then dividing by 4, since each tetrahedra is counted four times). Therefore, consider first any graph G with n vertices Xl, X2,· .. , Xn and e edges. Let us bound the number of triangles in terms of e. If d(Xi) is the degree of the vertex Xi, then there are at most (d(~i») :::; d(~i)2 triangles containing the vertex Xi. But the number of triangles containing the vertex Xi is also obviously bounded by the number of edges of G, that is e. Thus for triangles containing the vertex each vertex Xi, there are at most d(xd .

Xi. Summing over i and taking into account that we count each triangle three times in this way, we obtain that

f(G) < -. 1 - 3

~-. 2 L

d(Xi)

= -e 2 3

~-.2

Note that the constant appearing in this estimate in optimal, by taking for = Kn and then letting n --+ 00. This already shows that we are on the right path, since we solved the two dimensional version of the problem. To solve the original problem, all we have to do it to repeat some of the arguments in the previous paragraph. Namely, take a graph G with vertices XlJ . .. ,Xn and fix a vertex Xi. The number of tetrahedra containing Xi as a vertex is the number of triangles in the subgraph induced by the vertices connected to Xi. Let ei be the number of edges in this subgraph, then the previous paragraph shows that there are at most ~ei.JW tetrahedra containing Xi as vertex. Also, note that we have 3f(G) = L: ei, since for each vertex Xi there are ei triangles containing Xi as vertex (and we count each triangle three times this way). However, one still needs an observation to end the proof: we gave an estimate for the number of tetrahedra having Xi as vertex in terms of ei, but there is also an obvious estimate: this number is at most f(G).

Therefore, there are at most f/~f(G)ei tetrahedra containing Xi. Summing over i and taking into account that we count each tetrahedron four times, we finally obtain

Taking the cube of the last inequality yields the desired estimate

and ends the proof.

".17. Let k be a positive integer. A graph whose vertex set is the set of positive integers does not contain any complete k x k bipartite subgraph. Prove

250

Chapter 6. Some Classical Problems in Extremal Graph Theory

that there are arbitrarily long arithmetic progression of positive integers such that no two elements of the progression are joined by an edge in this graph.

6.1. Notes

251

Fix a large integer N and consider all arithmetic progressions with m terms, all among 1,2, ... , N. There are at least

~ [N - a] > ~ (N - a

KaMaL

a=i

m -1

m-l

1) _ -

N(N -1) 2(m _ 1) - N

Proof. We will first prove the following result, of independent interest: Lemma 6.3. Let a 2: 2 and let Ka,a be the complete bipartite graph with a vertices in each half of the partition. There is a constant c > 0 such that a graph with n vertices and without Ka,a has at most cn2-~ edges.

Proof. Let G be such a graph and call 1,2, ... , n its vertices. Let di be the degree of vertex i. Let us count pairs (V,{Vi, ... ,Va }), where {Vi, ... ,Va } is a set of a (distinct) vertices, all connected to v. For each v = i, there are precisely (~) sets {Vi, ... ,Va } sharing a pair with v. Thus we find L~=i (~) pairs in total. On the other hand, for each set of a vertices {Vi, ... , Va} there are at most a-I vertices V sharing a pair with {Vi, . .. , va}. Thus

such arithmetic progressions, since each is determined by a pair (a, d) of positive integers such that a + (m - l)d ::; N. This is greater than c(m)N2 for N sufficiently large, where c( m) > 0 is a constant depending only on m. Since each of these progressions contributes an edge to the graph G N (the subgraph of G induced by the vertices 1,2, ... , N), and since each edge is counted at 2 most m times, we deduce that G N has at least c~rr;;) N 2 edges for all sufficiently large N. Since c~rr;;) N 2 > cN 2-i for all sufficiently large N (and this for any constant c> 0), it follows by the previous lemma that G N contains a Kk,k for all sufficiently large N. Since this contradicts the hypothesis on G, the conclusion follows. 0

6.1 Let q be the number of edges in the graph and let f (x) = (~) if x > a-I and f(x) = 0 for 0 ::; x ::; a-I. Then f is convex and since L di = we deduce that

2q,

This yields the rough estimate

a (2q

aÂˇn 2:n

~-a+

from which the result follows immediately.

l)a , o

Coming back to the proof, fix an integer m > 1 and suppose that among any m vertices forming an arithmetic progression, at least two are connected.

Notes

We would like to thank the following people for providing solutions to some of the problems presented in this chapter: Alon Amit (problem 12), Alexandru Chirvasitu (problem 10), Xiangyi Huang (problem 3), Szymon Kubicius (problem 4), Joel Brewster Lewis (problems 2, 7, 15), Fedja Nazarov (problem 14), Richard Stong (problem 13), Gjergji Zaimi (problem 5).

6.A. Some Pearls of Extremal Graph Theory

Addendum 6.A

Some Pearls of Extremal Graph Theory

The purpose of this addendum is to present some beautiful theorems from extremal graph theory, that nicely complement the more elementary results discussed in chapter 6. The main result is the famous Szemeredi-Trotter theorem, a deep result in incidence geometry which has a lot of wonderful geometric consequences. For instance, it is the basic tool in dealing with natural (but very hard) questions such as: if we are given n points in the plane, how many distinct distances do they always determine, what is the greatest number of segments of length 1 (or triangles of area 1) they determine, etc. Thanks to a brilliant observation of Elekes, it also plays an important role in additive combinatorics, giving nontrivial bounds on the so-called sum-product problem, a major research problem in modern combinatorics. The results discussed in this addendum found wide ranges of applications and their extensions are a very hot topic of research. See the papers [1], [11], [15]' [22], [23], [36], [74], [72], [73], [75], [SO], [SI], and the excell:nt book [S2] for more details, as we will only scratch the surface of the subject (but hopefully that will be enough to convince the reader of the beauty of these results). Also, some proofs in this addendum use the probabilistic method, so the reader is invited to take a look at addendum 6.B for details on this method and for probabilistic vocabulary.

6.A.l

The Szemeredi-Trotter theorem

The famous Szemeredi-Trotter theorem deals with the number of intersections between geometric objects, giving a fairly nontrivial (and even sharp up to absolute constants) upper bound for the number of incidences between a family of points and a family of curves. Since we do not want to delve into subtle topological considerations, let us agree that curves will always mean arcs of circles or polygonal lines, as this appears in all applications we have in mind. Theorem 6.A.1. (Szemenidi-Trotter) Consider n points H, P2,"" Pn and m curves C l , C2, ... , Cm in the plane. Suppose that Ci and Cj have at most

253

one common point for all i i- j. Then the number of pairs (i, j) such that 2 Pi E Cj is at most m + 4max(n, (mn)3). In applications, the following immediate consequence is very handy:

Corollary 6.A.2. Let S be a set of n points in the plane, L a collection of curves such that any two curves have at most one common point. Suppose that each curve in L contains at least k 2: 2 points of S. Then the number of pairs of a point in S and a curve in L containing that point is at most 29 . max ( n, ~) .

Proof. Let 1 = ILl and let p be the number of such pairs. The theorem 2 1 2 gives p ::; 1 + 4n3 max(n 3 , l3). By hypothesis, we have p 2: lk, therefore " 2 1 2 p-l2:p (1-];1) 2: l!.2' Henceourmequahtybecomesp::;Sn3max(n3,(pjk)3). From this the result follows by considering two cases. 0 It is important to note that corollary 6.A.2 (and also Szemeredi-Trotter's theorem) is sharp up to a constant. Suppose that 2k2 ::; n and consider the set of points (x, y), where 1 ::; x ::; k,1 ::; y ::; f and x, yare integers. Also, consider the set of lines y = mx + b, where m, b are positive integers such that 1 ::; b ::; and 1 ::; m ::; ib. It is easy to check that any such line contains exactly k such points (namely all points (i, mi+b) with 1::; i ::; k). Moreover, 2 there are about n points and about ;ft:, lines. Also, if k 2: yn, one can simply pick about njk lines and put k points on each of them. The original proof [SI] of theorem 6.A.l was very intricate. In a beautiful paper [SO], Szekely gave an amazing proof using a graph-theoretic result on crossing numbers. We will follow this path, but in order to do that we first need some terminology. Let G be a graph and consider an injective map that sends vertices of G to points of the plane. Draw an arc (in the plane) between any two points that come from the endpoints of an edge of G, such that except for the endpoints this arc does not meet any vertex. Call such a map a drawing of G. A crossing (or crossing point) is simply the intersection of two arcs not at the image of a vertex. Such a point belongs therefore to at least two arcs, but it is not an image of a vertex of the graph (with a piece of paper and a pencil in front of you, all these definitions should be obvious!). The number

254

Chapter 6. Some Classical Problems in Extremal Graph Theory

of crossings is the total number of crossing points, counted with multiplicities (Le. if a crossing point belongs to k arcs, its multiplicity is (~)). It is also the number of pairs of edges with no common endpoints and whose associated arcs intersect each other. After this dry list of obvious definitions, let us glorify one that will be very helpful in what follows:

Definition 6.A.3. The crossing number of a graph G, denoted c(G), is the minimal number of crossings over all possible drawings of G. Note that, by definition, a graph is planar if and only if its crossing number is 0. The key ingredient in Szekely's proof of theorem 6.A.1 is the following deep theorem [1], that will be proved in the next section.

"Theorem 6.A.4. (Ajtai, Chvatal, Newborn, SzemenÂŁdi, Leighton) Let G be a simple graph with e edges and n vertices. If e ;:::: 4n, then c( G) ;:::: 6J~2' Let us see why theorem 6.A.4 implies theorem 6.A.1. We may assume that every curve Gi contains at least one of the points PI, P2 , â&#x20AC;˘.. , Pn . Consider the graph G whose vertices are PI, P2 , .. . , Pn and whose edges connect adjacent points on some Ci (i.e. two points Pj, Pk belonging to some Gi, such that there is no other point Pz between them on Gi ). Let e be the number of edges of G and let I be the number of incidences between points and curves (Le. the number of pairs (i,j) such that Pi E C j ). As every curve contains at least one point among PI, P2 , ... , Pn , we have e = 1- m (since if a curve contains s points, it yields s - 1 edges of G and edges coming from two different curves are distinct). Now, since two curves intersect in at most one point, it follows that c( G) ::; m 2. If e ::; 4n, we deduce that I ::; m + 4n and we are done. Otherwise, the previous theorem yields m 2 ;:::: 6J~2' thus e ::; 4( nm) ~ and so

I ::; m

+ 4( nm)"3.

6.A. Some Pearls of Extremal Graph Theory

255

Lemma 6.A.5. IfG is a graph with e edges and n vertices, then c(G) ;:::: e-3n. Proof. First, we reduce the proof to the case when G is planar. To do this, remove edges of G which cross other edges until this is no longer possible, so you end up with a graph G' which is planar. Suppose that you removed k edges of G. Each of them removes at least one crossing, so that if we accept the truth of the lemma for G', we can write

c( G) ;:::: c( G')

+ k ;:::: e( G') + k -

3n = e - 3n.

Now, assume that G is planar, so c(G) = 0. We need to prove that e ::; 3n. If n ::; 2, this is clear, so suppose that n ;:::: 3. Let f be the number of (possibly infinite) faces of G. By Euler's formula we have n + f = e + 2. As every face has at least three edges, we have 3f ::; 2e. We easily deduce that e ::; 3n - 6, finishing the proof of the lemma. Finally, let us recall the proof of Euler's formula: if you start with a single vertex and try to reconstruct the graph, then every time you add an edge, you either add a vertex or a face, so the quantity n + f - e - 2 does not change. Since initially it is obviously zero, it is zero all the time. 0 Now, we use the probabilistic method to improve the previous inequality. Take an arbitrary number p E (0,1] and consider a random induced subgraph H of G, by picking each vertex of G independently and with probability p. As the probability of a given vertex to be in H is p, by linearity of expectation we have E[v(H)] = np, where v(H) is the number of vertices of H. Also, since an edge appears in H with probability p2, we have E[e(H)] = p 2e, where e (respectively e(H)) is the number of edges of G (respectively H). The previous lemma and linearity of expectation yield

The result follows. E[c(H)] ;:::: E[e(H)]- 3E[v(H)].

6.A.2

Proof of theorem 6.A.4 and a generalization

The proof of theorem 6.A.4 is simply beautiful: we will start with a rather weak inequality obtained from Euler's formula and, using the probabilistic method, we will improve it drastically by averaging. Let us start with the weak inequality:

We cannot easily express E[c(H)] in terms of c(G), but we can at least say that E[c(H)] ::; p 4 c(G). Indeed, take a drawing of G with exactly c(G) crossings and observe that the probability that a crossing survives in H is p4. The conclusion is that

256

Chapter 6.

Some Classical Problems in Extremal Graph Theory

As p was arbitrary, it is tempting to choose a value that will optimize the previous inequality. This value is p = ~~, but unfortunately it is not necessarily smaller than 1. However, this is far from being a subtle issue: simply choose something a bit smaller, namely p = This finishes the proof of the theorem. In geometric applications, it is useful to have a more flexible version of theorem 6.A.4, which allows graphs with multiple edges. This is the objective of the following theorem of Szekely [80].

6.A. Some Pearls of Extremal Graph Theory

257

crossings. However, one has to pay attention to the fact that each crossing is counted in 2(z-1)(ei- 2 ) members of the family, so in total we obtain at least

4;:.

Theorem 6.A.6. Let G = (V, E) be a multigraph with n vertices and e edges. If the maximal multiplicity of the edges of G is m, then e < 32nm or c( G) 2':

crossings in Gi , proving the claim. Finally, by definition of the crossing number, it is clear that

e3 216 n 2 mÂˇ

Proof. The idea is to reduce everything to the case of simple graphs, where theorem 6.A.4 applies. The details are however a bit involved. For each 1 :::; i :::; log2 m + 1, let Gi = (V, E i ) be the subgraph of G using only those edges with multiplicity between 2i- 1 and 2i (more precisely, two vertices a, b are joined by k edges in G i if they are joined by k edges in G and k E [2 z- 1 , 2Z)). Let ei be the number of edges in G i , without multiplicity. In order to apply theorem 6.A.4, we will restrict to those i for which ei 2': 4n. Let S be the set of these i. As Gi has maximal multiplicity at most 2i , we have (under the assumption that e 2': 32mn, which will be made from now on)

Now, since

it is natural to use Holder's inequality in the form

Combining this with the easy estimate

IE(Gi)1 = e -

iES

1+ [log2 (m)]

IE(Gi)1 2': e -

irf-S

4nÂˇ 2i 2': e - 16nm 2':

1+ [log2 (m)]

L 2i/2 :::; L

i=l

iES

We claim that for i E S we have

Thus, over this whole family of graphs we obtain at least 2(i-l)ei

i=l

yields

Indeed, pick an arbitrary drawing of G i and for every pair of vertices of G i with a multi-edge between them, select 2i- 1 of these edges and then arbitrarily and independently only one of them. We get a family of (2 i- 1 )ei simple graphs, each having ei 2': 4n edges (as i E S) and so with crossing number at least

61t2.

2i / 2 < 4V2m

61n 2

finishing the proof.

6.A.3

An application to additive combinatorics

In [23] Elekes made a nice connection between theorem 6.A.1 and a famous problem of Erdos and Szemeredi, the sum-product problem. This asks for a

Chapter 6.

258

6.A. Some Pearls of Extremal Graph Theory

Some Classical Problems in Extremal Graph Theory

The variant of the sum-product problem in which instead of sets of real numbers one considers subsets of a finite field has generated a huge amount of work. The proofs are in general very technical, as the analogue of the Szemen§di-Trotter theorem over finite fields is wrong. One has nevertheless the following deep theorem [11].

sharp lower bound of the expression max(IA + AI, IA . AI) over all sets of real numbers A with n elements. Here A

+ A = {a + bla, bE A}

and A· A

= {a· bla, bE

A}.

In particular, they made the very deep conjecture that for any c > 0 there exists C > 0 such that for all sets A with sufficiently many elements we have

Theorem 6.A.8. (Bourgain, Katz, Tao, Glibichuk, Konyagin) Let c > O. There exist positive constants C, (j such that for any prime p and any A c lFp with IAI < p1-E we have

max(IA + AI, IA· AI) 2: CE ·IAI

2 -

E .

max(IA + AI, IA· AI) 2: CIAI Ho .

The following result treats the case when 2 - c = ~. In [72], Solymosi proves the case 2 - c

259

The analogue of the Erdos-Szemeredi conjecture for finite fields would be

11'

T h eorem 6 .A.7. (Elek es) For any finite set of real numbers A we have IA + AI·IA· AI 2: ;2IAI5/2.

for subsets A of lFq . This is far from being settled. The following theorem can be deduced from the sum-product theorem.

Proof. Consider P = (A + A) x (A· A) as a set of points in lR~. Let L be the 1 with equations y = a(x - b) ranging over all chOIces of elements · d' . t set 0 f 1mes a b a, b of A. N~te that all lines la,b with a E.A - {O} and b E A are Istmc, ~o ILl 2: IAI(IAI- 1). Also, la,b is incident wIth (b + c, a' c) E P, for any c E . Thus we have at least IAI . ILl incidences between Land P. Thus, by the

Theorem 6.A.9. (Bourgain, Katz, Tao) Let 0 < 0: < 2. There exists c > 0 3 and C > 0 such that there are at most Cn'2- E incidences between n ::; pO'. points and n ::; pO'. lines in IFp'

Szemeredi-Trotter theorem, we have IAI.ILI ::; ILl

+ 4max(IPI, (IPI·ILI)2/3).

Note that the previous theorem no longer holds if one considers subsets A of lF q , where q is not a prime number. It suffices to consider a nontrivial subfield of lF q , for which IA + AI = IAI and IA . AI = IAI. Also, it is necessary to have some upper bound on IAI, since otherwise A = IF; would be again a counterexample for p large enough.

If IFI 2: ILI2, then clearly

IPI2: IAI2 . (IAI- 1? 2:

6.AA

1~14

and we are done. Otherwise, the previous inequality yields IFI 2: (

IAI- 1)3/2 IjTILI> fATAl' (IAI- 1)2 > ~IAI5/2 4 Villi - V 1.1"1.1 8 - 32

and the result follows.

Some geometric applications

In this section, we present some nice geometric applications of SzemerediTrotter's theorem. The first result deals with the number of triangles of unit area spanned by n points in the plane. It is not difficult to construct examples in which there are at least cn 2 log n unit-area triangles, for an absolute constant c > 0, but it is rather hard to give nontrivial upper bounds for the number of such triangles. We will use an easy geometric argument combined with corollary 6.A.2 to give such an upper bound.

260

Chapter 6. Some Classical Problems in Extremal Graph Theory

Theorem 6.A.I0. There exists an absolute constant e > 0 with the following property: the number of triangles of area 1 with vertices among n points in the 7 plane is smaller than en 3" • .

6.A. Some Pearls of Extremal Graph Theory

261

points of S. Thus the number of pairs of points that lie on average lines is bounded by

Proof. Let P be a set of n points in the plane. The essential observation is the following: if a, b are points of P, then all points pEP such that the triangle abp has area 1 are on the union of two fixed lines La,b, L~,b' parallel to abo This is immediate. Now, consider those pairs (a, b) for which the lines La,b and L~,b contain at most n~ points of P apiece. Each such pair determines at most 2n 1/ 3 unit-area triangles and so these pairs contribute at most 2n 2 . n 1/ 3 = 2n7 / 3 unit-area triangles. On the other hand, by corollary 6.A.2, there are O(n 4 / 3 ) incidences between lines l containing at least n 1 / 3 points of P and points of P. However, given such a line l, there are O(n) pairs (a, b) such that l = La,b (as for any a and l there are at most two points b such that l = La,b). Thus pairs (a, b) for which at least one of the lines La,b and L~,b contains more than n 1/ 3 points of P also determine O(n·n 4/ 3 ) = O(n 7 / 3 ) unit-area triangles. The re~~~.

Using a lot more work, one can improve the previous result to O(n9/4+c) for any E > 0 as shown in [22]. The exact order of growth of the number of unitarea triangles is unknown. If instead one considers the number of triangles with perimeter 1, the same kind of reasoning (but working with incidences between ellipses and points) yields a bound O(n 16 / 7 ). The second application is a famous result of Beck. It is again an easy consequence of corollary 6.A.2.

Theorem 6.A.l1. (Beck) For any n points in the plane, either more than n· 2- 14 of them are on a line or these points determine at least 2- 29 n 2 different lines.

Proof. Let S be a set of n points in the plane. Call a line average if it contains between 214 and n/2 14 points of S. We claim that there are at most n 2 /2 pairs of points of S that determine average lines. Indeed, for each i E [14, log2 ih] there are at most 29 . max (~, ~n lines that contain between 2i and 2i+ 1 points of S, by corollary 6.A.2. Any such line contains at most 4i+1 pairs of

proving the claim. So we have at least n 2 /2 pairs of points of S that do not determine an average line. Now, we have two possibilities: either some line contains more th~n n/214. points of ~, in which case the result is proved, or actually the n 2 /2 paIrs of pomts determme only lines that contain less than 214 points of S. But then each such line contains at most 228 pairs of points of S, so there must be at least n 2 /2 29 such lines. S<? again the result is proved. 0 We continue with another rather natural question: what is the maximal number of unit-dis1~nc~s spanned by n points in the plane? Erdos proved the lower bound n loglogn and the upper bound en 3 / 2 and conjectured that the true order of growth should be about nl+logl~gn. This is far from being proved, but the following theorem gives a better upper bound. In dimension 3, CI~rkson ~roved that the number of unit distances is O(n~+C) for all E > 0, while Erdos, Pach and Hickerson gave examples with at least en 4 / 3 unit d~stances. Also, note that already in dimension 4 one might have en 2 unit distances, as shown by placing n points on + x§ = 1/2 and n points on x§ + xa = 1/2.

!heorem 6.A.12. (Spencer, Szemeredi, Trotter) Let S be a set of n points 4 2n the plane. Then there are at most 16n / 3 ordered pairs of points (P, Q) in S such that PQ = 1. Pro~f. Draw a unit circle around each point of S and consider the multigraph G with vertex set S and in which P, Q E S are joined if they are adjacent on one of these circles. Note that there might be more than one edge between two

262

Chapter 6. Some Classical Problems in Extremal Graph Theory

vertices and that G might have loops (if there are circles containing exactly one point of S). If q is the number of unit distances between points in S, then G has q edges, for if a circle centered at PES contains Xi points of S, these points contribute Xi edges to G. Now, remove all circles containing at most two points of S, so we get a new multigraph with at least q - 2n edges. We claim that the maximal multiplicity in this multigraph is at most 2. Indeed, if there are at least 3 edges between P -I- Q E S, then there are at least three circles containing P, Q, so the circles of radius 1 centered at P, Q have at least three common points, a contradiction. Next, for every pair of vertices with exactly two edges between them, remove one edge (arbitrarily), so that we end up with a simple graph having at least ~ - n edges. Now, if q > lOn, then this simple graph has more than max ( 4n) edges and so at least (qj4)3 j(64n 2 ) crossings by theorem 6.A.4. But since any two circles have at most two common points, there cannot be more than 2G) < n 2 crossings. The conclusion is that if q > lOn, then q3 ::; 4096n4 and the result follows (if q::; lOn, everything is clear). 0

We end this addendum with a rather technical result concerning a famous question of Erdos: what is the least number of distinct distances determined by n points in the plane? Though rather innocent-looking, this is an extremely difficult problem. Erdos gave examples of configurations which determine at most vl~~n distances, for some absolute constant c. For more than thirty years, the best result was Moser's bound cn 2 / 3 , which is a rather tricky, but very elementary argument. Note that this bound also follows from the previous 2 theorem: there are at least ~ ordered pairs of points and the previous theorem (combined with a rescaling argument) shows that each distance appears at most 16n4 / 3 times, thus the number of distinct distances is at least n~~3. It was only in 1984 that Chung [15] improved Moser's bound tocn 5 / 7 . The next big step was done in the paper [16], where bounds of the form (logn I n4/;c 1 are proved by rather difficult arguments. After a huge effort, a major breakthrough was made in [36], where Guth and Katz prove that n points always determine at least loC;n distinct distances. Here, we prove a much weaker estimate, but which is already highly nontrivial (following [80]).

6.A. Some Pearls of Extremal Graph Theory

263

Theorem 6.A.13. (Szekely) Any set of n points in the plane determines at least cn 4 / 5 distinct distances, for an absolute constant c > O.

Proof. From now on c is an absolute constant that will change throughout the proof without changing its name. Fix n distinct points PI, P2 , ... ,Pn in the plane and let t be the number of distinct distances am~ng them. Let d l , d2 , . .. , dt be these distances. We may assume that t < n 10, as otherwise we are done. Around each point Pi draw t circles, having radii d l , d2 , ..â&#x20AC;˘ ,dt . Consider now the multigraph whose vertices are PI, ... , Pn and whose edges are arcs connecting adjacent vertices on these circles. Note that there are at least n(n -1) edges in this multigraph (for each Pi, there are n - 1 points on the union of the circles centered at Pi and if a circle contains s points, it contributes sedges). As usual, we remove edges that come from circles containing at most two points. It is easy to see that this removes at most 2nt edges, so we still have n 2 + o(n 2 ) edges. Unfortunately, the maximal multiplicity might be too big for the previous theorems to yield the desired estimate. The crucial point is to show that there cannot be too many edges with very high multiplicity. This is the aim of the following lemma. Before stating it, call a pair of points k-rich if there are at least k edges between them. Lemma 6.A.14. There are at most c~~2 of points.

+ ctn log n

edges joining k-rich pairs

ProofÂˇ By definition, this number of edges is at most the number of pairs (d, e), where d is the symmetry axis of an edge e of G such that d contains at leas~ k vertices. If 2i ::; jn, there are at most c;;;2 lines containing at least 2i vertIces, by corollary 6.A.2. If such a line contains u points then, by definition of t, this line bisects at most 2tu edges. So, as long as we are counting pairs (d, e) such that d contains between k and 4jn vertices, the total number of such pairs is at most

264

Chapter 6. Some Classical Problems in Extremal Graph Theory

and an easy computation shows that this is bounded by ct12. On the other hand, for each a ~ 4y'7i", it is easy to see that there are at most lines containing between a and 2a vertices. These lines yield at most

c;:

pairs. Combining these two observations finishes the proof.

Finally, the previous lemma shows that there is an absolute constant Cl such that if we delete all Cl vt-rich edges, the resulting multigraph G 1 has at least cn 2 edges. As clearly G 1 has crossing number at most 2n 2 t 2 , it follows from theorem 6.A.6 that

and the result follows.

Addendum 6.B

Probabilities in Combinatorics

This addendum presents some applications of probabilities in comb inatorics, or what is commonly called the probabilistic method. This is one of the most powerful tools in modern combinatorics, even though, just as with the pigeonhole principle, the underlying idea is extremely easy. The probabilistic method was used at first by Erdos, who proved a great deal of fairly nontrivial results using rather simple probabilistic arguments. For a much deeper study of the method, a canonical reference is the excellent book [2] by Alon and Spencer. Many of the examples we are going to discuss are taken from this book (however, some of them are left as exercises in the book and are not really easy ... ). We begin by recalling some useful notions from probability theory. In discrete combinatorics, one works with finite probability spaces, which makes the discussion much more elementary, avoiding subtle issues from measure theory. A finite probability space is the data of a finite set 0 (called the sample space) and of a map (called the probability distribution) P : 0 ---+ [0,1] such that

I:: P(w) =

wEn

One obvious such map is the constant map It21' which is called the uniform distribution. One defines for any subset A of 0, which we call an event, its probability as

P(A) =

I:: P(x).

XEA

It is very easy to check that P(AUB) +p(AnB) = P(A) +P(B) and that the probability of the complement of A is 1 - P(A). Given a probability space, a random variable X is simply a map X : 0 ---+ 1Ft The expectation (or mean value) of X is

E[X] =

I:: X(w)P(w) = I:: xÂˇ P(X = x). xElR

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Chapter 6. Some Classical Problems in Extremal Graph Theory

Note that the second sum is finite, as the image of X is finite. Another extremely important notion is that of independence. Some events A I, A 2, ... , Ak are called independent if for all I c {1, 2, ... , k} we have p(niEIAd

IT P(Ad·

it follows that P(Bj ) ::; P(Aj ) (as B j C Aj and P takes nonnegative values) and P(UjA j ) = P(UjBj ) =

L P(Bj ) ::; L P(Aj ). o

2) is an easy consequence of the definition of expectation.

Two random variables X, Yare called independent if the events X = a and Y = b are independent for all a, b. It is an easy exercise to check that in this case E[XY] = E[X]E[YJ. The following theorem consists of two elementary, but incredibly powerful principles. The first one is basically the principle of the probabilistic method: if you want to show the existence of an object with properties PI, . .. ,Pk, it is enough to prove that there is a probability space (0, P) such that the sum of the probabilities that an object does not have property Pi is smaller than 1. Of course, the difficult point is constructing the good probability space (though in practice the construction is naturally imposed by the statement of the problem we are trying to solve) and estimating these probabilitiew.- The second part of the theorem is called linearity of expectation and it is also a very powerful result, as we will soon see.

Theorem 6.B.1. Let (0, P) be a finite probability space.

267

The rest is immediate.

iEI

1) For any subsets AI, A 2, . .. ,Ak of

6.B. Probabilities in Combinatorics

When using the probabilistic method, one is naturally confronted with estimating probabilities, which sometimes can be quite painful. A simple example is that if (0, P) is a probability space and if X is a random variable on 0, then. the defi.nition ~i~es that there exists wE such that X(w) 2: E[X]. The followmg two mequalItIes are also very basic, but useful tools in estimating probabilities:

Theorem 6.B.2. 1) (Markov's inequality) If X is a random variable taking nonnegative values and if a> 0, then P(X 2: a) ::;

~E[X]. a

2) (Chebyshev's inequality) Let X be a random variable and let a> 0. Then

we have k

P(U~=IAi) ::;

L P(Ai ),

Proof. For the first part, simply note that

i=l

with equality if the events are pairwise disjoint. So, if L:Z=1 P(Ai) then UAi =I=- 0.

< 1,

2) If Xl, X2, ... ,Xk are random variables, then

E[X]

P(X

x)x 2:

P(X

x)a

a).

For the second part, using Markov's inequality, we can write P(IX - E[X]I 2: a) = P((X - E[X])2 2: a 2) ::;

Proof. 1) Let BI = Al and let Bi be the complement of Uj:,i Aj in Ai for all i 2: 2. Then clearly UjB j = UjA j and the Bj's are pairwise disjoint. But then

= aP(X 2:

~E[(X a

E[X])2]

and an easy computation using linearity of expectation yields the result.

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Chapter 6.

Some Classical Problems in Extremal Graph Theory

As usual, knowledge comes with practice, so we will spend the remainder of this chapter by giving a lot of applications of the previous theorems. We start with some applications of the sub-additivity of probabilities. One can prove the following inequality using standard techniques, but there is also a very elegant probabilistic proof: B.2. Let Xij E [0,1] for 1 ::; i, j ::; n. Prove that

Proof. Consider a random binary matrix (aij) such that P(aij = 1) = Xij and all these events are independent. Then I1j=1 (1 - I1~1 Xij) is simply the probability that each column contains at least a zero, while the expression

I1~1 (1 - I1j=1 (1 - Xij)) is the probability that each row contains at least a 1. Since all binary matrices have either at least one zero in every column or at least one one in every row, the result follows. D

6.B. Probabilities in Combinatorics

269

Proof. It is enough to check the equality for x E (0,1), since a polynomial vanishing on (0,1) vanishes everywhere. Consider a random two-coloring of S such that the probability that a point is black is x and all these events are independent. For a given P, the quantity x a (P)(1 - x)b(P) is simply the probability that the vertices of P are black and the points outside P are white. As these events are disjoint, the sum over all P is the probability that there is a polygon with black vertices and such that all points outside it are white. But this probability is 1, since the convex hull of the black points is such a polygon (note that there might be 0,1 or 2 black points and this is why the sum is also taken over degenerate polygons). D We continue with two problems which use the principle of the probabilistic method, namely the fact that if the sum of probabilities of some events is (strictly) less than 1, then there is a point in the probability space not belonging to any of these events. B.4. Prove that there is a four-coloring of the set M = {I, 2, ... , 1987} such that M contains no monochromatic arithmetic progression with 10 terms. IMO Shortlist 1987

Another application of theorem 6.B.l is the following nice problem: B.3. Let S be a finite set of points in a plane, no three of which are collinear. For each convex polygon P with vertices in S, let a(P) be the number of vertices of pI and let b(P) be the number of points of S lying outside P (Le. outside its interior or border). Prove that for all real numbers x, L x a(P)(I- x)b(P)

Proof. Pick a random coloring and observe that the probability that M contains a monochromatic progression of length at least 10 is bounded by N /4 9 , where N is the number of progressions of length 10 contained in M. So the problem is solved if we prove that N < 49 . But there are lI98;-iJ progressions of length 10 contained in M and whose first term is i. So 1987

N<" - 6

1987 - . 9 z

< 18 . 2000 2 .

t=1

where here the sum is taken over possibly degenerate convex polygons(polygons with 2, 1, or 0 vertices), too

lWith the convention that a(P) = 0,1,2 if P is connecting two points of S.

So, it remains to check that 28 . 56

< 9 . 2 19 , which follows. from

IMO Shortlist 2006

0, a point of S, respectively a segment

Recall that the m-th Ramsey number R(m) is the smallest positive integer n such that in any coloring of the edges of Kn (complete graph with n vertices)

270

Chapter 6. Some Classical Problems in Extremal Graph Theory

with two colors, one can find a monochromatic Km subgraph. Note that it is not obvious that this is well defined, but one can prove without too much difficulty that any coloring of the edges of KC:::~12) with .two colors contains a

6.B. Probabilities in Combinatorics

B.6. Let AI, A 2, ... , An and B 1 , B2, . .. ,Bn be distinct subsets of N such that Ai n Bi = 0 for all i and (Ai n Bj) U (Aj n Bd -:j::. 0 for all i -:j::. j. Prove that for all p E [0, 1]

monochromatic K m , so that R(m) is well-defined and at most (2~1-=}) (which grows roughly as 4m). The following result of Erdos from 1947 is an absolute classic. Amazingly, even after more than 60 years, it remains close to the best known lower bound (and 4m remains close to the best known upper bound). B.5. If 2(';)-1 > (~), then R(m) > n. In particular, R(m) > 2~ for all m24. Erdos

Proof. By definition, saying that R( m) > n is the same as saying that we can find a coloring of the edges of Kn with no monochromatic Km. Consider a random coloring of the edges of Kn with two colors, each having probability 1/2. If S is an m-element subset of the vertices of K n , let As be the event that the corresponding subgraph of Kn is monochromatic. We want to prove that there is an element in the probability space (i.e. a 2-coloring) which belongs to none of the events As. It is enough to check that LP(As)

271

LplAil(l - p)IBil ::; l. i=l

Tusza

Proof. Let X be the union of all Ai and Bi and consider a random subset S of X such that the events xES for x E X are independent and of probability p (more formally, let 0 be the set of all subsets of X and define P(S)

= plsl(l

_ p)lxl-lsl,

which is a probability measure on 0, because 2:: sc x P(S) = 1 by the binomial theorem). Consider the eventEi: Ai eSc X - B i . By hypothesis, no two events occur at the same time, thus

s However, it is clear that we have (~) choices of sets S and for each of them P(As) = 21-(';).

On the other hand, we have P(Ei )

L plsl(l - p)IXI-lsl AiCSCX-Bi

= plAil(l

_ p)IBil,

Thus, as long as 2(';)-1 > (~), we can find the desired coloring. The second part of the theorem is now easy: taking n = {2m/2], we have (for m 2 4)

the last equality being an easy computation left to the reader. Inserting this in the previous inequality yields the desired result. 0

nm < (n/2r ::; 2~2 -m < 2(';)-1. 0 m!

Remark 6.B.3. If all Ai have a elements and all Bi have b elements, we deduce from Tusza's inequality (by taking p = a~b) that n ::; (a~:!b:+b.

Here is a more delicate problem using similar ideas, but for which it is more difficult to find a good probabilistic interpretation.

We continue with a series of applications of the linearity of expectation and Chebyshev's inequality. First, a very simple problem:

n) = n(n - 1)··· (n - m (m m!

+ 1) <

272

Chapter 6. Some Classical Problems in Extremal Graph Theory

B.7. Let Pn(k) be the number of permutations of {I, 2, ... , n} having exactly k fixed points. Prove that n

L kPn(k) = n!. k=O

IMO 1987

6.B. Probabilities in Combinatorics

273

For what follows, in a tournament there is exactly one match between each pair of players and there is no draw. A Hamiltonian path in a tournament is a permutation 0' of the players such that player O'(i) beats player O'(i + 1) for all i. B.9. There exists a tournament with n players which lias at least 2:::~1 Hamiltonian paths. Szele

Proof. Let 0' have the uniform distribution on the set of permutations of {I, 2, ... , n}. If X(O') is the number of fixed points of 0', then 1 n

- "kPn(k) n!~ k=O

On the other hand,

Proof. Pick a random tournament (in which the results of each match occur with equal probability), and let X be the number of Hamiltonian paths. If 0' is a permutation of the players, let Xa(T) be 1 if 0' induces a Hamiltonian path in the tournament T and 0 otherwise. It is clear that, as random variables, we have X = 2:.:17 Xa' It is equally clear that E[Xa] = 21 - n (since the result of the matches between O'(i) and 0'(i+1) is imposed for all i). Thus E[X] = 2:::~1 and the result follows. D

= E[X].

where Xi(O') = 1a(i)=i' Then

A good exercise for the reader is to prove that the following result implies (a weak form of) Turan's famous theorem on graphs without n-complete subgraphs.

E[Xi ] = P(O'(i) = i) = n

and the conclusion follows by linearity of expectation.

The next two applications are absolute classics. B.8. Any graph with q edges contains a bipartite subgraph with at least q/2 edges. Erdos Proof. Pick a random subset S of the set V of vertices, by including a vertex in S independently with probability 1/2 . Let X(S) be the number of edges xy for which exactly one of x, Y is in S. For a given edge xy, the probability that exactly one of x, y is in S is 1/2, so by linearity of expectation E[X] = q/2. Thus one can find S such that X(S) 2: q/2. By construction, the subgraph comprising all edges with exactly one vertex in S is a solution. D

B.1O. Let db d 2 , ... , dn be the degrees of the vertices of a graph. Prove that one can find a subset S of vertices such that

1) S has at least 2:.:~=1 di~l elements. 2) There are no edges between vertices in S. Proof. Consider a random permutation 0' of the vertices of our graph G, all permutations having equal probability Let Ai be the event: 0'( i) < O'(j) for any neighbor j of i. We claim that P(Ai) = di~l' Indeed, we need to find the number of permutations 0' of the vertices such that 0'( i) < O'(j) for any neighbor j of i. If Yl, ... ,Ydi are the neighbors of i, there are (di: 1 ) possibilities for the set {O'(i),O'(yt), ... ,O'(Ydi)}' di ! ways to permute the elements of this

rb.

274

Chapter 6. Some Classical Problems in Extremal Graph Theory

set (and not (d i + I)!, since a( i) is the smallest element of the set) and finally (n - di - I)! ways to permute the remaining vertices. So

, _ ( P(A) -

n di

) (n - di - 1)!# _ _ 1_' n! - di + 1 '

275

from which the result follows, as ([~]) is the largest among the binomial coefficients. 0 Actually, the proof shows the following more general result, known as the Lubell-Yamamoto-Meshalkin-inequality: let A!, A 2 , •.• ,Ak be subsets of {I, 2, ... , n} such that no Ai is a subset of any A j . Then

as claimed. If X is the random variable X (a) = L:~=l 1aE A i' then 1

E[X] = '"' P(Ai) ~ i=l

6.B. Probabilities in Combinatorics

= '~d'+l "'-. i=l

The following famous theorem of Bollobas generalizes this result further:

Hence one can find a such that

B.12. Let AI, A 2 , ..• , An and BI, B 2 , ... , Bn be distinct sets of positive integers such that Ai n Bi = 0 for all i, but Ai n B j i= 0 for all i i= j. Then

It is clear that the set of vertices i such that a E Ai satisfies both properties.

o We continue with a beautiful proof, due to Lubell, of a famous theorem of Sperner on maximal anti-chains. B.ll. Let F be a family of subsets of {I, 2, ... ,n} which does not contain two elements A, B such that A c B. Then IFI :::; ([ ~]) . Sperner's theorem

Proof. Consider a random permutation a of {I, 2, ... , n} (with uniform distribution) and let Ai be the event that {a(1),a(2), ... ,a(i)} E F. The hypothesis on F implies that the random variable X (a) = L:~=l 1aE A i satisfies X(a) :::; 1 for all a, so that its expectation E[X] :::; 1. On the other hand, E[X] = L:~=IP(Ai) and the probability that· {a(l), ... ,a(i)} is a given set with i elements is i!(:,i)!, thus P(Ai) = where ni is the number of subsets

with i elements in F. Putting this together yields the beautiful inequality n

L z=l

(:z)" :::; 1, z

Bollobas

Proof. We may assume that all Ai's and Bj's are subsets of {I, 2, ... , N}. Consider the uniform distribution on the set SN of all permutations of {I, 2, ... , N}. Let Ei be the event consisting of all permutations a with the following property: all elements of Ai come before all elements of Bi in the list 17(1),17(2), . .. ,a(n). It is easy to see that the hypothesis implies that the events Ei are pairwise disjoint. Thus L:i P(Ei ) :::; 1. It remains to notice that P(Ed = (lAil!IBil)' as any subset with IAil elements of Ai U Bi is equally IAil

likely to form the first follows.

IAil elements in the list 17(1),17(2), ... , a(n).

The result 0

We take a break here to discuss a very nice consequence of Bollobas' inequality. Recall that an r-uniform hypergraph is a pair (V, E), where V is a set and E is a collection of subsets of V, each subset having r elements. Let K[ be the complete r-uniform hypergraph on t vertices.

276

Chapter 6. 80me Classical Problems in Extremal Graph Theory

B.13. Let G = (V, E) be a r-uniform hypergraph on n vertices. Suppose that G contains no K[, but that if we add any r-element set to E, at least one K[ appears. Then

6.B. Probabilities in Combinatorics

On the other hand, the probability that i ¢ A + B is clearly the n-th power of the probability that a given integer is not in A, that is

P( Z E A

(IAI)n 1- n = 1 - (l)n 1- h

+ B) = 1 -

We deduce that

E[X] =

Bollobas Proof. Let [n] = {I, 2, ... , n} and let [n](r) be the collection of r-element subsets of In]. If A E [n](r) - E, consider a copy C(A) of K[ that appears by adding A to E. Then A E C(A), yet A' is not in C(A) if A' i=- A is in [n](r) -E. So, if {AI, ... , Am} = [n](r) -E and Bi is the complement of C(A i ), then (Ai)i and (Bj)j satisfy the hypotheses of Bollobas' inequality, yielding 0 m S (n~~+r). The result follows.

We continue with a very nice result from additive number theory. B.14. Let A c 7l/n 2 71 be a subset with n elements. Prove that there exists a subset B c 7l/n 2 71 with n elements such that IA + BI 2: ~2 • IMO Shortlist 1999 Proof. Pick a random collection of n elements of 7l/n 271, each of the n elements being taken with probability 1/n2 and all choices being independent. Consolidate the distinct elements among the n chosen ones in a set B, which may have less than n elements. Consider then the random variable X = IA + B). As

X =

1iE A+B,

iEZ/n 2 Z

we have by linearity of expectation

E[X]=

L iEZ/n 2 Z

277

(1 _(1 _~) n)

and the result follows from the inequality (1 - ~r

Often, the choices of the probability space and distribution are imposed by common sense. However, in order to get better quantitative results, one uses sometimes less natural probability distributions. The following problems illustrate this: B.15. Let V be a set with n elements and let F be a family of m subsets of V, each having three elements. Prove that if 3m 2: n, then there exists 8 c V having at least 2;~ elements and such that no element of F is contained in 8. Proof· Choose p E [0, 1] and pick a random subset 8 of V such that P( v E 8) = p for all v E V, all the events being independent. If n(8) is the number of elements of F contained in 8, then linearity of expectation yields

E[181- n(8)] = E[l81]- E[n(8)]

= np - mp3.

This is maximal for p = ~ E [0, 1] and is equal to ~. Thus, we can find 8 such that 181 - n(S) 2: ~. Choose such 8. For each e E F with all of its three elements in 8, delete arbitrarily one of these three elements. We thus end up with at least 181 - n(8) elements of V which obviously have the desired property. 0

B.16. The minimal degree of the vertices of a graph G with n vertices is d Prove that there exists a subset 8 of the vertices such that P(iEA+B).

1) 181

s n· 1+1;1~+1).

> 1.

278

Chapter 6. Some Classical Problems in Extremal Graph Theory

2) Any vertex of G is either in S or neighbor of a vertex in S.

6.B. Probabilities in Combinatorics

x, y, z E S. Pick x E

279

IF; randomly with uniform probability and consider the

set Noga Alon

Proof. Consider a random subset S of the set of vertices V such that each vertex is in S independently with probability p = Ini~~d). If S' is the set of vertices which do not belong to S and which have no neighbor in S, then SuS' satisfies the second condition. It is thus enough to show that

Ax = {a E Alax

(mod p) E S}.

This is easily seen to be a sum-free subset of A. To prove that one can choose x such that IAxl > it is enough to compute the expected value of IAxl. Linearity of expectation shows that

141,

E[lAxl] = L

P(ax

(mod p) E S)

aEA

and it is clear that for any a E A we have P(ax (mod p) E S) = ;~Il' as However,

(ax )XElF; is a permutation of IF;. Thus E[IAx I] 2:

+ E[IS'I] LP(v E S) + LP(v E S')

E[lS u S'I] ::; E[ISI] =

d+ 1

We end this chapter with a very beautiful application of Chebyshev's inequality. Before attacking this problem, let us recall a few basic properties of the variance. Note that for any constant c and any random variable X we have Var(cX) = c2 Var(X). Also, it is not difficult to check that if X, Yare independent variables, then Var(X + Y) = Var(X) + Var(Y). Finally, if X is o with probability p and 1 with probability 1 - p, then Var(X) = p(I - p).

::; pn + n(I _ p)d+1 I+ln(d+I) < n . ---'----'-

;k:t\ IAI and we are done.

the last inequality being equivalent (after dividing by n, canceling p and taking logarithms) to log(I - p) ::; -p, which is well-known. 0 A very beautiful but rather subtle application of the probabilistic method is the following gem due to Erdos: B.I7. Let A be a finite set of nonzero integers. Then A contains a subset B of cardinality greater than I~I such that the equation x + y = z has no solutions in B x B x B. Erdos

Proof. The proof is quite tricky. Choose a prime p = 3k + 2 large enough, say such that p > 21al for all a E A. The subset S = {k + 1, k + 2, ... , 2k + I} of lFp is a sum-free subset, i.e. the equation x + Y = z has no solutions with

B.I8. Let VI, V2," ., Vn be n vectors in the plane, whose coordinates are integers of absolute value less that 160 I, J of {I, 2, ... , n} such that

Iff. Prove that there are disjoint subsets

LVi = LVj. iEI

jEJ

Proof. Note first of all that it is enough to find two distinct such subsets I, J, for then it is enough to take out of each the common elements of I, J. Now, assume that we cannot find two such distinct subsets and write Vi = (Xi, Yi). Consider a random binary sequence (aI, a2," ., an), each ai being 0 or 1 with probability 1/2 (all these events being independent). Consider the random variables X = 'L~l aixi and Y = 'L~=l aiYi. The point is that the values taken by (X, Y) are all different and that a lot of values are concentrated near

280

Chapter 6. Some Classical Problems in Extremal Graph Theory

the expectation of (X, Y). To be more precise, let mx = E[X], my = E[Y] and observe that the discussion preceding the problem yields

by hypothesis. Let a 2

= 45;00. Using Chebyshev's inequality, we obtain

P(IX - mxl 2 2a)

:s: 4'

P(IY - my I 2 2a)

hence

P(IX - mxl

:s: 2a, IY -

my I :s: 2a) 2

Chapter 7

:s: 4'

Complex Combinatorics

"2.

This means that at least half of the (pairwise distinct) points (X, Y) are located in the square IX - mxl :s: 2a, IY - my I :s: 2a. But this square contains at most (1 + 4a)2 lattice points, so

There is a class of combinatorial problems, most of the times with numbertheoretic flavor, which have very elegant solutions using finite Fourier transforms or versions of it. The main observation is the identity k-l

which is easily seen to be impossible (as say 25a 2 2 (1 unless n = 1, in which case we are easily done.

+ 4a)2

1n=a (mod

for n 2 16) 0

" Z j(n-a) , -- ~ k 'L.....t j=O

which holds for any primitive root of unity z of order k. This gives a rather powerful approach to problems concerning the distribution mod k of the sums of subsets of a given set or to tiling problems. Actually, this relation is a very special case of a much broader theory, that of representations of finite groups. The case of finite abelian groups is particularly easy and useful in combinatorial problems and we refer the reader to addendum 7.A for a more detailed discussion. For the next problems the previous relation is actually sufficient.

7.1

Tiling and coloring problems

We start with two tiling problems which have rather elegant solutions using complex numbers. Of course, the idea is similar to the usual coloring method, but we think it is neater.

282

Chapter 7.

Complex Combinatorics

1. Let k be an integer greater than 2. For which odd positive integers n can we tile a n x n table by 1 x k or k x 1 rectangles such that only the

central unit square is uncovered? Gabriel Dospinescu Proof. Assign coordinates to the squares of the table, with (0,0) assigned to the upper left corner. Put the number zX+Y in square (x, y), where z is a primitive k-th root of unity. Then all 1 x k and k x 1 tiles will cover a total value of O. Thus the total value of the whole board must be O. On the other

hand, this total value is precisely 1 ') 2 ( L~:o z~

(L~:OI zi

zn-l. Thus we must have

n 1 n-l - zn-l = 0 so for some c: E {-1, 1} we have zz"::-l = c: . Z-2-. This

can be also written as (z n;-l - c:)(z ntl + c:) = 0, which immediately implies that n is congruent to 1 or -1 (mod k). The converse is easily seen to hold, for if n == 1 (mod k) then we can tile both the center column and the center row (excluding the center square) using the rectangles, leaving four rectangles of dimensions multiples of k. And if n == -1 (mod k), then we can split the table into four rectangles each with a dimension a multiple of k and so again the tiling is possible. Therefore the answer to the problem is: all n for which n == Âą 1 (mod k). 0 2. Let n 2:: 2 be an integer. At each point (i, j) having integer coordinates we write the number i + j (mod n). Find all pairs (a, b) of positive integers such that any residue modulo n appears the same number of times on the sides (except for the vertices) of the rectangle with vertices (0,0), (a,O), (a, b), (0, b) and also any residue modulo n appears the same number of times in the interior of this rectangle. Bulgaria 2001 Proof. First, let us get rid of the easy case n = 2. As the interior of the rectangle must have an even number of lattice points, clearly one of a, b must be odd. But if one of a, b is odd, then clearly all conditions are satisfied, so that in this case the solutions are all pairs (a, b) with a, b not simultaneously even.

7.1.

Tiling and coloring problems

283

Now, let us treat the more difficult case n > 2. Let z be a primitive root of unity of order n and put the number zi+j in the square (i,j). By hypothesis, the sum of the numbers associated to the squares of the rectangle is 0 so that ~a-l ~b-l i + j . ' L...i=1 L...j=1 Z = O. But thIS factors as

a-lb-l LLzi+j =

Z2.

i=1 j=1

a-I

b-l -1. _z__-_1 z- 1 z- 1 '

so that necessarily one of a, b is congruent to 1 modulo n. By symmetry, we may assume that a == 1 (mod n). Now, since every residue modulo n appears the same number of times on the sides of the rectangle, we also have

As a

1 (mod n), the previous equality becomes (z + 1) (L~=i zj) = O. But

L;=i

n. > 2,.so. that z+ 1 =I- 0 and so we must have zj = 0 and b == 1 (mod n). Smce It IS clear that for a == b == 1 (mod n) all conditions are satisfied, it follows that this is the solution if n > 2. 0 The next two problems have very beautiful solutions using the methods of this chapter and some classical algebraic identities. 1(1 3. Each element of M = {1, 2, ... ,n} is colored in either red, blue or yellow. Let A be the set of triples (x, y, z) E M x M x M such that n divides

x + y + z and x, y, z have the same color. Let B be the set of triples (x,y,z) EM x M x M such that n divides x + y + z and x,y,z have pairwise different colors. Prove that 21AI 2:: IBI.

Chinese TST 2010 ProofÂˇ For simplicity write 1,2,3 for the colors and let

fJ(X) =

LXi, iEM,c(i)=j

Chapter 7. Complex Combinatorics

284

zj(n-a)

j=O

IBI-IAI =

yields

n-1 2i L ~ Le :;k(x+y+z) = j=1 c(x)=c(y)=c(z)=j k=O 3

IAI = L and

n-1

IBI = ~ L n

285

Proof Let us write Zk = etr and c(j) for the color of number j. Define h(X) = Lc(a)=j xa. As in the previous solution, one ends up with the formula

k-1

k) =

(mod

Counting problems 2i7rk

where c( i) = j if i has color j. The identity

1n=a

7.2.

n-1

N L (2h(Zk)2 12(zk)2 + 212(zk)2 h(Zk)2 k=O

+ 2h(Zk)2 h(Zk)2

~ LLfj(e :;k)3 2i

k=O J=1

Readers familiar with Heron's formula have already noticed that

2(a2b2 + b2c2 + c2a2) _ (a 4

II h(e .

N-1

2i:;k).

k=OJ=1

Of course, this method has some limitations, because it is rather difficult to compare complex numbers. The miracle is in the formula

+ b4 + c4 ) (a + b + c)(b + c -

a)(c + a - b)(a + b - c).

Also, as in the previous solution we have N

x 3 + y3 + z3 _ 3xyz = !(x + y + z)((x - y)2 + (y - z)2

2 and especially in the fact that 3 ""'"

h (Zk)

x)2)

""'"

2i7fki

= 0

j=1

h(l)

unless k = 0, when it equals n. Putting these observations together, we deduce that

2IAI-IBI = (h(l)

+ 12(zk) + h(Zk) =

L e 2i;,r a=1

and this is nonzero if and only if k = 0, in which case it equals N. The result follows from these remarks and the hypothesis, which ensures that

n 2i7fk

~h(en-) = ~e

j=1

+ (z _

+ 12(1) 2: h(l)

and similar inequalities.

- 12(1»2 + (12(1) - h(1»2 + (h(l) - h(1»2 2: o.

7.2

Counting problems

The following problem is very similar, but more technically involved. 4. Color the numbers 1,2, ... , N using 3 colors such that there are at most ~ numbers of each color. Let A be the set of 4-tuples (a, b, c, d) E {I, 2, ... , N}4 such that a+b+c+d == 0 (mod N) and a, b, c, d have the same color. Let B be the set of 4-tuples (a, b, c, d) E {I, 2, ... , N}4 such that a + b + c + d == 0 (mod N), a, band c, d have the same color, but these colors are distinct. Prove that IAI ~ IBI.

KaMaL

In this section we combine the technique of generating functions with the fundamental relation

k-1

1 ""'" j(n-a) 1n=a (mod k) -_ k ~ Z . j=O

This mixture is very powerful and applies very well for counting problems with arithmetical flavor, as roots of unity can detect congruences very efficiently.

Chapter 7. Complex Combinatorics

286

7.2. Counting problems

287

5. Three persons A, B, C play the following game: a subset with k elements of the set {1, 2, ... , 1986} is selected randomly, all selections having the same probability. The winner is A, B, or C, according to whether the sum of the elements of the selected subset is congruent to 0, 1, or 2 modulo 3. Find all values of k for which A, B, C have equal chances of winning.

We present two solutions for the following problem: the first one is the standard proof using complex numbers, while the second one is a very elegant probabilistic proof.

IMO 1987 Shortlist

IMC 1999

Proof. Let z be a primitive third root of unity and consider the polynomial 1986

P(X) = (1 + Xz)(l + Xz 2) ... (1

+ Xz

1986

)

2:= ajXj.

6. We roll a regular die n times. What is the probability that the sum of the numbers shown is a multiple of 5?

Proof. We need to count the number of sequences (Xl, X2, ... , xn) with 1 :S Xi :S 6 and such that 5 divides Xl + X2 + ... + Xn- Let aj be the number of sequences (Xl, X2,' .. , xn) such that 1 :S Xi :S 6 and Xl + X2 + .,. + xn == j (mod 5). Then for each fifth root of unity z we have

j=O

Note that

P(X) =

This equals zn unless z = 1, in which case it equals 6n . Adding these relations for all fifth roots of unity z gives 5ao = 6n + zn + z2n + z3n + z4n, for some primitive root of unity of order 5. If n is a multiple of 5, zn + ... + z4n = 4 and the probability is ~ = ~~;t.4. If not, then

Ie {1,2, ... ,1986}

where m(I) is the sum of elements of I. Thus

aj =

2:= zm(I) =

III=j

2:=

III=j,3Im(I)

(

2:=

1) z

III=k,3Im(I)-1

2:=

1) z2

III=k,3Im(I)-2

so that the probability is 65~6nl.

and this is zero if and only if 1=

III=j,3Im(I)

zn +ÂˇÂˇÂˇ+z4n

1= III=j,3Im(I)-1

III=j,3Im(I)-2

Therefore, the problem asks precisely for those k such that ak = 0. Now, fortunately the polynomial P has a very simple form, since

so the nonzero coefficients are precisely those whose index is a multiple of 3. Thus the answer to the problem is: those k == 1,2 (mod 3). 0

zn _ z5n =

1- zn

= - 1,

Proof. We use a simple fact from probability. Suppose X is an integer-valued random variable which is uniformly distributed mod m (that is, takes on every value mod m with probability 11m) and suppose Y is any other integer-valued random variable independent of X. Then X + Y is also uniformly distributed modm. To use this we break a roll of a die into two steps. We first flip a coin with a 516 probability of giving a heads. If we get a heads, then we roll a fair 5-sided die with the numbers 1,2,3,4,5 on it. If we get a tails, then we just get the number 6. If anyone of our n coin flips gives a heads (which occurs with probability 1- (ir), then one of our summands is uniformly distributed mod

288

Chapter 7.

Complex Combinatorics

5 and hence the sum is uniformly distributed mod 5. Otherwise, all the coin flips were tails and hence all the dice were 6's, so that the total is n (mod 5). Thus the probability of getting a sum which is a multiple of 5 is ~ (1 - (~t) for n not a multiple of 5 and ~ (1 - (~t) + for n a multiple of 5. D

The following problem is a good opportunity to recall a very useful result: if p is a prime and if ao, al," ., ap-l are rational numbers such that ao + alZ + ... + ap_Iz p- 1 :;:::: 0 for some primitive root z of order p of unity, then ao = al = ... = ap-lÂˇ This is an immediate consequence of the irreducibility of 1 + X + ... + Xp-l over the rational numbers. 7. Let p > 2 be a prime. How many subsets of {I, 2, ... ,p - I} have the sum of their elements divisible by p?

7.2.

289

Counting problems

So Xo - 1 + XIZ + ... + Xp_IZp-1 = 0, which implies that Xo - 1 = Xl = ... = Xp-l = k for some k. Since Xo + Xl + ... + Xp-l is simply the number of subsets of {I, 2, ... ,p - I}, that is 2P- I , we deduce that kp + 1 = 2P - 1 and ~o k = 2P-~-I. Since Xo = 1 + k, the problem is solved. Note that we included the empty set when counting Xo (by convention the sum of the elements of the empty set is zero). D The following problem is technically more involved, but uses the same ideas as before. The special case when n is a prime was problem 6 of IMO

1996. 8. Prove that the number of subsets with n elements of {I, 2, ... ,2n} whose sum is a multiple of n is

Ivan Landjev, Bulgaria TST 2006 Proof. Let z be a primitive root of order p of unity and consider the sum

zm(A) ,

s= AC{I,2, ... ,p-l}

where meA) = LaEA a. If Xj is the number of subsets A C {I, 2, ... ,p - I} such that meA) == j (mod p), then clearly

+ XIZ + X2z2 + ... + Xp_IZ P- I .

Proof. Consider the polynomial I(X, Y)

= (1 + XY)(I + X2Y) ... (1 + x2ny).

If meA) is the sum of the elements of A, we also have

On the other hand, we can explicitly compute S, since I(X, Y)

p-l

S =

IT (1 + zi) = IT (1 + (), <:

i=l

the product being taken over all roots ( of the polynomial XP 1 = 1 + X X-I

+ ... + Xp-l .

f a=O

xm(A)) y a ,

IAI=a

where all sets A in this solution are subsets of {I, 2, ... ,2n}. Taking for X ~ the roots of unity of order n, say Zj = en, the fundamental relation implies that

We deduce that

IT<: (I + () =

(-I)P - 1 = 1. -1-1

where Xa is the number of subsets A with a elements and such that meA) == 0 (mod n). We deduce that X n , which is the number of sets A we are trying to

290

Chapter 7.

Complex Combinatorics

find, is also the coefficient of yn in the polynomial ~ (f (Zl' Y) + ... + f (zn, Y)). Now, fix j and let us compute

We can write j

f(zj, Y) = (1

+ zjY)(l + zJY) ... (1 + zry).

= dU,n = dv,

where d

f(Zj,Y)=((l+Y)(l+e

2i11"U

= gcd(n,j)

Y)···(l+e

Counting problems

291

Note that

and then we clearly have

2i11"(v-l)u

7.2.

Y))

=(l-(-Y

)v)2d

et).

So the coefficient of yn is (-1 )d+n Since for any din there are exactly cp(n/d) integers 1 :S j :S 2n such that gcd(j, n) = d, we deduce that the coefficient of yn in ~(f(Zl' Y) + ... + f(zn, Y)) is exactly

which finishes the solution.

We continue with a very nice problem concerning the number of solutions mod p of a quadratic equation in several variables. Actually, the same methods also apply in some other situations, but the general problem of estimating the number of solutions mod p of systems of polynomial equations mod p is very deep. See the addendum of this chapter for more details. 9. Let p be an odd prime. Find the number of 6-tuples (a, b, c, d, e, f) of integers between 0 and p - 1 such that

!~ ( p k=O

zka

2 )

3 . (

aEZ/pZ

L dEZ/pZ

In the previous sum, there is one obvious term: the one for k = 0, which gives us p6. Also, for each 1 :S k :S p - 1 we have

by the change of variable a - d = x, a + d = y (which recovers a, d uniquely because 2 is invertible modulo p). On the other hand, for a fixed x i= 0, we have

zkxy =

yEZ/pZ

zky = 0,

yEZ/pZ

as the map y -t kxy (mod p) is a bijection of Z/pZ. LX,YEZ/PZ zkxy = p, which shows that

(L (L 2

a2+b2+c2=d2+e2+f2

(modp).

ka aEZ/pZ Z

MOSP 1997

Proof. Let Z be a primitive root of order p of unity. Since L~:~ zkx = 0 if x is not a multiple of p and equals p otherwise, the desired number of 6-tuples is

z-kd2 ) 3

)

-kd dEZ/pZ Z

2 )

Therefore,

= p.

Combining the previous paragraphs yields the answer to the problem, namely p5+(p_1)p2. 0

Proof. First look at a2 - d2 = (a - d)(a + d) mod p. Since p is an odd prime, we can choose x = a - d and y = a + d arbitrarily and then solve uniquely for a and d via a = (x+y)/2, d = (y-x)/2. If x is nonzero, then since p is prime, xy takes on every value exactly once whereas if x = 0 then xy is always zero.

292

Chapter 7.

Complex Combinatorics

Thus a 2 - d2 = xy takes on the value zero 2p - 1 times and takes on every nonzero value p - 1 times. Therefore it is described by the generating function

2p - 1 + (p - l)X + (p - 1)X2+ ... + (p - 1)XP-1 = P + (p - 1)(1 + X + X2 + ... + Xp-1)

in Z[X] modulo XP - 1. Note that modulo XP - 1, the polynomial

7.2.

Counting problems

293

for some integers ai· Since ZX only depends on x (mod p), it is clear that ai is exactly the number of ways residue i is represented by the numbers ±1±2±· . .±P;1. Thus, the problem asks us to prove that a1 = a2 = ... = ap-1 and to find this common value. The point is that 8 has a nice closed expression, since it obviously factors as p-l

-2-

8 =

1 + X + X2 + ... + Xp-1

IT (zj + z-j). j=1

has the feature that

On the other hand,

f(X)(l + X + X2 + ... + XP-1) = f(l)(l + X + X2 + ... + XP-1)

p-1

IT (zj + z-j) = 8· IT

for any polynomial f(X) (as X -1 divides f(X) - f(l)). Thus the generating function for the values taken on by (a 2 - d2) + (b 2 - e2) + (c 2 - j2) is

j=1

_P;_l

(zj + z-j)

= 8· IT (zp-j

j=~

+ zj-P)

= 82

j=1

and

(p + (p - 1) (1 + X + X2 + ... + Xp-1 ) ) 3

p-1

== d2+e2+ f2 D

j=1

p; 1

represent each nonzero residue class mod p the same number of times. Compute this number.

P(;-l)·

p-1 (1 + z2 j ) =

j=1

IT (1 + zj) = 1.

j=1

The last relation uses the fact that x -+ 2x is a bijection of the nonzero remainders modulo p (as p is odd) and that

We continue with a very nice problem, in which one uses a mixture of the previous techniques, algebraic manipulations and a rather tricky numbertheoretic argument.

10. Let p be an odd prime. Prove that the 2E=.! 2 numbers ±1 ± 2 ± ... ±

p-1

IT (zj + z-j) =

= p3 + (p5 _ p2)(1 + X + X2+ ... + XP-1).

From this we read off the number of 6-tuples for which a 2+b2+c2 (mod p) as p5 + p3 - p2.

p-1

p-1 IT(l+z j ) = 1, j=1 which follows from

R. L. McFarland, AMM 6457 2i7r

Proof Let z = e p

by taking X and write

8= ciE{-1,1}

-1.

The previous computation shows that 8 2 = 1, so that 8 = ±1 is definitely an integer. But then then relation ao - 8 + a1Z + ... + ap_1 zp- 1 = 0 implies that ao - 8 = a1 = ... = ap-1. In particular, a1 = ... = ap-1 and the first part of problem is solved.

294

Chapter 7.

Complex Combinatorics

On the other hand, we clearly have ao

+ al + ... + ap-l =

which combined with ao - S =

S == 2

z=.! 2

= ... = ap-l and with S = ±1 shows that

(mod p) == (-1)

p2_1

since n is odd. But conversely, if we have a sequence bl , b2, ... , bn of elements of Z j nZ satisfying bi =1= 0 (mod n) and bl + b2 + ... + bn == 0 (mod n), then we can find precisely n admissible sequences giving rise to bl , b2, ... , bn . Indeed, choose any ao E {I, 2, ... , n} and take for ak the remainder modulo n in {I, 2, ... , n} of the number ao + 1 + 2 + ... + k + bl + b2 + ... + bk . Thus, if fen) is the number of sequences (b l , b2, ... , bn ) E (ZjnZ - {O} such that bl + b2 + ... + bn == 0 (mod n), then the desired answer is nf(n). 2i7r

Now, we evaluate fen) in the standard way: let z = en, so that

p2_1

so that S = (-1)

295

Counting problems

(mod p),

7.2.

and the value we are looking for is z=.!

al = a2 = ... = ap-l =

p2_1 2 8 -----''----~-

(-1) p

We used here a standard result in quadratic residues, saying that Legendre's symbol ( ~ ) = (-1)

£=.! 8- •

It is rather difficult to approach the following problem directly with the methods of this chapter. However, a small observation reduces the problem to a more familiar one. 11.

a) Let n be an odd integer. Find the number of sequences (ao,al, ... ,an ) such that ai E {1,2, ... ,n} for all i, an = ao and ai - ai-l ¢ i (mod n) for all i = 1,2, ... , n. b) Let n be an odd prime. Find the number of sequences (ao,al, ... ,an ) such that ai E {1,2, ... ,n} for all i, an = ao and ai - ai-l ¢ i, 2i (mod n) for all i = 1,2, ... , n. Reid Barton, USA TST 2004

Proof. a) Call a sequence as in the problem admissible. Considering

Since n-l

Lzkb

= n-l

b=l

for k

= 0 and

it equals \--=-zz:k - 1

fen) =

-1 for 1 :::; k :::; n - 1, we deduce that

(n - l)n - (n - 1) , n

so the answer to part (a) is (n _1)n - (n - 1). for 1 :::; i :::; n, the condition becomes bi =1= 0 (mod n). Note that bl

+ b2 + ... + bn == - (1 + 2 + ... + n) == 0

(mod n),

b) The same argument as in the first paragraph of (a) shows that it is enough to count the number g(n) of sequences (b l , b2, . .. , bn ) with bi E ZjnZ, bi ¢ 0, i (mod n) and bl + b2 + ... + bn == 0 (mod n). The desired answer will be ng(n).

296

Chapter 7.

Complex Combinatorics

We compute in the same way

7.2.

Counting problems

297

and of the fact that n is odd. We deduce that

for all 1 :; k :; n - 1 and so the answer to the problem is g(n)

= (n - l)(n - 2)n-1 - (n - 1) .

Â°

the corresponding product equals trivially (n - 2)n . (n - 1) (the For k = factor n - 1 comes from the sum associated to i = n). Now fix 1 :; k:; n - l. For all 1 :; i < n we have

L b~O,i

zkb

= 1 + zk + ... + z(n-l)k

+ zki) =

-(1

+ zki).

(mod n)

The following problem is hard, even though the first steps are rather clear. The difficulty lies in the algebraic technicalities. 12. Let p

> 3 be a prime number.

If X is a nonempty subset of {O, 1, ... ,p - I}, let f(X) be the number of sequences (aI, a2,Âˇ .. , ap-d such that aj E X for all j and p divides L~:Uaj. Prove that f({O,1,3}) ;:: f({O,1,2}), with equality if and only if p = 5.

For i = n the corresponding sum is - 1. Thus

II n

i=l

( b~O,i

zkb

)

(mod n)

IMO 1999 Shortlist

II (1 +

n-l

zki).

i=l

Proof. Let us first find a formula for f(X). argument yields

Now, since n is a prime and 1 :; k < n, we have gcd(k, n) = 1, so the numbers zki with 1 :; i :; n - 1 are precisely the n-th roots of unity different from l. Thus n-l

II (1 +

zki) =

+ u)

II un=l,u~l

(X - u)

xn-1 X-I

~ I: k=O

i=l

the last equality being a consequence of the equality

Letting z

zkX) .....

xEX

2i7r

= e P, the usual

zk(P-I)X) .

xEX

In particular,

p-l (P-I ) }](1+zi + z2i

f({O,1,2})=~t;

)

Chapter 7. Complex Combinatorics

298

= {z j ll :s: j :s: p -I} for

Since {zijll :s: j :s: p -I} since p > 3, it follows that

f( {O, 1, 2}) =

3P -

+ (p

(3 j _ (j _ 1

-1)}] p-l

all i =1= 0 (mod p) and

= 1+

P- 1 -

We also have

f( {O, 1, 3}) =

p-l (P-l

} ] (1

+ (ij + (3i

)

~ ~ (3"-1 + (p - 1) 3P -

+ (p -

((j - a)((j - {3)((i -

7))

l)ap

p where

and so the characteristic polynomial of the sequence (an)n is

Thus, there exists a constant C such that for all n we have

It is not difficult to compute that the first terms of the sequence (an)n are 0,1,1,3,1,1,3,8, ... and that the sequence is increasing after the sixth term (and C = -2). The result follows easily. 0

So, if a, (3, 'Yare the roots of x 3 + x + 1, we have (using again that for all i =1= 0 (mod p) the numbers (zij)j are a permutation of the numbers (zJ)j)

f( {O, 1, 3})

Remark 7.1. Except for the technical combinatorial part, which, as we have seen, is quite standard, the difficult point of the problem is establishing the inequality ap 2: 1, with equality precisely for p = 5. Proving that ap 2: 1 is however rather easy, without the computation of the characteristic polynomial. Indeed, note that ap is a symmetric polynomial with integer coefficients in the roots of X3 + X + 1, so it is an integer. On the other hand, the polynomial function x 3 + x + 1 is increasing on the whole real line, so among a, (3, 'Y precisely one is real, say a. Moreover, it is easy to check that -1 < a < O. Thus \-=..c; > o. Also,

(an - 1)((3n - 1)("(n - 1) an =

(a - 1)((3 - 1)("( - 1)

(1 - (3P)(1 - 'Y ) - 11-'-----'--'---'--'P

(1-(3)(I-'Y)

It is thus enough to prove that ap 2: 1, with equality precisely for p = 5. However, this sequence satisfies a linear recursive relation with characteristic

polynomial

(X - a)(X - (3)(X - 'Y)(X - a· (3) (X - (3. 'Y)(X - a· 'Y)(X - a· (3. 'Y)

= (X + 1)(X3 + X + 1)(X -

299

7.3. Miscellaneous problems

- (3P 12 > 0

1-(3

so that ap > o. It follows that ap 2: 1. However, we know no easy proof of the fact that ap > 1 for p > 5.

7.3

Miscellaneous problems

a· (3)(X - (3. 'Y)(X - a· 'Y).

We can easily compute that

(X - a· (3) (X - (3. 'Y)(X - a· 'Y) = (X + (X = X 3 - X2-1

+~) (X + *)

13. Let ak, bk , Ck be integers, k = 1,2, ... , n and let f(x) be the number of ordered triples (A, B, C) of disjoint subsets (not necessarily nonempty) of the set S = {I, 2, ... ,n} whose union is S and for which

iES\A

iES\B

iES\C

300

Chapter 7. Complex Combinatorics

Suppose that f(O) = f(l) = f(2). Prove that there exists i E S such that 3 I ai + bi + CiÂˇ

7.3. Miscellaneous problems

301

Proof. Consider the matrix A = (Xij), whose determinant is given by det A =

Gabriel Dospinescu

Proof. Observe that F(X) =

(xai+bi

+ Xbi+Ci + XCi+ai)

detA

(xj - Xi).

XL:iES/A ai+L:iES/B bi+L:iES/C Ci

The first formula and the definition of A j , B j imply th(;) following equality in Z[X]

{A,B,C}

= f(l) = f(2) =

== 3n -

II lS,i<jS,n

F(X)

xS(a).

a odd

On the other hand, we also have a simple closed form for det A, given by Vandermonde's formula:

i=l

Thus, the equality f(O)

xS(a) -

a even

3n -

+ X + X2)

becomes (mod X 3

p-l -

Since 1 + X + X 21X 3 - 1 this means in particular that 1 + X. + X2!F~~): So there exists i such that Zai+bi + zbi+Ci + zCi+ai = 0, where z IS a pnmItIve third root of unity. This means that {ai + bi , bi + Ci, Ci + ad is equivalent to a permutation of {O, 1, 2} (mod 3) and so 31ai + bi + Ci. The conclusio~ follows. Readers who are not familiar with basic linear algebra will have a rather hard time trying to solve the following problem. 14. Let p be an odd prime and n ;::: 2. For a permutation a of the set {I, 2, ... ,n} define

S(a) = a(l)

+ 2a(2) + ... + na(n).

Let A J. be the set of even permutations a such that S (a) == j (mod p) and let B be the set of odd permutations a for which S(a) == j (mod p). Prov: that n > p if and only if Aj and B j have the same number of elements for all j. Gabriel Dospinescu

L(Aj - Bj)Xj j=O

(xj - Xi)

(mod XP - 1).

lS,i<jS,n

Therefore, the problem comes down to showing that

n>p ~

(xj - Xi)

== 0

(mod XP -1).

lS,i<jS,n This is actually very easy. Indeed, if n > p then clearly

X P - llXp+l - XI

(xj - Xi) .

lS,i<jS,n For the other direction, take z any primitive root of order p of unity. Then by hypothesis I11S,i<jS,n(zj - zi) = 0, showing that for some i < j we have zj-i = 1. Since this implies that j - i is a multiple of p, thus at least p, we are done. 0 The following problem is very challenging. Here, it is very hard to find the correct approach, especially because the problem suggests number theory. 15. Is there a positive integer k such that p = 6k + 1 is a prime and (mod p)?

ekk) == 1

USA TST 2010

Chapter 7.

302

Complex Combinatorics

Proof. The answer is negative, but the proof is far from being obvious. Suppose that p = 6k+ 1 satisfies == 1 (mod p). Let g be a primitive root mod p and let z = g6, so that z has order k mod p. Consider the sum

e:)

7.3. Miscellaneous problems

Proof. Let r(x) E {O, 1, ... ,p - I} be the remainder of x mod p and observe that r(x) = p. {~}. Thus the hypothesis can also be written r(an)

Since z has order k mod p, the sum L::~~ zij is 0 unless j is a multiple of k, which happens if and only if j = 0, k, 2k, 3k. Hence

P-l (

e:) == 1 (mod p), we deduce that (1

+ i)3k ==

== 4k

(mod p). On the other hand,

+ zi)~ == -1,0,1

~jzmja'

p-l

)

= 2p L z j

-2p.

j=l

It is easy to check the identity

(mod p).

It is now immediate to check that we cannot have k remainders mod p, each of them -1,0 or 1, adding up to 4k modulo p. This contradiction shows that no such k can be found and solves the problem. 0

+ r(bn) + r(cn) + r(dn)

for any gcd( n, p) = 1. Let z be a primitive root of unity of order p and let a', b', e', d' be the inverses of a, b, e, d modulo p. For any m not a multiple of , 1 ' P we have zmnaa = zmn, so we have L: r(an)zmnaa = 2pzmn. Fix a number m relatively prime to p and let n run over a system of representatives of the nonzero classes mod p. Then r(mn) runs over 1,2, ... ,p-1 and so does r(an), as a and m are not multiples of p. Hence if we take the sum over n of the previous relation, we obtain

L Since

303

1 + 2X

+ ... + nXn-1 =

n 1 nX + - (n

(X -

+ l)xn + 1 1)2

which easily implies that p-l

"J'zma'j = ~

We present two difficult solutions for the following beautiful, but very challenging problem.

j=l

P aIm

and similarly for b, e, d. Thus, we can write "16. Let p be an odd prime and let a, b, c, d be integers not divisible by p such that

for all integers r not divisible by p (here {.} is the fractional part). Prove that at least two of the numbers a + b, a + c, a + d, b + c, b + d, c + dare divisible by p. Kiran Kedlaya, USAMO 1999

" 1~

1 za'm -2 -

and this for all m relatively prime to p. Clearing denominators and canceling equal terms yields (after a tedious computation) an equivalent equality 2+L 1 All

z(a'+b'+c')m = L

unspecified sums are over a, b, c, d.

za'm

+ 2z(a'+b'+c'+d')m,

Chapter 7.

304

Complex Combinatorics

7.3.

which continues to hold (trivially) when m is a multiple of p. Finally, we add these relations for 0 ~ m ~ p - 1. Note that

Miscellaneous problems

305

Proof. This second proof uses Lagrange's interpolation theorem, for which we refer the reader to chapter 11. As in the previous solution, r(x) E {O,l, ... ,p-l}

p-l

Lzma'

=0 is the remainder of x modulo p. Define

m=O

and similarly for b', c' , d' and that

Q(x) = 2r(x) - r(2x), p

p-l

zNm

so that Q(x) = 0 if 0 ~ r(x) ~ (p-l)/2 and Q(x) = 1 if (p-l)/2 < r(x) < p. Call (a, b, c, d) good if it satisfies the relation in the problem, which can also be written as r(ka) +r(kb) +r(kc) +r(kd) = 2p for alII ~ k < p. Note that if (a, b, c, d) is good, then so is (ka, kb, kc, kd) for any k which is not a multiple of p. Also, note that if (a, b, c, d) is good, then

m=O

for any N (simply because the corresponding sum equals p when zN = 1 and o otherwise). We deduce that p-l

Q(a)

zm(a'+b'+c'+d') 22p,

Combining these two observations, we deduce that

m=O

which implies that a' + b' + c' + d' is a multiple of p (otherwise the left-hand side would be 0). Therefore, by the previous key equality, we also have "\""' - a' m

L....t z

"\""'

= L....t

Q(ka)

+ Q(kb) + Q(kc) + Q(kd) =

for all 1 ~ k < p. By Lagrange's interpolation theorem, there exists a polynomial P(X) of degree at most p - 2 such that P(x) == Q(x) (mod p) for all x =t 0 (mod p). Note that if R(X) = P(X + 1) - P(X), then R(x) == 0 (mod p) when

a'm

for any m relatively prime to p. By multiplying the previous relation by z-ma' , we get l+zm(b'-a')

+ Q(b) + Q(c) + Q(d) = 2.

p-3 p+l x = l""'-2-'-2-, ... ,p-2

+ z(c'-a')m + zm(d'-a')

and R((p - 1)/2) =t 0 (mod p), so degR = p - 3 and degP = p - 2. On the other hand, the polynomial S(X) = P(Xa) + P(Xb) + P(Xc) + P(Xd)' is congruent to 2 mod p for p - 1 values of the variable. Since deg S ~ p - 2, S - 2 must be the zero polynomial mod p. Imposing that the coefficient of Xp-2 vanishes in S, we obtain the key relation

_ z -2a'm +z -(a'+b')m +z -(a'+c')m +z -(a'+d')m ,

and a similar argument (add the equations and observe that the left-hand side is at least p) shows that at least one of a' + b', a' + c', a' + d' is a multiple of p. Say pia' + b', then also pic' + d' (as pia' + b' + c' + d') and so pia + band pic + d, finishing the proof of this hard problem. D

a P-

, \

+ bP- 2 + cP- 2 + dP- 2 == 0

(mod p),

Chapter 7. Complex Combinatorics

306

which can be also written (by Fermat's little theorem)

III ~ + b+ ~

1 d = 0 E lFp .

+ Finally, since r(a) + r(b) + r(c) + r(d) = 2p,

we have a + b + c + d

= 0 in

7.3. Miscellaneous problems

307

the sum being taken over all characters X of F = IF3d. The whole point is to be able to estimate the previous quantity for each character X. Let g(d) = f(~) and take a subset 8 as in the theorem, but with 181 = f(d) = 3d g(d). N~te that

lF p . Combining the two relations yields

L(1xES - g(d xEF

1 1 1 1 +-=--abc a+b+c'

which readily becomes (after clearing denominators) (a + b) (b + c) (c + a) = O. By symmetry, we may assume that a + b = 0 in IFp' But then c + d = 0 also and we are done. 0 We end this chapter with a very deep result, whose proof is however elementary (and follows [52]). It improves previous results of Brown and Buhler [12], Frankl, Graham and Rodl [33] and finally Ruzsa [52]. For more details on some arguments concerning orthogonality relations and properties of characters, we refer the reader to addendum 7.A. 17. Let p be a prime and let 8 C (Z/3Z)d be a subset containing no line of the affine space (Z/3Z)d. Prove that 8 has at most elements.

2't

l))x(x) = LX(s) sES

if X is not trivial and it equals LSES X(s) - 3d g(d - 1) otherwise (again by orthogonality of characters). We deduce that

181

g(d - 1)181

+ 31d L x

( L X(S)) sES

Proof. Let f(d) be the largest cardinality of a set 8 C lF~ (we write lF3 for Z/3Z) containing no line. Note that three distinct points of lF~ adding up to the zero vector form a line and conversely, the sum of the points of a line is the zero vector. So, f(d) is also the largest cardinality of a set 8 containing no three elements that add up to O. Since lF~ and F = IF 3d (the field with 3d elements) are isomorphic as additive groups, we may work from now on with subsets 8 of F. In particular, if 8 is such a set, the orthogonality relations for characters of abelian groups (theorem 7.A.5) yield

(L(1XES - g(d - l))X(X)) xEF

2': g(d - 1)181

31d L

L X(s) x sES

L(1xES - g(d - l))X(x) xEF

Here comes the crucial estimate: Lemma 7.2. For all characters X we have

However, prove that we can find such a set with at least 3ÂĽ-1 elements. Meshulam-Roth theorem

s: 3d g(d -

L(1xES - g(d - l))x(x) xEF

1) -181.

Proof. If X is nontrivial, let Ho be its kernel, while if X is trivial, let Ho be any hyperplane of F (where F is seen as lF3 -vector space of dimension d). Take HI, H2 two affine hyperplanes parallel to Ho so that the Hi's form a partition of F. Note that by definition X is constant on each Hi, say X(x) = Zi for x E Hi (where of course IZil = 1). Then 2

L(1xES - g(d - l))x(x) xEF

L zi(18 n Hil- f(d -

1))

i=O

;d L X

(Lx(s))3 sES

s: L i=O

118 n Hi I - f (d - 1) I .

308

Chapter 7. Complex Combinatorics

But clearly S n Hi does not contain three elements adding up to 0 and since Hi is (d - 1)-dimensional, we deduce that IS n Hi I ~ f (d - 1) (by definition of f (d - 1)). Therefore, the previous estimate yields

7.4. Notes

On the other hand, since we work in characteristic 3, we have

(al

I ) l xE S

g(d - l))X(x) ~ 3d g(d - 1) -

IS n Hil

i=O

xEF

309

+ a2)(al + a2)3 d = (al + a2)(at + a~d) x-l = alx + a2x + ala 2x-l + a2al.

+ a2)X =

(al

We deduce that

which is precisely what we wanted.

Coming back to the proof and using the lemma and Plancherel's identity (theorem 7.A.5)

a contradiction.

L LX(s) x sES

7.4

3 1SI,

we deduce the estimate (do not forget that S was chosen such that lSI = 3d g(d) = f(d)) lSI 2:: g(d - 1)IS1 2 -ISI(3 d g(d - 1) -lSI), which implies that

(d) < g(d-1)+39 - g(d - 1) + 1

â&#x20AC;˘

Since g(l) = ~, it is immediate to check that the last estimate implies g(d) ::; ~, which is precisely what we wanted. Finally, let us show that f(3d) 2:: gd, which will trivially imply the remaining part of the theorem. Let x = 3d + 1 and consider

This has gd elements and we claim that it does not contain three elements d adding up to O. Note that aX E lF3d if a E lFgd, because (a x )3 -l = 1 if a =I- O. If (aj, aj) add up to 0, we obtain al + a2 + a3 = 0 and af + a2 + a3 = o. But then

Notes

We thank the following people for providing solutions to the problems discussed in this chapter: Mitchell Lee (problems 1, 2, 10, 15), Richard Stong (problems 6, g), Qiaochu Yuan (problems 5, 6, 8), Victor Wang (problem 16), Alex Zhu (problems 15, 16), Gjergji Zaimi (problems 12, 13, 14).

'l.A. Finite Fourier Analysis

Addendum 7.A

Finite Fourier Analysis

311

The identity n-l

1 """" 2i7rk(a-b) - ~e n = 1a =b n

(mod n)

k=O

(for all integers a, b and all positive integers n) is at the origin of a lot of beautiful mathematical results, as we could see in chapter 7, but it is just a special case of a broader theory. In the first part of this addendum we will see a vast generalization of this relation, through Fourier analysis on finite abelian groups. Though rather elementary, these results are extremely useful in number theory and combinatorics. On the other hand, they are the very first component of a much broader picture, that of harmonic analysis. Things are much easier for finite abelian groups, since one avoids quite a lot of technicalities that appear when dealing with harmonic analysis on other groups (such as JR, the unit circle in C or, much harder, non-abelian and non-compact topological groups). For instance, all convergence issues are automatic (as all sums we deal with are finite), while the integration theory has no subtlety (on the other hand, if one wants to do harmonic analysis on JR, one has to be very careful with these issues and one has to develop a rather powerful integration theory first). Also, the fact that the group is abelian simplifies the problem considerably, basically because all irreducible complex representations of such a group are one-dimensional. We will also recall the main features of Fourier analysis on finite general groups, without proving anything, since this is already the content of a course in representation theory. Next, we deal with a classical, but amazing application of these ideas, namely Dirichlet's theorem. This allows us to say a few words about L-functions of Dirichlet characters. This is again just the tip of a huge iceberg, which is far from being understood, even though progress is constantly being made.

7.A.l

Dual group

From now on, let G be a finite abelian group with n elements. We want to define the Fourier transform of a function f : G -+ C. Recall that if f is an integrable function on JR, with complex values, its Fourier transform is defined

The presence of the characters y -+ analysis on abelian groups.

e2i7rxy

of JR will be our guide to Fourier

Remark 7.A.1. When dealing with abelian groups (which will always be the case for us), one also denotes by + the internal operation of G. This is quite intuitive when dealing with groups such as ZjnZ, but definitely not suitable for groups such as (ZjnZ)*. Our convention will be the following: when dealing with an abstract abelian group G, we will use multiplicative notation for the internal operation of the group, while in concrete examples we will choose the most intuitive notation, depending on the situation. Hopefully, this will not create too much confusion ... A character of G is a morphism of groups X : G -+ C*. So, according to whether we use additive or multiplicative notation for the internal operation of G, a character satisfies X(x + y) = X(x) . X(y) for all x, y E G (respectively x(xyl = X(x) . X(y))Âˇ The character is called trivial if X(g) = 1 for all g E G. Let G be the set of all characters of G. It becomes a group with respect to the obvious multiplication (XlÂˇ X2) (g) = Xl (g )x2(g) and it is called the dual group of G. Note that for all X E G and g E G we have X(g)n = X(gn) = 1, because gn = 1 by Lagrange's theorem. In particular, Ix(g)1 = 1 and X-I (g) = X(g) for all g E G and X E G. The idea of harmonic analysis is t~at all information about the space of ([:::-valued functions on G is encoded in G.

Example 7.A.2. Take n 2': 2 and G = ZjnZ. What is the dual group of G? We saw that X(l) is an nth root of unity. And clearly X is uniquely determined by x(1), as G is generated by 1. Conversely, if z is an n-th root of unity, x -+ ZX defines a character of G (by ZX we mean za for any lifting a of x; this does not depend on the choice of a, as zn = 1). We deduce that G is isomorphic to ZjnZ, even though this isomorphism depends on the choice of a primitive nth root of 1, so it is not really canonical. It is an easy exercise to check that passing to duals is compatible with direct products (i.e. the dual of G x H is Gx H). Since any finite abelian group is a direct product of cyclic groups (this is a nontrivial, but absolutely classical

Chapter 7. Complex Combinatorics

312

7.A. Finite Fourier Analysis

result), it follows from this remark and the previous example that for any finite abelian group G, its dual is isomorphic to G. In particular, Gand its dual have the same number of elements, which is really not obvious at .first sight. We also deduce from this observation and example 7.A.2 that if x E G - {I}, then there exists X E such that X(x) #- 1 (if you think about this for a moment, you will realize that it is nontrivial!). Thus, the map g -+ (X -+ X(g)) is an injective

Theorem 7.A.3. If q is a finite abelian group, then

isomorphic to G and

e is canonically isomorphic to G.

e is an abelian group

Let F(G, q be the ~::-vector space of all maps f : G -+ C. It is a ((>vector space of dimension IGI, since the map F(G, q -+ C IGI sending f to (f(g))9EG is obviously a C-linear isomorphism. If f,g E F(G,q, let

(f,g) =

because x -+ gx is a permutation of G. Since X is not trivial, there is g such that X(g) #- 1 and the previous identity yields S = 0 and so (Xl, X2) = o. Step 2. We claim that for all x E G - {I} we have I:XEGX(x) = O. The same argument as in the first step shows that it is enough to prove that there exists X E such that X(x) #- 1. This is nontrivial, but has already been explained. Step 3. Since (X)XEG is an orthonormal set, it is linearly independent. But since it has the same cardinality as the dimension of the vector space F(G, q (namely IGI, by theorem 7.A.3), basic theory of vector spaces shows that it has to be a basis of F(G, q. The result follows. 0

homomorphism G -+ realizing G as subgroup of Since lei = IGI, the previous injection has to be an isomorphism. So G is canonically isomorphic to its double dual. Let us now glorify what we have just proved:

313

â&#x20AC;˘

In practice, one really uses the following consequence of the previous theorem:

Theorem 7.A.5. For any finite abelian group G, the following relations hold: 1) (orthogonality relations) For all X, Xl, X2 E

1 '""' fGI ~ f(x)g(x).

I~I ~ X(g)X(h) = I g=';.

xEG It is easy to check that this is an inner product on the C-vector space F(G, q. We are now ready to prove the main theorems of Fourier analysis on G.

Theorem 7.A.4. The elements of

e and g, h E G

e form an orthonormal basis of F(G, q.

Proof. We split the proof in several steps. Step 1. We prove that (Xl, X2) = lX1=x2' i.e. that (X)XEG is an orthonormal set. If Xl = X2, everything follows from the fact that X(g) has magnitude 1 for any g and any character X. Assume that X = is not trivial. Then

Let S = I:xEG X(x). Then for all g E G we have

X(g)S = L X(gx) = L X(x) = S, xEG xEG

XEG 2) (Fourier inversion) For all f

F(G, q we have f = I:XEGU, xh.

3) (Plancherel's identity) For all f E F(G, q 1 '""' fGI ~ If(x)1 2 =

XEG

'""' ~ IU, X}I 2 . XEG

Proof. 1) has already been proved while proving the previous theorem. For 2), note that for any x E G we have, by the orthogonality relations ( LU,X). X) (x) x 1

fGI L Y

I~I LX(x) Lf(y)x(y) x

f(y) L

X(x/y) = f(x),

Chapter 7.

314

Complex Combinatorics

from which the result follows. Finally, using 2) and the previous theorem, we can write

I~I

L xEG

If(x)12 = (1,j) = LL(1,XI I' (1,x21' (XI,X2)= L Xl

1(1,xI1

XEG

Remark 7.A.6. More generally, we have (1, gl = L

(1, Xl (g, Xl

XEG

for all f,g E F(G, q, as can be deduced by an obvious adaptation of the proof of Plancherel's identity. By analogy with the usual formula in classical Fourier analysis

i(n)

2~ 127r f(y)e-inYdy,

we write i(x) = (1, Xl and call it the x-Fourier coefficient of

7.A.2

A glimpse of the non-abelian case

One may ask whether the above theory can be adapted to the case when the group G is still finite, but no longer abelian. The answer is positive, but things are much more complicated than in the abelian case. In order to develop Fourier analysis on any finite group G, one has to consider all complex finite dimensional representations of G. By definition, such a representation consists of a finite dimensional C-vector space Von which the group G acts linearly, i.e. for each element g E G one has an automorphism still denoted g of V such that gh = go h (in the left-hand side gh is the automorphism associated to gh E G, while in the right-hand side we have a composition of automorphisms). In more down to earth terms, a representation of G is a morphism 2 P : G ----t GLn(C) for some n 2: 1. Two representations PI, P2 : G ----t GLn(C) are called isomorphic if there exists P E GLn(q such that P2(g) = PpI(g)P- 1 for all g E G. 2Recall that GLn(1C) is the group of matrices A E Mn(1C) with nonzero determinant.

7. A. Finite Fourier Analysis

315

If n = 1, we call P a character (this is compatible with the definition of a character given in the previous section). Such a representation V is called irreducible if no proper subspace of V is stable under all automorphisms g, with g E G. For instance, a character is an irreducible representation, as there is no proper subspace at all! Moreover, the only irreducible representations of a finite abelian group are the characters, so the new theory will be compatible with the theory developed in the previous section. To prove the previous assertion, one has to use a basic result from linear algebra, stating that any commutative family of endomorphisms of a finite dimensional vector space over an algebraically closed field has a common eigenvector. So, if G is an abelian group acting irreducibly on V, there is a common eigenvector v for all g E G. But then Cv is a nonzero subspace of V stable under all g E G and so we must have Cv = V; hence V is a character. We have the following very basic theorem due to Maschke:

Theorem 7.A.7. Any finite dimensional C-representation V of a finite group is a direct sum of iT'reducible repT'esentations, that is theT'e exist sub-vectorspaces VI," . ,Vk of V stable under G, which are iT'reducible repT'esentations and such that V = EBi Vi.

~ Now, the role of the dual G of G in the abelian case is played by the set G of (isomorphism classes of) irreducible representations of G. It turns out that this is a finite set and by the previous discussion its definition agrees with the usual definition if G is abelian. The set of maps F(G, q is naturally a Calgebra and as such it is isomorphic to the group algebra qG]. By definition, the elements of the last object can be uniquely written L9EG a g Âˇ g with a g E C (so the elements of G form a C-basis of qG]) and multiplication is defined by

Note that qG] itself is a representation of G (each element of G acting as multiplication by g). The following theorem summarizes the properties of C[G].

316

Chapter 7.

Complex Combinatorics

1) G is a finite set and qG] is isomorphic as a representation ofG to EBvEG(dim V)V, where nV = VEBVEB路路路EBV (n times). In particular,

Theorem 7.A.8.

IGI

L (dim V)2.

7.A. Finite Fourier Analysis

7.A.3

317

Some concrete examples

Let us specialize the previously quite abstract (at first ... ) theory to a very concrete situation that appears frequently in number theory. Let N be an integer greater than 1 and let

VEG

2) The center of qG] consists of all maps f : G --t C such t!!:..at f(g) f(hgh- I ) for all g, hE G. Its dimension over C is equal to IGI and also to the number of conjugacy classes3 of G.

In the abelian case, we defined an inner product on qG] and we showed that the characters of G formed an orthonormal basis of qG]. All this can be done in the non-abelian case, though things are usually more difficult to prove. Namely, define for !I, 12 E qG]

(!I, h) =

1" TGI ~ !I(g)h(g)路 gEG

Now, to any representation p : G --t GLn(C) one can associate its character, which is the element of qG] defined by Xp(g) = Tr(p(g)). The main theorem of Fourier theory on finite groups is then the following:

1) If VI, V2 are two representations ofG such that XVI = XV2 as elements ofqG], then VI and V2 are isomorphic (and conversely).

Theorem 7.A.9.

2) (XV)VEG is an orthonormal basis of the center of qG] for the previously defined inner product. Moreover, a representation V is irreducible if and only if (Xv, Xv) = 1. 3) If g, h E G, then LVEG Xv(g)Xv(h) is equal to the cardinality of the centralizer of 9 in G if g, h are conjugate in G and it is equal to 0 otherwise. 3The conjugacy class of an element 9 EGis {hgh-Ilh E G}.

G = (71jN71)*

be the abelian group of invertible residue classes mod N. A character of Gis called a Dirichlet character of modulus N or simply a Dirichlet character mod N. If X is such a character, we set

x(n) = l~cd(n,N)=1 . x(n) for all integers n, obtaining in this wayan N-periodic function. We will focus on a more restricted class of characters modulo N, namely the primitive ones. Let us recall what that means. If d divides N and if Xd is a character mod d, then Xd yields a character mod N simply by composing it with the natural map (71j N71)* --t (71j d71)*. We say that a character mod N is primitive. if it is not obtained in this way, for any proper divisor d of N and any Xd. A more practical and useful criterion is the following: a character X mod N is primitive if and only if for any proper divisor d of N there exists n == 1 (mod d) such that gcd(n, N) = 1 and x(n) =I- 1. We leave to the reader the easy task of checking this. Let us see what happens when we apply the abstract theory to this situati~n. Let a be an integer prime to N and let f be the map fen) = 1n =a (mod N). It IS naturally a map on G and it is clear from the definitions that for all characters X of G we have icx) = x(a). So, using Fourier's inversion formula, we obtain

1n =a

(mod N) =

1,,-

!.p(N) ~ x(a)x(n),

the sum being taken over all Dirichlet characters mod N. This relation holds for all n prime to N and plays a crucial role in the proof of the famous Dirichlet's theorem, to be discussed in the next section.

Chapter 7.

318

Complex Combinatorics

Gauss sums and Fourier coefficients of Dirichlet characters To any Dirichlet character X mod N we attached an N-periodic function on Z, defined by x(n) = 19cd (n,N)=l . x(n). Any N-periodic function on Z induces a map on ZINZ and we can look at its Fourier coefficients with respect to this finite abelian group. As we have already seen, the characters of ZI NZ are identified with ZINZ (we identify a and the character x ---+ e 2i'lvax). Via this identification, the Fourier coefficients of a map f on ZI NZ will be denoted

l(r)

7.A. Finite Fourier Analysis

and the result follows from Ix(a)1 = 1. 2) Write N = dv and a = du with d > 1 and gcd(u, v)

= 1.

Let

2i7rU

(= e--v- ,

a primitive root of unity of order v. Note that

x(a) =

f(x)e -2~rx.

319

X(x)C

x (mod N)

xEG

N-l

The following result gives further information about these coefficients when comes from a Dirichlet character:

1 ~ '"' = N

X(j)(J.

j=O d-l v-l

Proposition 7.A.I0. Let X be a Dirichlet character mod N.

~ LLX(j + kv)(j k=Oj=O

1) For any a relatively prime to N we have

x(a) 2) If X is primitive, then x(a)

x(a)X(l).

0 whenever gcd(a, N) > 1 and we have Ix(a)1 =

(

j (mod v)

X(j

+ kV))

(j.

k(mod d)

It is thus enough to prove that

Sj =

L k

X(j

+ kv)

(mod d)

for gcd( a, N) = 1.

Proof. 1) Since X(x)

= 0 whenever gcd(x,

x(a) =

N) > 1, we have

x(x)(ax,

xEG 2i1f

where ( = e- N

x(n)Sj =

As x ---+ ax is a permutation of G, we have

x(a)x(a) =

~L x(ax)(ax = ~L X(X)C = X(l) xEG

vanishes for all j. Now, since X is primitive, there exists n == 1 (mod v) such that gcd( n, N) = 1 and X( n) =f- 1. It is easy to see that (n(j + kv) (mod N)) k (mod d) is simply a permutation of (j + kv (mod N)) k (mod d). Thus

xEG

x(n(j

k (mod d)

+ kv)) =

X(j

+ kv) = Sj

k (mod d)

and since x(n) =f- 1, we have Sj = O. This proves the first part of 2). Finally, using Plancherel's identity (theorem 7.A.5) for ZINZ and the

Chapter 'l.

320

Complex Combinatorics

321

nality relations we can write

information we have already gathered, we deduce that 'P(N) N

'l.A. Finite Fourier Analysis

Ix(x)12

x (mod N)

Ix(a)12

a (mod N)

Ix(a)x(1)1

gcd(a,N)=1

= 'P(N)lx(1)1 ,

from where we obtain

7.A.4

IX(l)1

JR. Using again part 1), the result follows.

Some applications

We have developed enough theory in the previous sections to be able to prove quite a few nontrivial results. The first one is very elementary, but tricky and taken from [71]. The interested reader will find in loc.cit many beautiful applications of finite Fourier analysis.

Proposition 7.A.l1. Let A be a finite set of integers and let f : A ----t Z/pZ be a map. Then for any positive integer k there exist at least IA~2k (2k )-tuples (al,' .. , a2k) E A2k such that

But it is apparent in this last expression that N(j) < N(O), finishing the proof. 0 The method used to prove the following result is extremely useful in additive combinatorics.

Theorem 7.A.12. (D. Hart, A.Iosevich) Let q be a prime power, d a positive integer and let A C lF d be a set with 4±l

more than q 2 elements. Then for any x E lF~ one can find a, b E A such that a· b = x (here· is the standard inner product in lFg). Proof. The idea is to express the number of solutions of the equation a· b = x using the orthogonality relations, then analyze the error term. Let X be a nontrivial additive character of IF q and let n( x) be the number of solutions of the equation a . b = x with a, b E A. The orthogonality relations yield

Proof. Let N(j) be the number of 2k-tuples (al,'" ,a2k) E A2k such that

IAI2

n(x) = L

qLx(c(a.b-x))=-q-+R,

a,bEA

cElFq

where Clearly L,~:6 N(j) = IAI2k. We will prove that N(O) 2: N(j) for all j, which will be enough to deduce the desired result. Next, note that by the orthogo-

R= - L L

q aEA bEA cElF;j

X( c( a . b - x)).

322

Chapter 'l.

Complex Combinatorics

Now, using the Cauchy-Schwarz inequality and once again the orthogonality relations, we obtain

~ I~I

L LLx(c(a.b-x)) q aEA bEA cioO

~ I~I q

L aEFg

323

Proof. Apply the previous theorem to the subset A x A x ... x A of lF~.

The following bound on character sums is a basic tool in analytic number theory.

'l.A. Finite Fourier Analysis

Theorem 7.A.14. (Polya- Vinogradov) Let X be a primitive character modulo N. Then for all positive integers m, n we have

L L x(a· (c 1b1 - C2 b2))x(X(C2 b1,b2EA C1,C2EF~

cd)

I~I

L L X(X(C2 - q)) L x(a· (c 1b1 - C2 b2)) q bl,b2EA C1,C2EF~ aEFg

Using the orthogonality relations and the substitution Sl = we can write

X(j) ~ ViilogN.

m5,j<n

Before giving the proof, let us mention that Schur proved that if X is a primitive character mod N, then

m~x

S2 = C2,

x(n)

2~ Vii,

n5,M

so the Polya-Vinogradov inequality is not far from being optimal. L L X(X(C2 - c1))1 C1 b1=C2bz b1 ,b2 E A C1 ,C2 EF~ =

Proof. Using Fourier's inversion formula and proposition 7.A.10, we can write

x(j) =

L L 181 b1=b2 L X( xs 2(1 - Sl)) bl,b2EA 81 EF~ 82EF~

X( a)e 2i;.a

j •

(a,N)=l

L (q - 1)h1=b2 - L L 181 b1=b2 bl,b2EA bl,b2EA 81 io1

~ (q - 1) L

Adding this over j, using the triangle inequality and proposition 7.A.1O, we deduce that

1b1=b2

2i7ran

2i1rarn.

e r::r - e-r:r- <_1_ 1 1 x(j)1 = L.. x(a)· 2i11"a m5,j<n (a,N)=l etr - 1 - VJV (a,N)=l 1sin ~ I·

b1,b2EA

< qlAI· Combining this with the first formula for n(x), we deduce that n(x) d+1 if IAI > q-2-, from where the result follows. 1

>0 0

An easy convexity argument shows that sinx :::: ~x for 0 ~ x ~ Applying this to x = ~ (respectively x = 7l"(~-a)) if a ~ If (respectively a > If), we deduce that

Corollary 7.A.13. Let A C lFq be a subset with more than q'j+2ri elements. Then for any x E lF~ there exist a1, ... , ad E A and bl, ... ,bd E A such that x = a1b1 + a2b2 + ... + adbd.

m5,J<n

and the result follows.

X(j) 1 ~

Vii·

L a5,N/2

~a ~ Vii log N

324

Chapter 1.

Complex Combinatorics

Here is a nice application of the previous theorem, taken from [59]. We refer the reader to that paper for many refinements and further discussion:

7.A. Finite Fourier Analysis

325

Now, everything follows from the simple observation that if S =I 0, then there exists 1 ::; n ::; T such that n q == a (mod p), while if S = 0, then

Theorem 7.A.15. (Murty) Let p be a prime and let q be a prime divisor of

p2logp T < -1. q

E-=l

p - 1. Let a be an integer such that a q == 1 (mod p). Then there exists an integer x such that Ixl < p3/2:0gp and x q == a (mod p).

In [59], Murty proves that the hypothesis that q is prime is not necessary and that we can strengthen the conclusion by asking that Ixl < cp3/2 jq for a suitable absolute constant c. This requires some stronger estimates on character sums.

If p is a prime, let n2 (p) be the smallest positive integer a such that a is a quadratic non-residue mod p. It is an easy exercise to check that n2(p) < 1 + yIP, but it is much more challenging to find nontrivial bounds for n2(p), The next result gives such an upper bound, by combining the PolyaVinogradov inequality with Mertens' theorems.

Proof. Consider a parameter p > T > 1 and look at the number of solutions of the congruence x q == a (mod p) with x E [1, T]. Using the orthogonality relations, we write this as

Theorem 7.A.16. (Vinogradov) If p is a sufficiently large prime number,

1nq =a

then there exists 1 ::; a

< p2.,je log2 p such that

(~) =-1.

(mod p)

n"S.T =

L p n"S.T

= p

~1 X

(mod p)

~ 1 Lx(a) L X

q x(a)x(n )

Proof. Let m = [ylPlog2 p] and suppose that

(~) =

xq(n).

n"S.T

In the previous sum we will distinguish those characters X such that Xq = 1 from the others. Indeed, if Xq = 1, then Ln<T xq(n) = T and x(a) = 1 (since the hypothesis on a implies that a is a qth power in IF;), while if Xq =I 1, Polya-Vinogradov's theorem gives

for all 1 ::; a ::; X =

[p 2:re log2 p].

Thus, since there are precisely q characters X such that Xq = 1, the previous remarks yield

qT p-1-q S - p _ 1::; p _ 1 y'plogp

2NI =

f (~) : ; x=l

ylPlogp,

so N > !Jj - !ylPlogp. On the other hand, any quadratic non-residue mod p must have a prime factor greater than X. Thus m

< y'plogp.

Let N be the number of quadratic non-

residues among {1, 2, ... , m}. By Polya-Vinogradov's inequality

1m L xq(n) ::; ylPlogp. n"S.T

1 2y'plogp

< N::;

,,",m ~

X<q"S.m q

326

Chapter 7.

Complex Combinatorics

The proof of theorem 7.A.17 combines Fourier analysis on G = (ZjNZ)* and very subtle estimates of the L function of a Dirichlet character near 1. To start with, define the complex L-function of a Dirichlet character X mod N by (recall that x(n) = 0 for gcd(n, N) > 1)

Mertens' theorem (theorem 3.A.5 and the remark following it) yields '"'

log m

= m log 10 X

X<qs"m q

Note that

lo;X

= O(JP .logp)

(m),

327

7.A. Finite Fourier Analysis

L(s, X)

and

= '"' x(n) . ~

n:?:l

log m log-- = log log X 1

= -

! log p + 2 log log p

2}e logp + 2 log logp 1 + 4 log logp

+ 10

logp

g 1 +4 vfe~ co logp

! _ 4( y'e 2

1) loglogp logp

As Ix(n)1 ~ 1 for all n, this series converges uniformly on compact subsets of Re(s) > 1, so L(s,X) defines a holomorphic function in this region. Moreover, a simple argument going back to Euler and using the unique factorization theorem and the fact that 1~x = I:n>O xn for Ix I < 1 shows that for Re( s) > 1 we have 1 L(s, X) = 1 _ X(p)p-s'

(1) X logp

(1)

(_1_) . logp

Xlogp

II p

We easily deduce from this that L( s, X) does not vanish if Re( s) if we choose a branch of the complex logarithm, we can write

Combining these estimates yields

m 1 - - -JPlogp

<- 2

loglogp 4(v'e - l)mÂˇ logp

7.A.5

10gL(s,X) = -

+ O(y'Plogp),

Dirichlet's theorem

We will use the previous tools and a bit of complex analysis to give a proof of the famous

L log(l- X(p)p-S) = LL X(pn)p-ns n

Dirichlet's key insight was to use the Fourier transform to express the condition that n == a (mod N) in an analytic way. More precisely, multiplying the previous relation by x(a), summing over X and using the orthogonality relations, we obtain for s > 1 1

rp(N)

x(a) log L(s, X) =

X (mod N)

p=a

1 = rp(N) 1 . log (s1 pS _ 1)

+ 0(1).

(mod N)

In particular, there are infinitely many primes p

(mod N).

1 npns

n>l

pn=a(~od N)

-pS + 0(1), p=a(mod N)

the last estimate being a consequence of the inequalities

o < '"' '"' ~ ~ == a

L p

Theorem 7.A.17. Let a and N be relatively prime positive integers. For s -+ 1+ we have

> 1. Moreover,

which is not possible for p large enough.

1 ~~~pn '"' '"' 1 = '"' 1 npns ~p(p-l) <00.

p n:?:2 pn=a(mod N)

p n:?:2

Chapter 7. Complex Combinatorics

328

The crucial point is that log L(s, X) remains bounded as s -t 1+ precisely when X is nontrivial and that log L( s, 1) (where 1 is the trivial character) is relatively easy to handle and yields the factor log S~1. If the second part is rather easy to prove, the first part is a very deep result, equivalent to the non-vanishing of L(s, X) at s = 1 whenever X is nontrivial. Dirichlet's proof was very roundabout, but we have quite a few different ways to prove this nowadays. First, let us deal with the "easy" part, which will also be important in the proof of the hard part.

Theorem 7.A.18. L(s, X) extends to a function on Re(s) > 0, which is holomorphic except possibly at s = 1. If X is nontrivial, this function is holomorphic at s = 1, but if X is trivial, we have lim (s - I)L(s, 1) =

s-+1+

I). II ( 1 P piN

Proof. The key ingredient is Abel summation. Suppose that X is nontrivial and note that for all n we have IX(I) + X(2) + ... + x(n)1 ::::; N, which follows easily from the orthogonality relations (theorem 7.A.5). An easy computation shows that

n -s - (n

+ 1)-S =

sn -s-1

+ O( n -S-2) ,

+ X(2) + ... + x(n)) (n- S -

+ 1)-S)

n2:1

we get

But for t E [n, n + 1] we have

le s

n-sl

II (1 - Is) . ((s), P

Since

S )

dt = - S n

((s)

inrt -SX-S-1dxl -< int

lsi

xRe(s)+1

dx < lsi - n1+Re(s) '

yielding the uniform convergence of the series

on all compact subsets of Re( s) > 0. Thus g is holomorphic on Re( s) > since ((s) = S~1 + g(s), the result follows.

Â°

and 0

Taking into account the previous theorem and discussion, theorem 7.A.17 is a consequence of the difficult

Proof. We saw that for all s > 1 we have 1

converges uniformly on compact subsets of Re(s) > 0. Moreover, by Abel summation, the sum of the series is L(s, X) if Re(s) > 1, which yields the holomorphic extension of L(s,X) to Re(s) > 0. On the other hand, if X is trivial, the inclusion-exclusion principle yields

L(s, X) =

329

Theorem 7.A.19. If X is nontrivial, then L(I, X) =1= 0, so log L(s, X) remains bounded for s -t 1+ .

which is uniform for s in compact sets. Thus the series

I)X(I)

7.A. Finite Fourier Analysis

1 = ' "L.... nS. n2:1

(n + 1)1-S - n 1- s , s-1

<p(N)

x(a) 10gL(s, X) =

x (mod N)

L L p

n>1

- >- 0 , npns

pn=a (;nod N)

so by taking a = 1 we obtain

L(s, X) 2 1. x (mod N) But by the previous theorem all L(s, X) with X nontrivial are holomorphic at s = 1 and L(s,l) has a simple pole at s = 1. Combining this observation with the previous inequality, we deduce that there is at most one nontrivial

Chapter 7. Complex Combinatorics

330

character X such that L(I,X) = 0 (otherwise the product of L(s,X) would vanish for s -t 1+). Assume that X is such a character, then clearly

L(I,X)

7.A. Finite Fourier Analysis

331

(bn(x))n is nondecreasing, as one easily checks using the AM-GM inequality that

= 0,

so that we must have X = X, that is X takes only values ±1 and O. Thus, it remains to prove that L(I, X) =f:. 0 whenever X is a nontrivial real character. We will present Paul Monsky's elementary proof from [54]. Consider the function

defined for x E [0,1) (the series is obviously absolutely convergent for such x). We claim that f is unbounded as x -t 1-. Indeed, we can write

1 ( 1 xn ) = I-x n(n+l) - (l+x+···+x n- 1 )(I+x+···+x n )

is nonnegative. Next, using Abel's summation formula, the monotonicity of the bn(x) and the fact that I 'L:~=l x(i)1 are bounded, it is easy to see that f is bounded. But this contradicts the result of the previous paragraph, ending the proof of Dirichlet's theorem. 0 Once we know that L(I, X) =f:. 0 for all nontrivial characters X, it is a natural question to ask how far it is from being zero. If we look carefully at the proof of the previous theorem, we see that it gives L(I, X) > 0 for any nontrivial real Dirichlet character. The following deep theorem gives much more. The proof is much more involved:

Theorem 7.A.20. (Siegel) For any E > 0 there exists C(E) > 0 such that L(I, X) > c;:J for any real primitive Dirichlet character modulo N. all manipulations of the series being justified by the absolute convergence. Let Cn = 'L:dln X(d). If p divides N, then Cpk = 1 for all k ~ 1, as X(p) = O. Also, it is easy to see that Cmn = CmCn for gcd(m, n) = 1. An.immediate computation shows that Clk ~ 0 for all primes I and all k and using the previous observation we deduce that Cn ~ 0 for all n. As infinitely many en's are equal to 1 and all Cn are nonnegative, 'L:n enxn is not bounded as x -t 1- and the claim is proved. Now, assuming that L(I, X) = 0, we will prove that f is bounded as x -t 1-, which will finish the proof. If

xn bn(x) = n(l-x) - l-xn' 1

then the hypothesis L(I, X) = 0 reduces the boundedness of f(x) as x -t 1 to the boundedness of 'L:n bn(x)x(n) as x -t 1. The miracle is that the sequence

7.A.6

A useful consequence

The purpose of this section is to give an analogue of Mertens' theorems for primes in arithmetic progressions. This uses the proof of Dirichlet's theorem and the standard technique of Abel summation. Recall that A is Von Mangoldt's function, defined by A(n) = logp if there is a prime p and k ~ 1 such that n = pk, and A(n) = 0 otherwise. Its usefulness in number theory comes from the relation 'L:dln A( d) = log n, which will be heavily used in the proof of the following result.

Theorem 7.A.21. Let a, N be relatively prime positive integers. Then '"

~ p=a(mod p:-S;x

logp = logx (N) p cp N)

O() 1.

Chapter 7. Complex Combinatorics

332

Proof. Since " L...J " L...J -logp kk?2 p P we have

p:S;x p=a (mod N)

A(n)

n<x n=a (~od N)

333

Another Abel summation (using the fact that the partial sums L~=l X(k) are bounded) shows that

< 00,

logp p

7.A. Finite Fourier Analysis

+ 0(1).

xC!)

_ J <x

= L(l, X) -

xC!)

J'> x

A(n) =

n<x n=a(~od N)

_1_" x(a) "

'(!'l(N) L...J

L...J

n<x -

We deduce that E(x) = F(X, x)L(l, X)

x(n) A(n),

(~L A(d)) . d:S;x

Since by theorem 7.A.19 we have L(l, X) =I- 0, in order to prove that

so everything comes down to the study of

F(X,x)

+ 0 (~) x

Using the orthogonality relations, we can write

" L...J

= L(l, X)

F(X, x) = 0(1)

L x(n) A(n). n

it is enough to prove that

n:S;x

-x1 L A(d) = 0(1).

If X is trivial, Mertens' theorem 3.A.4 yields

F(l,x)

L pk,:s;x

lo~p = L p

p:S;x

d:S;x

logp p

+ 0(1) = log x + 0(1).

But

gcd(p,N)=l

Suppose now that X is nontrivial. We will prove that F(X, x) is bounded as x --+ 00, which will finish the proof of the theorem. Exploiting the relation Ldln A(d) = logn, let us consider the expression E(x) =

L n:S;x

x(n) logn. n

10;

L...J

dld2:S;X

X(d 1 d2 ) A(d 1 ) = "X(dI)A(dI) d1 d2 L...J d1 dl:S;X

" L...J d2:S;X/dl

X(d 2 ). d2

pk:S;x

logp :s;

L logpÂˇ ~Og x = log x . 1f(x) = O(x), p:S;x

ogp

using theorem 3.A.2. The result follows.

7.A.7

Since x --+ x is decreasing for x 2 3 and since the partial sums of the sequence (x(n))n are bounded, Abel summation shows that E(x) = 0(1). On the other hand

E(x) =

L A(d) = L d:S;x

An "elementary proof" of Dirichlet's theorem

The proof of Dirichlet's theorem that we presented used rather heavily basic properties of holomorphic functions, but it turns out that the ideas used in the proof of theorem 7.A.21 may be combined with careful estimates to obtain an entirely "elementary" proof of Dirichlet's theorem, i.e. completely avoiding the use of complex analysis. We will sketch such a proof in this section, but we must emphasize that the complex analytic approach is much more powerful and conceptual.

Chapter 7.

334

Complex Combinatorics

We will actually give an elementary proof of theorem 7.A.21, which obviously implies Dirichlet's theorem. We will use the notations of arguments of the proof of theorem 7.A.21. The only "non-elementary" result used in the proof of this theorem is the fact that £(1, X) =I- 0 for all nontrivial characters X. Monsky's proof (see the proof of theorem 7.A.19) deals with the case when X is real in an elementary (but very tricky) way, so it remains to deal with the case when X is not real. Let k be the number of nontrivial characters X for which £(1, X) = 0 and assume that k :2: 1. Then Monsky's result implies that k :2: 2 (since if L(l, X) = 0 then £(1, X) = 0). Either a direct computation or a simple application of Mobius' inversion formula yields the identity

L JL(d) . log ~ = 1n=1 . log x + A(n), from which a standard computation yields

F(X, x) =

L n:S;x

x(n) A(n) n

The argument used in the proof of theorem 7.A.21 shows that F(X,x) is bounded if L(l, X) =I- O. Assume that £(1, X) = o. Then Abel's summation formula shows that L:d>x x~d) = O(l/x), so the previous formula yields

F(X,x) = -logx + 0

(~L log ~) dl:S;X

= - 1ogx+ 0 ( = -log x

[xl log x -lOg[Xl x

+ 0 (1).

The conclusion is that

LF(X,x)

= (1- k)logx +

0(1).

x But this contradicts the fact that k :2: 2 and (again by the orthogonality relations)

L x

F(X, x) = <p(N)·

'" ~

A(n) -:;;:-:2: o.

n<x n=l(mod N)

The reader has probably noticed that the crucial part of .thi~ argument was an "elementary" version of the argument used in the begmnmg of the proof of theorem 7.A.19.

din

335

'l.A. Finite Fourier Analysis

Chapter 8

Formal Series Revisited In this chapter, we discuss applications of generating functions in combinatorics or combinatorial number theory. This is a rather vast subject and we will only be able to scratch its surface through some nice examples. The idea is very simple (even though in practice things are less clear): given a sequence (an)n of complex numbers (but we may allow this sequence to take values in any ring, in theory), one is sometimes able to extract information about the sequence from its generating function Ln>o anX n in an easier way than by dealing with the sequence itself. For instance, if the sequence satisfies a rather complicated recursive relation, it is very hard to study the sequence directly, but quite often it turns out that its generating function satisfies some differential equations or some regularity properties that allow us to study the sequence. Also, in counting problems it is very often much easier to find the generating function of the desired number of objects to count than to find directly that number. Of course, one sometimes needs quite a lot of imagination to extract the information from the generating function, but it should be a principle that if one knows the generating function, then one knows a lot of things about the sequence. It is important to note that when dealing with generating functions one neglects all convergence and analytic issues. However, once one knows the generating function, one usually uses analytic arguments to study the sequence.

338

Chapter B. Formal Series Revisited

Before passing to concrete problems, let us recall a few things about operations with formal series. Addition and multiplication are defined just as for polynomials. It is an easy exercise to check that a formal series has an inverse (for multiplication) if and only if the constant term is nonzero. A more subtle problem is the composition of formal series. Here, one has to be a bit careful, as simple things such as L:n (x!?)n do not make sense formally. To do things properly, the best way is to introduce the X -adic valuation on formal series: if I = ao + alX + ... is a formal series with complex coefficients (more generally, any field), let vx(f) be the least nonnegative integer n such that an =I- O. Thus I . X-vx (I) is an invertible power series and one easily checks using this that vx(fg) = vx(f) + vx(g) and vx(f + g) 2 min(vx (f) , vx(g)). Thus Vx behaves as the p-adic valuation.

Definition 8.1. Say a sequence of formal series In converges towards a formal series I if vx(f - In) tends to 00. If I = ao + alX + .. " this means that for all i, the coefficient of Xi in In is ai for all sufficiently large n. One says that I = L:n>O In, respectively I = TIn>O In if the sequence (fo + h + ... + In)n (respectIVely (foh ... In)n) converges to I· It is an easy exercise for the reader to check that L:n>O In converges to a formal series if and only if vx(fn) tends to infinity (the analogous result holds for p-adic numbers). Also, if In are formal series such that In(O) = 0, then TIn>O(l + In) converges if and only if vx(fn) tends to 00. We can now explain the -composition of two formal series. Assume that I, 9 are formal series such that g(O) = 0 and I = ao + alX + a2X2 + .... The composition log is then by definition the formal series

B.l.

Counting problems

339

a(a-I)···(a-n+1) A f h ( a) were n = n! . very use ul special case is a = -m, where m is a positive integer. In this case the binomial formula becomes

1 = " (m + 1)

(1 - x)m

L..;

n m _ 1

n~O

Finally, we will sometimes use the classical notation [XnlJ for the coeffiin I. cient of

8.1

Counting problems

The first exercise is quite technical, but there is no serious difficulty in it and the result is quite beautiful. Recall that if Xn is a sequence of complex numbers, an equivalent for it is a sequence Yn in closed form such that :En. Yn converges to 1. 1. Let aI, a2,···, an be relatively prime positive integers. Find an equivalent as k --+ 00 for the number of positive integral solutions of the equation alXI +a2x2 + ... + anXn = k.

Proof. The generating function of the sequence Yk, the number of positive integral solutions of the equation alXI + a2X2 + ... + anxn = k, is

log = ao + alg + a2g 2 + .... The series converges, because vx(gn) 2 n. All other operations on formal series (such as differentiation, integration) do not involve any difficulty and have the properties that we imagine. Before passing to problems, let us recall the following useful extension of the binomial formula: for any rational number a we have This is a rational function and in order to analyze sequence Yk we will decompose it into simpler pieces. Namely, if 1 = Zl, Z2, .• . , Zr are the distinct

Chapter B. Formal Series Revisited

340

roots of the polynomial n~=l (1 - xa;), with multiplicities then by general theory there exist constants Cij such that r

f(X) = Coo

ml,

m2,' .. ,mr ,

Counting problems

341

The following problem is a bit trickier, but it shows how formal series in several variables appear quite naturally. 2. For positive integers m and n, let f (m, n) denote the number of n-tuples (Xl, x2,···, xn) of integers such that IXll + IX21 + ... + IXnl :::; m. Show that f (m, n) = f (n, m).

+ LL (X - zy' ij

i=l j=l

B.l.

Using the binomial formula, we obtain

Putnam 2005 Proof We will use a two-variable generating function in this problem: with the convention f(m,O) = f(O, n) = 1, we have the chain of equalities

Inserting this in the formula for

Yk -

f and re-arranging terms yields

~ ~(-I)j zi-(k+j)G1J.. ~~

F(X, Y) =

(j + k 1)

f(m,n)xmyn =

m,n~O

k -

i=l j=l

kl, ... ,knEZ

= _I_,",yn

I-X~

n~O

We claim that the dominant term in the previous sum corresponds to i = 1 and j = ml = n. First, note that mi < n for all i =f. 1. Indeed, each of Xaj - 1 has simple roots, so the only possibility for mi 2: n would be that Zi is a root of all Xak -1, thus also of X -1 and so i = 1. Thus the contribution to Yk of all terms with i =f. 1 is dominated by ckn - 2 for some constant c. A similar argument shows that the contribution of the terms with i = 1 and j < n is also dominated by a constant times k n - 2 . Thus the main contribution is that of the term with i = 1 and j = n, which is (~2G (k + 1)··· (k + n -I)Gl n ' To find GIn, note that GIn

lim f(x)(x

x-+l

I-x a

-It = (_1)n x-+l lim TI 1 . 1=1

Ikll+"'+lknl::om

Xlkll+'+lknl yn ( )n = '"' _ . '"' Xlkl I-X ~I-X ~ n~O

kEZ

(1+ ~)n = _1_,", (y, I-X

I-X~

n~O

I+x)n I-X

Since F(X, Y)

= F(Y, X), it is clear that f(m, n) = f(n, m) for all m, n.

o 3. For n 2: 3 and A C {I, 2, ... , n}, say A is even if the sum of the elements of A is an even number. Otherwise, we say that A is odd. By convention, the empty set is even. a) Find the number of even, respectively odd subsets of {I, 2, ... , n}.

X '

Combining the previous observations yields the equivalent (n-~~:;

Xmyn

m,n~O

= Lyn n~O

b) Find the sum of the elements of the even, respectively odd subsets of{I,2, ... ,n}. an

for Yk·

Romanian TST 1994

342

Chapter B. Formal Series Revisited

Proof. The generating function for the sums of the elements of the subsets of {I, 2, ... , n} is n~=l (1 + Xi), because of the obvious identity n

II (1 + Xi) = L

B.l.

343

for even n. By induction, we deduce that an = bn for all n and then solving the previous recursive relation (which becomes an = 2a n -l + n . 2n- 2) yields an = 2n- 3 n(n + 1). We continue with a very nice problem, though quite simple.

xm(A) ,

i=l

Counting problems

where meA) = I:aEA a. Let E(n), respectively O(n) be the sets of even, respectively odd subsets of {I, 2, ... ,n}. Taking X = -1 in the previous identity yields = IE(n)I-IO(n)l. Since we clearly have IE(n)1 + IO(n)1 = 2n , we deduce that the number of odd, respectively even subsets is 2n-l. To answer the second question, we differentiate the identity, to get

4. How many polynomials P with coefficients 0, 1, 2, or 3 satisfy P(2) = n, where n is a given positive integer?

Â°

L iXi- II(1 + xj) = L m(A)Xm(A)-l.

Romanian TST 1994

Proof. Let an be the number of such polynomials. Then an is also the number of solutions of the equation XO+2Xl + .. ,+xk2k = n in integers Xi E {O, 1,2, 3}, for varying k. Thus, the generating function is

L anXn = II (1 + X2k + X 2.2k + X

j#i

i=l

n20

Taking X = -1 yields

) .

k20

This is also equal to

meA)

AEO(n)

meA).

II n20

AEE(n)

On the other hand, taking X = 1 gives

meA)

AEO(n)

1_ X

2n +2

I_X4

1- X2 n

I-X

I_X8 I_X2

which simplifies drastically to

meA) = n(n + I)2n-2.

~ anX = (1 _ X)(I- X2)"

'"'

AEE(n)

We deduce that the answer to the second question is n(n quantities involved.

+ I)2 n- 3

n20 for both 0

Now, we can decompose the last rational fraction in simple elements, following the standard procedure. A simple computation yields

Remark 8.2. Here is an alternative proof for the second part. If an = I:AEO(n) meA) and bn = I:AEE(n) meA), then

1 X)(I - X2)

="4

1 1- X

+ 1+X

+ 2(1 -

X)2'

for n odd, as the odd subsets of {I, 2, ... ,n} are either odd subsets of {I, 2, ... , n -I} or the union of {n} and of an even subset of {I, 2, ... , n -I}. A similar analysis shows that

2an-l

+ n . 2n- 2,

bn = 2bn- 1 + n . 2n- 2

Expanding this once more, we deduce that the answer is ~ J + l. Note that we could have avoided the simple elements decomposition, because (1-X)(1-X2) is simply the generating function for the number of ways to express n as a + 2b, where a, b are nonnegative integers. There are ~ J + 1 possible choices for b, each of which leaves one choice for a. So there are ~ J + 1 ways to express n as a + 2b and the conclusion follows. 0

Chapter B. Formal Series Revisited

344

Remark 8.3. Here is another nice solution: if P(X) = ao + a1X + ... + akXk satisfies P(2) = n, define f(P) =

l~o J l

J ·2+··· + l~k J ·2k.

In the next two problems, one combines an easy combinatorial argument with a generating functions argument. Such mixtures appear very often in combinatorial problems and in this case generating functions do the dirty work of solving quite complicated recursive relations. Let us start with an absolute classic.

5. In how many different ways can we parenthesize a non-associative product X1X2··· xn? Catalan

Proof. Let an be the desired answer. Note that in a product of k, respectively n - k factors, we can put parentheses in ak, respectively an-k ways. Looking at the position at which the first parenthesis ends, we deduce the recursive relation a1 = 1 and an = L~:i akan-k· Next, define f(X) = Ln>1 anxn and observe that the recursive relation implies the chain of equalities-

= X2 + L n2:3

(I: VI -

0) that

tV 6.

2 - 2(1 -

~ - ~. L(-I)nC~2)4nxn.

4X)2

e:-=-n.

n- 1

L. Lovasz, Miklos Schweitzer Competition

Proof. The key point to obtain a recursive relation for the sequence F(n) is to 1 look at If- (1)1. If If- 1 (1)1 = j, then the j elements of f-1(1) can be chosen in (j) ways and f can be defined in F(n - j) ways on the remaining elements. Thus, there are (j)F(n - j) maps f satisfying the property in the statement of the problem and for which If- 1 (1)1 = j. Summing over all possible values of j, we cover all functions f satisfying the property in the statement, so

J=1

(~)F(n -

j) =

I: (~)F(j). j=O

Here we took by convention F(O) = 1. Considering the exponential generating function f(X) = L~=o F~~) X n , the previous relation yields

= 1 + ;(e X

1·3··· . (2n - 3) (-It. 2n = _~ (2n - 2)

Let F(n) be the number of functions f : {I, 2, ... , n} -+ {I, 2, ... ,n} with the property that if i is in the range of f, then so is j, for all j ::; i. Prove that kn F(n) = 2k+1 . k2:0

f(X)

(_I)n-1 .

and so, using the previous relation yields an = ~

aka n- k ) Xn = f(X) - X,

k=1

345

Now,

F(n) =

which implies (taking into account that f(O)

f(X) = 1 -

Counting problems

(-It. 4n = ( 1/2) n

We obtain a map f from the set of polynomials P as in the statement of the problem, to values in {O, 1, ... , ~ J}. It is easy to check that this map is a bijection.

f(X)2

B.l.

Next, we can write

n2:0

l)f(X)

f(X)

2 _1 eX.

Chapter B. Formal Series Revisited

346

Expanding now e

B.l.

Counting problems

347

On the other hand, the unique factorization of monic polynomials into products of irreducible monic polynomials yields

nX (nX)2 =1+-+--+··· I! 2!

and identifying coefficients in the previous equality yields the desired formula for F(n). Note l that we cheated a bit, since En>o does not converge X -adically, but this can be easily fixed: consider the function J(z) = E~=o F~7) zn, defined in a neighborhood of in C. This series converges absolutely and the previous computations show that J(z) = 2!e' if z is close enough to 0. But then we can expand

e;:

= II(1 + Xdeg h + X 2deg h + ... ) = II 1- xdegh' 1 h

the product being taken over the irreducible monic polynomials h. Taking the formal logarithm yields log

1 1 1 1- pX = Llog 1- Xdegh = LNnlog 1- xn' h

n~O

The desired formula is easily obtained from this equality and the classical expansion where this time the series converges in <C. Expanding enz , collecting terms and identifying coefficients yields the result. 0 We end this section with two more challenging problems. The first one is a very classical result in the theory of finite fields. 7. Let N n be the number of irreducible monic polynomials of degree n with coefficients in Z/pZ. Then for all n we have Edln dNd = pn. Proof. Write IFp = Z/pZ and consider the generating function F =

xdegJ,

JElFp[X]

the sum being taken over monic polynomials J E IFp[X]. As there are pn monic polynomials of degree n with coefficients in IFp, we have

log

1 -X =

LT' k~1

Remark 8.4. Using Mobius' inversion formula and the result of the previous problem, we obtain I",

N n = - L/-l(d)pd. n din

It is fairly easy to see from here that N n > 0, so there exists at least one irreducible polynomial of degree n over IFp' This result is absolutely not trivial. There is also an arithmetic solution of the previous problem, the idea being to prove the stronger result

II J =

Xpn - X,

JElFp[XJ, degJln

where the product is taken only over those irreducible monic polynomials J E IFp[X] whose degree divides n. 8. Let x and y be noncommutative variables. Express in terms of n the constant term of the expression (x + y + x-I + y-I)n.

1 We

thank Richard Stong for pointing this out.

M. Haiman, D. Richman, AMM 6458

348

Chapter B. Formal Series Revisited

Proof. Note that variables cancel in variable/inverse pairs so the constant term is zero for n odd. Let am,n be the number of products UI U2 ... Un where each Ui is one of {x,y,x-I,y-I} and after cancellation we are left with a word of length m. Note that aO,n is the desired constant term. By convention we set a-I,n = O. Starting from a word of length m > 0 there are three ways we can add one more term on the right and produce a word of length m + 1 and one way we can add a cancelling term and produce a word of length m - 1. Therefore we have

B.2. Proving identities using generating functions

we can write this as 1

6TJl-12T

An = (3X ,

+ X-It + L

1 )

+ 18T

a(T).

2J1- 12T - 1 1-16T

a(T) = :-1-:-+-:::2-v'-;=;;1=-==1;::::::2:;::::T

X-I).

aO,2r =

This is a first order linear difference equation in An and it can be solved by standard techniques to give n-I

1 1 - 6T ( J1 - 12T - 18TJ1 - 12T

Extracting the coefficient of T r using the binomial formula gives

Letting An = L:~=o am,nxm we can rewrite this as

+ X-I )An + ao,n(X -

Solving for a(T) and simplifying gives

3am-1 n + am+! n m =I- 1 amn+1 = { ' , , 4aO,n + a2,n m = 1

An+1 = (3X

349

2k=O t 24(r-k) (-12)k (1/2) _24r k r 24(r-k)+23k (2k - 2) 4r 2 - L . k=1

aO,n-m-I(X - X- I )(3X + X-I)m.

k- 1

m=O

By definition it is clear that An is a polynomial in X, therefore the coefficient of any negative power of X in this series must be zero. In particular, setting n = 2s + 1 and looking at the coefficient of X-I gives

Forming generating functions of this identity gives

8.2

Proving identities using generating functions

Generating functions are a very powerful method for proving combinatorial identities. Namely, we compute the generating functions of both sides of the identity we are given and we prove that they are the same. Here are a few examples. 9. Prove that for all positive integers n,

~ (n+k-1)

Let a(T) = L:~o ao,2rTr. Recognizing the other sums via

L k=1

2k-1= F2n ,

where Fn is the Fibonacci sequence (with

H = F2 = 1). Iranian Olympiad 2008

Chapter B. Formal Series Revisited

350

n2':l

~ (n + k ~ 2k-1 k=l

= " " Xn (n

k2':l

n2':k

+k 2k-1

351

Proof. Let f(n, k) be the desired sum of products and consider the generating function

Proof. The generating function of the left-hand side is "

B.2. Proving identities using generating functions

1) .

g(X) = Lf(n,k)X n = L

On the other hand, we have

L a1Xal ..... ak xak = al,a2, ... ,ak2':O

n(n+k-1) =Xk,,(n+2k-1)Xn=Xk(1_X)-2k ~X 2k _ 1 ~ 2k - 1 ' n2':k n2':O

_ X - ( (1 _ X)2 )

by the binomial formula. So the generating function of the left-hand side is

"k -2k X ~ X (1 - X) = X2 _ 3X

_ k -2k - X (1 - X) .

Expanding (1- X)-2k thanks to the binomial formula and using the previous relation, we deduce that

+ 1.

k2':l

It remains to compute the generating function of the sequence F2n' This is very easy, since we know that Pn = x:=f, where x, yare the roots of the equation t 2

t - 1 = O. Thus

f(n,k) = (_1)n-k(-2k) == (n+k-1). n- k 2k-1 On the other hand, it is easy to check that

2 " X np2 _ _ 1_ ( x X _ y2X ) ~ n- x-v 1-x2X 1_y2X n2':l

from where the result follows.

(x -y)(l - x 2X)(1 - y2X)' Since x

+ y = 1, xy = -1, we easily deduce that

Remark 8.5. Here are a few other such identities that can be easily proved using generating functions:

(1 - x 2X)(l - y2 X) = X2 - 3X + 1 D

and the result follows.

10. Let nand k be positive integers. For any sequence of nonnegative integers (al, a2,"" ak) adding up to n, compute the product ala2'" akÂˇ Prove that the sum of all these products is n(n 2 _ 12)(n 2 - 22 ) ... (n 2 - (k _1)2)

(2k - 1)!

a) Lal +a2+ .. + ak=n ala2 ... ak = P2n , where the sum is taken over all ordered partitions of nand Fn is the nth Fibonacci number.

We continue with a very nice combinatorial identity, for which we give two proofs: a natural one using generating functions and a more subtle, purely combinatorial one.

Chapter B. Formal Series Revisited

352

11. Let m, n be positive integers with m 2: n, and let S be the set of all sequences of positive integers (al, a2, . .. , an) such that al +a2+·· ·+an = m. Show that

l a1 2a2 .. . nan

Proving identities using generating functions

353

Multiplying this by 1 - iX and taking X ---+ yields the expression of Ai and a trivial computation shows that An = (-1 )n-i (7) in . 0 Proof. Define k = m - n and let T be the set of sequences of nonnegative integers (b l , b2, ... , bn ) such that bl + b2 + ... + bn = k. The substitution bi = ai - 1 transforms the desired identity into

= i)-lt-i(:}m.

(al, ... ,an)ES

B.2.

t=l

Palmer Mebane, USA TST 2010 Proof. The solution using generating functions is rather straightforward. Indeed, note that

~ Xm. C+.~,~m 1"'2"'·· 'n".)

~",x"'(2x)a, ... (nX)a.

a ...

X 1- X

2X 1- 2X

=--.---

nX I-nX

On the other hand, we have

Thus, it is enough to prove that n! (1 - X)(1 - 2X)··· (1

nX) =

~(-1)

().n

n-i n z i ' l-iX'

But the theory of simple fractions decomposition shows that there exist A l , A2"'" An such that n! (1 - X)(1 - 2X)··· (1- nX)

To prove this, we will count in two ways the colorings of k + n objects with n colors such that each color is used at least once. The first method of counting is to line up the objects in some order and to consider the first appearance of each color. Suppose object Ci is the first appearance of the ith color (in order of occurrence). Let bi = Ci+l - Ci - 1 (with Cn +! = n + k + 1) be the number of objects between Ci and Ci+l. Any choice of (b l b2, ... , bn )' with bl + b2 + ... + bn = k also gives a valid choice of the Ci'S via Ci = i + L:~=l bj. Now for the bi objects in between the first appearance of color i and color i + 1, there are i ways to color each one since only i colors have appeared so far. Therefore, for each choice of (b l , b2 ,· .. bn ) the number of ways is 1b12b2 ... n bn . Summing over all possible cases and taking into account that the order the colors appear in can be rearranged in n! ways, we get n! L:T 1b12b2 ... n bn ways to color our objects. On the other hand, there are nk+n ways to color k + n objects with n colors and there are (n~l)(n - 1)k+n ways to do the coloring with n - 1 of the n colors, (n~2) (n - 2)k+n ways with at most n - 2 colors and so on. A standard inclusion exclusion argument shows that the number of admissible colorings is j

(~)nk+n -

(n:

1) (n -

l)k+n

+ ... + (-It- 2 (~) 2k+n + (-It- l

(7)

~ 1- iX·

and we are done.

Chapter B. Formal Series Revisited

354

8.3

Recurrence relations

Generating functions are an extremely powerful tool to deal with the complicated recurrence relations that appear quite often in enumerative combinatorics. The reader is warned that the problems in this section are rather technical and difficult. 12. Let Al =

0, BI

{O} and

An+l = {I + xl

x E B n },

Bn+! = (An \ Bn) U (Bn \ An).

Find all positive integers n such that Bn = {O}. Chinese Olympiad

Proof. Using the relations given in the statement of the problem, it is immediate to check that

B.3. Recurrence relations

355

Explicitly finding Pn might be tedious, however we do not need to do so since we only care about when Bn = {O} or equivalently Pn(X) = 1 or Qn(T) = l. It is easy to see from the formula above that this occurs if and only if n is a power of 2. More explicitly, from Legendre's formula it follows that the smallest k > 0 for which (~) is odd is k = 2v2 (n). Therefore Qn(T) has degree n - 2v2 (n) as a polynomial in T and Pn(X) has degree! (n - 2v2 (nÂť) as a polynomial in X. Therefore Qn(T) = 1 if and only if n is a power of 2. 0

Proof. As above, we consider the sequence of polynomials defined by Po = 0, PI = 1 and Pn(X) = Pn-I(X) + XPn-2(X) and we observe that, modulo 2, the coefficient of Xk in Pn(X) is simply IBn (k). We consider the generating function P(X, Y) = L:~=o Pn(X)yn and observe that (1 - Y - Xy2)p(X, Y)

= Po(X) + (PI (X) - Po(X))Y

+ I)Pn(X) - Pn-I(X) - XPn_2(X))y n = Y, n=2

where lA(X) = 1 if x E A and 0 otherwise. This relation suggests considering the sequence of polynomials defined by so we get

Looking at the two recursive relations, it is clear that, modulo 2, the coefficient of Xk in Pn(X) is simply 1Bn(k). To solve this recursion, it is convenient to introduce a new variable T and set X = T + T2. To avoid confusion let Qn(T) = Pn(T + T2). Then this relation becomes

P(X, Y) = 1 _ Y _ Xy2 00

:L ym+l (1 + xy)m m=O

I: f

m=Oj=O This relation is easy to solve mod 2 giving

I::L (n -~ -1)

n=O It is obvious either from the original recursion or because Qn(T) == Qn(T + 1) (mod 2), that Qn(T) can actually be written mod 2 as a polynomial Pn(T+T2).

(~)Xjym+j+l

Xjy n.

Thus the problem is reduced to finding all n such that (n-;-l) is even for all n > j > O. We will prove that this happens if and only if n is a power of 2.

Chapter B. Formal Series Revisited

356

Consider the representation of n in the form o2 r where 0 is an odd number. If n is not a power of 2, then 0 2:: 3 and we have (n-~~-l) is odd. If n is a power of 2, then we need to prove that (n-j-l) is even 'no matter which j we choose. Using the Legendre formula we are done if we can find an m for

J l

J l-im J.

which n-;:tn- 1 > n-ij,-l + We can choose one more than the binary logarithm of the highest power of 2 that divides j. Thus the conditions are that n is a power of 2. 0 The following problem is very technical and its solution is too. But these kind of problems appear quite often in real life and we think it is rather important to be able to deal with them. 4113. Suppose that ao = al = 1 and (n + 3)an+1 = (2n + 3)an n 2:: 1. Prove that all terms of this sequence are integers.

+ 3nan-l

for

KaMaL

Proof. The first step is to consider the generating function

B.3. Recurrence relations

Replacing these expressions in the previous equality and collecting similar terms, we end up with the differential equation

!'(X)(X - 2X2 - 3X 3 )

n~O

v1-

And now? It is absolutely not c~ear that the coefficients in the Taylor expansion of f are integers! The tricky point is to write (1 - X - 2X2 f(X))2 = 1 - 2X - 3X 2 ~ 1 + (X - 1)f(X)

n~l

Now, since f'(X) Ln>l nanX n- 1 , it is very easy 'to express each of the previous sums in terms orf and f'. More precisely, a straightforward computation yields

= !'(X) +

an+2 = an+l

2 X(f(X) - 1) - 3,

+ 3f(X) -

3 and

f(X) =

1- X ( 2X2 1-

n~l

akan-kÂˇ

~ 00

4X2 ) 1 - (1 _ X)2

2k (2k) X t; k + 1 k (1 - X)2k+l

n~l

= 3Xf(X) + 3X 2!'(X).

And since the first terms are integers, the previous relation shows inductively that all terms of the sequence are integers. Another elegant way, found by Richard Stong, is to note that

L3nan- 1 X n

= O.

k=O

n~l

L(2n + 3)an Xn = 2X!'(X)

+ X2 f(X)2

= L(2n + 3)an xn + L 3nan _ 1 X n .

L(n + 3)an+1Xn

3X - 3X2) - 2 = O.

Now comes the technical part, which we will skip: solving this differential equation. There are standard methods to solve this, but unfortunately when one uses them in this case, one obtains rather horrendous expressions. Thus, we leave to the reader to convince himself that the resolution of this equation yields 1- X 2X - 3X2 f(X) = 2X2

Multiplying by xn the recursive relation and adding these relations yields

n~l

+ f(X)(2 -

And now we are saved, because if we identify the coefficients in the previous relation, we obtain the following recursive relation

f(X) = LanX n .

L(n + 3)an+1Xn

357

t; + 1(2k) (2:)]

[LnI2J

Xn.

358

Chapter B. Formal Series Revisited

e:) e:) -

and that -d:-r = (k~l) is an integer (this is the famous Catalan number). We leave as a challenge to the reader to prove directly from the recursive relation that

B.3.

Recurrence relations

Proof. Considering the generating functions A(X) = l:anX n ,

B(X) = l:bnX n ,

n~2

an+2 = an+l

359

+ l: akan-k

n~2

the recursive relation can also be written in the form

k=O A(X) = XB'(X)

or that

Ln/2J

an =

+ A(X)B(X),

from where we obtain the fundamental identity

(2k) ( n ) k +1 k 2k' 1

He will probably appreciate the power of generating functions (as far as we know, the proofs by induction are hideous, to say the least ... ). 0

Remark 8.6. The numbers appearing in the previous problem are called Motzkin numbers and they have nice combinatorial interpretations. Here is one of them: a Motzkin path is a lattice path from (0,0) to (n,O) with steps (1,0), (1,1) and (1, -1), never going below the x-axis. Then anis the number of Motzkin paths of length n. This is a beautiful and not obvious exercise that we leave to the reader. Another good exercise is to prove that an is also the number of paths on N with n steps, each step being -1,0 or 1, starting and ending at O. To make everything precise, let us recall that if A is a subset of zn, then a lattice path of length l from x E zn to y E zn, with steps in A is a sequence Vo = x, VI, ... ,VI = Y of elements of zn such that Vi - Vi-l E A for all i. We give three different solutions for the following beautiful problem. The first two proofs are a bit technical, but quite natural. The third one is very short and elegant, but not very natural. "14. Consider (bn )n::?.1 a sequence of integers such that b1 = 0 and define al = 0 and an = nbn + albn- l + ... + an-Ib l for all n 2': 2. Prove that pJap for any prime number p.

KaMaL

A(X) _ XB'(X) _ d - 1 - B(X) - -X dX log(l - B(X)). Now, we have the following very useful result:

Lemma 8.7. Any fEZ [[X]] with constant term 1 can be written in the form 00

f =

II (1- aiXi)-1 i=l

with ai E Z. Proof. Write f(X) = 1 + iIX + hX 2 + ... for some integers Ii- ~ooking at the coefficient of X we obtain that al = iI. In general, we find an III terms of al, ... , an-l by imposing the condition that

This expresses an as a polynomial with integer c~efficients in al, ... , an- l and iI,ÂˇÂˇ., fn, from where the conclusion follows immediately by induction. 0 Using this result, let us write 00

1 - B(X)

II (1 i=1

CiXi)-1

I . b '=(-")C1i

Chapter 8. Formal Series Revisited

360

- J -

8.4. Additive properties

L Q'W\ . 'h'\.

r• ~., I """

1:(

with

361

we deduce that

E Z. Then

f i=2

= - L(iCiX

+ ic;X2i + ... ).

i2:1

Since a p is the coefficient of XP in A(X), the only contribution comes from the terms with i = 1 and i = p and since the contribution for i = p is clearly a multiple of p, it is enough to check that CI = O. But this is clear, since 1 - B(X) == 1 (mod X2). D Proof. Let us modify the definitions of the generating functions a bit, by imposing different initial terms. Namely, define ao = al = O,bo = -l,bl = 0 and 00 00 A(X)

LaiXi,

B(X)

-biXi.

A(X)B(X) = -XB'(X), -A(X) X

= log ( 1 -

(b2 X2

i=l

+ b3.X3 + ... )Z Z

Proof. Let Xl, X2, ... , xp be the roots of the polynomial

Comparing the identities al = bl and an = nbn + albn- l n 2: 2 with Newton's identities, we easily deduce that

for every i E {I, 2, ... ,p}. On the other hand, bl = Xl

(b2 X

+ b3X 3 + ... )) .

X2 X3 10g(1- X) = -X - - - 2

+ X2 + ... + xp =

+ ... + an-Ib l

for all

implies that

The desired result follows then from corollary 9.15.

8.4

B'(X) B(X)'

Integrating both sides we get 00 Xt' ""'" -ai L.. - ,i=2 z

Looking at the coefficients of XP in both sides shows that a p is the coefficient E 23' S·Illce the denominator ofP this coefficient is 2 J (b2X +b3 X +,.)' of X P'III ",L L..i=l i . D clearly relatively prime to p, it follows that plap .

Then the recursive relation satisfied by the sequences implies that

so that

az~z

i=O

Since

O~_I

Additive properties

When studying additive properties of sets A c Z, it is very useful to consider the generating function f = LaEA xa. For instance, the square of f encodes the number of solutions to the equation a + b = n, with a, b E A. This rather innocent-looking observation yields quite a lot of nontrivial results and the purpose of this section is to present a few of them. Quite often, it is more convenient to reduce the generating function modulo suitable primes, as this simplifies a lot the computations: it is a very fortunate feature of lFp that (1 + X)P = 1 + XP in lFp(X].

362

Chapter 8. Formal Series Revisited

8.4Âˇ Additive properties

363

lV15. Let A be a finite set of nonnegative integers. Define a sequence of sets by: Ao = A and for all n ~ 0, an integer a is in An+l if and only if exactly one of the integers a -1 and a is in An. Prove that for infinitely many positive integers k, Ak is the union of A with the set of numbers of the form k + a with a E A.

Suppose that this is not the case and note that the partition hypothesis yields an identity of formal series

Putnam Competition 2000

However, the right-hand side has a pole at a primitive a~h root of unity, while the left-hand side definitely does not have such a pole. 0

Proof. The key point is to note that the definition of An+l in terms of An can be expressed algebraically by

1 n n _=~~xaik+bi=~

1- X

L..t L..t

Xbi

L..t 1 - Xai

i=l k2:0

i=l

The following problem is much trickier.

~ 17. Let p be a prime and let n ~ p and aI, a2,"" an be integers. Define 10 = 1 and Ik the number of subsets B C {I, 2, ... ,n} having k elements for all n, the equality being in lF2[X], We deduce that

and such that p divides L:iEB ai. Show that 10 is a multiple of p.

II + h - ... + (-l)n In Saint Petersburg 2003

for all n. On the other hand, the condition that An is the union of A and A + n can be also written (when n > max A) in the form

L aEAn

X a = (1 +xn)

Proof. Adding to all of the ai's a suitable large multiple of p (which does not affect the hypothesis or the conclusion), we may assume that the ai's are positive. The point is to consider the remainder of

L xa .

aEA

Thus, it suffices to find infinitely many n such that (1 + x)n = 1 + xn in lF 2 [X]. Simply take for n a power of 2 greater than maxA and we will have An=AU(A+n). 0 The following problem is an absolute classic. 16. Prove that if we partition the set of nonnegative integers into a finite number of infinite arithmetic progressions, then there will be two of them having the same common difference.

k=O

xm(B)

E lFp[X]

BcA IBI=k

modulo XP - 1, where m(B) is the sum of the elements of B. On the one hand, the hypothesis n ~ p and the fact that 1 - X divides 1 - Xai imply that 1 - XP = (1 - X)P (remember that we are working in lFp[X]) divides I(X). On the other hand, we have XN == XM (mod XP - 1) if N == M (mod p), thus

I(X) ==

I: j=O

Proof. Suppose that N is partitioned into the arithmetic progressions aiN + bi , with ai, bi ~ 0 and al :S a2 :S ... :S an. We will actually prove that an-l = anÂˇ

L (_l)k L

(t(-l)k k=O

L IBI=k,m(B)=j

(mod p)

(mod XP - 1).

Since this polynomial is 0, it follows that its constant term is 0 in lFp . But this is precisely saying that p divides 10 - II + h - ... + (_1)n In. 0

Chapter 8. Formal Series Revisited

364

The following is also a rather tricky problem. We found it in the wonderful little book [62]. It was also proposed in a Chinese Team Selection Test in 2002.

,"18. For which positive integers n can we find real numbers such that

all

a2,· .. , an

8·4· Additive properties

365

Proof. It is easy to build examples with n = 1,2,3, and 4. Suppose n 2: 5 and without loss of generality that al < a2 < ... < an. Then the largest difference must be an - al = (;) and since an-al

n-l n-l L(ak+l-ak) 2: Lk= (;) k=l

Proof. It is clear that if ai are such numbers, then all their differences are integers, thus we may actually ass,-!me that they are integers. Let f(X) = Xal + X a2 + ... + X an , so that the hypothesis can be written

G) G) XG)+1 _ x-G) f(X)f (~=n+t;xi+t;x-i=n-l+ X-I . ) The point is to look at values of f at points on the unit circle, since for 1 we have f (~) = fez) = fez), thus f(z)f (~) = If(z)j2 2: O. We deduce that for any z = eix we have

Izl =

n -1

Z(~)+l - z-(~)

z-1

> O. -

However, an easy computation shows that this is equivalent to

n - 1+

sin(n 2

. x sm

+ 1)

k=l

we see that the differences ak+l - ak for 1 ::; k ::; n -1 must be a permutation of 1, 2, ... , n 1. In particular, all differences of ai's with indices two apart are at least n. Suppose ak+l - ak = 1, then if they are present, both ak+2 - ak = 1 + (ak+2 - ak+l) and ak+l - ak-l = 1 + (ak - ak-I) would be at most n, a contradiction since they are at least n and distinct. Thus the difference of 1 can only occur at one of the ends a2 - al or an - an-l and it must be adjacent to the difference of n - 1. Since this accounts for the difference of n all remaining differences of ai's two apart must be at least n + 1. Therefore repeating the above argument shows that the difference of 2 must also occur at one of the ends and can only be adjacent to the difference of n - 1. But this is impossible since the ends are at least n - 1 2: 4 apart. 0 The following gem is one of our favorite problems .. 19. Find all positive integers n with the following property: for any real numbers aI, a2,···, an, knowing the numbers ai + aj, i < j (but not knowing which number corresponds to which sum) determines the values aI, a2,·.· ,an uniquely.

2: O.

Selfridge and Straus

We will take x such that (n 2

+ IH

3; to deduce that

1 2(n2-n+l) n - 1 >. 37r > 31f . sm 2(n L n+l)

However, it is immediate to check that the last inequality cannot hold unless n ::; 4. And indeed, for n ::; 4 such sequences exist. For n = 1,2 things are clear, while for n = 3 and n = 4 one can take the sequences 1,2,4, respectively, 1,2,5,7. 0

Proof. We will eventually prove that the answer is: all positive integers but the powers of 2. First, let us give a nice counter-example when n = 2k. Let Al = {0,3} and Bl = {I, 2} and define inductively Aj+l = Aj U (2)+1

+ B j ),

. 1 Bj+1 = B j U (2)+ + Aj

It is an amusing exercise left to the reader to check that A j , B j have 2j elements and that

366

Chapter 8. Formal Series Revisited

for all j. Actually, Aj and Bj are precisely the sets of numbers having at most j+l digits when written in base 2 and whose sum of digits is even (respectively odd). All this can be easily proved by induction and shows that Ak, Bk are a counter-example for n = 2k. Let us prove now that if n is not a power of 2 and if the collections (ai + aj)i<j and (b i + bj)i<j are identical, then the collections (ai)i and (bj)j are identical. This is the hard part. We may always assume that ai, bi are positive real numbers, by adding to each of them the same large integer. Assume first that the ai's and bi's are integers and consider the polynomials n n I(X) = LXai,

g(X) = LXbi.

i=l

Â°

(X2 - l)kp(X2),

which, after division by (X - l)k, can be written as p(X)(f(X)

+ g(X)) =

inequality, any equality ai have qi

+ qj

(r i

+ r j)

+ aj

= bi

= qi - pai

+ bj forces

+ qj -

paj -

+ qj

h -

= ri

+ rj (indeed, we

pbi ) - (r j - pb j )

and each term has magnitude smaller than 1/4). Hence the two collections (qi + qj)i<j and (ri + rj)i<j are the same and so (by what has already been done) we have an equality of polynomials rr~l (X - qd = rr~=l (X - rd. This can also be written as

IT (X i=l

+ pai -

qi) =

IT (x _ + bi

pbi - ri) .

i=l

Â°

If 1 = g, then we are done, so assume that h = 1 - 9 is not the zero polynomial. Write h(X) = (X - l)kp(x) for some polynomial p with p(l) f. and some k ~ 1 (note that 1(1) = g(l) = n). Then

+ g(X)) =

367

Additive properties

By taking E smaller and smaller, we obtain sequences Un, Vn which converge to and such that rr~l (X - ai + un) = rr~=l (X - bi + v n ) for all n. It is clear that this forces Il7=1 (X - ad = rr~=l (X - bd, which is the desired result. 0

i=l

Then the hypothesis implies the equality

(X - l)kp(X)(f(X)

8.4.

+ l)kp(X2).

Taking X = 1 in this identity and dividing by p(l) f. 0, we obtain that n = 2k-l. Since this is a contradiction, it follows that 1 = 9 and the result is proved for integers. It follows trivially that the result also holds for rational numbers (by multiplying them by a suitable positive integer we can make all of them integers). In order to prove the general case, we will use Dirichlet's approximation theorem. Suppose that n is not a power of 2 and that the two collections of numbers (ai + aj)i<j and (bi + bj)i<j are the same. Pick any E < ~. By Dirichlet's approximation theorem, we can find integers p, qi, ri such that p f. 0, Ipai - qil < E and Ipbi - ril < E for all 1 ::::; i ::::; n. By the triangle

The following alternative proof is a very elegant argument due to Selfridge and Straus ([70]): Proof Assume that n is not a power of 2. Let us prove that the knowledge of the multiset (ai + aj )i<j uniquely determines the symmetric elementary sums of the ai's (this in turns uniquely determines the multiset (ai)i, yielding the desired result). Using Newton's formulae, it is enough to prove that we can recover the sums Sk = a} + a~ + '.. + a~ from the multiset (ai + aj)i<j, and this for every k (it would be enough to take k ::::; n). Note that L(ai

+ aj)k =

i=h

(ai

+ aj)k -

2k Sk

l:<:;i,j:<:;n

t (~)a:-laj

k - 2 Sk

i,j=ll=O

t (~)Sk-ISI 1=0

k - 2 Sk

368

Chapter 8. Formal 8eries Revisited

As 2n-2 k i- 0 for all k, the previous relations show (by induction on k) that 8k can be uniquely determined as a linear combination of the sums ~i>j(ai+ajt and 80,81 , ... , 8k-1' This is precisely what we needed. . 0 The last two problems of this chapter are very hard. The first one requires a few facts about polynomials over finite fields, for which we refer the reader to the addendum 9.A.

8.4. Additive properties

369

Now, consider the permutation z -+ z2 of the roots of x2n-1 - 1. If C 1 , ..• ,Cl are the cycles of this permutation, the irreducible factors of 2n 1 X - - 1 are the polynomials I1z EC i (X - z). We deduce that the existence of S is equivalent to: for any root z i- 1 of X 2n -1 - 1, there is no k such that k z2 = 1/ z. This is equivalent to (x 2n-1 - 1, X2k+1 - 1) = X-I for all k 2': 0 and so therefore gcd(2n - 1, 2k + 1) = 1 for all k 2': 1. Hence the least s such that 2n - 1128 - 1 is odd and writing s = 2k + 1 we are done. 0

'\f20. Prove that there exists a subset S of {I, 2, ... ,n} such that 0, 1, 2, ... , n-l all have an odd number of representations as x - y with x, y E 8, if and only if 2n - 1 has a multiple of the form 2· 4k - 1.

Rather strong analytic skills are required to prove the following result, which we found in chapter III of the excellent book [62]. It is there considered an "appetizer."

Miklos Schweitzer Competition

21. Let A be an infinite set of positive integers. Let Xn be the number of pairs (a, b) E A x A such that a < b and a + b = n. Prove that the sequence (xn)n is not eventually constant.

Proof. Define f(X) = ~8ES X8-1. The fact that S satisfies the property in the statement is equivalent to the following equality in IF 2 [X]

Donald J. Newman

Indeed, the coefficient of xa in f(X)f (*") is the number of representations of a as a difference of two elements of 8. Conversely, if we can find a polynomial f E lF2[X] satisfying the previous equality, we can define a set 8 with the desired property simply by saying that s E 8 if and only if X 8-1 has coefficient 1 in f. Thus, we must find n for which one can find f E lF 2 [X] satisfying the previous relation. A first crucial remark is that 1 + X + ... + X 2n - 2 has no multiple root in the algebraic closure of lF2. Indeed, this is already the case with X 2n - 1 - 1, since this polynomial is relatively prime to its derivative. Next, the roots of f are precisely the inverses of the roots of X n - 1 f (*"). We deduce that no irreducible factor of 1 + X + ... + X 2n - 2 can vanish at z and l/z for any root z of 1 + X + ... + X 2n - 2. But conversely, if this property holds, we can find such f. Indeed, we can then pair the irreducible factors of 1 + X + ... + X 2n - 2 such that in each pair the roots of the first polynomial are the inverses of the roots of the second polynomial. One can then take for f the product of the first components of these pairs.

Proof. Consider the generating function f(X) = ~aEA xa. Note that fez) converges for alllzl < 1 and z -+ fez) is a continuous function on the open unit disk. Moreover, the hypothesis that Xn is eventually constant is equivalent to the existence of a constant c and of a polynomial P such that f(X)2 - f(X2) = 1 ~ X

+ P(X).

We will prove that this cannot happen. The idea is to look at the behavior of f on the real axis, close to 1 and then at its average behavior on circles of radius tending to 1. To avoid introducing too many functions, we will write f » g (when f,g are positive functions on (0,1)) if there exists a constant c > 0 such that fer) > cg(r) for all r in a neighborhood of 1. In particular, we have IPI « 1 and since f(x2) > 0 for 0 < x < 1, we deduce tha;t f(x)22': + P(x) and so f(x) » So f grows quite fast on the real axis, near 1. Now we integrate the relation satisfied by f on circles of radius close to 1. Using the triangle inequality, we deduce that for all r E (0,1) 27r 1 r27r If(re ix )1 2dx ::; 21 r If(r 2eiX )ldx + sup IP(z)1 + cF(r), 27r 7r Izl9

l-=x

vLx'

Chapter 8. Formal Series Revisited

370

where F(r) = 2~

8.5. Miscellaneous problems

371

Theorem 8.9. Let A be a set oj positive integers and let r(n) be the number oj pairs (a, b) with a, bE A such that a + b = n. Suppose that Jor some E > 0

II_dr~iXI' Using Parseval's identity we obtain

we have r(O)

+ r(l) + ... + r(n) n +1

Jor some constant C. Then C

On the other hand, Cauchy-Schwarz inequality and Parseval's identity yield

= C

(---i-) n"4+ E

= O.

For the origin of this result and a complete proof, see [62].

8.5

Miscellaneous problems

Putting these remarks together, we obtain the estimate:

J(r 2) ::;

The solution of the following problem is quite short, but the problem is actually quite tricky.

vi J(r 2) + CI + C2F(r)

for some constants C I , C 2 > 0 independent of r. This immediately yields J(r 2) ÂŤ 1 + F(r). If we combine this with the result established in the first paragraph, we deduce that VI~r2 ÂŤ 1 + F(r). It remains to prove that this is not the case. Note that for 0 ::; x ::; 7f we have

22. Is it possible to partition the set of all 12-digit numbers (leading zeroes are allowed) into groups of four numbers such that the numbers in each group have the same digits in 11 places and four consecutive digits in the remaining place? St. Petersburg Olympiad

Proof. The answer is negative. Assume that we found such a partition and let A be the set of all 12-digit numbers. Consider J(X) = LaEA xs(a), where s(a) is the sum of digits of a, and let G I , ... , Gk be the groups appearing in

hence

117T 11 dx . I re dx 1 r -Jr(=1=-=r=:)=;;;:2=+=4=r=x=;;;:2=::/7f~2

F(r) =::;;: "

the partition. By the condition imposed on the structure of each group we deduce that LaEGi xs(a) is a multiple of 1 + X + X2 + X 3 for any i. Thus

::; 7f } 0

= .

2~arcsinh (:~r ) .

F(r)

It follows that hmr--+I- -In(l-r) ::; "2 an

d h

1 +X +X2 +X IJ(X)

xs(a).

i=l aEG i

But we can actually compute

. ence

lim v:r=rF(r) = O.

J in a

closed form, since

r--+l-

Remark 8.8. With basically the same techniques, but with more work, one can prove the following beautiful theorem of Erdos and Fuchs:

Since this is clearly not a multiple of 1 + X + X2 it does not vanish at i), the problem is solved.

+X3

(for instance, because 0

372

Chapter B.

Formal Series Revisited

B.S.

The following problems are quite challenging. They establish congruences using generating functions. The first one is taken from [64J. See also [77J, [78J, [79J for other references and delicate congruences.

Miscellaneous problems

The following problem is similar in nature to problem 23, but more complicated. 24. Let p > 3 be a prime. Prove that

23. Let p be a prime and let dE {O, 1, ... ,p}. Prove that

2::

p-l (

2k ) d +

== r

373

(mod p),

c:)

,,0

(mod p2)

k=O where r == p - d (mod 3), r E {-I, 0,1}.

David Callan, AMM 11292 H. Pan, Z.W. Sun

Proof. The key point is to observe that

2k ) = [X- 2dJ ( k+d

(x + ~)2k X

~( L..,

k=O

2k ) = [X-2dJ

~ (x + ~)2k X

L..,

k=O

+ ~) 2p =

( XP

(1 - X) (L:2~ (2p2)Xk _ XP2) k-O k 1- X3 p2_1 2 = [XP2_ 1J 1 - X "" )Xk 1- X3 L.., k . k=O The second key ingredient is the following nice

S = [XP2_ 1 J

ElFp((X)).

Lemma 8.10. For all 1 :::::; k

< p2

C~2)

F = XP+X-p+ 1 X +X-l + 1 1- X3

(2P

+ ~p) 2 ,

we deduce that S (mod p) is the coefficient of X- 2d in X;~!~-~:il So S (mod p) is also the coefficient of X- d in

= X(1- X) (XP

1 (mod p2). The

thus

P = [X- 2d J (X + )2 -1. X2 + 1 + X-2

Now, since we are only interested in S (mod p), it is enough to understand the previous rational function taken mod p. Since in lFp((X)) we have

S =

ef),

Proof. Let S = L:~:-;/ so we need to prove that S first key point is to observe that

we have

== 2

(P:)

(mod p2).

+ X)2p2. But since 2 2 1 + X ) p == 1 + XP (mod p) (this follows by raising to the p-th power the equality (1 + X)P == 1 + XP (mod p)), we can write Proof. The left-hand side is the coefficient of Xk in (1

= (XP + X-p + 1)

X-p

(2:: X n2:0

1) 3n 1

(

2:: X

+2) .

n2:0

By inspection of the last product, the result follows.

+ X)p2 = 1 + Xp2 + pA(X)

for some A E Z[XJ. The lemma follows then by taking the square of the previous relation and comparing the coefficients of Xk. 0

Chapter 8. Formal Series Revisited

374

Coming back to the proof, we deduce that S - 1 is the same (mod p2) as twice the coefficient of Xp2 -1 in

8.6. Notes

375

Proof. Observe first of all that

which is precisely

where n = p;l. We will prove that for all nonnegative integers n we have We need to prove that Sl is a multiple of p2. Well, the good news is that 2i1T Sl = 0, since if z = eT , then by computing the generating function of the left-hand side. We have

On the other hand, since p2

== 1

(mod 6), we have

~ (~(-l)kC:) (n~k)) X" = ~(-1)'C:) ~xt~k) =L(-1)kC:)XkLxn(n~k2k) = L(-l)kC:)X k. (1_~)2k+1 n?O k?O X =1 ~ xL C:) (- (1_ X)2) k= 1 ~ X. (1- 4Âˇ (1 =~)2) ~ k?O k?O

which combined with the previous relation shows that Sl = O. The result follows. 0

The following result is very difficult. It was conjectured by RodriguezVillegas and proved in a rather difficult way by E. Mortenson. The following beautiful proof is due to R. Tauraso, taken from [83]. 25. Prove that for any p p-1

k=O

l+X Combining the previous two paragraphs yields the desired result.

> 2 we have 1

16 k

(2k) k

8.6 2

== (-1)

E.::.l 2

Notes

The following people provided solutions for the problems in this chapter: Robin Chapman (problem 24), Prasad Chebolu (problem 14), Darij Grinberg

376

Chapter 8. Formal Series Revisited

(problem 14), Xiangyi Huang (problems 18, 22), Mitchell Lee (problems 4, 10, 12), Palmer Mebane (problem 11), Greg Martin (problem 2), Fedja Nazarov (problem 21), Fedor Petrov (problem 23), Richard Stong (problems 8, 12, 18), Qiaochu Yuan (problems 1, 6, 16), Victor Wang (problem 12).

Addendum B.A

Lagrange's Inversion Theorem

In many counting problems, we start by finding a recursive relation for the number of objects we are trying to count, then we consider the generating function of that sequence (or the exponential generating function, according to the context) and establish a functional equation for it, based on the recursive relation. For instance, when counting the number of ways to put parantheses in a product of n terms, the associated generating function turns out to satisfy the easy functional equation J(z) = 1 + zJ(z)2. Of course, in this case one can solve this quadratic equation and find a formula for J(z), which in turn allows us to find the desired number of ways (which is the classical nth Catalan number). But what if the equation was J(z) = 1+zJ(z)5? Then surely such a method would not work, simply because it is impossible to solve this equation using radicals. Of course, one might argue that such an equation is unlikely to come up in enumerative combinatorics, but this is also wrong! The purpose of this addendum is to present a basic and very powerful tool for dealing with such problems (and many more), the Lagrange inversion formula. After discussing the very beautiful proof of this result, we will turn to applications, which will hopefully show its power.

B.A.1

Statement and proof of Lagrange's inversion formula

Before stating Lagrange's inversion formula, let us recall a very basic result on compositional inverses. Proposition 8.A.!' Let K be afield and let F(T) = alT+a2T2+ ... E K[[TJ] be a formal series with al -I- O. Then there exists a unique J E K[[T]] such that F(f(T)) = T and it also satisfies J(F(T)) = T. ProoJ. Let us look for solutions of the equation F(f(T)) = T having the form J(T) = blT + b2T2 + ... (it is clear that if J is a solution, then J has zero constant term). By comparing the coefficient of T on both sides, we obtain bl = all' Doing the same for the coefficient of T2 yields a l b2 + a2br = 0 and, in general, looking at the coefficient of Tn we obtain an equation of the

378

Chapter B. Formal Series Revisited

form aIbn + Gn(aI, a2,"" an, bI , ... , bn-d = 0 for some polynomial G n . This already shows that if the solution exists, then it is unique. But it also shows that a solution exists, since the previous equations can be-solved recursively. In order to prove that f(F(T)) = T, observe that f has the same property as F (namely its constant term vanishes and its linear term is nonzero). So, by the previous paragraph there is a unique g such that f(g(T)) = T. But then F(T) = F(f(g(T))) = geT) and so F = g and f(F(T)) = T. 0 We are now ready to state and prove the main theoretical result of this addendum. The proof given here (due to Hofbauer) is an adaptation of an analytic proof that would work over complex numbers. However, using the purely algebraic notion of residue of a Laurent series, one can obtain a proof over any field of characteristic zero. Recall that if f E K[[T]J, then [Tnlf(T) denotes the coefficient of Tn in f.

Theorem 8.A.2. (Lagrange's inversion formula) Let K be a field of characteristic 0 and let e be a series in K[[TJ] whose constant term is nonzero. If f E K[[T]] satisfies f(T) = Te(f(T)), then for any g E K[[T]] and any n > 0 we have [rn]g(f(T)) = ~[rn-I](g'(T)e(Tt). n

Proof. Let F (T) = e(;)' so F (T) = 0:' T + .. , for some 0: E K*, as the constant term of e is nonzero. The hypothesis f(T) = Te(f(T)) becomes F(f(T)) = T. Thus f is the (compositional) inverse of F and so f(F(T)) = T, too. If h(T) = I::nEZ anTn E K[[T]] [l/T] is a Laurent series with coefficients in K (so an = 0 for all n small enough), let res(hdT) = a-I. Obviously, for all h we have res(h'dT) = O. The key point is the following "change of variable formula" Lemma 8.A.3. If G E K[[T]][l/T], we have

res(G(T)dT) = res(G(F(T))F'(T)dT). Proof. By linearity, we may assume that G(T) = Tk, with k E Z. If k -I- -1, both G(T) and G(F(T))F'(T) are of the form g' with g E K[[T]][l/T] and

B.A. Lagrange's Inversion Theorem

379

so both terms of the equality we want to establish are zero. So, assume that k = -1. Then 2

F'(T) _ 1 F(T) - T so res (

;g:] dT)

res

(d!f)

e'(T) _ 1 ( e(T))' e(T) - T - log e(O)

and we are done again.

Corning back to the proof of the theorem and using the lemma, we can write

[Tn]g(f(T)) = res (g(f(T)) dT) Tn+I res (g(f(F(T))) F'(T)dT)

F(T)n+I

= --res(g(T)(F(T)-n)'dT). n

As u'v+uv' = (uv)', we have res(u'v) with the previous equality yields

= -res(uv') for all u, v, which combined

The result follows.

Here is an easy application of the inversion formula, which is not so easy to prove by other means: 2Note that log e(T) = ,,(_l)n-l (e(T) _ e(O) L.....t n e(O) n2: 1

exists in K[[T]], as :i~?

TK[[TJJ.

l)n

Chapter B. Formal Series Revisited

380

Example 8.A.4. Suppose that two sequences (an)n, (bn)n of complex numbers satisfy

B.A. Lagrange's Inversion Theorem

8.A.2

Two variations and some applications

In applications, one also encounters the following versions of the inversion formula. For instance, we will use the following result to prove a very nice combinatorial identity due to Abel.

Theorem 8.A.5. Let K be a field of characteristic 0 and let e be a series in K[[T]] whose constant term is nonzero. Then for all f E K[[T]] we have

for all n 2: O. Prove that

f(T) = f(O) for all n 2: 1.

Proof. Let A(X) = Ln>o anxn and B(X) = Ln>o bnxn be the generating functions of the two sequences. An easy computation shows that

B(X)

xn . ( L (n:k)ak)

= LakXkL (n:k)x

= LakXkL k

381

n=O

n k Now, substitute T = Y/e(Y) to obtain the desired result.

ak Xk (l +X)k = A(X + X2).

= Y'1+}(Y)"

Here is the promised application, which is really not easy to prove by direct computational means.

Let Y = X +X2, so that X = fey) with fey) inversion formula, we obtain

Proof. Let X = X(T) be a formal series such that X = Te(X) (it exists by proposition 8.A.1 applied to peT) = e(;Âť)' Then Lagrange's inversion formula yields

(~)xn =

+ ~ (e~)) n . ~[Tn-ll(f'(T)en(T)).

Using Lagrange's

Theorem 8.A.6. (Abel's identity) For all complex numbers a, x, y and all positive integers n,

an = [Yn](A(Y)) = [Yn](B(X)) = [yn](B(f(Y)) =

~[yn-l](B'(Y)(l + y)-n)

n =

kb k [yn-l](yk-l(l

L k

"kbk [yn-k]((l L n

+ y)-n) =

+ y)-n)

Proof. The desired equality is equivalent to

kb k ( -n ) n n k

and the result follows from the easily checked equality

( -n ) = n

(_l)n-k

(2n - 1) . kn-k

+ y)n =

~ x(x

k=O

+ ak)k-l (y k!

ak)n-k (n-k)!

Let us consider the generating functions of the two sides of the previous equality. The generating function of the left-hand side is e(x+y)T. The generating

382

Chapter B.

Formal Series Revisited

function of the right-hand side is

~ ~ x(x + ak)k-l (y - ak)n-k Tn = ~ TkX(x + ak)k-l T(y-ak) k! (n-k)! ~ . k! e . n O~k~n k Thus, by dividing by e yT , it suffices to prove that

= 1+

383

Here is a quite exotic problem: suppose that e E K[[T]J and consider the sequence an = [Tn](en(T)). What is the generating function of the sequence an? The following result answers this question in a more general context:

~ ~

exT

B.A. Lagrange's Inversion Theorem

Theorem S.A.S. Let X, Y be variables satisfying Y = X e(Y), where e E K[[T]J has nonzero constant term. Then for any F E K[[T]J we have

1 2: k!x(x + ak)k-le-akTTk. k::O:l

But this is a consequence of theorem 8.A.5 with

f(T) = exT and e(T) = eaT. Let us apply this result to prove the following nice-looking identity.

Example 8.A.7. Prove that for all n :::: 1 we have

(7)(i+1)i-l(j+1)j-l =2(n+2)n-l.

Proof. Apply theorem 8.A.5 to f(T) = JoT ~(~] du (recall that this is formal integration, so f(T) is the unique formal series vanishing at 0 and such that 1'(u)e(u) = F(u)). Using that l' = ~, we obtain feY)

i,j::O:O,i+j=n

= 2: ~(Y/e(Y)tÂˇ n::O:l

AMM E. 2828

Proof. Take a

= 1, x = 1 and y = n + 1 in the

2: z+J=n

(7) (i + l)i-l(j +

previous theorem. We obtain

l)j = (n

+ 2)n.

=2:

(7)(i+1)i-l j (j+1)j-l.

.2: z+J=n

+1-

2: ~[Tn-l](F(T)e(T)n-l)xn. n

n::O:l

F(Y) e(Y)

= .!ÂŁ fey) = '"'[Tn-l](F(T)e(T)n-l)xn-l dX. dY

n::O:l

Finally, differentiating Y = X e(Y) we obtain dY = dX . e(Y) + X e' (Y)dY, so

dX dY

z+J=n Now, observing that j = j

[yn-l](F(T)e(T)n-l)

Differentiating this equality with respect to Y we obtain

Unfortunately, this is not really what we want to prove, but it shows that we are on the right track. To get rid of the extra 1 in the exponent of j + 1, differentiate with respect to y the equality in the previous theorem and then take a = 1, x = 1 and y = n + 1. This time we end up with

n(n+2t- 1

1- Xe'(Y) e(Y)

1, we can write the last sum as Replacing this in the previous equality yields the desired equality.

(7)(i+1)i-l(j+1)j-.2: (7)(i+1)i-l(j+1)j-l. z+J=n

Combining this with the first relation, the result follows.

Example 8.A.9. Find a closed form for the generating function of the sequence an, where an is the constant term of (1 + X + n.

"* )

384

Chapter B.

Proof. Note that an previous theorem

Formal Series Revisited

[Tn](en(T)), where e(T) = T2 ~

+ T + 1.

So, by the

~ anX = 1 - X e' (Y) ,

= X e(Y).

Solving the equation in Y yields 1- X -

VI -

Once such a distribution is made, we have a n1 a n2 ... a nk ways to label the elements of the forest and so the contribution to the total number is

2X - 3X2

(n - I)! ,anlan2···ank· nl·n2···· nk. " The total contribution coming from all partitions of n - 1 is

and then an easy computation shows that ~

~anX = Vl- 2X n

8.A.3

n (n nl) (n - I)! ( ni' n2 ···=nl!n2!···nk!·

where Y

385

vertices, so ni + n2 + ... + nk = n - 1. The number of ways to distribute the n - 1 vertices different from the root in these k subtrees is

B.A. Lagrange's Inversion Theorem

3X2'

Examples from enumerative combinatorics

In this section we consider applications of the inversion formula in counting problems. We start with an absolutely classical and beautiful theorem of Cayley, but we need a series of definitions before stating it. Recall that a tree is a connected graph with no cycle. A labeled tree on the set {I, 2, ... , n} is a tree whose set of vertices is {I, 2, ... , n}. A rooted tree is a tree in which one of the vertices (called the root) is distinguished. There is a unique (nonbacktracking) path between any two vertices of a tree. The parent of a vertex in a rooted tree is the vertex connected to it on the unique path to the root. A child of a vertex v is a vertex whose parent is v. A tree is called ordered if one is given an ordering of the children of each vertex.

Theorem 8.A.IO. There are nn-2 unordered labeled trees on the set {1,2, ... ,n}. Proof. Let an be the number of unordered rooted labeled trees on {I, 2, ... , n}, with the natural convention that ao = O. It is enough to prove that an = nn-l, as clearly the number of unordered labeled trees is an/no The point is that giving such a tree is the same as giving its root and a forest of subtrees whose roots are the children of the root. Suppose that the root has k children and that the corresponding subtrees have nl, n2, ... , nk

however each configuration is counted k! times (since we do not care about the order of the children of the root) and the main root can be chosen in n ways. We finally deduce the very complicated recurrence relation

This simplifies drastically if we consider the exponential generating function T(X) = Ln2':O an -;:~, since

[xn-I]T(X)k =

an1 ... ank .

~ nl!···nk! nl+,,·+nk=n-I

Hence the recurrence relation can also be written

an = [xn-I] ~ T(X)k = [xn-I] T(X) ~ k! e, n! k

that is XeT(X)

= T(X).

Using Lagrange's inversion formula, we obtain

an = [xn](T(X)) n! from where the result follows.

= ~[Xn-l]enX = n

nn-I, n!

Chapter B. Formal Series Revisited

386

We end this addendum with two more difficult examples. The following beautiful result is taken from [65J.

Example 8.A.11. An intransitive tree on the set of vertices {I, 2, ... ,n} is a tree such that for alII::; i < j < k::; n, {i,j} and {j,k} are not simultaneously edges. Prove that the number of such trees is 1 n路 2n -

~ (n)kn-l . 1 ~ k k=l

Note that it is absolutely not clear that the above quantity is an integer!

Proof. Let Fn be the number of such trees and let F(T) = L n2':O

Tn Fn+1 - , n.

be an associated exponential generating function. Call a vertex i left if all of its neighbors are greater than i. Let Ln be the number of rooted intransitive trees on the set of vertices {I, 2, ... ,n}, whose root is a left vertex. Then Ll = 1 and we clearly have Ln = ~ Fn for n 2: 2 (n comes from the choice of the root, division by 2 takes into account the fact that the probability that the root is left is 1/2). If Tn L(T) = LL n - , n. n2':l

is the exponential generating function associated to L n , then Ln = ~Fn for n 2: 2 yields L(T) = ~(1 + F(T)). But the exponential formula implies that F(T) = eL(T), as an intransitive tree on the set of vertices {I, 2, ... ,n + I} is obtained from a forest of left-rooted trees on {I, 2, ... , n}, by connecting n + 1 to each root. Thus, we obtain F(T) = e~(HF(T)). If f(T) = T(1 + F(T)), we deduce that f(T) = T(1 + e f (T)/2). An application of Lagrange's inversion formula yields

B.A. Lagrange's Inversion Theorem

387

If we expand e kT /2 and collect terms according to the exponent of T, we finally deduce that Fn = 1 (n)k n- 1 . o n路 2n - 1 ~ k

k=l

Finally, a question by James Propp with a nice proof from [69J.

Example 8.A.12. The vertices of a polygon P with N + 2 vertices are labeled 1,2,1,2, ... in order (stopping when the end is reached). Let aN be the number of triangulations of P with no monochromatic triangle. Then aN = 2~:1 e~) n 1 2 + (3n+l) 1路f N = 2n an d aN = 2n+2 n 1路f N = 2n + 1. Proof. Define ao = 1. Suppose that N = 2n+ 1 and call a triangulation proper if it contains no monochromatic triangle. Consider a proper triangulation 7r of P. Note that P has an edge labeled 1, 1. This edge must be a side of a triangle with a vertex labeled 2. If this vertex is the ith vertex labeled 2, with i 2: 0, then the two sides of the triangle split P into a 2i + 2-gon and a 2n - 2i + 2-gon and both of these polygons are properly triangulated by 7r. By adding over all i we obtain a2n+l = L:::~o a2ia2n-2i and a similar argument for N even yields a2n = L:::;::!Ol aia2n-l-i. This suggests considering the two generating functions A(X) = La2nxn,

B(X) = La2n+lxn.

n2':l

n2':O

Then the previous relations can be written in a compact form

A(X) = 2X(1 Indeed, we have

+ A(X))B(X),

B(X) = (1

+ A(X))2.

388

Chapter B. Formal Series Revisited

and

389

where

A(X) =

L n2:1

=X . =

(2t1 L (t

aia2n-1-i) Xn = X .

aia2h+1-i) Xn

n2:0

i=O

a2ia2n-2i+1) Xn n2:0 i=O 2X(1 + A(X))B(X).

(2f1 L (t

h(IXI) =

t=O

n2:0

a2i+1a2n-2i) Xn

t=O

We deduce that A(X) = 2X(1 + A(X))3 and an easy application of Lagrange's inversion formula finishes the proof. D

8.A.4

B.A. Lagrange's Inversion Theorem

the sum being taken again over ordered partitions of X. For instance, consider the problem of counting the number of partitions (Sl, 82, ... , 8 k ) of {I, 2, ... ,n}, where each 8 i is nonempty. By taking f(n) = 1 if n 2: 1 and f(O) = 0, we deduce that the exponential generating function for the number of such partitions is

Composition of generating functions

One of the key points in the proof of Cayley's theorem 8.A.1O is to establish the functional equation XeT(X) = T(X) for the exponential generating function of the number of labeled trees. We would like to give a more abstract and general context for this kind of argument, which appears very often in counting problems. We follow rather closely the wonderful book [76] and we strongly advise the reader to take a look at the first chapter of it, which contains an impressive number of examples and problems on this topic. Let K be a field of characteristic 0 and let f, 9 : N -+ K be two sequences of elements of K. Let E f = Ln>o f (n) .:;;~ be the generating function of f. We would like to give a combinatorial interpretation of the generating functions EfÂˇ Eg and Ef 0 E g. Note that EfÂˇ Eg = Eh, where

h(n) =

t, (~)f(k)g(n

- k).

and expanding ejX yields the desired number of partitions. Suppose now that f(O) = o. We would like to understand the generating function Eg 0 E f. '

Theorem 8.A.13. Let f, 9 : N -+ K be sequences such that f(O) = 0 and g(O) = 1. Then Eg 0 Ef = E h, where h : N -+ K is a sequence such that h(O) = 1 and for all finite sets X we have

the sum being taken over all unordered partitions (81 ,82 , .. " 8 k ) (with arbitrary k) of X into nonempty subsets. Proof. We clearly have h

Lk2:0 g(k)hk' where

We deduce that for any finite set X we have

h(IXI) =

f(181)g(IX - 81) =

SeX

f(18 1I)g(1821),

(~,~)

the second sum being taken over all ordered partitions (81 ,82) of X (and 8 1 , 8 2 may be empty). By an obvious induction, we deduce that Eh . Eh ..... Efs

Eh,

11:

the sum being taken over unordered partitions with k classes. Hence it is enough to prove that Ehk = ~{. But this follows from the previous discussion and the fact that we are only considering unordered partitions here (thus the k classes may be permuted in k! ways and yield the same unordered partition). D

390

Chapter 8.

Formal Series Revisited

Example 8.A.14. Let us consider again the problem of finding the functional equation for the exponential generating function of the number of unordered rooted trees on {I, 2, ... , n}. Let T be this generating function. Then by the previous theorem X eT(X) is the generating function for the number of pairs (r, F), where r is a root and F is a forest of unordered rooted trees starting from this root. But it is clear that any unordered rooted tree arises in this way, so we actually have X eT(X) = T(X). Example 8.A.15. Suppose that we want to count rooted unordered labeled trees such that the number of children of each node is in a fixed set S, containing O. The argument used in the previous example yields the functional equation

f(X) = XL sES

cycles of (J. Then theorem 8.A.13 yields 3

Example 8.A.18. For nonnegative integers CI, C2,"" let an(cl, C2,"') be the number of permutations (J E Sn having Ci cycles of length i for all i ::::: n. Consider indeterminates X I, X 2, . . .. Then the previous example yields the following cycle-index formula

Example 8.A.16. Let E be the generating function for the number of connected graphs with vertices 1,2, ... ,n. Giving a graph with vertices 1,2, .. , ,n is the same as giving a family of disjoint connected graphs (its connected components), thus by theorem 8.A.13 the generating function for the graphs with vertices 1,2, ... ,n is eE But since there are 2(;) such graphs, we deduce that

Example 8.A.19. Let us count the number of permutations of odd order, i.e. for which all cycles have odd length. By taking Xi = 1 when i is odd and 0 otherwise, we deduce that the exponential generating function for this counting problem is

n>l

n -odd

= exp

(L2(;). ~~) .

Example 8.A.17. Consider two sequences f, g such that f(O) = 0, g(O) and define h(O) = 1 and =

~n)

Ej = exp (

Og (l

+ T)

2 10g(1 - T) )

=V +

1 T 1-T

n::::O

h(lX!)

391

f(~)S .

Using Lagrange's inversion formula, we obtain a formula for the number of such trees.

E = log

8.A. Lagrange's Inversion Theorem

(1 + T)(l - T2)-~

3Note that we also have

g(k) . f(ICII) . f(IC2 !) ..... f(ICk!) ,

CTESym(X)

where Sym(X) is the set of permutations of X and G I , G2 , ... , Gk are the

the sum being taken over unordered partitions 7r of X, since the cycle decomposition of a permutation yields a partition of X and since one can cyclically permute in (IS;I I)! ways the elements of a class with ISil elements.

Chapter B.

392

Formal Series Revisited

and an easy application of the binomial formula yields the number of permutations f(n) = (1·3··· .. (n-I))2 when n is even and f(n) = (1·3··· ·.(n-2))2· n when n is odd.

B.A. Lagrange's Inversion Theorem

393

Cayley's theorem is a direct consequence of the previous theorem and of the multinomial formula: the number of trees on {I, 2, ... ,n} is

Example 8.A.20. Let k be a positive integer and let f(n) be the number of permutations a E Sn such that a k = 1. This is equivalent to the fact that the length of each cycle of a divides k. Thus by the previous examples

E f = exp

= (1

(L :d) .

nn-2.

Proposition 8.A.22. There are (~=i). (n_I)n-k-1 labeled trees with vertices 1,2, ... ,n in which vertex 1 has degree k.

dlk

8.A.5

+ 1 + ... + 1t- 2 =

More tree-counting problems

In this section we present another proof of Cayley's theorem as well as some similar counting problems, all related to trees. The following general result is quite useful in problems concerning trees.

Theorem 8.A.21. Let dI, d2, ... ,dn be positive integers such that d 1 + d 2 + ... + d n = 2n - 2. Then the number of trees on the set {V1' V2, . .. ,vn } such . (n-2)! that vertex Vi has degree di zs (dl-1)!.(d2- 1)! ..... (dn -1)!.

Proof. This is also an easy consequence of the previous theorem. The desired number of trees is

2) .

_---.,--_---.C....(n_------"-2)_!_ _ _ = (n k+d2+ .. +dn=2(n-1) (k - I)! . (d2 - I)! ..... (dn - I)! k- 1

(n _I)n-k-1

the second equality being a consequence of the multinomial formula.

Let us introduce a very useful notion in graph theory. Proof. We will prove the result by induction on n. We may assume that d n = 1, by permuting the di's if necessary. Consider a tree on {VI, V2, ... , v n } such that deg( Vi) = d i and remove vertex Vn and the unique edge whose endpoint is Vn . We obtain a tree on {V1' V2, ... ,vn-I} whose degrees are db' .. ,dj - 1, d j - 1, d j +1, ... ,dn - 1 if Vn is connected to Vj. Conversely, any such tree on {V1' ... , Vn-1} yields a tree on {V1' ... , v n } simply by connecting Vn with Vj. It follows that there are n-1

L(d1

J=l

(n - 3)! I)! ... , . (d j - 2)! ..... (d n n-1

(

Ldk -n+I k=l

such trees and the result follows.

)

(n - 3)! Il(dk - I)!

1 -

I)!

(n - 2)! Il(dk - I)!

Definition 8.A.23. Let G be a loopless graph. A spanning forest is a subgraph without cycles and having the same vertices as G. A spanning tree is a connected spanning forest. . Here is a nice application of Abel's identity. Example 8.A.24. There are (n - 2) . nn-3 spanning trees of Kn which do not contain a fixed edge of Kn. Proof. CallI, 2, ... , n the vertices of Kn and assume without loss of generality that the fixed edge is e = 12. Let f(n) be the number of spanning trees that contain e. Such a tree appears uniquely as a result of the following process: consider two trees T 1, T2 whose vertices form a partition of 1,2, ... , n with 1 a vertex of T1 and 2 a vertex of T 2 , and then join these two trees by the edge e. If T1 has k vertices different from 1, these vertices can be chosen in (nk2)

Chapter 8. Formal Series Revisited

394

ways. Once these vertices are chosen, we have (k and (n - k - l)n-k-3 possibilities for T 2 . Thus

+ l)k-l

possibilities for Tl

n-2 ( n:2 ) (k+1)k-l.(n_k_l)n-k-3=2.n n- 3, f(n)=L

k=O

the last equality being an easy consequence of example 8.A.7. The result

Chapter 9

follows.

Remark 8.A.25. Here is another approach, suggested by Richard Stong. Let Xe be the probability that a randomly chosen spanning subtree of Kn contains the edge e. Then by symmetry it is clear that E[Xe] is the same for all edges e. Since any spanning subtree of Kn has exactly n - 1 edges we have

A Little Introduction to Algebraic Number Theory

LE[Xe] = n-1. e

Therefore since all

terms in this sum are equal,

E[Xe] =

n -1

2 =~.

Hence by Cayley's formula, there are 2Âˇ nn-3 spanning subtrees containing e and (n - 2)nn-3 that do not contain e.

This rather long chapter is concerned with elementary algebraic number theory. The techniques are rather diverse: basic linear algebra, algebraic numbers and symmetric polynomials, cyclotomy and p-adic analysis are some of the topics discussed in this chapter. Since we will use the notion of algebraic number quite often in this chapter, we end this introduction with a few recollections. For more details and some proofs, the reader is referred to the addendum 9.B. A complex number z is called algebraic if it is root of some nonzero polynomial with rational coefficients. In this case, there exists a unique monic polynomial with rational coefficients, called the minimal polynomial of z, which vanishes at z and has minimal degree. The roots of this polynomial are called the conjugates of z. The crucial property of the minimal polynomial is that it is irreducible over Q and divides any polynomial with rational coefficients that vanishes at z. A fundamental theorem in algebraic number theory states that the algebraic numbers form an algebraically closed subfield of Q, thus an algebraic closure of Q. If z is an algebraic number, we let Q(z) (or Q[z]) be the subfield of <C generated by z. It consists of all numbers of the form f (z), with f E Q[X] (or equivalently f E Q(X)). This is a finite extension of Q, of degree equal to the degree of the minimal polynomial of z. The primitive

396

Chapter 9. A Little Introduction to Algebraic Number Theory

element theorem ensures that all finite extensions of Ql are of the form Ql(z) for some algebraic number z. We call such extensions number fields. We will frequently use the notation [L : K] to denote the dimension of L as K-vector space, as well as the fundamental tower relation [M : K] = [M : L] . [L : K] for any finite extensions M / L / K. A more refined notion is that of algebraic integer. This is a complex number that is killed by some monic polynomial with integer coefficients. By Gauss' lemma, we can characterize algebraic integers as those algebraic numbers whose minimal polynomial has integer coefficients. An easy but fundamental result is that a rational number which is also an algebraic integer is necessarily a rational integer. Another important result is that the algebraic integers form a subring of the field of algebraic numbers.

9.1

Tools from linear algebra

9.1.

Tools from linear algebra

implies the existence of integers A, B, C such that Au + Bv + Cw = 1. We deduce that Aua+Bva == a (mod w). Since au+bv == 0 (mod w), we deduce that (Ab - Ba)v == -a (mod w), so that we can take p = Ab - Ba to get up == b (mod w). Also, we can immediately check that

u(Ab - Ba)

a = nw - pv, b = pu - mw, c = mv - nu. Octavian Stana§ila, Romanian TST 1989

Proof. Consider the linear system in the variables m, n, p a = nw - pv, b = pu

mw, c = mv - nu.

Trivially, the determinant of this system is 0 and the rank of its associated matrix is 2. It is thus enough to solve in integers the system a = nw - pv, b = pu-mw. This system has integer solutions if and only if there is an integer p such that vp == -a (mod w) and up == b (mod w). Now, the hypothesis

== Aub - Bua == b(Au + Bv) == b

(mod w).

Proof. . We will actually prove a stronger result: for any integers abc , , and an! I~tegers u, v, w such that au + bv + cw = 0 and gcd (u, v, w) = 1, there eXist mtegers A, B, C such that a = Bw-Cv, b = Cu-Aw and c = Av-Bu. Indeed, since gcd(u, v, w) = 1, a standard application of Bezout's lemma yields the existence of integers X, Y, Z such that Xu + Yv + Zw = 1. Let us define A=cY-bZ, B=aZ-cX, C=bX-aY. Then,

In this section we consider a few applications of linear algebra to number theory. These concern especially divisibility issues and linear diophantine equations. 1. Let a, b, c be relatively prime nonzero integers. Prove that for any relatively prime integers u, v, w satisfying au+bv+cw = 0, there are integers m, n, p such that

397

Bw - Cv = (aZ - cX)w - (bX - aY)v = a(Xu + Yv =a. Thus a = Bw follows.

+ Zw) -

X(au

+ bv + cw)

Cv and similarly b = Cu - Aw and c = Av - Bu. The result

A very nice and classical result is that ill<" "< ai-aj is an integer for . _t<J_n t-J any mtegers aI, a2,.··, an. There are many proofs of this result, at least two of them being presented in [3]. The following problem is a variation on this topic. 2. Prove that for any integers aI, a2, . .. ,an the number lcm(al' a2, ... ,an) ala2 " . an

(aj-ai)

l:Si<j:Sn

is an integer divisible by 1!2! ... (n- 2)1. Moreover, we cannot replace 1!2!··· (n 2)! by any other multiple of 1!2!·· . (n - 2)1.

Chapter 9.

398

A Little Introduction to Algebraic Number Theory

9.1.

i" Proof. Consider the matrix A = {ai,j h:S;i,j:S;n with ai,j = c-~r::i) for j ai,l = Icrn(al,a2, ai ... ,an) . "IX! vve WI'11 prove th a t

2: 2 and

",.1

det(A)

lcm(al' a2,···, an) ala2" . an

1!2!··· (n - 2)!

t:t

..•

Since the entries of A are integers, it is clear that its determinant is an integer, from which the first part of the problem will follow. Factoring an L = lcm(al, a2, ... , an) out of the first column and multiplying the i-th row by ai, it follows that 1:... 1 (a\-l) (a\'-l) a2 1

l-1) (an-l (a 2-1) n-l

1 (anI-I)

(an-I) n-l

•

lcm(al,a2, ... ,an )=n!na l,,·a n -l

_because the numbers an, ¥, i = 1,2, ... , n - 1 are pairwise relatively prime. . The result follows easily from this. 0

1 al al(al - 1) 1 a2 a2(a2 - 1)

a (aI-I) I n-2 2-1) a 2 (an-2

1 an an(a n - 1)

(an-I) an n-2

Remark 9.2. A similar result is proved in the beautiful paper [6]: if . , t hen Il i<j (2 . 1 . 1 f 2!·4!2....+·(2n)! an d ao, al,"" an are mtegers, ai - a 2) n 1 j IS a mu tIp eo

Taking out the numbers J. ~ that appear at the denominators of the binomials in each column shows that

L detA = ala2 ... a n 1!2! .. , (n - 2)!

Remark 9.1. Quantities such as Ili<j a~=;j and the one in the previous problem have natural combinatorial interpretations: they are the dimensions of some irreducible representations of special unitary groups. Of course, explaining this is beyond the scope of this modest book, but the reader should know that these are not "just some r·andom problems."

ala2'" an L

399

To see that the result is optimal, simply choose an = (n!)2 and ai = (n!)2 for 1 ::; i < n. Then

. :1:~' # .... (i .

IlI:S;i<j:S;n(aj - ai)

Tools from linear algebra

1 al 1 a2 1 an

ar - al a 22 a2 an2 - an

a n-l l a n-l 2 a nn - l

+ ... + ... + ...

Note that the (i,j) entry of this matrix can be written as Pj(ad, where Pj, o ::; j ::; n - 1 is a monic polynomial of degree j. For each column j add a suitable linear combination of the previous columns to reduce the previous determinant to 1 al a 2I a n-l l 1 a2 a 22 a 2n-l L al ... a n 1!2!··· (n - 2)! n-l 1 an a n2 an Since the last determinant is Vandermonde, the identity stated above is proved.

this result is optimal. This is also related to the dimension of some irreducible representations of the symplectic group. We continue with a nice application of linear-algebraic arguments. The ideas used in the following solutions are very useful in other contexts, too. Let p be a prime and let aI, a2, . . , , ap+l be real numbers such that no matter how we eliminate one of them, the remaining numbers can be divided into at least two nonempty pairwise disjoint subsets each having the same arithmetic mean. Prove that al = a2 = '" = ap+l. Marius Radulescu, Romanian TST 1994 Proof. Subtracting from the ai's their arithmetic mean (observe that the new numbers have the same property), we may assume that al +a2+" ·+ap+l = O. Fix some 1 ::; j ::; p + 1 and let GI, ... , Gr be the classes of a partition of {al,' .. ,ap+l} - {aj}, such that d LXEC1 x does not depend on l. Since the sum of the ai's is zero and since

Chapter 9. A Little Introductio~ to Algebraic Number Theory

400

we deduce that we have a linear relation of the form

9.1.

Let

Tools from linear algebra

ISi,ll = m

and observe that

401

L:j;i1ISi,j1 = p ==}

m and

LX==O

(mod p)

XEAi

with

IC11 < p.

Now, consider all such linear relations, obtained by making j run over all 1,2, ... ,p + 1. This gives us a linear system with p + 1 equations and p + 1 unknowns (the ai's), whose matrix has ~ on the main diagonal and numbers of the form -k with k < P elsewhere. But then the determinant of the matrix will be of the form p:t-l for some rational m == 1 (mod p). Thus, the determinant is nonzero and since the system is homogeneous, the only solution is the trivial one. This implies that all ai's are zero and the conclusion follows. 0

Proof. . First, we will reduce the problem to the case when all ai are integers. The following method is classical and very useful in a whole variety of situations: consider the vector space spanned over Q by the ai's. This is a finite dimensional Q-vector space and if we take a basis of it and write each ai as a linear combination with rational coefficients of the elements of the basis , we easily see that the coordinates of the ai's also satisfy the conditions of the problem (because by definition the elements of the basis are linearly independent over Q). Working coordinate by coordinate reduces therefore the problem to the case when all ai are rational. Multiplying all ai's by N! for some sufficiently large N reduces then the problem to the case when all ai are integers. Assume now that all ai are integers and let us prove the result by induction on max lail. The base case is obvious, so let us focus on the inductive step. Removing every element ai gives sets Si,j C {a-1' a2, ... ap +1} - {ai} = Ai nonempty, pairwise disjoint so that Isll L:xES'/"J. x is independent of j, say equal to k. Then 'tIJ

Summing over all choices of i we obtain p+1

Lai==r

(modp)

==}

ai==r

(modp)ViE{1,2, ... ,p+1}

i=l

Thus, we can write ai = pbi + r for some integers bi . But clearly {b 1 , b2, ... , bp +1 } also satisfy the conditions of the problem and moreover max Ibil < max lail. By the inductive hypothesis all bi's are equal and so all ai's are equal 0

Proof. Here's a "no formula" proof, which uses the same kind of argument, but replaces the choice of a basis in a vector space with an approximation argument: first, we reduce to the case when all ai are integers in the following way. By Dirichlet's approximation theorem there exists a large integer M such that all M ai are very close to some integer Ai. The linear equations deduced from the fact that the ai's satisfy the conditions of the problem become approximate linear equations for the Ai's. But there are finitely many such equations and each has rational coefficients. Thus, if at the beginning we ensured that M ai are sufficiently close to the Ai's, the approximate equations in Ai are actually exact. Thus the Ai's are integers satisfying the conditions of the problem. If we solved the problem over the integers, it follows that all Ai are equal. But then any two ai's are less than 2/ M apart and since M is arbitrarily large, this implies that all ai are equal. Now, let us assume that the ai's are integers. If we remove one number, the common rational arithmetic mean for the rest of the numbers cannot have p in the denominator, so the sum of all other numbers is rational with p in the numerator and, thereby, an integer divisible by p. Hence all numbers have the same remainder modulo p as their sum. Now continue as in the end of the previous solution. 0

402

9.2

Chapter 9. A Little Introduction to Algebraic Number Theory

Cyclotomy

9.2.

Cyclotomy

403

Theorem 9.6. For all n there are infinitely many primes p 2i7rk

There are cp( n) primitive nth roots of unity, namely e ----:n:-, where k relatively prime to n. Hence the nth cyclotomic polynomial

IT (X -e l<k<n

2i7rk) n

gcd(k,;:;-)=l

has degree cp( n). The splitting field Q( e 2~7r) of rPn is called the nth cyclotomic extension of Q. These polynomials and their splitting fields playa very important role in many areas of mathematics and gave rise to a whole series of very deep results. Their study would require a whole book by itself, so we decided to focus only on some very elementary and classical applications. Since any nth root of unity in C is primitive of order d for a unique din, we get the:

Proposition 9.3. (Fundamental identity) We have

Xn - 1 =

1 (mod n).

Proof. For k > n large enough (it is actually enough to take k > 2) we have rPn(k!) > 1 and so we can choose some PklrPn(k!). Since rPn(O) is 1 or 1, we have rPn(k!) == 1, -1 (mod k!), which obviously implies that gcd(pk, k!) = 1. As k > n we get Pk > k > n and by the previous theorem we deduce that Pk == 1 (mod n). The result follows. 0 For the next three problems we will use a very useful rationality result: if rand COS(T1f) are both rational numbers, then COS(T1f) E {±1,±~,0}. Let us recall the argument: 2 COS(T1f) = e Zr7r + e- Zr7r and the numbers e ir7r , e- ir7r are algebraic integers (they are roots of unity), so 2 cos( rn) is an algebraic integer. Thus, if it is rational, it has to be a rational integer and the result follows. Before passing to the next problem, let us discuss a beautiful consequence of the previous observation. We will prove that the only regular n-gons all of whose vertices are lattice points are the squares. Indeed, let A, B, C be three consecutive vertices of the polygon and observe that 1 + cos ~ = cos 2 2n = (AB2 + BC 2 - AC2)2 2 n 4AB2 . BC2 E Q.

IT rPd(X). din

This easily implies (by strong induction) that rPn(X) E Z[X] for all n. The following result is not trivial and plays an important role in many proofs concerning cyclotomic polynomials. We'll also see that a weak form of Dirichlet's theorem follows very easily from it. For a proof of Dirichlet's theorem in full generality, see addendum 7.A.

Theorem 9.4. Let a be an integer and let P be a prime divisor of rPn (a). Then either the order of a modulo P is n (and so P == 1 (mod n)) or P divides n.

Proof. By assumption, P divides rPn(a), and so if a has order k (mod p) then kin. If k < n, then P divides both a k - 1 and ~:=i (the second because of the fundament~l identity and the fact that P divides rPn(a)). As gcd (a k - 1, ~:=i) divides ~ by the Euclidean algorithm, pin and we're done.

Remark 9.5. Note that the proof also works for prime powers p.

Using the previous observation, the result follows easily. We strongly advise the reader to look for a geometric proof in order to appreciate the power of algebraic numbers! 4. Let A, B, C be lattice points such that the angles of triangle ABC are rational multiples of n. Prove that triangle ABC is right and isosceles.

Proof. Note that any angle () = LABC with A, B, and C lattice points must have tan () rational or infinity. To see this note that all lines between lattice points have rational or infinite slopes and if tan 0: and tan (3 are rational (or infinite) then so is tan( 0: - (3) = l~~a~-;;,t:~,6. This implies that tan 2 ()

= sec2 () _

1 - cos 2() 1 + cos 2()

is rational and hence cos 2() is rational. Combining this with the discussion preceding the problem shows that cos 2A, cos 2B, cos 2C are all equal to ± 1,

Chapter 9. A Little Introduction to Algebraic Number Theory

404

±~ and O. It is immediate to check that the condition tanA, tanB, tanC rational or infinite says A, B, C must be integer multiples of 7r / 4. Hence the only possibility is when ABC is right and isosceles. 0

5. Let a be a rational number with 0 cos(37ra) Prove that a

< a < 1 and

+ 2 cos(27ra)

IMO Shortlist 1991

Proof. Let x = cos 7ra and observe that the equation satisfied by a can be written as ===?-

(2x

+ 1) (2x2 + X -

Of course, if x = - ~, we must have a = ~ and we are done. The difficult point is to prove that we cannot have 2x2 + x - 2 = O. If this is the case, then x = -l~vTI, because Ixl ::; 1. We will then prove that cos(2n7ra) takes infinitely many values as n runs over the positive integers. This will clearly contradict the hypothesis that a is rational. But since cos(2n7ra) = 2 cos 2 (2n-17ra) - 1, it is easy to prove that we can write n

)

cos ( 2 a7r =

+ 4bn VI7 '

The previous relations yield by induction that an, bn are odd integers and that an+! > an. Thus cos(2n7ra) takes infinitely many values. 0

Remark 9.7. In general, let us choose relatively prime integers m, n with n > 2 and find the degree of the algebraic number x = cos (2:m). Define z = e 2i~rn , a primitive n-th root of unity. The irreducibility of the cyclotomic polynomials (which is a very nontrivial theorem) implies that z has degree !.p(n) as an algebraic number. On the other hand, we have

[Q(z) : Q] = [Q(z) : Q(x)]· [Q(x) : Q]

405

and we have [Q(z) : Q(x)] = 2. Indeed, 2x = z + z-l, which implies that z satisfies a quadratic equation with coefficients in Q(x), so [Q(z) : Q(x)] ::; 2. On the other hand, we cannot have Q(z) = Q(x), because z is not a real number. Putting these observations together, we deduce that x has degree 'P~n). Using the previous result and the fact that cos C~ x) = sinx, we can compute the degree of sin The answer is a bit complicated: if n -=I- 4, the degree of sin is 'P~n) if 8 divides n, 'P~n) if gcd(n, 8) = 4 and !.p(n) if gcd(n, 8) < 4.

= O.

4x3 + 4x 2 - 3x - 2 = 0

9.2. Cyclotomy

2:.

6. Prove that none of the numbers Vn+1- Vn for positive integers n can be the written in the form 2 cos e!1r) for some integers k, m. Chinese Olympiad

Proof. First, we will find a polynomial with x = Vn+1 - Vn as a root. 2 We have x = 2n + 1 - 2Vn 2 + n and so (x 2 - 2n - 1)2 = 4n 2 + 4n, from where we easily find that x4 - 2(2n + 1)x 2 + 1 = O. Note that the other roots of the polynomial f(X) = X4 - 2(2n + 1)X2 + 1 are x = ±vn + 1 ± Vn. Next, we will find a polynomial with roots x = 2 cos 2!1r. Let Tm be the mth Chebyshev polynomial, defined by the equality Tm (cos x) = cos mx for all x. Then Tm (~) = cos 2k7r = 1. Thus the numbers 2 cos 2!1r for k = 0, ... , m - 1 are roots of g(X) = Tm (~) - 1. These m numbers are not distinct, but 2 2k1r = 2 cos 2(m-k)1r C cos m m lor 1::; k < m / 2 are double roots of this polynomial since 9 achieves a local maximum at these points. Thus these are the only roots of g(X).

If Vn+1 - Vn = 2 cos 2!1r, then f(X) and g(X) have a common factor in Z[X]. The only roots of f which lie in the interval [-2,2] (which contains all roots of g) are vn + 1 - Vn and Vn - vn + 1. Therefore this common factor is either X - (Vn + 1 - Vn) or X2 - (Vn+1 - Vn)2. In either case we see that (vn + 1 - Vn)2 = 2n + 1 - 2y'n(n + 1) is an integer and hence n( n + 1) is a square. But this would make 4n( n + 1) and (2n + 1)2 consecutive positive squares, a contradiction. 0 We continue with a very beautiful and classical result ([50]) concerning linear equations in roots of unity.

406

Chapter 9.

A Little Introduction to Algebraic Number Theory

a) Suppose that aI, a2,···, ak are rational numbers and (1, (2, ... , (k are roots of unity such that al (1 + a2(2 + ... + ak(k = o. Moreover, suppose that LiE! ai(i -=J 0 for any proper subset I of {I, 2, ... ,k}. Prove that (i = (j for all i, j, where m is the product of primes smaller than or equal to k. b) Let Z be a complex number. Prove that there are at most 24k2 . kk ktuples (1, (2, ... , (k) of roots of unity with the following property: there exist rational numbers aI, a2, ... , ak such that Z = ai(i and Z -=J LiE! ai(i for any proper subset I of {l, 2, ... , k}.

L7=1

Mann's theorem Proof a) We may assume that al = (1 = 1. Let m be the least positive integer such that (i = 1 for all i and choose a prime factor p of m. If m = p1 n with gcd(n,p) = 1, we will prove that j = 1 and p S k. This will imply that m divides TIp::;k p and the first part of the theorem will follow. Proving this is however not a simple task. 2i7r

We start with an observation: let Z = e J.;J and let ( be an m-th root of unity. We claim that there exists 0 S r < p and x such that x ~ = 1 and ( = ZT . x. This is very easy: if ( = e 2::;1, simply choose 0 S r < p such that Tn == l (mod p). Applying this observation to each (i, we can write (i = ZTiXi with Xi, ri as above. We have Xl = 1 and rl = o. The equation ai(i = 0 can be . ""p-l b Z h b "" 2i7rp wntten ~Z=O ZZ, were Z = ~Ti=Z aiXi· Note that b E Q(ern). On the

L7=1 z

2i7rp

other hand,we can compute the degree of Z over Q(ern). Indeed, observe ~ 2i1T that Q(z, em) = Q(e-m), so that 2i7r

[Q(e~)(z) : Q(e 2:;:P)] = [Q(e2~): Q] = lj?(m) = lj?(P') [Q(e-r;!') : Q]

lj?(m/p)

lj?(pJ-l)

and the last quantity is p - 1 for j = 1 and p otherwise. Note that Lf~Ol bzXz is not the zero polynomial, since otherwise we obtain the relation LTi=Z ai(i = 0 for all 0 S l < p. But the hypothesis yields then {ilri = l} = 0 or {I, 2, ... , k} for alll. As rl = 0, this gives ri = 0 for all i !!!

and so (i P = 1 for all i, contradicting the minimality of m.

9.2.

407

Cyclotomy

If we combine the results of the previous two paragraphs, we see that we must have j = 1, as z is killed by the nonzero polynomial Lf~~ bzX z, of degree ~

at most p 1. But then z has degree plover Q( em) (as follows from the previous computation) and so Lf~~ bzXz is the minimal polynomial of z over

Q( e 2:-:? ). As z is also killed by 1 + X + ... + Xp-l, we deduce that these two polynomials differ by a constant. In particular, all bz are nonzero. So for all

Os l < p one can find i such that ri = l. Clearly, this implies that p S k and the proof is finished. ai(i of the equation and consider another b) Fix a solution z = solution, say z = biZi· Thus

L7=1

LbiZi = 0, i=l but one has to be a little bit careful, as this relation does not necessarily satisfy the conditions of (a). However, if we fix 1 SiS k, we can find a minimal sub-relation of the previous relation which contains Zi. By hypothesis, such a sub-relation must contain some (j. As the length of this sub-relation is at most 2k and as it clearly satisfies the hypothesis of (a), we deduce that zi = (j for all (j in this sub-relation. Here m = TI p::;2k p. So, for any i, Zi can take at most km values and so the number of solutions of the equation in Zl, Z2, ... ,Zk is at most (km)k. It remains to use Erdos's famous inequality (theorem 3.A.3) TIp::;n p S 4 n to conclude. 0 Remark 9.S. Let aI, a2, ... , an be nonzero complex numbers and consider the equation alZl + a2z2 + ... + anZn = 1. A non-degenerate solution is an n-tuple (Zl' Z2, ... ,zn) of roots of unity which satisfies the equation and such that LiE! aizi -=J 0 for any nonempty subset I of {I, 2, ... , n}. Conway and Jones [20] improved Mann's theorem by proving that if ai E Q, then for any nondegenerate solution we have = z~ = ... = z~ = 1 where d is the product of primes Pl,P2, ... ,Ps such that (pi 2) S n - 1. Also, in [30] the author proves using rather elementary and very beautiful arguments that there are at most (n + 1)3(n+1)2 non-degenerate solutions of the equation.

L:=l

408

9.3

Chapter 9.

A Little Introduction to Algebraic Number Theory

The gcd trick

8. Let a, b be two positive rational numbers such that for some n 2: 2 the number y'a + V'b is rational. Prove that y'a is also rational. Marius Cavachi, Gazeta Matematidl Proof. Let us write y'a+ V'b = c for some (positive) rational number c. Then y'a is a root of xn - a and also of (c - x)n - b. The key point is that it is the unique common root of these polynomials. Indeed, if Z is a common root, then we can write z = y'aZI and c - Z = V'bz2 for some nth roots of unity Zl, Z2. We deduce that y'a + V'b = y'aZI + V'bz2. Since IZil = 1, the real parts of Zl, Z2 are at most 1. Passing to real parts in the previous equality then implies that Zl = Z2 = 1 and the claim is proved. Now, since the two polynomials don't have multiple roots, it follows that gcd(xn - a, (c - x)n - b) = X - y'a. The result follows now from the gcd trick. 0

9. Let m, n be relatively prime numbers and, let x > 1 be a real number such that xm + x ;,. and xn + ~ x are integers. Prove that x + -xl is also an integer.

+ x;,.

and b = xn

The gcd trick

409

and

The division algorithm shows that if K cLare fields and if f, 9 E K[X] are two polynomials, then their gcd is the same if we see f, 9 as polynomials with coefficients in K or with coefficients in L. That is, the greatest common divisor of two polynomials is not sensitive to the field in which the coefficients of these polynomials live. Combining this observation with Gauss' lemma, we also obtain that if f and 9 are monic polynomials with integer coefficients, then their gcd computed in Q[X] has integer coefficients. This gives a very indirect, but sometimes very useful way to prove the rationality or integrality of a real number x: it is enough to exhibit X - x as the gcd of two polynomials with rational coefficients (respectively of two monic polynomials with integer coefficients). The next problems in this section illustrate this trick.

Proof. Let a = xm

9.3.

+ xIn

and consider the polynomials

The crucial claim is that gcd(p, q) = X2 - (x

+ x-I)X + 1 =

(X - x)(X - x-I).

Assuming this for a moment, we can conclude that x + X-I is an integer by the gcd trick. It remains to establish the claim and for that it is enough to prove that x and x-I are the only common zeros of p, q (since clearly p, q have no multiple root). But if Z is a common zero, we have zm = xm or zm = x- m and similarly zn = xn or zn = x-no We may assume (by changing z and z-l) that zm = x m , so that Izl > 1. Then clearly we must have zn = xn. But then z/x is a root of unity whose order divides both m and n. Since gcd(m, n) = 1, it follows that z = x and we are done. 0 The following problem is very similar to the previous problem, but a bit more difficult. 10. Let 0 E (0,11"/2) be an angle such that cos 0 is irrational. Suppose that coskO and cos[(k + 1)0] are rational for some positive integer k. Prove that 0 = 11"/6. USA TST 2007 Proof. We will actually prove more: it is enough to replace k+ 1 by any integer l which is relatively prime to k. The key point is the following

Lemma 9.9. If cos kO and cos are rational for relatively prime positive integers k, l, then either cos 0 is rational or 0 is a rational multiple of 11".

Proof. If cos kO = p and cos = q, tnen eiB is a common root of the polynomials f(X) = X2k - 2pXk + 1, g(X) = X21 - 2qXl + 1.

On the other hand, it is not difficult to check that if 0 is not a rational multiple of 11", then eiB and e- iB are the only common roots of f and g. Indeed, all roots

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Chapter 9. A Little Introduction to Algebraic Number Theory

of f are e±iB+~ for 0 ::; ] < k and all roots of 9 are e±iB+~ with 0 ::; ] < l. On the other hand, since gcd(k, l) = 1, the only solution of the equation e±iB+ 21r~h = e±iB+ 21r;h with 0 ::; ]1 < k and 0 ::; ]2 < l (for some choices of signs) is]l = 12 = O. This proves that the greatest common divisor of f and 9 is precisely (X - eiB)(X - e- iB ) = X2 - 2 cos OX + 1, thus cos 0 is rational. 0 Coming back to the proof, the previous lemma shows that 0 is a rational multiple of 7r. On the other hand, we saw in section 9.2 that the only rational numbers r E [0, 1] such that cos r7r is rational are r = 0, ~, ~, ~, 1. We deduce that kO and to are integer multiples of~. Since gcd(k, l) = 1, Bezout's lemma implies that 0 is an integer multiple of ~. Since cos 0 is irrational, we deduce that 0 = ~. 0

9.4

9·4·

The theorem of symmetric polynomials

411

Proof We will use induction on n and inside the induction step an induction on deg(f). For n = 1 everything is clear, so assume the result holds for n _ 1. We now prove by induction on deg(f) the assertion of the theorem with n variables. If deg(f) = 0 or 1, everything is clear. It is clear that the polynomial g(X I , ... , X n- l ) = f(X I , ... , Xn-l, 0) is still symmetric, so by (the first) induction it is a polynomial of the form h(Xl + ... + X n- l , ... ,Xl· .. X - ) n l for some h E R[Xl , ... ,Xn- l ]. Note that the difference

vanishes when Xn = 0 and is a symmetric polynomial. Therefore this polynomial is a multiple of Xl ... X n . Applying the inductive hypothesis to the quotient between this polynomial and Xl·· . Xn (which has degree less than deg /), the result follows. 0

The theorem of symmetric polynomials

The proof of the following result is quite elementary, but the result itself is incredibly powerful and useful. If R is a commutative ring and if f E R[Xl' X 2, ... ,Xn ] is a polynomial, we say that f is symmetric if for all permutations (T of {I, 2, ... ,n} we have

Remark 9.11. It is not difficult to prove that the polynomial 9 is unique. This means that there are no algebraic relations between the polynomials (Tl, (T2, ... ,(Tn. Remark 9.12. The theorem also implies that any symmetric rational function f E R(Xl, X 2, ... , Xn) is a rational function in the (Ti'S. Indeed, let (T. P(XI, X 2, ... , Xn) = P(Xo-(l) , X<T(2) , ... , X<T(n»)

Recall that the fundamental symmetric polynomials are

(Tk =

for P E R[XI, X 2, ... , Xn]. Then we can write

Xi1Xi2 ..... X ik ,

l:Sil <i2<··-<ik:Sn

for 1 ::; k ::; n. We have the equality

Theorem 9.10. (Fundamental theorem of symmetric polynomials.) Let R be a commutative ring and let f E R[Xl , ... ,XnJ be a symmetric polynomial. Then there is 9 E R[Xl , ... , Xn] such that f(X l , ... , Xn) = g((TI, (T2,···, (Tn).

for some polynomials P, Q, Pl. Since ! is symmetric, so is Pl. The result follows from the theorem of symmetric polynomials applied to PI and to n<T (T . Q.

Remark 9.13. We refer the reader to [66], chapter 5 for the proof of the following theorem of Lagrange: let K be a field of characteristic O. If f E K(X I ,X2, ... ,Xn ), let Gf be the set (actually group) of those permutations (T E Sn such that f(X I,X2, ... ,Xn ) = !(X<T(1),X<T(2), ... ,X<T(n»). If

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A Little Introduction to Algebraic Number Theory

9·4·

The theorem of symmetric polynomials

413

f,g E K(X I ,X2,··· ,Xn ) satisfy Gf eGg, then one can find a rational function h whose coefficients are symmetric polynomials in Xl, X2, ... ,Xn such that g = h(f).

11. Let a, b, c be integers. Define the sequence (xn)n~O by Xo = 4, Xl = 0, X2 = 2c, X3 = 3b and x n+3 = aXn-1 + bXn + cXn +1. Prove that for any prime p and any positive integer m, the number XpTn is divisible by p.

A very important consequence of theorem 9.10 is the following result, that will be constantly used in this section.

Calin Popescu, Romanian TST 2004

Corollary 9.14. a) Let f E Q[XI, X2, ... , Xn] be a symmetric polynomial and let g E Q[X] be a polynomial of degree n, with complex roots ZI,Z2,· ",Zn' Then f(zl,z2"",zn) E Q. b) If f has integer coefficients and if g is monic with integer coefficients, then f(zl, Z2,· .. , zn) is an integer. Proof. Using theorem 9}10, we can write

Proof. Let rl, r2, r3, r4 be the roots of the characteristic polynomial of the recurrence relation, namely X4 - cX2 - bX - a. The crucial point and by far the hardest step in the proof is to realize that Xn =

for all n. This is suggested by Xo = 4 and by the fact that problem creators tend to try to be sneaky. I Proving the previous formula is immediate by induction, once we prove it for n = 0,1,2,3. For n = 0,1 this is trivial, for n = 2 follows from the identity

f(Xl, ... , Xn) = h(al, a2,···, an)

for some h E Q[XI , ... , Xn] (resp Z[XI,"" Xn]). The result follows from the fact that ai(zl, Z2, . .. , zn) are rational (respectively integers), because the coefficients of g are so. 0 Another very useful result is the following generalization of Fermat's little theorem. Corollary 9.15. Let f E Z[X] be a monic polynomial with complex roots Zl, Z2,.' ., Zn (multiplicities counted) and let p be a prime number. Then

Zi + z~ + ... + z~ == (Zl + Z2 + ... + zn)P

rr + r'2 + r!3 + r'4

L: r; = (L: ri)

2 -

L: rirj = 2c i<j

and for n = 3 we can use the recursive relation (since it is easy to see that Yn = r 1 + r'2 + r!3 + r'4 together with Y-I = -b/a satisfies the same recursive formula for the general term of the sequence , relation as xn). With this closed Tn we need to prove that L is a multiple of p. Since L ri = 0, the result follows from corollary 9.15 and by induction on m. 0

Let us consider now a few more or less direct applications of theorem 9.10 and of corollary 9.14.

(mod p). 12.

Proof. Corollary 9.14 implies that both sides are integers. Consider the quotient by p of the difference between the left-hand side and the right-hand side. Using the multinomial formula, it is easy to see that this quotient is a symmetric polynomial with integer coefficients in Zl, Z2, ... ,Zn, thus the result fqllows from corollary 9.14. 0

Here is a nice application of the previous corollary. It was one of the difficult problems given in the Romanian IMO Team Selection Tests in 2004.

a) Let P, R be polynomials with rational coefficients such that P =I 0. Prove that there exists a non-zero polynomial Q E Q[X] such that P(X)IQ(R(X)) b) Let P, R be polynomials with integer coefficients and suppose that P is monic. Prove that there exists a monic polynomial Q E Z[X] such that P(X)IQ(R(X)) Iranian Olympiad 2006

1 As

Richard Stong kindly remarks ...

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Chapter 9. A Little Introduction to Algebraic Number Theory

9.4.

The theorem of symmetric polynomials

Proof The idea is very natural: the first condition that should be satisfied in order to have P(X)/Q(R(X)) is that for each root z of P we have Q(R(z)) = o. Therefore, if Xl, X2, ... ,Xn are the roots of P (some of the X'i'S may be equal), then we would like to have Q(R(Xi)) = o. The most natural choice is to take

i=l

II h(ai) . II 92(ad

415

= 1.

Q(X) =

II(X -

R(Xi)).

i=l

Note that it satisfies P(X)IQ(R(X)), because X - Xi divides R(X) - R(xd for all i. It remains to check that Q has rational (respectively integer, for the second part of the problem) coefficients. This foUows from coroUary 9.14. 0

~ 13.

a) Let aI, a2,.·., am, b1, b2,··· ,bn E <C be such that m

h(X) = II(X - ai),

h(X) =

II(X -

bi ) E Z[X].

i=l

Suppose that there exist 91,92 E Z[X] such that h91 Prove that: m

i=l

+ 1292

= 1.

)

= 1.

j=l

b) If ai, bi are integers and m

{a -l,a + I}.

Thus, by symmetry in A and B and by making a translation of the variable X -+ X - a, it is enough to consider the cases when A = {O}, B = {I} and A = {O}, B = {-I, I}. In each case h divides som~ xn and 12 divides some k (X2 -l)m. Thus h divides X2 and 12 divides (X2 _1)n for k, n sufficiently large. It is thus enough to find Bezout relations with integral coefficients for k the polynomials X2 and (X2k - l)n. But this is immediate. 0

IIII(ai - bj ) = 1, i=l j=l prove that there exist polynomials 91, 92 E Z [X] such that 1292 = 1.

cp: Z[X]deg(!2)-l

h 91 +

Ibero-American Olympiad

Proof a) Note that the relation to be proved can also be written as m

ITh(ai) = 1. i=l

Remark 9.16. The assumption that a~ and bi are integers is useless. Here is a proof, due to Richard Stong. We advise the reader not familiar with the notion of resultant to read the discussion before problem 27 in chapter 12. It is not too difficult to check that the resultant of hand 12 is n~l nj=l (b j -ai) = ±1. But then the map

II II (ai -

On the other hand, n~l h(ai) and n~192(ai) are integers, by corollary 9.14. These two observations are enough to conclude. b) Note that /ai - bj I = 1 for all i, j, because ai, bj are integers. It is then immediate that we have only two possible cases: 1) A, B are singletons of the form {a}, {a + I} or {a + I}, {a}. 2) A = {a} and B = {a -l,a + I} for some integer a or B = {a} and

Z[X]deg(!Il-1 -+ Z[X]deg(!Il+deg(!2l-1

defined by cp(g1,92) = 91(X)h(X) + 92(X)h(X) is invertible, thus we can find 91,92 E Z[X] such that h91 + 1292 = 1. A classical problem is to prove that 1

la + bv'21 2': v'2(l a l + Ibl) for any integers a, b, not both of them equal to o. The idea is that it is not clear how to deal with la + bv'21 directly, but it is very easy to say something

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Chapter 9.

A Little Introduction to Algebraic Number Theory

about the product of this number and its conjugate la - bV2l. Indeed, this product is a nonzero integer, thus at least 1. The result follows immediately. With a similar idea, it is not difficult to prove the following' absolutely classical theorem of Liouville: if Z is an algebraic irrational number of degree d, then

Iz - I

there exists c > such that for all integers p and q we have ~ > Iqld' That is, irrational algebraic numbers are badly approximable with rational numbers. A much deeper result, for which Roth won the Fields medal, is that we can improve the previous inequality to Iz - ~ I > 1:I~cL for all E > 0. The following problems use this trick of multiplying by conjugates and estimating the conjugates, but they are much more challenging than the very simple example discussed above. .~ 14.

Let k, n be positive integers and let P(X) be a polynomial of degree n with all coefficients in the set {-1,0, I}. Suppose that (X - l)kIP(X) and that there exists a prime q such that r!q < In(:+l)' Prove that the primitive complex roots of unity of order q are roots of P. IMC 2001

9.4·

The theorem of symmetric polynomials

417

we have I1i~}(l - Zi) = q and so

q-1

i=l

II P(Zi) = l II Q(Zi).

But the same argument as before shows that I1i~; Q(Zi) is a nonzero integer, therefore I1i~; IP(Zi) I 2:: qk. This is however impossible, since by assumption we have IP(zdl ::; n + 1, therefore

q-1

II IP(Zi)1 ::; (n + 1)q-1 < l. i=l

The previous arguments show that P vanishes at one of the primitive roots of unity of order q. But since the polynomial 1 + X + ... + Xq-1 is irreducible over the rational numbers, if P vanishes at a primitive root of unity of order q, then it also vanishes at all the other roots. This ends the proof. 0 One needs some gymnastics if one wants to avoid the use of Galois theory for the following problem. <.(115. Let p be a prime and let nI, n2, .. . ,nk be integers. Define

Proof. The problem looks rather complicated because of the strange inequality imposed on q. Let us forget first about that and consider the product of all values of P at the primitive qth roots of unity, I1i~; P(Zi). This is an integer, by corollary 9.14. If it is not 0, then I1i~; IP(zdl 2:: 1. However, by assumption there exists a polynomial Q (necessarily with integer coefficients) such that P(X) = (1 - X)kQ(X). Therefore

q-1

II P(Zi) = II (1 i=l

i=l

L cos 27rnj j=l

Prove that either S

or S 2:: k

(A) Holden Lee

2i7T

q-1 Zi)k .

S =

Proof. Let

II Q(Zi).

= e p. The crucial ingredient in the proof is the following:

Lemma 9.17. The number

i=l

P.=.! 2

Since

q-1 Xq-1 + Xq-2 + ... + X + 1 =

II (X i=l

II 1=1

zd, is an integer. Moreover, N

if and only if S

= 0.

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Chapter 9.

A Little Introduction to Algebraic Number Theory

Let us admit this for a moment and see how we can finish the proof. Assume that 8 -I- O. Then INI 2: 1. So

< - ~ Ie.!

Ie.!

181 . II2 1=2

'" cos 27rlnj ~

181· k~-l

and the conclusion follows. Now, let us prove the lemma. As is well-known (and easily proved by induction) there are polynomials F j E Z[X] of degree j such that xj + x-j = Ie.!

Fj(X + X-I). Let F = L:J=l F nj . Then N = I11~1 F(zl will be proved if we prove the following result:

Lemma 9.18. The minimal polynomial of z

+ z-l

+ z-l).

The lemma

II (X -

(zl

+ z-l)) = F~ +

The theorem of symmetric polynomials

419

zs.!

Now, assume that N = O. Thus, there exists 1 :S l :S such that F(zl + z-l) = O. By the lemma, F is a multiple of f and so F vanishes at z + z-l. But this means that 8 = 0, a contradiction. Thus, we have proved the crucial claim and the result follows. 0 The following result is certainly classical, but it is rather difficult to find an elementary proof in the literature. We follow one of the approaches proposed in the beautiful article [10]. \116. Let aI, a2, ... , an be positive rational numbers and let k1, k 2 , . .. ,kn be integers greater than 1. If a~/kl +a~/k2 + ... +a;!kn is a rational number, then any term of the previous sum is also a rational number. Proof. It is clearly enough to prove the following result: let k > 1 and suppose that the positive rational numbers al, ... , an, bl, ... ,bn satisfy

Ie.!

f(X)

9.4.

+ ...

1=1

Then {;'bi E Q for all i. Let Ai = {roots of Xk - a7bd

Proof. Note that f(X

+ X-I)

II (X -

zl)lX - z-l)

p-l -2-

II (X -

zl)(x - zp-l)

P(X) =

by definition of the polynomials F j . In particular,

2 2 .

has integer coefficients and so N is an integer (by the fundamental theorem of symmetric polynomials). Moreover, f has degree and vanishes at z + z -1. But -1 [Q(z) : Q] p- 1 [ Q(z + z ): Q] = [Q(z) : Q(z + z-l )] 2: -2-' so f must be the minimal polynomial of z

zs.!

+ z-l.

L:ai\lb:

and

+ F p-3 + ...

i=l

1=1 I+X + ···+Xp-1

Thus f = Fp-l

k},

where w is a primitive root of order k of 1. Also, let

1=1 = X-~

= {wjai \Ib: 11 :S j :S

(8-X-

X2-"·- X

n).

By theorem 9.14 we have P E Q[X]. Note that P(a1 ijbl) = O. Let d be the least positive divisor of k for which

fjbi E Q (it exists, as ifb'i E Q).

If we

manage to prove that d = 1, it will follow that ijbl E Q, so we can delete the first term of 8 and conclude by induction on n. So, let us prove that d = 1. By definition, we can write a1 ijbl = {IX with x E Q+. The crucial fact is the following:

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Chapter 9. A Little Introduction to Algebraic Number Theory

Lemma 9.19. X d

9.5. Ideal theory and local methods

~ 17. Suppose that (a n )n21 is a linear recurrence sequence of integers such

x is irreducible in Q[X].

that n divides an for all positive integers n. Prove that linear recurrence sequence.

Proof. If F is a monic polynomial with rational coefficients, of degree between 1 and d - 1 that divides X d x, all roots of F have absolute value rx and so

IF(O) I =

(rx)deg(F) is a rational number, that is

Proof. By the general theory of linear recurrence sequences, we can find distinct nonzero algebraic numbers Zl, Z2, . .. , Zm and polynomials iI, 12,Âˇ .. ,fm with algebraic coefficients such that

Since P( rx) = 0, the previous lemma yields Xd - x I Pin Q[X]. Thus, is a primitive root of order d of 1, we have P(zrx) = 0 and so there are (X2, ... , xn) E A2 X A3 X ..â&#x20AC;˘ x An with 8 - z{/k = X2 + ... +x n . If d 2: 2, then Re(z) < 1, so

Re(z\YX + X2

an = iI(n)zl

+ ... + xn) n

i=2 n

i=2

< \YX +

2: ai Vb:

ajpf ==

i=2

1 and al ijbl E Q.

2: fi(O)zl

(mod 1),

i=l

=8, a contradiction. So d

+ 12(n)z2 + ... + fm(n)z?;;"

for all n. We will prove that if nlan for all n, then fi(O) = 0 for all i, from which the result follows easily. Let K be the field obtained by adjoining to Q all z/s and all coefficients of the polynomials k Choose a prime p which does not divide any of the norms of the (nonzero) coefficients of fi's or the norms of one of the Zi'S. All sufficiently large primes satisfy this property. Fix such a prime p and consider a prime ideal I of Kover p, with norm N(1) = pl. Impose the condition that jpl divides ajpf. Note that

=8 =

is also a Polya

Re(8)

(a;)

b~eg(F) E Q, contradicting

the minimality of d. if

421

pi == Zi (mod 1) and since pEl. Thus, we must have

SInce Zi

9.5

Ideal theory and local methods

We strongly advise the reader not familiar with algebraic number theory to read the appendices on number fields and p-adic numbers before reading this section, which is short but rather challenging. We start with a beautiful result otI~olya concerning linear recurrence sequences. Recall that a sequence (an)n .alled a linear recurrence sequence if one can find d and XI, ... , Xd such thai'an+d + Xlan+d-l + ... + Xdan = 0 for all n.

2: fi(O)zl == 0

(mod 1)

i=O

for all j = 0,1, ... , m - 1. Seeing this as a linear system in the fi(O)'S, it follows that fi(O) E I for all i, unless I divides the determinant of the matrix associated to this system. However, this is a Vandermonde determinant in the Zi'S and so, if we ensure that I and nih (Zi - Zj) are relatively prime, we will be able to conclude that fi(O) E I. But to ensure the last property, it is enough to choose a prime p which does not divide the norm of the algebraic number nih (Zi - Zj). Again, all sufficiently large primes have this property.

Chapter 9. A Little Introduction to Algebraic Number Theory

422

The previous paragraph shows that we can find infinitely many primes p and for each such prime an ideal I over p such that fiCO) E I for all i. But then p will divide the norm of fiCO) for infinitely many primes p and so fiCO) = 0 for all i. It is then clear that is still a linearly recurrence sequence. 0

'II

Proof. If d i is the degree of the algebraic number ai, it is clear that d2k :2: d2k+l (because Q(a 2k+l) C Q(a 2k )). Thus there exists an integer j and a positive 2J integer d such that d2k = d for all k :2: j. Let al = a , a2,"" ad denote the conjugates of a 2J and choose a positive integer c such that fo = c(X Z[X] is primitive. Then go = c(X + a1)'" (X + ad) E Z[X] is also primitive, so by an easy application of Gauss' lemma f1 = c2(~ aD' .. (X - a~) E Z[X] is also primitive. Since ha~ deg~ee d and Slllce

a1)'" (X

We present two approaches for the following challenging problem: a rather exotic elementary one and a more advanced approach which uses standard facts about number fields and their p-adic completions.

423

9.5. Ideal theory and local methods

ad)

deg h = d, it follows that h is irreducible over Q. Repeatlllg thls argument.' we obtain that fr = c2 (x -ar)··· (X -ar) E Z[X] is primitive and ~;reduclble. Next, since nar is an algebraic integer, we have hr = nd(X - da1 ) ... (X 2T n '71 S' ar) = (nX - nar) ... (nX - nad ) E Z[X], so we must have c2T E /LJ •. lllce T

18. Let al, a2, ... , an be complex numbers such that af + ar + ... + a~ is an integer for all positive integers m. Prove that (X - al)(X - a2)" . (X -

an) l-oc...t -

Z[X]. Michael Larsen, AMM E 2993

this happens for all sufficiently large r, it follows that c = ±1 and so a is an algebraic integer. As a result, a is an algebraic integer and we're done. 0

6l(Jb...!

Proof. Let

O'k =

ail ai2 ... aik

l::;il <i2<··-<ik::;n

and Pk = a~ l:S;k:S;n)

+ ... +

Pk - O'lP~-l

a~, so Newton's identities2 can be written (for

+ (J'2 Pk-2 -

.. , + (-I)kkO'k = O.

It follows immediately from these relations that if Pk E Z for all k then O'k E 1 niZ for all 1 :S; k :S; n. In particular, (X - al)(X - a2) ... (X - an)' E ~Z[X] and so n! . ai are algebraic integers. Observe that if al, a2, ... , an satisfy the conditiEms of the problem, then so do aI' a2, ... , a~ for all r :2: 1. We deduce that n!ai is an algebraic integer for all r. The next lemma shows that all ai's are algebraic integers, so the coefficients of (X - al)(X - a2)'" (X - an) are algebraic integers. Since these coefficients are rational numbers (this has already been established), they must be integer~ and the result follows.

Lemma 9.20. Let n be a positive integer and a be an algebraic number. ]f na k is an algebraic integer for all positive integers k, then a is an algebraic integer. 2See the remark 9.22 for a proof of these.

Proof. This proof uses rather heavy material, but it is much ~or~ concept~al than the previous one. Namely, we will use a local-global prmclple, statlllg that an algebraic number x is an algebraic integer if and only if v( x) :2: 0 for any valuation v on Q.This follows easily from the relations between a number field and its completions (see the addendum on number fields), but the result is not obvious at all. Anyway, once we have this, the lemma is immediate: if v is a valuation, then we know that v(n) + kv(a) :2: 0 for all k. Dividing by k and making k -t 00 yields v(a) :2: 0, which is enough to ensure that a is an algebraic integer. The lemma is proven, and so we are done.

Remark 9.21. The case when all ai's are rational numbers is much easier and is a rather folklore problem. In this case, the problem reduces easily to the following: if p is a prime number and if a1, a2,'" ,ak are integers such that pn divides al + a2: + ... + ak for all n, then p divides all ai's. This follows easily from Euler's theorem, by choosing n = r.p(pN) with N sufficiently large. Actually, using ideal theory as in the previous problem and imitating the proof for rational numbers, one can give yet another solution of the problem. We leave this as a nice exercise for the reader.

Chapter 9. A Little Introduction to Algebraic Number Theory

424

Remark 9.22. Let us recall the proof of Newton's relations. Let ai be elements of a field K of characteristic 0 and define n

i=l

i=O

9.5. Ideal theory and local methods

Assume that no ai is a multiple of p and let Zi q

+ a~ + ... + a~.

Observe that

!'(X) = - f(X) . L Pk X k - 1 k2':l

yields Newton's relations

for 1 :::;m:::;

The following is also a very tricky problem. We use a p-adic approach to solve it and we refer the reader to the appendices on p-adic numbers and number fields for more details.

i :::; q). Since

p-l

E.=!

there exists C7(i) E {I, 2, ... , q} such that vp(a i have vp ( 2:::i

Identifying coefficients in the equality

= zi-1 (1 :::;

L vp(a~ - Zj) = vp(af-1 - 1) > 0, j=l

f(X) = I1(1- ai X ) = LbiX i . Let Pk = a~

425

a:~l)

Za(i)) > O. Since we

> 0, it follows that vp (2::: Za(i)) > O. So, if f(X)

2:::~=1 Xa(i)-\ we have vp(f(z)) > O. Since f(Zi) is an algebraic integer for all i, it follows that vp(f(l) rrZ=2 f(Zi)) > O. Let N = rrZ=2 f(Zi). N is an integer, because it is a symmetric polynomial expression with integer coefficients in the roots of the polynomial Xq-1 + ... + X + 1. We claim that N is nonzero. Otherwise, there exists 2 :::; i :::; q such that f(Zi) = O. The irreducibility of the polynomial 1 + X + ... + Xq-1 over the rational numbers implies that f is a multiple of 1 + X + ... + Xq-1. But then r = f(l) is a multiple of q, a contradiction with the hypothesis. So N is a nonzero integer and vp(rN) > 0, so that vp(N) > 0 (clearly p does not divide r). Thus INI 2: p. On the other hand, we have If(Zi)1 :::; r, so that INI :::; r q - 1 < p. This contradiction finishes the proof. 0 Proof. Here is a more elementary, but still very tricky solution, based on the theorem of symmetric polynomials. Suppose that none of the ai's is a p-l

~ 19. Let p, q be prime numbers and let r be a positive integer such that qlp-1,

q does not divide rand p-l

> r q- 1. Let aI, a2, ... ,aT be integers such that

p-l

a~ + a-;:- + ... + aT q is a multiple of p. Prove that at least one of the ai's is a multiple of p.

multiple of p and let h

= g-q-,

where 9 is a primitive root mod p.

p-l

can therefore find positive integers mi such that a i q f = Xml + X m 2 + ... + xmr and let 9 E Z[X] satisfy

== h mi

(mod p). Let

J. Borosh, D.A. Hensley, J. Zinn, AMM 10748

Proof. Let Z = e 2~7r and let K = Ql(z). This is an. extension of degree q - 1 of

Ql. By choosing a prime dividing p in the ring of algebraic integers of K and completing K with respect to this prime, we obtain an extension of the p-adic valuation vp on K: Moreover, if x is an algebraic integer in K, then vp(x) 2: O.

where C7i are the symmetric fundamental sums. Let Zl, Z2, ... , Zq-1 be the complex roots of iq~ll. Note that

xq -

1 2 1 X-I = (X - h)(X - h ) ..... (X - h q- ) E lFp[XJ,

Chapter 9. A Little Introduction to Algebraic Number Theory

426

as h, h 2 , ... ,hq -

are distinct qth roots of unity in IFp. This implies that

ai(zl, Z2, ... , zq-d

== ai(h, h 2, ... , h q- 1)

(mod p)

9.6. Miscellaneous problems

427

and first compute the minimal polynomial of

cos~.

In order to do this, we

will first find a rational equation of low degree satisfied by cos ~. Let so that z7 = -1 and

Z =

eT ,

for all i. Therefore q-l

II f(Zi) == g(al (h, h 2, .. . , hq-

), ... , aq-l (h,

h 2, . .. ,hq- 1 ))

Dividing this by z3 and rearranging terms yields

i=l

q-l

II f(h

)

== 0

3+ -z31 - (2Z + -z21) + Z + -Z1 - 1 = O.

(mod p),

i=l

that is p divides the integer N = f(Zl) ..... f(Zq-l). As in the previous solution we obtain INI ::; r q - 1 < p and so N = O. We conclude as in the previous solution. D

9.6

Miscellaneous problems

It is really not easy to solve the following problem without the use of minimal polynomials. However, once the yoga of minimal polynomials is understood, the argument is rather standard.

20. Find the least positive integer n such that cos ~ cannot be written in the form p + 01 + ijr with p, q, r E Q.

O. Mushkarov, N. Nikolov, Bulgaria Proof. For n ::; 6, explicit computations show that cos ~ can be written in the

desired form (the argument is a bit tricky for n = 5, but note that Z = e W- is a solution of the equation z4 - z3 + z2 - Z + 1 = 0" which can also be written as (z + z-l? - (z + Z-l) - 1 = 0.) The question is whether we can write cos ~ in the form p + 01 + ijr with p, q, r E Q and the answer turns out to be negative, implying that the answer to the problem is n = 7. Let us assume that 7r

cos 7

= P + y'q + ijr

Thus, if x = cos ~ = z~z , then the previous relation gives 8x 3

that is 8x 3

4x 2

4x + 1

6x - (4x 2

+ 2x -

= O. Since the polynomial

f(X) = X

1 2 1 1 - -X - -X + 2

is trivially irreducible over the rational numbers (it has degree 3, so we only have to look for rational roots), this is the minimal polynomial of x. Therefore x and x - p have degree 3 over Q. But then (observe that the identity (( y'q + ijr) - 01)3 = r easily implies that 01 E Q(01 + ijr))

[Q(y'q + ijr) : Q(y'q)]. [Q(y'q) : Q] = [Q(y'q + ijr) : Q] = 3 and since [Q( 01) : Q] is 1 or 2, it follows that 01 is a rational number. Thus x - p - 01 = ijr and u = p + 01 is a rational number. Now, since x is irrational, we must have ijr irrational and so X3 - r is irreducible over the rational numbers. Since f(u + ijr) = f(x) = 0, it follows that X 3 r divides f(u + X) and so (for degree reasons) we must have f(u + X) = X 3 - r. It is trivial now, by identifying coefficients, to see that this is not possible. The result follows. D

Chapter 9. A Little Introduction to Algebraic Number Theory

428

9.6. Miscellaneous problems

Now, by assumption sntn is an integer for all n. But then

Proof. As in the previous solution it is enough to show that we cannot have

Sitj As before we compute that cos ~ satisfies 8x 3 - 4x 2 - 4x + 1 and that this polynomial is irreducible since it has no rational roots. Also either by noting that the other two roots are cos and cos or by plugging in a few values, we see that this polynomial has three real roots. Now let z = e 27ri / 3 and suppose x = p Âą Vii + zk ijr. Then

[(x - p)3

+ 3q(x _

p) _ r]2 = (3(x _ p)2

+ q)2q.

Thus

g(X)

= [(X - p)3 + 3q(X - p) - r]2 - (3(X - p)2 + q)2q E Q[X]

is a sixth degree polynomial with roots p Âą Vii + zk ijr. If the equality above holds, then this polynomial must be a multiple of f(X), the minimal polynomial of cos ~ = p + Vii + ijr. However f(X) has three real roots and g(X) has only two real roots (the ones with k = 0). Thus this cannot occur. D We continue with a very beautiful problem and a very elegant solution. 21. Let Sl, S2, ... and t1, t2,'" be two infinite nonconstant sequences of rational numbers such that (Si - Sj)(ti - tj) is an integer for all i,j 2': 1. Prove that there exists a rational number r such that (Si sj)r and ti~tj are integers for all i, j.

USAMO 2009

Proof. We start with some useful reductions: first of all, by working with the sequences (Si - Sdi and (ti - tdi' we may assume that Sl = t1 = O. Secondly, there is u such that Su =I- 0 and, by working with the sequences (::) nand

(su . tn)n, we may assume that Su

429

+ Sjti = siti + Sjtj -

(Si - Sj)(ti - tj)

is also an integer for all i,j. Since sitj + Sjti and (Sitj) . (Sjti) = (Siti)(Sjtj) are integers, sitj and Sjti are algebraic integers. Since they are also rational numbers, they must be rational integers. Thus sitj is an integer for all i, j. For i = u, we obtain that all tj are integers. Let d be their greatest common divisor. Then clearly ~ is an integer for all i. We claim that dS i is also an integer for all i, which will solve the problem. But since d is a linear combination with integer coefficients of some ti's (by Bezout's lemma) and since sitj E Z for all i, j, it is clear that dS i E Z for all i. The conclusion follows. D In order to motivate the next problem, we will discuss first a very classical and nontrivial result in elementary number theory. The reader is advised to read the addendum 9.A before reading the proof. Theorem 9.23. (Lucas-Lehmer) Define a sequence by ao = 4 and an+1 = a~ - 2 for n 2': O. Let m be an odd positive integer and let n = 2m 1. Then n is a prime if and only if nlam -2.

Proof. The first difficulty is to actually find a manageable formula for the general term of the sequence. We use the identity x 2 + x- 2 = (x + x- 1)2 - 2 and set an = Xn + x;;-l for a sequence Xn > 1 (note that an > 2, so Xn exists). Then Xn+1 = x~, so Xn = and we easily conclude that

Suppose that n

= p is a prime and

2': 3. Since p == 1 (mod 3) and

-1 (mod 8), the quadratic reciprocity law implies that

(~)

= -1. Pick some

in an algebraic closure3 of lFp such that

= 1 and 0:

= 3.

. 30 ne does not need the existence of an algebraic closure to prove the existence of a: if 3 is a quadratic residue mod p, it is clear what we have to do; otherwise, it is easy to check that JFp [Xl/(X2 3) is a field with p2 elements and we can take for a the image of X in this field.

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Chapter 9. A Little Introduction to Algebraic Number Theory

Note that 0: is actually an element of IFp2 and that we can define a map f : Z[V3] -+ lFp2 by f(a + bV3) = a + bo:, where a = a (mod p) (seen as an element of IFp2). Since 0: 2 = 3, it is immediate to check that this is a ring homomorphism. Trivially, f vanishes on pZ. Let x = f(2 + V3) = 2 + 0: and y = f(2 - V3) = 2 - 0:. Thus x, y E lFp2 and they are nonzero, since xy = f(l) = 1. We want to prove that f(a m -2) = or equivalently that 2m - 2 2m - 2 l±! x + x= 0, i.e. x 2 = -1. Since 2x = (1 + 0:)2, we obtain the following equality in IFp:

2x l±! 2 = 2

(2)P .

9.6. Miscellaneous problems

If x ElF;, everything is easy, since then Lagrange's theorem for this subgroup yields 2n+ 2 1p -1 and so trivially 2n +3 1p2 -1. So, assume that x is not in IF;. Then x, yare roots of the irreducible polynomial X2 - 4X + 1 E lFp[X], so that we must have x P = y. Indeed, since x 2 - 4x + 1 = 0, we also have (by raising the previous equality to the p-th power and by using the formula (x + y)P = x P + yP, valid in fields of characteristic p) x2p - 4xP + 1 = 0, so that x P is also a root of X2 - 4X + 1, which cannot be x (because otherwise x P = x and x ElF;). Thus x P = y and so x p+l = 1. But then 2n+2, which is the order of x, must divide p + 1 and we are done again. 0

X l±! 2 = (2x) l±! 2 = (1 + o:)pH = (1 + 0:) (1 + o:P).

G) .

Since 0: 2 = 3, we have o:p = 0: = -0:, which combined with the previous equality yields the desired result. Let us prove the converse now. Suppose that nlam -2, we need that n is a prime. It is enough to check that for all pin we have p > .;n. Since p divides a m -2, the previous arguments yield the equality (2+0:) = -1 in IF*2. Thus p 2 + 0: E IF;2 has order n + 1 and Lagrange's theorem yields n + 11p2 - 1. The result follows. 0

There is really no obvious approach to the following rather exotic problem. 23. Let k be a positive integer and let al,a2, ... ,ak and bl ,b2 , ... ,b k be two sequences of rational numbers with the property: for any irrational numbers XI,X2, ... ,Xk > 1 there exist positive integers nl,n2, ... ,nk and ml, m2, ... , mk such that adx~l] + a2[x~2] + ... + ak[x~k]

22. The sequence ao, aI, a2, ... is defined by ao = 2 and ak+l = 2a~ - 1 for k 2: 0. Prove that if an odd prime p divides an, then 2n+3 divides p2 - 1. IMO Shortlist 2003 Proof Note that 2a n is precisely the sequence studied in theorem 9.23, so

Let now p > 2 be a prime factor of an and let 0: E lFp be such that 0: 2 = 3. Define f, x, y as in the proof of the previous theorem. Since plan, we have n 2n X + y 2 = 0, thus x 2n +1 = -1. Hence x has order 2n +2 in the group IF*P2 and so by Lagrange's theorem 2n+2 divides p2 - 1. Unfortunately, this is not enough, but we are close.

431

= bdx;'H] +

b2[x~2] + ... + bk[X~k].

Prove that ai = bi for all i. Gabriel Dospinescu, Mathlinks Contest Proof The key point is the following result:

Lemma 9.24. For any integer N 2: 2 we can find irrational numbers a, b > 1 such that for every positive integer m we have [am] == -1 (mod N) and [b~ == (mod N).

Proof We will choose a, b to be algebraic integers of degree 2. Let us show how to construct a and leave to the reader the details for the construction of b. We want to find a polynomial with integer coefficients

f(X) = X2

+ uX + v =

(X - a)(X - e)

for some irrational numbers a > 1, 0< e < 1. In this case, since am + em is an integer for all positive integers m, it follows that [am] = am + em 1 for all m. Thus, we need to ensure that am + em == (mod N) for all m. Since a m+l + em+ l = -u(am + em) _ v(a m- l + em-I)

432

Chapter 9. A Little Introduction to Algebraic Number Theory

for all m, it is enough to ensure that N divides u, v. Also, to ensure that For instance, we can take u = -2N,v = N, yielding a = N + JN2 - N. Similarly for b, we will choose u = -(2N + 1) and v = N, so

9.7.

Notes

Let us prove the base case: assume that n

o < c < 1 we will choose v > 0 and 1 + u + v < O.

2N + 1 + J 4N2 2

Coming back to the proof, choose a positive integer N and a, b irrational numbers as in the lemma. Set Xl = a and X2 = ... = Xk = b. By hypothesis, we can find positive integers nl, n2, ... ,nk and ml, m2, ... ,mk such that

433

= 1 and that

for some rational numbers a, b. Squaring this relation and using that y'Pl is irrational, we deduce that ab = O. But then either ql q2 ... qm or ql q2 ... qmPI is a perfect square, which is clearly not possible. Now, assume that the result holds for n and let us prove it for n + 1. Let F = Q(y'Pl, $2, ... , ffn) and assume that ..jqlq2··· qm = a + bJPn+1 for some a, bE F. Again, we square this relation to deduce that

By the properties of a and b we deduce that al == bl (mod N). Since N was arbitrary, it follows that al = bl . Since we can do the same with the other pairs (ai, bi), the result follows. 0

However, by the inductive hypothesis we have ~ rJ. F, so we must have ab = O. If a = 0, we obtain that JPn+lqlq2'" qm E F, contradicting the inductive hypothesis. If b = 0, we get a similar contradiction. In all cases, the inductive step is proved and the conclusion follows. 0

The following result is really a mathematical gem, taken from [5]. It is quite difficult and has a very elementary proof.

Remark 9.25. In [58], Mordell proved the following generalization:

24. Prove that if PI,P2, ... ,Pn are distinct primes and if

for some rational numbers a!, a2, . .. ,an, then ai = 0 for all i.

Theorem 9.26. Let K c L be fields of characteristic 0 and let Xl, X2, ... , Xr be elements of L such that for all i there exists a least positive integer ni such that x7 i E K. Suppose that for all integers el, e2, ... , e r , if X~l . X~2 ..... x;r E K, then ni divides ei for all i. Finally, suppose that L c lR. or that K contains all nith roots of unity, for all i. Then (XiI. X;2 ..... x~r)O:s'ij<ni is a linearly independent set. In particular, [K(xl, X2," ., x r ) : K] = nl ..... n r .

Besicovitch's theorem Proof. We will prove by induction on n the following statement: for any m :2:: 1 and any distinct primes ql, q2, ... , qm, PI, P2, ... , Pn we have4

4If Kj F is an extension of fields and if Xl, X2, ... , Xn E K, we let F(XI, X2, .•. , Xn) be the smallest subfield of K which contains F and Xl, X2, .•. , X n . It is also the set of elements of the form f (Xl, X2, ... , x n ), where f is a rational function in n variables with coefficients in

9.7

Notes

We thank the following people for providing solutions: Amol Aggarwal (problem 18), Darij Grinberg (problem 1), Daniel Harrer (problem 9), Holden Lee (problems 8, 10), Thanasin Nampaisarn (problem 13), Fedja Nazarov (problem 3), Richard Stong (problems 4, 6, 20), Qiaochu Yuan (problems 4, 17), Victor Wang (problems 9,19), Gjergji Zaimi (problems 2, 3).

9.A. Equations over Finite Fields

Addendum 9.A

Proof. First, let us check that IFq is a field. It is clearly stable by multiplication and stability under addition follows from the previous proposition. IFq has q elements since xq - X splits into linear factors over IFp (because IFp is alge-

Equations over Finite Fields

This addendum is a modest introduction to finite fields and polynomial equations over finite fields. There are some very beautiful and extremely deep results on the subject, which are far beyond the scope of this book. But the fact that their proofs are very difficult should not be a reason for not presenting them. We highly recommend the introductory text [43] for more details. To avoid spending too much time on preliminaries, we will fix a prime number p and an algebraic closure IFp of the field IFp = 'iZ/p'iZ. Recall that this means that any x E IFp is a root of some nonzero polynomial f E IFp[X] and that any f E IFp[X] has at least one root in IFp (which actually implies that it splits into linear factors over IFp). It is a rather nontrivial theorem of Steinitz that any field has an algebraic closure and any two algebraic closures are isomorphic. We take this approach when introducing finite fields since it is pretty rapid, though not very elegant ... Before proving the first fundamental result, let us glorify the following easy result, which will be constantly used in this chapter: Proposition 9.A.1. Let p be a prime and let A be a ring such thatf5 pa = 0 for all a E A. Then for all powers q of p and for all aI, a2, ... ,an E A we have

Proof. By induction on n, we may assume that n

435

= 2. Then everything follows

from the usual binomial formula, the hypothesis on A and the fact that (mod p) for any 1 ::; i < q.

m== 0

If q is a power of p, let

q IFq = {x E IFplx = x}.

braically closed) and all of these linear factors are distinct (because xq - X is prime to its derivative -1). Let us consider now a subfield L of IFp with q elements. As L * is a group with q - 1 elements, Lagrange's theorem yields x q - 1 = 1 for all x E L *. Thus x q = x for all x ELand so L C IFq. A cardinality argument finishes the ~~

A more subtle result is the following generalization of Gauss' classical theorem on primitive roots modulo prime numbers. Theorem 9.A.3. IF~ is a cyclic group of order q - 1. More generally, if K is any field and G is a finite subgroup of K*, then G is cyclic. Proof. Let d be the maximal order of the elements of G. It is a general property of finite abelian groups that if x, y E G have orders m, n, then one can find z E G with order lcm( m, n) (the reader can take this as an easy exercise). Using this, we deduce that the order of any element of G divides d. Thus for all g E G we have gd = 1. But the polynomial X d - 1 E K[X] vanishes at all elements of G, so d ~ IGI. On the other hand, d is the order of some element of G, so dllGI by Lagrange's theorem. Therefore d = IGI and G is cyclic. 0

There is a trap concerning finite fields: it is not true that if n ~ m, then IFpm C IFpn. Actually, this inclusion takes place if and only if Xpm -1 -1 divides Xpn-1 - 1 (this follows immediately from the definition and the fact that the roots of xq - X are simple) and this happens if and only if pm - 1 divides pn _ 1, which in turns happens if and only if m divides n. A fundamental object in the theory of finite fields is the Frobenius map

We have the following easy, but crucial result: Theorem 9.A.2. IFq is the unique field with q elements contained in IFp. 5We say that A has characteristic p.

an automorphism of IFqn which acts as identity on IF q. Moreover, any such automorphism is an iterate of the Frobenius map and there are precisely n such

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Chapter 9. A Little Introduction to Algebraic Number Theory

automorphisms. 6 All these results would be pretty hard to prove without theorem 9.A.3, but they become easy exercises once we have it. The following result is fundamental. It says that if you know a root of an irreducible polynomial over lFq, then the other roots are obtained by successively applying the Frobenius map to that root.

Theorem 9.A.4. Let f E lFq[X] be a monic irreducible polynomial of degree 2 n-l n and let x E lFp be a root of f. Then the roots of fare x,xq,x q , ... ,xq . In other words, f(X) = I1~==-Ol(X - xqi). Proof. The key point is that xqn = x. Indeed, the field generated by x over IFq (inside ifp) has qn elements, because x has degree n over IF q, so this field is IFqn. But in IFqn all elements are roots of xqn - X. Having done this, define the polynomial G(X) = I1~==-Ol(X - xqi). The key point and proposition 9.A.1 yield n-l n-l G(X)q = (xq - xqi+l) = (xq - xqi) = G(xq).

i=O

Thus, if we write G(X) = go + 91X + ... 9.A.1 96 + 9iX q + ... + gi Xql = go

+ 91XI,

then again by proposition

+ 91Xq + ... + glX ql ,

which implies that g; = gi for all i and so gi E lFq. Thus G E lFq[X]. Since G vanishes at x and f is irreducible, we deduce that f divides G. A degree argument finishes the proof. 0

9.A.1

Norm and trace maps

Consider a finite field IF q and a finite extension IF qn. Define the norm and trace maps by

n-l NFqn/Fq : lFqn ----+ lFqn,

II x j=O

The following result summarizes the basic properties of these maps, that will be used in future sections.

Proposition 9.A.5. The norm and trace maps are surjective maps from lFqn to lFq. The norm map is multiplicative and the trace map is additive. Proof. To avoid complicated notations, write Nand T for the norm, respectively trace map. First, let us check that N(x), T(x) E lFq for all x E lFqn. It is enough to see that N(x)q = N(x) and T(x)q = T(x). For N(x), this is clear since xqn = x, while for T(x), this follows from proposition 9.A.1 and the equality xqn = x. It is clear that N is multiplicative and proposition 9.A.1 shows that T is additive. It remains to prove the surjectivity of these maps. Let ~ be a generator of lF~ and let u be a generator of lF~n. There exists a E Z such that ~ = u a . As ~q-l = 1, we have ua(q-l) = 1 and so there is an integer b such that a = b· q~/. But then ~ = N(u b ) and the surjectivity of N follows. For the trace map, this argument does not work, however we note that T(ax) = aT(x) for any a E lFq and any x E lFqn. Thus, it is enough to prove that there exists x such that T(x) -=I- O. But this is clear, as the polynomial X + xq + ... + xqn-l has at most qn-l roots and so it cannot vanish on all of IFqn . 0

9.A.2

Characters of finite fields

As lFpn is an n-dimensional vector space over lFp, the choice of a basis yields a group isomorphism lFpn c:::: lFp x ... x lFp . Now, basic properties of the dual of a group discussed in section 7.A.1 yield the following result.

Proposition 9.A.6. Let q be a power of p. There is an isomorphism of groups a ----+ 'lj;a between IF q and its character group, where

n-l ,

TrFqn/Fq : lFqn ----+ lFqn,

2i1fTr

'lj;a(x)=e p

L xqj. j=O

6In fancy terms, the Galois group of the extension IF qn IlF q is cyclic of order n and generated by Fr q •

437

9.A. Equations over Finite Fields

Also,

lF~

IFq/lFp

(ax)

•

is a cyclic group, so its group of characters is also cyclic of order

q -1. The following result will play an important role in the following sections,

when we will compute the zeta function of a diagonal hypersurface.

438

Chapter 9. A Little Introduction to Algebraic Number Theory

439

9.A. Equations over Finite Fields

Proposition 9.A.7. Let d be a divisor of q - l. The map X ----t Xn, where Xn(x) = X(N'l" q n/'l"q (x)) induces a bijection between characters of order d of'Fq* , and characters of order d of 'F~n. '

Finally, the following result will be used in the proof of the DavenportHasse relation, to which the next section is devoted. It is by no means specific to finite fields, but the short proof we are going to give uses properties of finite fields developed in the previous sections.

Proof. The fact that Xn is a character of order dividing d is a consequence of the multiplicativity of the norm map. The fact that it has order exactly d and that X ----t Xn is injective is a consequence of the surjectivity of the norm map (proposition 9.A.5). It remains to check the surjectivity of X ----t Xn. Let X be

Proposition 9.A.9. Let x E 'F qn and let f = X d - alX d- 1 + ... + (-I)d ad E 'Fq[X]

qn_l

a character of order d of 'F~n and let u be a generator of 'F~n. Then,;- = u q-l is a generator of 'F~ and since X(u)q-l = 1 (because Xd = 1 and dlq -1), there is a unique character X of 'F~ such that X(.;-) = X(u). By construction, X = Xn and the result follows. 0 Just as for Dirichlet characters, it is convenient to extend the definition of a multiplicative character X of 'F~ to 'Fq, by defining X(O) = 0 if X is nontrivial and X(O) = 1 if X is trivial. The following innocent-looking identity will play a crucial role in future arguments and is constantly used when dealing with equations over finite fields:

Proposition 9.A.B. Let d be a divisor of q - 1 and let x E 'Fq. The number of solutions of the equation yd = x with y E 'F q, denoted N (yd = x) is L:xd=l X( x), the sum being taken over all multiplicative characters whose order divides d.

Proof. If x = 0, this is clear, as both sides are equal to l. Assume that x =I O. If the equation yd = x has a solution in 'F q' then it has exactly d such solutions, as the equation yd = 1 has precisely d solutions in 'F~ (because dlq - 1 and 'F~ is cyclic of order q - 1). On the other hand, the dual group of 'F~ is also cyclic of order q - 1, so the equation Xd = 1 has d solutions and for each of them X(x) = X(yd) = X(y)d = 1, so both sides of the equality we want to prove are equal to d and we are done. Finally, if the equation has no solution, the result is a consequence of the orthogonality relations (theorem 7.A.5) for the abelian group 'F~/{xdlx E 'F~}, whose dual group is precisely the subgroup of those multiplicative characters X such that Xd = 1 (actually, this argument also covers the previous case ... ). 0

qj be its minimal polynomial over 'Fq. Then din and n:j~t (X - x ) = f~. In particular, N'l"qn/'l"q(x)

= aJ and Tr'l"qn/'l"q(x) = Jal.

Proof. Since ['Fq(x) : 'Fq] = deg(j) = d, we have 'Fq(x) = 'Fqd (this uses theorem 9.A.2). But then 'Fqd C 'Fqn and, as we have already remarked, this implies qj that din. Next, for degree reasons it is enough to prove that 9 = TIj~t (X -x ) has only one irreducible monic factor, namely f. But if h is such a factor, then h has some root xqj. But proposition 9.A.4 implies that f also vanishes at xqJ d (note that the cited proposition applies only for j < d, but we have x q = x anyway, since we have seen that 'Fq(x) = 'Fqd). Thus gcd(j,h) is nonconstant and by irreducibility f = h. This finishes the proof. 0

9.A.3

Gauss and Jacobi sums, the Davenport-Hasse relation

Gauss and Jacobi sums playa fundamental role in the theory of equations over finite fields and in number theory, in general. We give here their basic properties, that we will need in the following sections. But before doing that, it is convenient to define them ...

Definition 9.A.IO. 1) If 'l/J and X are characters of 'Fq, respectively 'F~, the associated Gauss sum is

9(X, 'l/J) =

X(x)'l/J(x).

XE'l"~

2) If Xl and X2 are characters of 'F~, the associated Jacobi sum is

J(XI,X2) = XI(X)X2(Y)Âˇ x,yE'l"q,x+y=1

Chapter 9. A Little Introduction to Algebraic Number Theory

440

where

Theorem 9.A.H. If X and'lj; are nontrivial, then Ig(x, 'lj;)1 = ~.

r(x) = Proof. The orthogonality relations (theorem 7.A.5) yield (using also the substitution ~ = t) Ig(x, 'lj;)12

X(x/y)'lj;(x - y)

L X(t) (L 'lj;(y(t tElF:;

= q- 1-

e-ttX-ldt,

B(x,y) =

10 t X- l (l_ t)y-ldt,

the integrals being convergent for Re(x) , Re(y) > 0.

Theorem 9.A.l3. If Xl, X2 are nontrivial characters of such that Xl . X2 is nontrivial, then for all nontrivial characters 'lj; of Wq we have

X(t)'lj;(y(t - 1))

t,yElF:;

X,YElF:;

441

9.A. Equations over Finite Fields

1)) - 1) =

yElF q

X(t) = q -

L X(t)(q· I

t =l -

tElF:;

Proof. This is a rather tricky computation:

X(t) = q.

t#O,l xElFq-{O,l} YElF:;

Using the substitution a = xy and b = y(1 - x), this becomes

Corollary 9.A.l2. If X and 'lj; are nontrivial, then

g(X, 'lj;) . g(X-l, 'lj;) = X( -1)q. Proof. This is just a long string of obvious computations, using the previous theorem and the fact that g(X,'lj;(-·)) = X(-I)g(X,'lj;) (which is immediate by definition and the fact that x -+ -x is a permutation of W;). More precisely, we have

One has the following beautiful result which connects Gauss and Jacobi sums. Note the striking similarity with Euler's famous formula

+ b) =

g(XI, 'lj;)g(X2' 'lj;) -

XI(a)X2( -a).

aElF:;

As XIX2 is nontrivial, the orthogonality relations (theorem 7.A.5) yield the desired result. D Here is a striking application. Assume that p == 1 (mod 4) is a prime. As 1, there exists a nontrivial character Xl of order 4 of

W; is cyclic of order p -

W;.

g(X-l, 'lj;) = g(X, 'lj;) = g(X, 'lj;) = g(X, 'lj;( -.)) q = X( -1)g(X, 'lj;) = X( -1) g(X, 'lj;)"

B(x y) = r(x)r(y) , r(x+y)'

Xl (a)X2(b)'lj;(a

a,bElF:; ,a+b#O

Let X2(X) =

(~) be Legendre's symbol. The previous two theorems imply

that IJ(XI, X2W = p. On the other hand, it is clear that Xl takes only the values 0, ±1, ±i, thus J(XI, X2) E Z[i]. In particular, IJ(XI, X2W is the sum of the squares of two integers. We recovered thus Fermat's celebrated theorem that any prime of the form 4k + 1 is the sum of the squares of two integers. We end this section with a much deeper result, the famous DavenportHasse relation. This is a quite strong identity between Gauss sums, which is crucial for the proof of Weil's theorem 9.A.24 that we will see a bit later on. It also has relations to the Langlands program, but that is really beyond the scope of this book. The proof is very ingenious.

442

Chapter 9. A Little Introduction to Algebraic Number Theory

Theorem 9.A.14. (Davenport-Hasse) Let X and'ljJ be nontrivial characters ofF~,

respectively Fq. Then -gn(X, 'ljJ) = (-g(X, 'ljJ))n, where gn(X, 'ljJ) = L

9.A. Equations over Finite Fields

differentiating, we obtain

L'(T) T· - L(T)

X(NlFqnllF q(X))'ljJ(TrlFqn IlFq (x)).

XElFqn

Proof. As NlF qn IlF q (x) and TrlF qn IlF q (x) only depend on the minimal poly~omial of x and not on its roots, we will partition F qn into collections of conjugate elements over F q' Define

for any bi E F q' It is easy to check that A(J g) = A(J) . A(g) for all monic polynomials f,9 E F q[X]. Combining this with proposition 9.A.9 shows that if x E Fqn has minimal polynomial P = X d - a1X d- 1 + ... + (-l)d ad over Fq, then

Summing over all conjugates of x and then over collections of conjugate elements in F qn yields the crucial identity

gn(X, 'ljJ) =

L dA(Pt ld , d=degPln

ITl P

_ "

A(p)Tdegp - L

A(f)Tdegf

d T· -logL(T) dT

deg(P)A(P)nTndeg(P)

(

L dA(p)nld) Tn. d=deg(P)ln

Combining this with the key relation of the previous paragraph, we conclude that L'(T) n T· L(T) = Lgn(X,'ljJ)T . n

Finally, we will show that L(T) has a very simple expression. Note that

On the other hand,

L A(J) = L A(X - a) = L x(a)'ljJ(a) = g(X, 'ljJ), degf=l a a

the sum being taken over all irreducible monic polynomials P E Fq[X] whose degree divides n. To exploit this relation, consider the L-function

L(T)

443

Here the product is taken over all P E F q[X] monic irreducible, while the sum is over all f E Fq[X] monic. The second equality follows from multiplicativity of A and the unique factorization theorem in Fq[X]. Taking log and

while for n 2: 2

by the orthogonality relations (theorem 7.A.5). We deduce that

L(T) = 1 + g(X, 'ljJ)T and the result follows immediately from this and the last equality of the pre0 vious paragraph.

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Chapter 9. A Little Introduction to Algebraic Number Theory

We end this section by stating a very deep result of Dwork, a consequence of his proof of the rationality of the zeta function of algebraic varieties. It is a vast and very difficult generalization of the Davenport-Hasse relation. Theorem 9.A.I5. Let f,9 E IF q[X] and let X, 'ljJ be multiplicative, respectively additive characters of IF q' If

Sn = L

9.A. Equations over Finite Fields

the sum being taken over all nontrivial characters Xi of IF~ such that and XOXI ... Xl = 1.

X (NlFqn/lFq(f(X))) . 'ljJ (TrlFqn/lFq(9(X))),

'"' Sn n) exp ( L..t -:;T

= Q(O) = 1 and

and let 9(Xi)

L la.u=o L lxO'=uo····· lxl'=u/ L N(xO' = uo) ..... N(xl = UI),

a·u=O

Diagonal equations and a theorem of Wei!

IF q /lFp

laoxO'+ ... +a/xl'=o

xo, .. ·,xz

P(T) Q(T)'

Using almost everything we have done so far, we can prove the following beautiful theorem of Weil. The true beauty of the result will be revealed in a next section, when we will use this result and the Davenport-Hasse relation to compute the zeta function of a diagonal hypersurface. Before stating the theorem, we need some notation. Let ao, aI, ... ,al E IF~ and let XO, Xl, ... ,Xl be multiplicative characters of IF~. Consider the additive character 'ljJ(x)

= =

n2:l

(x)

xo, .. ·,x/ElFq

then there exist polynomials P, Q E qT] such that P(O)

2i7rTr

xi =

Proof. Let M be the number of solutions (xo, Xl, ... ,Xl) E IF~+1 of the equation aoxO' + ... + alxl = O. Since we work with a projective variety, we have IX(IFq)1 = ~~/, so it remains to find M. Note that

XElFqn

9.A.4

445

where we wrote for simplicity a . u = 0 for aouo + al Ul + ... + alul = O. Using proposition 9.A.8, then expanding the product and re-arranging terms, we obtain

Next, note that

= g(Xi,'ljJ). Finally, define

9(Xo) g(Xl) g(XI) Wq(Xo,XI,··· ,Xl) = - (-) . - (-) ..... - (-)' Xo ao Xl al Xl al Note that this quantity depends on the ai's, but we suppress the dependence from the notation, as we will consider the ai's as fixed elements, while the characters Xi will vary. We are now ready to state and prove: Theorem 9.A.I6. (Weil) Let ao, al, ... , al E IF~ and let X be the projective variety defined by aoxO' + alx 1 + ... + alx l = O. Then

XO(UO)xI(UI)'" XI(UI) =

a·u=O

j=O

where

So, we end up with the pretty complicated formula

1-1

" ,qJ' IX(IFq)1 = ,L..t

.°

-1

'"' L..t

q XO,Xl",·,Xl

Wq(Xo, Xl, ... ,Xl),

II Xj(aj)-l Jo(XO, ... , Xl),

II Xj(aj)-l Jo(XO, ... , Xl).

XO'='''=xl'=l j=o

446

Chapter 9. A Little Introduction to Algebraic Number Theory

It is convenient to study in more detail these sums Jo(Xo, Xl,·· ., Xl), which are generalizations of the Jacobi sums discussed in section 9.A.3. The following lemma deals with those terms in the sum defining M for which some character is trivial.

447

9.A. Equations over Finite Fields

Lemma 9.A.IB. If Xo, Xl, X2,· .. , Xl are nontrivial characters and

=I 1, Xl = 1,

xo . Xl ..... Xl then JO(XO,XI, ... ,Xl) = O. Ifxo' Xl·····

then

Jo(xo, Xl, ... ,Xl) = q - 1 9(XO)g(XI) ... g(XI). q

Lemma 9.A.17. Suppose that the trivial character appears in the list Xo, Xl, ... ,Xl· Then either Xj = 1 for all j, in which case

Proof. Let Jo = Jo(xo, Xl,· . . , Xl) and JO(XO,XI, ... ,Xl) = ql, or JO(XO,XI,'" ,Xl) = O. Proof. If all Xj = 1, it is clear that JO(XO,XI, ... ,Xl) = ql (don't forget that the trivial character evaluated at 0 yields 1 by convention). Assume that not all Xi are trivial, say (without loss of generality) xo = Xl = ... = Xk = 1 and Xj =I 1 for j > k. Then Jo(Xo, Xl, ... ,Xl)

Xk+I(Uk+I)'" XI(UI)

JI = JI(XI,X2,···,XI) =

XI(UJ)X2(U2)···XI(UI).

Ul +U2+",+uz=1

Then

L = L

Jo=

tElF'~ Xl

+·+xz=-t

XO(t)(XI ... Xl)( -t)JI

tElF'~

= (Xl'" Xl)( -1)

XOXI ... Xl(t)) J I .

tElF';;

Uk+l,""UZ

If XOXI ... Xl =I 1, we are done since LtElF'*q XOXI ... Xl(t) = 0 in this case (orthogonality relations). If XOXI .,. Xl = 1, the previous equality becomes Jo = XO( -l)(q - l)JI. Since g(XO)g(XOI) = Xo( -l)q (by corollary 9.A.12), it remains to prove that

=0, the last equality being a consequence of the orthogonality relations (theorem 7.A.5). D Using this, we obtain

M=ql+

J = g(XJ)g(X2) ... g(XI) I g(XIX2 ... Xl) . Now, by definition we have

g(XI)g(X2)'" g(XI) =

IIXj(aj)-IJo(Xo,XI, ... ,XI)' Xi=l,xdl j=O

XI(XI)x2(X2)'" Xl (XI)1/J(XI

+ X2 + ... + Xl)

Xl,X2,···,Xl

= JO(XI, X2,·· ., Xl)

1/J(t)XIX2 .. , Xl(t)) JI

t~O

The crucial step is the following lemma, whose proof uses a generalization of theorem 9.A.13.

= JO(XI, X2,···, Xl)

+ g(XIX2'"

XI)JI .

448

Chapter 9. A Little Introduction to Algebraic Number Theory

As XIX2路 .. Xl -I- 1, we have JO(Xl, X2,路路路 ,Xl) proposition and the conclusion follows.

o by

the first part of the

Finally, using the previous lemma, we obtain

449

9.A. Equations over Finite Fields

is exp (Ln2: 1 nC;n--\)Tn ), as this time two solutions that differ by a nonzero element of lFpn are the same element of the projective space.

Remark 9.A.20. Suppose that

IX (IFqn ) I = zl + Zz + ... + z~ - ul - ... - uf the sum being taken over all nontrivial characters Xj such that xo ... Xl = 1. The result follows.

9.A.5

xj =

1 and 0

for all n ~ 1 and some complex numbers Zi, Uj (as we will see, this always happens, but this is very difficult to prove). Then

The zeta function of an algebraic variety

In essence, an affine variety over a field k is the locus in kN of a bunch of polynomial equations in N variables with coefficients in k. A projective variety is the locus in the projective space lpm(k) of a bunch of homogeneous polynomial equations in n + 1 variables and coefficients in k. If X is an algebraic variety over IF q, it is natural to consider the number of points of X over the various finite extensions IF qn. The zeta function of the variety X is (up to a convenient normalization) the generating function of the sequence obtained in this way, i.e.

essentially by definition of Zx(T) and by the equality of formal series

an Tn =

n2:1

10g(1 - aT).

For instance, if X = pr, the projective space, then

IX(lFqn)1 = qnr

+ q(r-l)n + ... + qn + 1,

so r

Zx(T) =

II 1- qJT..

j=O

Clearly, Zx(T) is a formal series in T with rational coefficients.

Example 9.A.19. Consider finitely many polynomials h, 12, ... , is in k variables with coefficients in lFp, and let an be the number of solutions in lFpn of the system of equations

The zeta function of the affine variety defined by the polynomials

h, ... ,is

exp (Ln2: 1 ~ Tn ). On the other hand, if h, ... ,is are homogeneous polynomials , the zeta function of the projective variety defined by these polynomials

Another trivial example is the variety defined by the equation xyz = 1 in three-dimensional affine space. Then clearly IX (IF qn) I = (qn - 1)2, so that

(1 - qT)2 Zx(T) = (1 _ T)(l _ q2T)" Let Z[[T]] be the set of formal series in T with integer coefficients. Note that in all previous examples we have Zx(T) E Z[[T]]. It turns out that this is always the case. The following result discusses more generally when exp (Ln Tn) E Z[[T]].

450

Chapter 9. A Little Introduction to Algebraic Number Theory

Proposition 9.A.21. Let an be a sequence of integers. There exist unique sequences of rational numbers bn and en such that

9.A. Equations over Finite Fields

451

Considering the coefficient of Tn in the left-hand side of the previous equality, we obtain that Cn E Z. Using also the explicit formula of the cn's in terms of the an's yields the last statement of the proposition and finishes the proof. 0 Let X be an algebraic variety over IF q, say defined by some polynomials in n variables with coefficients in lF q . If

iI, 12,路路路, fd Moreover, all bn are integers if and only if all en are integers, if and only if n divides Ldln J-l (~) ad for all n (where J-l is Mobius' function). Proof. The existence and uniqueness of bn is clear. As for to consider

= -

L: (L

n:2:1

define Fr(x)

en, the key point is

(xi,x~, ... ,x~).

It is easy to see that this is again an element of X (IF qrn ). Let f be the smallest positive integer such that x E X (IFqf) and associate to x the cycle (x, Fr( x), ... , Fr f -1 (x)), of length f. Then X (IF qrn) is the disjoint union of the cycles of length dividing m, so if an is the number of cycles of length n, then IX (IF qrn ) I = Ldlm d . ad. Combining this and the previous proposition yields

Zx(T) =

II (1 -

Tn)-a n E Z[[T]].

n:2:1 dC d ) .

din

Remark 9.A.22. The previous proposition is powerful in other contexts, too. For instance, it immediately yields the equality of formal series

Thus we need to find the sequence Cn for which Ldln dCd = -an for all n. But the Mobius inversion formula yields the explicit form

exp(X)

/L(n)

II (1- Xn)--n , n:2:1

where J-l is Mobius' function. The proposition also easily implies the following equality showing the existence and uniqueness of cn 路 It is clear that if C n E Z, then bn E Z for all n. The converse is proved by induction. Since C1 = -b 1 , we have C1 E Z. Suppose that C1,' .. ,Cn -1 E Z and observe that f = TIi<n(l - Ti)Ci is invertible in Z[[T]], as its constant term is 1. Thus

exp

'""' xpn) ~ -n( n>O p -

/L(n) (1 - X n )-----;;;-,

n>l gcd(n-;-p)=l

X;n)

showing that the Artin-Hasse exponential exp (Ln:2: 0 has coefficients in Zp. This is absolutely not clear from the definition and plays a major role in padic analysis and also in Dwork's proof of the rationality of the zeta functions attached to algebraic varieties over finite fields.

Chapter 9. A Little Introduction to Algebraic Number Theory

452

In general, it is a very deep problem to compute the zeta function of a given variety. Yet even without computing the zeta function, one can say a great deal of things about it! This was conjectured by Weil in the wonderful paper [84] and proved after a gigantic work by Deligne and Grothendieck. The following theorem is really one of the most difficult and beautiful results of modern mathematics:

9.A. Equations over Finite Fields

453

This last estimate was obtained by Lang and Weil in 1954 for arbitrary varieties, before the proof of the previous deep theorem. For an even more concrete example, consider integers a, b and a prime p > 3. The condition that the curve y2 = f (x) be non-singular is that p does not divide 4a 3 + 27b 2 . In this case the curve y2 = f(x) has genus 1 and one point at infinity. Hence

Theorem 9.A.23. (Deligne-Grothendieck) Let X be a non-singular projective variety of dimension n over IFq' Then the zeta function of X is a rational function. More precisely, there are polynomials Po, PI, ... ,P2n E Z[T] such that

and a) Po(T) = 1 - T and P 2n (T) = 1 - qnT.

b) We can write Pi = I1;~I(I-WijT), where Wij are algebraic integers such that IWij = qi/2 for all i, j. 1

c) If X =

2:::;:0 (-1) bi , then the zeta function satisfies the functional equai

tion

If X is a smooth projective curve of genus g (this is an important invariant attached to curves; in the notations of the theorem, we have b1 = 2g) over IF q,

Weil proved in 1940 that its zeta function can be written in the form (1-~~~~qt) for some polynomial P E 1 + tZ[t]. Moreover, the roots of P have absolute value 1/ Vii, So in this case bo = b2 = 1, b1 = 2g and moreover we can write

IX (IFqn )1 = 1 + qn - (WI + w2 + ... + W29) IWil = Vii,

In particular, we

This reproves a famous theorem of Hasse (there are however easier proofs, but they require a good knowledge of the theory of elliptic curves and quite a lot of algebraic geometry). Dwork's p-adic proof (1960) of the rationality of zeta functions works for affine or projective varieties, be they non-singular or not. For a nonsingular projective hypersurface defined by a polynomial f E IF q [XI, ... ,Xn ], homogeneous of degree d, Dwork proved that its zeta function is of the form P(t)(-I)n+1 (l-t) ... (l-qn 2t)

the very nontrivial estimate

(d-l)n+(-I)n(d-l).

for some P E l+tZ[t] of degree d . We WIll prove this in the next section, in the much easier case of diagonal hypersurfaces (a famous theorem of Weil).

9.A.6

for some sign Âą.

with

and the bound above becomes

Zeta function of diagonal hypersurfaces

In this part, we show how to compute the zeta function of a diagonal hypersurface. This beautiful result, due to Weil was also the starting point of the famous Weil conjectures.

Theorem 9.A.24. (Weil) Let l 2:: 1, mlq - 1 and let X be the projective hypersurface of equation aoxo + a1xl + ... + azx'l of degree d = (m - 1)1+1+(m- l)I+I(m - 1) such that

= O.

There exists P E Z[T]

p(T)(-I)1

a) Zx(T)

= (I-T)(I-qT) ... (l-q I IT)

b) If P(z)

= 0,

â&#x20AC;˘

I-I

then 1/ z is an algebraic integer of absolute value q2"" .

454

Chapter 9. A Little Introduction to Algebraic Number Theory

c) Ther~ exists an explicit integer X such that ZX

( 1) = ± (

1-1 )X qZ-T Zx(T).

ql-IT

Proof. Recall that by theorem 9.A.I6 we have for any q an equality 1-1

IX(lF q)I = " L . qJ + -1 j=o

" L

Wq(Xo, Xl,···, Xl),

XO,Xl,,,,,XI

the sum being taken over all nontrivial characters Xi of lF~ such that xi = 1 and XOXl'" Xl = 1. The main point is to study how the numbers Wqn (XO, ... ,Xl) vary and this is accomplished by the Davenport-Hasse relation. More precisely, recall that for a character X of lF~ we have a character Xn(x) = X(NlFqn/lFq(X)) and that X -+ Xn induces a bijection between characters of lF~ of a given order and characters of lF~n of the same order (proposition 9.A.7). So, if 8 is the set of l + I-tuples (XO, . .. ,Xl) of nontrivial characters of lF~ such that xi = 1 and XOXI ... Xl = 1, then

9.A. Equations over Finite Fields

455

Note that Wq(Xo, . .. , Xl) =I- 0 for all (Xo, . .. , Xl) E 8, as the Gauss sum associated to a nontrivial character is nonzero. Thus deg(p) = 181. It remains to find this number. See 181 as a function f(l) of l. Let g(l) be the number of l + I-tuples (Xo, ... , Xl) of nontrivial characters such that xi = 1 and Xo'" Xl =I- 1. Clearly, g(l) = f(l + 1). But f(l) + g(l) is just the number of tuples of nontrivial characters such that xi = 1. As there are m 1 nontrivial characters of order dividing m, we deduce that f(l) + g(l) = (m - 1)1+1. One immediately deduces that

f(l) =

(m - 1)1+1

Wqn(XO,n, Xl,n,"" Xl,n) = (_I)l+l( _I)n(l+l)(Wq(Xo, ... , Xl))n. We deduce that

and so by the previous paragraph we finally obtain the first part of the theorem, with (_1)1+1 ) P(T) = q Wq(Xo, ... ,XI)T . (1-

(xo,,,,,XI)ES

finishing therefore the computation of deg(P). By definition of Wq(Xo, .. . ,Xl) and by the fact that Ig(xi)1 =

Vii,

deduce that IWq(xo, . .. ,xZ)1 = q I t . This yields part 2) of the theorem. Next, the fact that g(Xi)g(X;:I) = Xi(-I)q and XO"'Xl = 1 implies that -1

ql+l

-1

Wq (Xo , ... , Xl )

On the other hand, by the Davenport-Hasse relation (theorem 9.A.I4) we can write

+ (-I)I+1(m -

w:q (X 0,···, XI )

As clearly the set 8 is stable by inversion, we deduce that the map z -+ ql!lz is a permutation of the roots of P. From here, it is an easy but tedious exercise to deduce the third part of the theorem. Finally, it remains to prove that P E Z[TJ. ~s Zx(T) E Q[[TJ], w.e must have P E Q[TJ. We will prove that the coefficIents of P are algebraIc integers, which will be enough to conclude. It is enough (taking into account the definition of P) to check that Wq (XO, ... , Xl) / q is an algebraic integer. As Xi(ai) are roots of unity, it will therefore suffice to check t h at g(xo)···g(xd· q IS a.n algebraic integer. But this is an obvious consequence of lemma 9.A.IS. ThIs finally proves the theorem! D

9.B. A Glimpse of Algebraic Number Theory

Addendum 9.B

A Glimpse of Algebraic Number Theory

This addendum recalls the basic properties of number fields. Of course, one would need a whole book (and actually much more ... ) to properly develop the theory of number fields, as even proving the basic properties requires a lot of commutative algebra. We will try to stay as elementary as possible, while still giving some proofs. We warn the reader that a long part of this addendum is very abstract. To see the power of the notions and theorems discussed, we advise the reader to start with the last part of the addendum, which discusses applications to problems with very elementary statements and very non-elementary solutions ...

9.B.1

Ideals and quotient rings

Let R be a commutative ring. An ideal of R is a nonempty subset I of R which is stable under addition and such that ax E I for all a E R and x E I. Note that this is far stronger than the stability of I under multiplication. It is fairly easy to construct ideals of R: if Xl, X2," ., Xn E R then

is obviously an ideal, called the ideal generated by Xl, X2, ... , X n . Once we have an ideal I in a ring R, we can naturally construct a quotient ring, whose elements are coset classes a = a + I with a E R and addition, multiplication are defined by a + b = a + b and a . b = abo It is an easy exercise to check that it is well-defined (the issue is that we may have a = a' even if a =1= a' and one needs to check that if a = a' and b = b', then a + b = a' + b', similarly for multiplication).

Definition 9.B.1. 1) An ideal I of R is called maximal if 1=1= R and if I is not contained in any ideal different from I and R. 2) An ideal I of R is called prime if 1=1= Rand ab E R-I for any a, bE R-I.

457

There is a very nice characterization of prime and maximal ideals of a ring in terms of quotient rings. The proof is essentially trivial unwinding of definitions, but the result is crucial:

Proposition 9.B.2. field.

1) An ideal I of R is maximal if and only if Rj I is a

2) An ideal I of R is prime if and only if Rj I has no zero divisors. As a field has no zero divisors, this proposition implies that any maximal ideal is a prime ideal. There are however prime ideals which are not maximal: the ideal (2) in Z[X] is prime and not maximal, as the quotient ring is lF 2 [X], which has no nonzero zero divisors but is not a field. There are natural operations on ideals: if I, J are ideals of a ring R, one defines their sum I + J = {i + j liE I, j E J}. It is easy to check that this is an ideal. The analogous definition for multiplication would fail (in general) to yield an ideal, so one defines the product of ideals I, J as the ideal generated by all products iÂˇ j with (i,j) E I x J.

9.B.2

Field extensions

We say that L is a field extension of K if both K, L are fields and K c L. The extension 7 Lj K is called finite if L is a finite dimensional K-vector space. In this case, we define the degree of the extension to be [L : K]

= dimK(L).

For instance, the extension CjJR has degree 2, as 1, i is a basis of C over JR. On the other hand, the extension CjQ is infinite (for example, because C is uncountable and Q is countable). We will mostly be interested in finite extensions, for which the following result is of constant use:

Proposition 9.B.3. Let L j K and M j L be finite extensions of fields. Then M j K is finite and

[M: K] = [M: L]Âˇ [L: K]. 7This notation should not be confused with the quotient ring previously discussed, simply because K is not an ideal in L unless K = L.

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Chapter 9. A Little Introduction to Algebraic Number Theory

Proof. One can easily check that if (Xi)i is a basis of M as L-vector space and (Yi)i is a basis of L as K-vector space, then (XiYj)i,j is a basis of M as K-vector space. 0

9.B.3

Algebraic numbers and algebraic integers

If L/ K is an extension of fields and if I E L, we say that l is algebraic over K if there is a nonzero polynomial f E K[X] such that f(l) = O. In this case, there is a unique monic polynomial trz E K[X] of least degree which vanishes at I. It is called the minimal polynomial of I and it is irreducible in K[X], by minimality. The division algorithm shows that the only polynomials f E K[X] such that f(l) = 0 are the multiples of 7rz. Recall that K(l) is the smallest field containing K and l and it can also be described as

K(l)

{;~:? If, g E K[X],g(l) i

K[l]

U(l)lf E K[X]}.

To prove this equality, it suffices to show that if fE K[X] does not vanish at l, then is of the form A(l) for some A E K[X]. But since f(l) i 0 and trz is irreducible, f and trz are relatively prime, so there are polynomials A, B E K[X] such that Af + B7rz = 1. Evaluation at I yields the result. The following proposition is easy, but fundamental.

AI)

9.B. A Glimpse of Algebraic Number Theory

459

then (the classes of) 1,X, ... ,Xd-l form a K-basis of K[X]/(7rz) and so [K[Xl!(7rz) : K] = d. Combining this with the previous isomorphism finishes the proof. 2) This is clear: if d = [K(l) : K], then 1,1,12, ... ,ld E K(l) cannot be linearly independent over K. This forces a nonzero polynomial equation with coefficients in K satisfied by I and so I is algebraic over K.

o Combining the previous results, we can now prove the following nontrivial result (which can also be obtained using the theorem on symmetric polynomials):

Theorem 9.B.5. Let L/ K be any field extension. The set of elements of L which are algebraic over K forms a subfield of L. This subfield is equal to L if L / K is finite.

Proof. If h,b E L are algebraic over K, then by proposition 9.B.4 K(h)/K and K(h)(b)/ K(h) are finite extensions (note that 12 is also algebraic over K(h)). Thus by proposition 9.B.3, K(h)(b)/ K is finite. But K(h)(12) contains K(h +12), K(hI2) and K(h/12). We deduce that if x E {h +12, h12, h/12}, then K(x)/K is finite and the result follows by proposition 9.B.4. The second part is also a trivial consequence of proposition 9.B.4. 0

Proposition 9.B.4. Let L/ K be any extension of fields. 1) Let l E L be algebraic over K. Then there is an isomorphism of Kalgebras8 between K(l) and K[Xl!(7rz). Moreover, [K(l) : K] = deg7rZ < 00.

2) Conversely, if I ELand [K(l) : K] <

00,

then I is algebraic over K.

Proof. 1) Consider the map sending f E K[X] to f(l). It is a map of K -algebras, vanishing precisely on the ideal (7rz), by definition of 7rz. It is easy to check that it induces an isomorphism between K[Xl!(7rz) and K[l], obtained by sending f + (7rz) to f(l). Next, if d = deg7rz, 8This means an isomorphism of rings which is K-linear.

Definition 9.B.6. 1) A number z E C is called algebraic if it is algebraic over Q. It is called an algebraic integer if its minimal polynomial over Q has integer coefficients. By Gauss' lemma, this is equivalent to the fact that z is root of some monic polynomial with integer coefficients. 2) We denote by Q (respectively Z) the set of algebraic numbers (respectively algebraic integers). The following result is an easy consequence of the theorem of symmetric polynomials 9.10.

Theorem 9.B.7. Q is an algebraically closed field and x E Q there exists n 2': 1 such that nx E Z.

Z is a ring. For any

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Chapter 9.

A Little Introduction to Algebraic Number Theory

Proof. The previous theorem shows that Ql is a field. Suppose that x E C satisfies xn + an_IX n - 1 + ... + ao = 0 for some ai E Ql. We want to prove

that x E Ql. Let a~l be the conjugates of ak (i.e. the roots of the minimal polynomial of ak, including ak). The theorem of symmetric polynomials easily implies that f =

(xn

461

2) Let (n be a primitive nth root of unity in C and consider K = Ql((n). The irreducibility of cyclotomic polynomials (a fairly nontrivial theorem) implies that the nth cyclotomic polynomial is the minimal polynomial of (n, so that K has degree <p(n) over Ql. One can prove with quite a lot of effort that OK = Z[(n], i.e. there are no algebraic integers in K except for the obvious ones.

+ a~k~11l X n - l + ... + a6kOl ) 9.B.4

ko, ... ,kn-l

has rational coefficients and vanishes at x, from where the result follows. To prove that Z is a ring, consider for instance x, y E Z and let Xl, . .. ,Xn and YI, Y2, ... ,Ym be all roots of the minimal polynomials of x and y. Another application of the theorem of symmetric polynomials shows that TIi,j(X Xi Yj), respectively TIi,j(X Xi' Yj) have integer coefficients and vanish at x + y, respectively x . y. Finally, to prove the last statement, take x E Ql and choose integers ao, aI, ... ,an such that anx n + an_IX n - 1 + ... + ao = 0 and an i= O. Then

so an' x E

9.B. A Glimpse of Algebraic Number Theory

Let us introduce now the main object of this addendum.

Factorization in OK, the fundamental theorems

Since the proofs of the theorems stated in this section are rather long and technical, we will simply state them without proof, referring the reader to any basic number theory book. We prefer to focus on their arithmetic applications. Let K be a number field of degree dover Ql and let OK be the subring of K consisting of algebraic integers.

Theorem 9.B.IO. If I is a nonzero ideal of OK, then OK/I is a finite ring. Remark 9.B.1l. Actually, one can prove that if [K : Ql] = d, then there exist Xl, X2,···, Xd E OK such that the map Zd -+ OK sending (nl' n2, ... , nd) to nlxl + ... + ndxd is a bijection. This easily implies the previous theorem: let x E I be nonzero, then the norm n of x is again in I and is nonzero. Hence I contains nO K. But the previous result shows that OK/nO K is in bijection with Zd /nZ d , which is finite, with Inl d elements.

Definition 9.B.8. A number field is a finite extension of Ql. If K is a number field, then we let OK = K n Z.

Corollary 9.B.12. If ~ is a nonzero prime ideal, then OK / ~ is a finite field

Example 9.B.9. 1) Let d i= ±1 be a square free integer and consider K = Ql( Vd) (as usual, if d < 0, Vd = iH). Then K has degree 2 over Ql, but the structure of OK depends on the residue class modulo 4 of d: if

Proof. The ring R = OK / ~ is finite with no zero divisors. Let x i= 0 be any element of R. As R is finite, there must be i < j such that xi = xj. Then j xi(x - i - 1) = 0 and as there are no zero divisors in R, we obtain x j - i = l. Thus x is a unit. This proves that R is a field and the result follows. 0

== 1 (mod 4), then OK = Z [1+2Vd ] , while if d == 2 or 3 mod 4, then OK = Z[Vd]. This is not difficult to prove: imagine that x E OK and d

write x = a+bVd. If b = 0, then a must be an integer (since it is rational and algebraic integer), so we are done. Otherwise, the conjugates of x are x and Y = a - bVd. Thus 2a and a 2 - db 2 must be integers, which easily implies the desired result.

and so ~ is a maximal ideal.

Definition 9.B.13. If I is a nonzero ideal of I, we define its norm

This is an integer which lies in I, by Lagrange's theorem in the group OK / I.

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We cannot emphasize enough the importance of the following result for the theory of algebraic number fields. Suffice it to say that it plays exactly the ~am.e role as the fundamental theorem of arithmetic (i.e. the unique factorIzatIOn theorem) and we wi11leave the reader recall that basically all results of elementary number theory follow from it.

Theorem 9.B.14. (Kummer, Dedekind) Let K be a number field. a) Any ideal of OK, different from 0 and OK, can be uniquely (up to permutation) written in the form PI . P2 ..... Pn for some n > 1 and some prime ideals Pi, not necessarily distinct. b) If I and J are two nonzero ideals of OK, then N(I . J) = N(I) . N(J). Moreover, I C J if and only if there exists an ideal J' such that I = J. J'. Remark 9.B.15. One can also prove that for all x E OK we have N(XOK) = IN(x)l. Remark 9.B.16. Suppose that K has degree dover Q and that p is a rational prime. Let pOK = f~l p~2 ..... P~g be the factorization of the ideal pOK. ~8 pOK has 1norm p (by the first part of the remark), theorem 9.B.14 yields g p = N(pI)e ••••• N(pg)e and so there are nonnegative integers fi such that N(Pi) = p/;o We have the fundamental relation

9.B. A Glimpse of Algebraic Number Theory

which is easily seen to be a nonzero ideal of OL. Moreover, pOL f- OL, as otherwise we would have 1 = aIxI + ... + anX n for some ai E p and some Xi E OL and by taking norms, we would get 1 E p. Hence, by the previous theorem, there exists a prime /3 of OL dividing pOL. By the same theorem, this simply means that pc /3. Note that /3 is not unique, but there are only finitely many possibilities for it. On the other hand, given a nonzero prime ideal /3 of OL, there is a unique prime p of OK such that pc /3. Indeed, the existence is obtained by taking p = /3nOK (it is an easy exercise for the reader to check that this is indeed a nonzero prime ideal of OK) and the uniqueness follows from the fact that any nonzero prime ideal of OK is maximal (hence, if PI C /3 and P2 C /3 for some different pri~es Pj of OK, then 1 E PI + P2 C /3, a contradiction).

Definition 9.B.IS. Let L / K be a finite extension of number fields and let p be a nonzero prime ideal of OK. A prime ideal /3 of OL is said to lie above p if P c /3. We also write /3lp in this case. We can resume the previous discussion by saying that any prime /3 of OL has a unique prime p of OK below it, namely /3 n OK and any prime p of OK has at least one (but finitely many) prime /3 above it. Note that if /3lp, then the inclusion OK C OL induces an injective morphism OK / p -t OL/ /3, realizing therefore 0 L / /3 as a finite extension of OK / p.

Definition 9.B.19. Let

/3lp be as above.

1) The residual degree of /3/p is defined by f(/3/p) Remark 9.B.17. Let K be a number field, let P be a nonzero prime ideal of OK a~d ~et x E OK be prime to P (i.e. P does not appear in the prime fa.ctonzatlOn of xOK or, equivalently, XOK + P = OK). Since OK/p is a field wIth N(p) elements, we obtain from Lagrallge's theorem the following useful analogue of Fermat's little theorem: xN(p)-I == 1 (mod p).

Consider now a finite extension L/ K of number fields and let p be a nonzero prime ideal of OK. Let

463

= [0£//3: OK/pl·

2) The ramification index of /3/ p is the exponent e(/3 / p) of /3 in the prime factorization of pOL. Note that by definition we have pOL =

II /3e(/J/p). /Jlp

Using this and proposition 9.B.3, we easily obtain the following useful property of the ramification index and residual degree in a tower of number fields.

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Chapter 9. A Little Introduction to Algebraic Number Theory

Proposition 9.B.20. Let M / L / K be a tower of finite extensions of number fields and let pl{3lp be primes of OM, OL and OK. Then

e(p/{3)· e({3/p) = e(p/p),

f(p/{3)· f({3/p) = f(p/p)·

Definition 9.B.21. Let L/ K be a finite extension of number fields. A prime p of OK is called a) unramified in L if e({3/ p) = 1 for all (3lp in L. b) totally split in L if e({3/p) = 1 and f({3/p) = 1 for all {3lp in L. This is equivalent 9 to: pOL is the product of [L : K] different prime ideals of OL. The following theorem gives a practical way to factor a prime in a number field. The precise statement is a bit complicated, but the message is very simple: for most primes p, factoring pO K in OK comes down to factoring 1 E IFp[X], where f is the minimal polynomial of any primitive element of K that lives in OK.

Theorem 9.B.22. (Dedekind, Kummer) Let K = Ql(x) be a number field, where x E OK has minimal polynomial f. Let p be a prime which does not divide 10 IOK/Z[X] I and let 9

1 = IIf/

i=l

be the prime factorization of1 E IFp[X]. Then

9.B. A Glimpse of Algebraic Number Theory

465

Proof. Lift arbitrarily fi to some monic polynomials fi E Z[X]. The key point is to prove that the natural map Z[X]/(p, fi) -+ Z[x]/(p, fi(x)) sending (the class of) h to (the class of) h(x) is an isomorphism of rings. Assume for a moment that we proved this. Let a = IOK/Z[X] I, so gcd(p,a) = 1 and aOK C Z[x]. A standard argument using Bezout's lemma shows that OK = pOK + Z[x], thus OK/Pi is naturally isomorphic to Z[X]/(p,li(X)). Since Z[X]/(p, fi(X)) is clearly isomorphic to IFp[Xl/(fi) , which is a field with pdeg li elements, it follows from the previous discussion that the Pi are different prime ideals with N(Pi) = pdegli . Since i C pOK + fi(XYiO K , f(x) = 0 and f == I1f=l Iti (mod pZ[X]) , it follows that I1f=l p:i C pOK and so by theorem 9.B.14 we have pOKI I1f=l p:i. So, if pOK = I1f=l pfi, we must have Si ::; ei· We conclude by noting that both sums L:i ei . deg fi and L:i Si' deg fi are equal to [K : Ql] (see remark 9.B.16), so we must have Si = ei for all i. Let us prove now that Z[X]/(p, fi) -+ Z[x]/(p, fi(X)) is bijective. Since surjectivity is clear, it remains to prove the following assertion: if h E Z[X] satisfies h(x) E pZ[x] + fi(X)Z[X]' then h E pZ[X] + fi(X) . Z[X]. Write h(x) = pA(x) + fi(x)B(x) for some A, B E Z[X]. Since f is the minimal polynomial of x, there is r E Z[X] such that h = pA + fiB + r f. It suffices to use again that f E pZ[X] + fi . Z[X] to finish the proof. D

Remark 9.B.23. It is not very difficult to prove that if p does not divide the discriminant of f, then p does not divide IOK/Z[x]l, so the theorem can be applied.

9.B.5

Two classical examples

pOK = where Pi = pOK

+ fi(X)OK

II P:\ i=l

are different prime ideals and N(Pi) = pdegli .

9We have I:l'Jlp e(f3!r.;)· f(f3/r.;) = [L: K], by an argument similar to that used in remark 9.B.16. lONote that the result stated in remark 9.B.11 shows that OK/Z[X] is a finite set.

Consider a squarefree integer d =I- ±1 and let K = Ql( y'd). We saw that OK = Z[x], where x = 1+2# if d == 1 (mod 4) and x = y'd otherwise. The minimal polynomial of x is X2 - X + 14: d in the first case and X2 - d in the second case. Theorem 9.B.22 shows that in order to understand the prime factorization of pOK, we need to understand the prime factorization of these polynomials modulo p. Since a quadratic polynomial modulo p is irreducible if and only if it has a root in IFp, we easily deduce the following

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Proposition 9.B.24. Let d =I- ±1 be a squarefree integer and let K Let p be a prime. a) If p

> 2,

then pOK is a prime ideal if

different prime ideals if

(~) =

= Q( Jd).

(~) = -1, a product of two

1 and the square of a prime ideal if pld.

9.B. A Glimpse of Algebraic Number Theory

b) If pin and n where s

9.B.6

= pk· m

with gcd(m,p)

= 1,

467

then pOK

= (Pl ... ··Ps)pk_pk-l,

ord(::2d m) and each Pi is of degree ord(p mod m).

The primitive element theorem and embeddings of number fields

b) Ifp = 2, thenpOK is the square ofaprime ideal ifd == 2,6,3,7 (mod 8), a prime ideal if d == 5 (mod 8) and a product of two different prime ideals if d == 1 (mod 8).

When working ~ith subfields of C, the following result is very handy. We will use it constantly to shorten proofs of results which actually hold in much greater generality.

Consider now n > 1 and let K = Q((n), where (n is a primitive nth root of unity. As we have already said, it is rather difficult to prove that OK = Z[(n] and we will take this for granted (actually, it is easier to use remark 9.B.23). In this case, theorem 9.B.22 reduces the prime factorization of pOK to that of tPn E lFp[X], where tPn is the nth cyclotomic polynomial (whose roots are precisely the primitive nth roots of unity). Assume that p does not divide n and let 9 be an irreducible factor of degree f of tPn E lFp[X]. Let x be a root of 9 in an algebraic closure of lF p . Theorem 9.A.4 shows that g(X) = n{==-~ (X - x p } Let d be the order of p modulo n. Since n is the least positive power of x equal to 1 (because x is a root of tPn), it follows that the sequence x p' is periodic with period d, so we must have f = d. That is, tPn factors mod p as a product of irreducible polynomials of degree d, the order of p modulo n. Since xn - 1 is squarefree modulo p (because the hypothesis that p does not divide n implies that xn - 1 is relatively prime to its derivative), we deduce that in theorem 9.B.22 we have el = e2 = ... = e g = 1 and fi = d for all i, so 9 = <p~n). Assume now that pin and let n = pk . m for some m relatively prime to p. It is easy to see that tPn(X) = tPm(XPk)/¢m(XPk-l), so

Theorem 9.B.26. (primitive element theorem) Let L/ K be a finite extension of subfields ,ofe. Then there exists l E L such that L = K(l).

k-l

modulo p we have tPn = tPfn -p . This reduces the problem of factoring pOK to the previous case. All in all, we have the following useful result: Proposition 9.B.25. Let n be an integer greater than 1, let K = Q((n) and let p be a prime. a) If gcd(n,p) = 1, then pOK is a product of or d(p<pen) rno d ideals, each of degree ord(p mod n).

different prime

Proof. As L / K is finite, there are elements Xl, . .. ,xn E L such that L K(Xl, X2,···, x n ) (for instance, the elements of a K-basis of Lover K). Thus, by induction on n it is enough to prove that if x, yare algebraic over K, then there exists l E L = K(x, y) such that L = K(l). Let f, 9 E K[X] be the minimal polynomials of x, y respectively and let Xl = x, X2,·· . ,Xn and Yl = y, Y2,· .. ,Ym be their roots. Clearly, there exists c E K such that Xi + eYj =I- x + ey for all (i,j) =I- (1,1) (each of the previous linear equations in c has at most one solution in K and K is infinite). We claim that l = x + ey works. Clearly K(l) C L, so it is enough to check that x, y E K(l). As l = x + ey, it is enough to do it for y. Now, we know f(l- ey) = 0, so the polynomial f(l - eX) E K(I)[X] has a common root y with g. But by construction this is the only common root of these polynomials. Moreover, it is a simple root, as 9 is irreducible over Q, so it has simple roots. Finally, we conclude that the greatest common divisor of these two polynomials is X - y. As these two polynomials have coefficients in K(l), so does their greatest common divisor and so y E K(l). The result follows. D

We will use the primitive element theorem to prove some basic results on the structure of number fields. The first one concerns the embeddings of a number field in Co Note that Q( V2) has two such embeddings, namely the identity map and a + bV2 r----t a - bV2. It turns out that a number field with degree d has exactly d embeddings in Co

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Theorem 9.B.27. Let L/ K be an extension of number fields (L/ K is automatically finite). Then any embedding K -+ C extends to exactly [L : Kl embeddings L -+ C.

Proof. Fix an embedding a : K -+ C and use the primitive element theorem to write L = K(l) for some I E L. Then I is algebraic over K, of degree d = [L : Kl, with minimal polynomial f E K[Xl· Suppose that a' : L -+ C is an embedding that extends a, so a'(x) = a(x) for all x E K. Thus, if 9 E K[XJ, then a'(g(l)) = gO'(a'(l)) (where gO' is the polynomial obtained from 9 by applying a to its coefficients) and so a' is determined by a' (I), which has to be a root of (use the previous equality with 9 = J). Conversely, if I' is a root of we can define an embedding a'(g(l)) = gO'(I') for all 9 E K[XJ, well-defined because any two polynomials 9 and h with g(l) = h(l) differ by a multiple of f. Thus the embeddings of L into C that extend a are in bijection with roots of and the result follows. 0

Taking K = Q in the previous theorem, we deduce that any number field L of degree dover Q embeds in exactly d ways in <C.

9.B.7

A bit of Galois theory

Let L, M be two extensions of a field K and suppose for simplicity that they are contained in <C. A K - morphism L -+ M is a K -linear map from L to M which is also a ring homomorphism. Stated differently (but equivalently), it is an additive and multiplicative map f : L -+ M such that f(x) = x for all xEK.

Definition 9.B.28. Let f E K[Xl. The splitting field of f is the field K(XI' X2,.··, x n ), where Xi E C are the roots of f. Note that a K-morphisrp f : K(XI, ... , xn) -+ M (where M/ K is an extension) is uniquely determined by the values f(Xi), as any element of K(XI' .. . , xn) is a polynomial with coefficients in K in the Xi'S. However, f(Xi) cannot be any element of M, since if P E K[X] kills Xi, then P also kills f(xd: note that P(f(Xi)) = f(P(Xi)) = 0, as f is K-linear and multiplicative. Moreover, there might be algebraic relations with coefficients in K

9.B. A Glimpse of Algebraic Number Theory

469

between the Xi'S and the same argument shows that the numbers f(Xi) must satisfy these relations. Thus, it is a fairly delicate issue to understand these K-morphisms. This is the content of Galois theory. Let us make an important definition first:

Definition 9.B.29. Let L be an extension of a field K. The Galois group of Lover K, denoted Gal(L/ K) is the set of bijective K-morphisms f: L -+ L. We can now prove one basic result of Galois theory, which is also of constant use:

Theorem 9.B.30. Let

L/ K

be an extension of number fields. Then

IGal(L/ K)I ::; [L : K],

with equality if and only if L is the splitting field of some polynomial f E K[Xl. Proof. Let L = K(l) for some I E L. Each element a E Gal(L/ K) is uniquely determined by a(l), which must be a conjugate of I. Thus we have a natural injection of Gal(L/ K) in the set of conjugates of I, which has [L : Kl elements. The inequality follows. Suppose that we have equality and let f E K[Xl be the minimal polynomial of I. We will prove that L is the splitting field of f. It is enough to prove that if x is a conjugate of I, then x E L. But the first paragraph and the equality case implies that if iI, ... ,In are the conjugates of I, then {a(l)la E Gal(L/ K)} = {iI, ... , In}. Thus there exists a E Gal(L/ K) such that x = a(l). Since a(L) C L, we have x ELand we are done. Conversely, suppose that L is the splitting field of some f E K[X], with roots Xl, X2,· .. , Xn · Theorem 9.B.27 applied to the natural inclusion map K -+ C shows that there are [L : Kl K-linear morphisms of rings L -+ <C. But for any such morphism a : L -+ C we have a(xd E L, as a(xi) is just some root of f and L contains all these roots. Thus the image of any such morphism is a subset of L, i.e. any such morphism is an element of Gal(L/ K). The result follows. 0 Definition 9.B.31. We say that a finite extension L/ K of number fields is Galois if IGal(L/ K)I = [L : Kl, or, equivalently, if L is the splitting field of a polynomial with coefficients in K.

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We are now able to prove the main theorem of Galois theory for number fields. The result holds in much greater generality, but we will not need it and we prefer to use all the extra data in order to shorten the proof rather dramatically. ~

Theorem 9.B.32. Let L/ K be a finite Galois extension of number fields. Sending a subfield M of L containing K to Gal(L/M) yields a bijection between subfields of L which contain K and subgroups of Gal(L/K). The inverse of this bijection is the map sending the subgroup H of Gal(L/ K) to

{x E LJdx)

x, VO" E H}.

Moreover, we have H1 C H2 if and only if LH1 contains LH2. Proof. Let us prove first that LGa1(L/M) = M for any intermediate field M between Land K. Note that L/M is again finite and Galois (as L is the splitting field of a polynomial with coefficients in K, thus also in M). Write L = M(l) for some primitive element l E L, with conjugates h,l2, ... ,zn over M. Suppose that x E LGal(L/M) is not in M. Thus, we can find f E M[X] nonconstant of degree less than n and such that x = f(l). We saw in the proof of the previous theorem that {O"(l)JO" E Gal(L/M)} = {h, ... , In}. So, for any i we can find O"i E Gal(L/M) such that O"i(l) = Ii. Then x = O"i(X) = f(li). Hence f(h) = f(12) = ... = f(ln), contradicting the fact that f is nonconstant of degree less than n. The result follows. Next, we need to prove that if H is any subgroup of Gal(L/ K), then Gal(L/ LH) = H. By the very definition of LH we have an inclusion H c Gal(L/ L H ). It is thus enough to check that JGal(L/ LH)J ::; JHJ and, using the previous theorem, it is enough to check that [L: LH] ::; JHJ. Write L = LH (I) for some I E L. Consider f(X) = ITaEH(X - 0"(1)). This is a polynomial of degree JHJ whose coefficients are clearly in LH. Hence I has degree at most JHJ over LH and the result follows. It remains to check that H1 c H2 if and only if LH1 contains LH2. It is clear that if H1 c H 2, then LH1 contains LH2 (any element of L fixed by H2 is also fixed by H1)' Assume that LH1 contains L H2, so LH1 n LH2 = LH2. But it is clear that the left-hand side of this equality is simply L H , where H

9.B. A Glimpse of Algebraic Number Theory

is the subgroup of Gal(L/ K) generated by H1 and H2. Hence LH as we have seen, this forces H = H2 and so H1 C H 2.

471

= LH2 and, D

Remark 9.B.33. Using similar arguments, it is not difficult to show that His • normal in Gal(L/K) (i.e. gHg- 1 = H for any g E Gal(L/K)) if and only if LH / K is Galois.

9.B.8

Prime factorization in a Galois extension

The results in this section will be crucially used in the applications that will be presented at the end of this addendum. They are absolutely fundamental in algebraic number theory. Theorem 9.B.34. Let L/ K be a Galois extension of number fields, with G = Gal( L / K). Let p be a nonzero prime ideal of OK and let 131 and 132 be two prime ideals of OL which lie over p. Then there exists 0" E G such that ll

132 = 0"(131)' Proof. Suppose that 132 -=J. O"(13d for all 0" E G. Using the general version 12 of the Chinese Remainder Theorem, we obtain the existence of a E 132 such that a t/: 0"(131) for all 0" E G. Then ITaEG O"(a) E OK n 132 = pC 131 and this contradicts the fact that 131 is a prime ideal and a t/: 0"-1 (13d for any 0" E G. The result follows. D Corollary 9.B.35. Let L/ K be a Galois extension of number fields and let p be a nonzero prime ideal of OK. Then all primes above p have the same ramification index and the same residual degree.

Proof. For the residual degrees, note that any 0" E G induces a bijection between OL/13 and OL/O"(13) , so these two sets have the same number of elements. We conclude by the previous theorem. Next, if e is a positive integer such that llNote that by definition a((3r) is the set of all a(x), with x E (31. It is easy to check that this is again a prime ideal of OL and that a((31) n OK = gJ. 12This is stated as follows: let A be a commutative ring and let h, h, ... ,In be ideals such that Ii + I j = A for all i #- j (which is satisfied if Ii are different maximal ideals of A, for instance). Then for any Xl, .•• ,X n E A there exists X E A such that x - Xi E Ii for all i.

Chapter 9. A Little Introduction to Algebraic Number Theory

472

9.B. A Glimpse of Algebraic Number Theory

473

(3e divides SJOL, then O"((3)e divides O"(SJOL) easily fr~m the previous theorem.

= SJOL. The result follows again 0

Definition 9.B.36. Let L / K be a Galois extension of number fields, with Galois group G. Let (3 be a prime of OL. The decomposition group of (3 is

Note that D(3 is indeed a group and that any 0" E D(3 induces an automorphism 0" : OL/(3 -+ 0L/(3, which is trivial on OK/SJ for SJ = (3 n OK. Hence, we have a natural map D(3 -+ Gal((OL/(3)/(OK/SJ)). Note that Gal((OL/(3)/(OK/SJ)) is a cyclic group, generated by the automorphism x -+ xN(p), since (OL/(3)/(OK/SJ) is a finite extension of finite fields (see the addendum 9.A for the structure of finite fields). The following result is rather tricky, but fundamental.

Theorem 9.B.37. With the previous notations, the map

is surjective. Proof. We may assume that OL / (3 -# OK / SJ, as otherwise the statement is trivial. Let a be a generator of the cyclic group (0 L / (3) *, so that clearly oL / (3 = (OK / SJ )[a]. Using the general form of the Chinese Remainder Theorem (see the proof of theorem 9.B.34), we can find a E OL such that a (mod (3) = a and a E (3' for any (3' -# (3 above SJ. Let F be the minimal polynomial of a over K. Then F E OK[X] and F(a) = 0, hence F(a) = 0 in OL/ (3. But then 13 F(aN(p)) = 0, so that F(aN(p)) E (3. So, we can find a conjugate b of a so that b - aN(p) E (3. It is not difficult 14 to see that there 13Recall that if lFq is the field with q elements, then for all f E lFq[X] we have f(xq) = f(X)q, so if a is a root of f in some extension of lFq, then a q is also a root of f· 14Since L/ K is Galois, L contains all conjugates of a over K, i.e. all roots of F. Let M be the field generated over K by these conjugates, i.e. the splitting field of K. Then M / K is a Galois subextension of L / K, thus any element of Gal( M / K) extends to an element of Gal(L/K). But the isomorphisms K[X]/(F) -+ M sending X to a and b respectively yield <7 E Gal(M/K) such that <7(a) = b.

exists 0" E Gal(L/ K) such that O"(a) = b. We claim that 0" is in D(3 and that it induces the automorphism x -+ xN(p) of 0L/(3. First, if 0"((3) #= (3, then 0"-1((3) -# (3 and so by our choice of a we have a E 0"-1((3), i.e. O"(a) E (3. But O"(a) - aN(p) E (3, forcing aN(p) E (3 and then a E (3, a contradiction with a (mod (3) = a -# O. Thus 0" E D (3. The automorphism induced by 0" on 0 L/ (3 is uniquely determined by its action on a (generator of (OL/(3)*) and by our choice this action is aN(p). The result follows. 0 Let us fix a prime (3 above SJ in L. By theorem 9.B.34, all primes above SJ are of the form 0"((3) with 0" E Gal(L/ K), i.e. Gal(L/ K) permutes transitively the different primes (31, ... ,(3g over SJ· Since there are precisely 1D (31 elements of Gal(L/ K) which fix (3, we obtain [L : K] = IGal(L/ K)I = g ·ID(3I. But we also have [L : K] = e((3/ SJ)· f((3 / SJ ).g, since all ramification indices and residual ~ degrees of the (3i' S are the same. Hence 1D (31 = e ((3/ SJ) . f ((3/ SJ). In particular, if SJ is unramified in L, i.e. if e((3/SJ) = 1, then ID(31 = f((3/SJ), which is the same as the cardinality of Gal((OL/(3)/(OK/SJ)). We obtain therefore the following crucial result:

Theorem 9.B.38. Let L/ K be a Galois extension of number fields and let SJ be a prime of OK unramified in OL. Let (3ISJ be a prime of OL over SJ. Then the map D(3 -+ Gal( (OL/ (3) / (OK / SJ)) is a bijection. In particular, there exists a unique ((3,L/K) E D(3 such that ((3,L/K)(x) == xN(p) for all x E OL. We call ((3, L/ K) the Frobenius substitution of (3 in L/ K.

Remark 9.B.39. Assume that we are only given SJ. The choice of some (3 over SJ yields a Frobenius substitution ((3, L/ K), but this depends on the choice of (3. However, any other prime above SJ is of the form 0"((3) for some 0" E Gal(L/ K) and it is immediate to check that (0"((3), L/ K) = 0" 0 ((3, L/ K) 00"-1 (indeed, the right-hand side is in Da«(3) , which is trivially equal to 0" D(3O"-l, and satisfies the congruence that uniquely characterizes it). Hence, the conjugacy class of ((3, L/ K) in Gal(L/ K) does not depend on the choice of (3 and is called the Frobenius conjugacy class of SJ. Note that if Gal(L/ K) is abelian, then this conjugacy class is reduced to an element, which we denote (SJ, L/ K): it is ((3, L/ K) for any (3 over SJ· We leave to the reader to check the following easy results:

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Proposition 9.B.40. Let M / L / K be a tower of Galois extensions of number fields and let pl,8lp be a tower of prime ideals. Assume that p is unramified in M (thus,8 is unramified in M and p is unramified in L). Then we have the following relations.

a) (p,M/L) = (p,M/K)f({3/pl. Actually, this relation still holds if M/K is nor necessarily Galois.

:;:::::=

b) (p, M/ K)I L

9.B.9

= (,8, L/ K).

Bauer's theorem and Chebotarev's density theorem

475

9.B. A Glimpse of Algebraic Number Theory

We will also need the following easy consequence of Chebotarev's density theorem: Proposition 9.B.43. Let 0" E Gal(L/ K). There are infinitely many prime ideals p of degree 1 of K such that one can find ,8lp in L with (,8, L/ K) = 0". Proof. Take a Galois extension E over Q which contains L. Since L/ K is Galois, we can extend 0" to an automorphism again denoted 0" of Gal(E/ K). By Chebotarev, we can find infinitely many primes p such that there is pip in E with (p, E /Q) = 0". Then Dp C Gal(E / K), so p = p n K has degree 1 (lemma 9.B.49). Let ,8 = p n OL. Then

Before stating the following fundamental result, we need one more definition. If S is a set of primes, its Dirichlet density is

(,8,L/K)

(p,E/K)I L

(p,E/Q)IL

and we are done.

if the limit exists. The next very deep theorem was conjectured (and proved in some special cases) by Frobenius. It is a vast generalization of Dirichlet's theorem. Theorem 9.B.41. (Chebotarev) Let K be a finite Galois extension of Q of degree n and let 9 E Gal(K/Q). Let S be the set of primes p such that (g--J, K/Q) is conjugate to 9 for all prime divisors p ofp. Then deS) = ~·I{hgh-llh E G}I. Here is a typical application of this deep theorem. Example 9.B.42. Let l be a prime and consider the set S of those primes p == 1 (mod l) such that 2Y == 1 (mod p). These two conditions are equivalent to the statement that Xl 2 splits into distinct linear factors in IF'p[X] (as

2Y == 1 (mod p) is equivalent to the existence bf y E IF'p such that yl = 2). This is also equivalent to p being a product of different primes in OK, where K = Q( e 2;71" ,-1'2) is the splitting field of Xl - 2. Using again Chebotarev's 2 theorem, we deduce that deS) = [K~Q}. But K contains (Q(e ;71") and Q(-1'2), which have degrees l - 1, respectively l. Thus [K : Q] is a multiple of l(l - 1) and so necessarily [K : Q] = l(l - 1) and deS) = 1(1~1).

Definition 9.B.44. If K is a number field, let Pl (K) be the set of rational primes p for which pO K has at least one prime ideal factor p of residual degree 1 (i.e. such that OK / p is the field with p elements). The following result explains why Pl(K) is an interesting object: Proposition 9.B.45. Let f E Z[X] be a monic irreducible polynomial, let B be a root of f and let K = Q( B). Then there is c such that Pl(K)

n [c, 00) = {p 2 cl:3x

E Zsuch that

plf(x)}.

Proof. If p is a sufficiently large prime, then the residual degrees of the prime divisors of pO K are given by the degrees of the irreducible factors of 1 E IF'p[X], by theorem 9.B.14. Hence, for such p we have p E Pl (K) if and only if has a linear factor in IF'p[X], i.e. if and only if there exists x E Z such that plf(x). 0

Here is a rather nice application of this proposition: . Theorem 9.B.46. (Nag ell) If f E Z[XJ, let P(f) be the set of primes p for which the congruence f (x) == o· (mod p) has at least one solution. If !1, 12,·· ., fk are nonconstant polynomials with integer coefficients, then P(!1) np(12) n··· n P(fk) is infinite.

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Proof. The case k = 1 is a classical result due to Schur, but for the reader's convenience let us recall the proof. If II (0) = 0, everything is clear. Consider the numbers Xn = J((2f~~r(OÂť. We have IXnl > 1 and Xn == 1 (mod n!) if n is large enough, so there exists Pnlxn with Pn > n and the result follows. The general case is more difficult. We may assume that fi are irreducible monic. Let Zi be roots of fi and let Ki = <Q(Zi). Let K be the least number field containing all K/s and write K = <Q(x) for some x E OK with minimal polynomial f. Applying the case k = 1 to f and using the previous proposition, we see that we can find infinitely many p which have a prime p of degree 1 in K. Let f3i = p n OKi Âˇ Since f(p/p) = 1, we also have f(f3dp) = 1. Applying once more theorem 9.B.22 (this time for each Kd we see that all but finitely many primes p (among the infinitely many we have just found) belong to n~=1 P(Jd, which is therefore infinite. D

The following beautiful theorem due to Bauer was discovered before Chebotarev proved his theorem. It is however more conceptual nowadays to see Bauer's theorem as a consequence of Chebotarev's density theorem, which is what we will do.

'{ Theorem 9.B.47. (Bauer) Let KI be a number field and let K2 be a number field which is Galois over <Q. Suppose that there exists a set of primes S of Dirichlet density 0, with the following property: if p tJ- S is in PI(KI), then p is completely split in K2. Then K2 c KI. Proof. We start with two easy lemmas, which express the condition p E PI (K) in a group-theoretic way. They are also results of independent interest and will be used in other applications.

Lemma 9.B.48. Let L be a number field which is Galois over <Q, with group G. Let <Q eKe L be a subfield and let H = Gal(L/K). Suppose that f3lplp is a chain of prime ideals in L, K, <Q such that p is unramified in L. Then f(p/p) = 1 if and only if D(3 c H. Proof. Note that the decomposition group of 13 with respect to the extension L/ K is simply D(3 n H. But since p is unramified in L, p is unramified in L / K and so the decomposition group of 13 with respect to L / K has f (13 / p)

9.B. A Glimpse of Algebraic Number Theory

elements. That is, c 11 The resuIt 10 ows.

IH n D(31

477

= f(f3/p). But this equals also J((3/p) = ~IDf:l1. J(p/p)

J(p/p

Lemma 9.B.49. Assume that Land K are as in the previous lemma and let p be a prime which is unramified in L. Then 1) p has a prime factor p in K with f(p/p) = 1 if and only if there is f3lp in L such that D(3 C H.

2) p is completely split in K if and only if for all f3/p in L, we have D (3

Proof. 1) is an immediate consequence of the previous lemma. For 2), it is clear that if p is completely split in K, then for any f3lp we have D(3 C H (as if p = f3nOK, then necessarily f(p/p) = 1). Conversely, if D(3 c H for any f3lp, then the previous lemma implies that f(p/p) = 1 for any pip in K. But note that p is unramified in K, as it is already in L. Thus p must be completely split in K. D

Let us prove now the theorem. ChooseI 5 a finite Galois extension L of <Q containing KI and K2 and let H j = Gal(L/Kj). By the main theorem of Galois theory (theorem 9.B.32), it is enough to prove that HI c H2. Choose any a E H l . By Chebotarev's theorem, there is p tJ- S such that p has a prime factor 13 in L with D(3 = (a) (cyclic subgroup of HI generated by a). By lemma 9.B.49, we have p E PI (K) and so p is totally split in K 2 , which forces, by the same lemma, D(3 C H 2. Hence a E H2 and we are done. D Remark 9.B.50. Here is another useful consequence of lemma 9.B.49. Let n > 1 be an integer and let H be a subgroup of (Z/nZ)*. We claim that a prime p not dividing n is totally split in <Q((n)H if and only if p (mod n) E H. Indeed, such a prime is unramified in <Q( (n) and by lemma 9.B.49 it is totally split in <Q((n)H if and only if for any f3lp in <Q((n) we have D(3 C H. Since Gal(<Q((n)/<Q) is abelian, this is also equivalent to (p, <Q((n)/<Q) E H, which is saying that p (mod n) E H (we naturally identified H with a subgroup of Gal(<Q( (n) /<Q)). 15For instance, if Kl = Q(x) and K2 all conjugates of x and y.

= Q(y),

consider the extension generated over Q by

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Finally, a reward: applications to "elementary-looking" problems

In this section we show how the deep results in the previous sections can be combined to yield some fairly nontrivial theorems with "elementary" aspects. The first one uses a generalization of Bertrand's postulate to obtain the following difficult irreducibility result.

I Theorem 9.B.5L (Schur) IrtJ n > . _ 2 an d aI, a2, ... ,an-I are zntegers, th en the polynomial xn

-n! + an-I'

xn-I (n - I)!

+ ... + a2 . -2! + alX + 1

is irreducible in Q. Proof It is enough to prove that f = xn + nan_IXn-1 + ... + n! is irreducible in Z[XJ. Suppose that this is not the case and let g be a monic irreducible factor of f of degree m, with m ~ ~. Choose any prime divisor p of n( n _ 1) ..... (n - m + 1) and cons~er the reduction of mod p. The choice of p im~lies that xn-m~1 divides f. If f = g. h, it follows that xn-m+1 divides ?] . h and since deg h = n - m, we must have XI?] and so plg(O). Let z be a root of g and let K = Q[z], a number field of degree m. Since plg(O), it follows that p divides N(z· OK) and so we can choose a prime p of OK dividing ZOK and such that pip. Let e = e(p/p) ~ [K : QJ = m be the ramification index of p. Since f(z) = 0, we obtain (set an = 1)

so we can find i such that vp

(z.") >. _ zVp () z

vp(z) = z. . e

i -.

Using the inequalities vp ( i!) < P~I and e ~ m, we deduce that p ~ m. We have therefore proved that all prime factors of n( n -1) ..... (n - m + 1) are less than or equal to m. This contradicts a famous theorem of Sylvester

9.B. A Glimpse of Algebraic Number Theory

479

and Schur, which generalizes Bertrand's postulate and for a proof of which we refer the reader to [26J. 0

,Theorem 9.B.52. Ifn ~ 2k, then (~) has a prime divisor greater than k. Let us consider now the second application.

Theorem 9.B.53. (Davenport, Lewis, Schinzel) Let f E Q[XJ be a polynomial with the following property: any arithmetic progression P c Z contains an integer x such that f(x) is the sum of squares of two rational numbers. Then there exist polynomials g, hE Q[XJ such that f = g2 + h 2. Proof. We may assume that f is not constant. Using Gauss' lemma, we can write f = Cf~l ..... f!/nm for some c E Q*, some primitive, distinct, irreducible polynomials fi E Z[XJ and some ei E N*. Fix an index j for which ej is an odd number and let () be a root of fj. We claim that L = Q( ()) contains Q( i). Using part a) of Bauer's theorem 9.B.47, it is enough to prove that if q is a sufficiently large prime in PI (Q( ())), then q E PI (Q( i)). We will need the following standard argument:

Lemma 9.B.54. There exists a prime qo with the following property: for all q E PI(L) n [qO, (0), one can find an integer x such that vq(Jj(x)) = 1 and Vq(Jk(X)) = for all k i- j.

Proof First, choose a prime ql > [0£ : Z[()JJ. Take a prime number q2 greater than all prime factors of (the numerator or denominator of) c and such that q does not divide gcd(fJ(x), fj(x) . IIk~j fk(X)) for any x E Z and any q ~ q2· To prove the existence of q2, note that fJ is relatively prime to fj . IIkh lk, write a Bezout relation over Q for fj and fj· IIk~j fk and clear denominators. We claim that qo = max(ql, q2) works. Indeed, let q E PI (L) n [qo, ooJ. By Dedekind-Kummer's theorem 9.B.22, we can find y E Z such that q divides fJ (y). Then q does not divide fj (y) or IIkh fk (y), so one of y or y + q satisfies the desired conditions. 0 Now, fix qo as in the previous lemma and let q E PI (L) n [qo, (0) and x as in the lemma. By hypothesis we can find y == x (mod q2) such that f(y) = a 2 + b2 for some rational numbers a, b. Note that vq(Jj(Y)) = 1 and

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Chapter 9. A Little Introduction to Algebraic Number Theory

Vq(Jk(y)) = 0 for k -=1= j. Then ej = vq(J(y)) = vq(a 2 + b2) is odd by our choice of y. We deduce that q == 1 (mod 4) and so q is split in Ql(i), hence q E P l (Ql( i)). This proves the claim made in the previous paragraph and we conclude that i = A E L. Hence we can write i = h(O) for some h E Ql[X]. Then h must divide h 2 + l. If G = gcd(h - i, fj) E Ql(i)[X], it is easy to see that NQ(i)/Q(G) = gcd(h - i, fj) . gcd(h + i, h) is a polynomial with rational coefficients dividing h. We deduce that fj is a constant times NQ(i)/Q(G), and this last polynomial is obviously the sum of squares of two polynomials with rational coefficients. Applying the previous arguments for each h such that ej is odd, we deduce that f is of the form c(u 2 + v 2 ) for a constant c and some u, v E Ql[X]. But using once more the hypothesis on c we deduce that c is the sum of squares of two rational numbers and the result follows. 0 Remark 9.B.55. Actually, Davenport, Lewis and Schinzel prove a more general result: let K/Ql be a Galois extension of degree n and let Vl, V2,"" Vn be such that OK = .!ZVl + .!ZV2 + ... + .!Zvn . Suppose that f E Ql[X] has the property that any arithmetic progression of integers contains an integer x for which the equation f(x) = NK/Q(UlVl + U2V2 + ... + unvn ) is solvable in (Ul' U2,"" un) E Qln. If either Gal(K/Ql) is a cyclic group or the multiplicity of any zero of f is prime to n, then one can find Ul, U2,"" Un E Ql[X] such that f(X) = NK/Q(Ul(X), U2(X), ... , un(X)). The proof follows precisely the same ideas, but is a bit more technical. 'Theorem 9.B.56. (Schinzel) Let f E .!Z[X] be an irreducible polynomial over Ql and let n be an integer greater than 1. Suppose that there is c with the following property: if p 2': c divides f(x) for some x E .!Z, then p == 1 (mod n). Then there are a E Ql and a polynomial 9 E Ql( (n) [X] such that

Proof. Let 0 be a root of f an let K = Ql(O). We claim that Ql((n) C K. By Bauer's theorem, it is enough to prove that if p is a large prime having a factor of degree 1 in K, then p is completely split in Ql( (n). But if p is such a prime,

9.B. A Glimpse of Algebraic Number Theory

481

greater than c and n, then p divides some f(x), so p == 1 (mod n) and p is completely split in Ql( (n). Now, write (n = u(O) for some u E Ql[X]. Then, since f is irreducible and <Pn(u(O)) = 0, the polynomial f divides <Pn 0 u in Ql[X] (recall that <Pn is the nth cyclotomic polynomial, minimal polynomial of (n over Ql). For j relatively prime to n, define h = gcd(J, u - (~) E Ql((n)[X]. Let (Tj E Gal(Ql((n)/Ql) be the automorphism sending (n to (~. Then clearly fj = (Tj(!1), so

h = N Q((n)/Q(!1)Âˇ

gcd(j,n)=l

Clearly, the polynomials h are pairwise relatively prime, so their product divides f. But, as we have seen, their product has rational coefficients, hence by irreducibility of f we must have f = a . NQ((n)/Q(JI) for some constant a. The result follows. 0

Remark 9.B.57. It is much easier to check that the converse of the theorem also holds. "Theorem 9.B.58. (Murty) Let f E .!Z[X] be a polynomial and let l, n be relatively prime positive integers. Let S be the set of primes p for which the congruence f(x) == 0 (mod p) has solutions in.!Z. Suppose that there exists c such that any pES greater than c satisfies p == 1 (mod n) or p == l (mod n). Also, suppose that there are infinitely many primes p == l (mod n) in S. Then l2 == 1 (mod n).

Proof. Let 0 be a root of f and let K = Ql(O) and L = K((n), so L/K is a finite Galois extension (splitting field of xn - 1). Let H be the subgroup of Gal(Ql((n)/Ql) generated by the automorphism (Tt, which sends (n to (~. Note that the hypothesis combined with Bauer's theorem and with remark 9.B.50 yield an inclusion M c K, where M = Ql((n)H. Here is the crucial technical result: Lemma 9.B.59. Restriction to Ql( (n) yields an isomorphism between Gal(L/ K) and H.

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Chapter 9. A Little Introduction to Algebraic Number Theory

Proof. Since M c K, the restriction of any a E Gal(L/ K) yields an element of H = Gal(Q(n)/M). It is clear that if the restriction of a is trivial, then a is trivial, as L = K(n). Surjectivity is more delicate. Choose a prime p'-= I (mod n) which is unramified in L (this excludes only finitely many primes) and for which K has a prime pip of degree 1. The existence of p follows from the second hypothesis. Let (3 be a prime of L lying over p and let p = (3nQ(n), a prime ofQ(n) lying over p. Let a = ((3,L/K) be a Frob~nius automorphism associated to (3. Since p has degree 1, we have a( x) -= x P (mod (3) for all x E OL. We claim that the restriction of a to Q(n) is at, which will prove the surjectivity and end the proof of the lemma. But this restriction, call it T, is an element of Gal(Q(n)/Q) which preserves p and such that T(X) -= x P (mod p) for all x E Z[(n], thus it has to be (p,Q(n)/Q) = ap (it is easy to see that a p satisfies the properties which uniquely characterize (p, Q(n)/Q)). It remains to observe that a p = at, as P -= I (mod n). D

It is now easy to finish the proof of the theorem: consider at2 = E H. By lemma 9.B.59 there is a E Gal(L/ K) restricting to at2. By proposition 9.B.43 we can find infinitely many primes p of degree 1 of K such that ((3, L/ K) = a for some (3lp. These primes p correspond therefore to primes p such that p -= 12 (mod n) and for which plf(x) for some x E Z. The first hypothesis yields therefore 12 1 (mod n) or [2 -= I (mod n). The result follows. D Using the tools developed in this addendum, we are also able to prove theorem 2.A.6. For the reader's convenience, let us recall the statement. The proof is adapted from the beautiful paper [32], where a much more general "'result is proved.

'V Theorem

9.B.60. Let n > 1 and a be integers such that a is an nth power modulo p for all sufficiently large prime numbers p. Then either a is the nth power of an integer or 81n and a = 2~bn for some integer b.

Proof. We will use without comment the following well-known, but nontrivial result: if an integer a is a perfect square mod p for all sufficiently large primes

9.B. A Glimpse of Algebraic Number Theory

483

p, then a is a perfect square. While elementary proofs exist,16 this statement is also an immediate consequence of Chebotarev's density theorem 9.B.41 applied to the splitting field of X2 - a. So, if.n is even, we already know that a is a perfect square, in particular positive (the case a = 0 is trivial and excluded in this proof). Hence, by replacing a by -a if n is odd, we may and will assume that a > O. The crucial ingredient of the proof is the following application of Bauer's theorem 9.B.47:

Lemma 9.B.61. If z is a complex root of xn - a, then Q(z) 2i7r (n = en.

Q(n), where

Proof. Let Kl = Q( (n) and let K2 be the splitting field of xn - a over Q. It is enough to prove that K2 c K 1 . Using Bauer's theorem, it suffices to prove that for all sufficiently large primes p E PI (K1 ), p is completely split in K2. If p does not divide n (which certainly happens if p is large enough), the condition p E PI (KJ) is equivalent to nip - 1 by proposition 9.B.25. Also, if p is large enough, then a is an nth power modulo p. These two conditions imply that xn - a is a product of n distinct linear factors in IFp[X]. Using the Dedekind-Kummer theorem 9.B.22,17 we deduce that p is totally split in K2, D which finishes the proof of the lemma. Coming back to the proof of the theorem, let d be the least positive divisor of n for which \(;;ji E Q. Write = {!C for some c E Q~. By lemma 9.19, the polynomial X d - c is irreducible over Q and by lemma 9.B.61 we have Q( {!C) C Q(n). Since Q(n)/Q is Galois with commutative Galois group, any subfield of Q(n) is again Galois over Q and so Q( {!C)/Q is Galois. In particular, (d . {!C (which is a conjugate of {!C) is again in Q( {!C) and so it is a real number. We deduce that (d E JR., which implies that d = 1 or d = 2. If

16The Jacobi symbol (~) for odd m 2': 3, defined as the product over the prime factors of m with multiplicity of the corresponding Legendre symbols, detects quadratic nonresidues mod m and is easily seen to obey quadratic reciprocity. The Chinese Remainder Theorem then permits the construction of a "bad" value of m. 17To be fair, we need the obvious version of this theorem in which Q is replaced by Kl and K by K2 = K 1 ( y'a). We leave to the reader to convince himself that the statement and proof of this general version are exactly the same.

484

Chapter 9. A Little Introduction to Algebraic Number Theory

d = 1, then a is the nth power of a rational number and we are done (as a is an integer, this rational number is actually an integer). Suppose now that d = 2 and write n = 2m so that a is an mth power. Note that a is also a perfect square by the discussion in the first paragraph. If m is odd, then a is an mth power and a perfect square, hence it is an n = 2mth power and we are done. Suppose that m is even, so 41n. Then a is a 4th power modulo all sufficiently large primes and, by what we have already seen, we can write a = c2 , with Ql(JC) c Ql((4) = Ql(i). Since c > 0, we deduce that Ql( JC) c lR n Ql( i) = Ql, so c is a square and a is a fourth power. If m = 2k with k odd, this implies that a is a lcm(m,4) = nth power and we are done again. Finally, assume that n = 2i . U, with i :::: 3 and U odd. Since a is an 2ith power modulo all sufficiently large primes, lemma 9.B.61 implies that Ql( JC) c Ql( (2i). The extension Ql( (2i) /Ql is Galois, with Galois group (Z/2iZ)*. Its sub-extensions which are quadratic over Ql are classified by the subgroups with two elements of (Z/2iZ)*, which in turn are classified by the nontrivial solutions to the equation x 2 == 1 (mod 2i). There are three such solutions, giving rise to the quadratic sub-extensions Ql( A), Ql( )2), Ql( A). Since JC is real, we deduce that Ql( JC) C 9!()2) and soc is either a perfect square or twice a perfect square. As a = c 2 , this finally yields the desired result. 0

Chapter 10

Arithmetic Properties of Polynomials This chapter deals with elementary, but rather challenging problems concerning the arithmetic of polynomials with integer coefficients. The techniques are very diverse, going from the very basic fact that a - b divides f (a) - f (b) for f E Z[X] and a, bE Z to Mahler expansions, finite differences, p-adic analysis, finite fields, etc.

10.1

The a - blf(a) - f(b) trick

One can extract a lot of arithmetic information from the fact that a - b divides f(a) - f(b) for all polynomials with integer coefficients f, but often one needs a lot of inventiveness to do this. The next problems are all quite tricky, even though they hide very simple ideas. The first problem is a version of a very classical result, stating that there are no nonconstant polynomials all of whose values are prime numbers. 1. Let

h, 12, ... ,fk be nonconstant polynomials with integer coefficients.

Prove that for infinitely many n all numbers h (n), 12 (n), ... , fk (n) are composite.

486

Chapter 10. Arithmetic Properties of Polynomials

Proof. Let g = hh ..... fk and fix a positive integer no. Since none of the polynomials g and fi is constant, there are a, b > no such that min(lg(a)l, Ih(a)I,···, Ifk(a)1) > 1 and Ifi(a

+ Ig(a)lb)1 >

Ig(a)1

for all i. Since Ifi(a)1 divides fi(a+ Ig(a)lb) and is less than Ifi(a+ Ig(a)lb)l, it follows that all numbers fi(a + Ig(a)lb) are composite. The result follows. 0 The following two problems deal with periodic points of polynomials with integer coefficients. 2. Let

10.1.

487

The a - blf(a) - f(b) trick

multiple of Xi - Xj. But Xi - Xj = f(f(Xi)) - f(f(xj)) is also a multiple of f(Xi) - f(xj). Thus, we must have If(Xi) - f(xj)1 = IXi - xjl. But then n

L i=l

If(xi+d - f(xdl = L

IXi+1 -

Xii

i=l

= Xn+1 - Xl n

= L(f(Xi+1) - f(Xi)) , i=l

Z[X] and let

n 2: 3. Prove that one cannot find distinct integers Xl, X2,···, Xn such that f(Xi) = Xi-1, i = 1,2, ... , n, indices being taken E

mod n.

Proof. If such a polynomial f and integers Xi existed,

would be divisible by Xi+1 - Xi. In particular, IXi - Xi-11 2: IXi+1 - Xii for all i, which forces the numbers IXi - Xi-11 to be equal. Since

which implies that all numbers f(Xi+1) - f(Xi) have the same sign. Combined with the equality If (Xi) - f (Xi+ 1) I = /xi - Xi+ 11 this shows that either all numbers f(Xi) + Xi are equal or that f(Xi) - Xi are all equal. This is however 0 impossible, as f has degree n.

Remark 10.1. A very similar problem was given at the IMO 2006: let P(X) be a polynomial of degree n > 1 with integer coefficients and let k be a positive integer. Consider the polynomial Q(X) = P(P(··· P(P(X)))). "-----v-----"

L(Xi+1 - xd

k times

= 0,

i=l

there exists i such that Xi+1 - Xi and Xi+2 - Xi+1 have different signs. But then Xi = Xi+2, which contradicts the hypothesis that Xi are all distinct. The result follows. 0 3. Let

Z[X] be a polynomial of degree n 2:·2. Prove that the polynomial

f(f(X)) - X has at most n integer zeros. Gh. Eckstein, Romanian TST

Proof. Suppose that Xl < X2 < ... < Xn+1 are distinct integers such that f(f(xd) = Xi· Then f(Xi) are also pairwise distinct and f(Xi) - f(xj) is a

Prove that there are at most n integers such that Q(t) = t. It follows from problem 2 that any integral solution of the equation Q(t) = t is a solution of the equation P(P(t)) = t, so this new problem is actually equivalent to the previously discussed one. The following two problems are a bit trickier than the previous ones. 4. Find all integers n > 1 for which there is a polynomial f E Z[X] with the property: for any integer k one has f (k) == 0 (mod n) or f (k) == 1 (mod n) and both these equations have solutions.

Proof. The answer is: exactly the powers of a prime number. If n is a prime power, one can see from Euler's theorem that the polynomial xcp(n) is a solution. Conversely, suppose there exists a polynomial f with the properties

488

Chapter 10. Arithmetic Properties of Polynomials

in the statement and that n has at least two different prime factors p, q. Changing f to f(X - a) for a suitable a, we may assume that f(O) == 0 (mod n). Thus f(O) == 0 (mod q), which implies that for all b E Z we also have f(bq) == 0 (mod q). Thus, we cannot have f(bq) == 1 (mod n) and so f(bq) == 0 (mod n). In particular, f(bq) == 0 (mod p). But then for all a E Z we also have f(ap + bq) == f(bq) == 0 (mod p) and so f(ap + bq) == 0 (mod n). Since any integer x E Z can be written ap + bq for suitable a, b, it follows that the equation f(x) == 1 (mod n) has no solutions, a contradiction. 0 5. Find all polynomials f with integer coefficients such that f(n)12n -1 for all positive integers n. Polish Olympiad

Proof. Of course, if f is constant, then f has to be either 1 or -1 and these are solutions. So, assume that f is nonconstant. We may assume that the leading coefficient of f is positive. Take n such that f (n) > 1 and let p be a prime factor of f(n). Then pl2n - 1 and also p/f(n + p)12 n+p -1. But then pl2P - I, which is certainly impossible. So any such f is constant and the solutions are I, -1. 0 6. Let f E Z[X] be a nonconstant polynomial. Prove that the sequence f(3 n ) (mod n) is not bounded.

Proof. Changing f with its opposite, we may assume that the leading coefficient of f is positive. So, given N, there exists m such that f(3 m ) > N. Choose a prime p > 2f(3 m ) and observe that f(3 pm ) == f(3 m ) (mod p) by Fermat's little theorem. In particular, if r = f(3 mp ) (mod mp), then r == f(3 m ) (mod p) and so r 2': f (3 m ) > N, finishing the proof. 0 7. Is there a nonconstant polynomial f E Z[X] and an integer a > 1 such that the numbers f(a), f(a 2), f(a 3 ), ..â&#x20AC;˘ are pairwise relatively prime? Saint Petersburg 1998

Proof. Assume that f is such a polynomial and a is as in the statement. Let 9 = gcd(a, f(a k )) = gcd(a, f(O)). If 9 > I, then 9 is a common factor of all f(a j ). Thus 9 = I, so a and f(a i ) are relatively prime for all

10.1.

The a

blf(a) - f(b) trick

489

i. Choose an i with If(ai)1 > 1 and choose j = i + <p(lf(ai )!), where <P is Euler's totient function. Then f(a i ) divides aj - ai by Euler's Theorem, so f(a i ) divides f(a j ) - f(a i ) and f(ai)lf(a j ). But this contradicts the fact that gcd(f(a i ), f(a j )) = 1 and shows that the answer to the problem is negatiw. 0

The next problem is a variation on the following very classical (but nontrivial) problem: what are the polynomials f E Z[X] such that f(p) is a prime for all prime numbers p? 8. Find all polynomials f with integer coefficients, having the following property: there exists k such that for all primes p, f(p) has at most k prime factors.

Proof. All polynomials of the form axm work, for obvious reasons. Let f be a solution of the problem and suppose f is not of this form. So we can write f(X) = xmg(X) for some polynomial 9 with integer coefficients such that g(O) =1= 0 and 9 is not constant. By Schur's theorem, there are infinitely many primes dividing at least one ofthe numbers g(l), g(2), .... In particular, we can choose PI,P2,'" ,Pk+l distinct primes greater than Ig(O)1 and we can choose positive integers aI, a2, ... , ak+l such that g(ai) == 0 (mod Pi)' Using the Chinese Remainder Theorem, we can find an integer a such that a == ai (mod Pi) for all i. Note that a is relatively prime to PIP2 ... PHI (otherwise some Pi would divide 'a, so Pi/ai and then Pilg(O), a contradiction). Thus, by Dirichlet's theorem, there exist infinitely many primes P == a (mod PIP2 ... PHd. In particular, we can choose such a prime P with f(p) =1= O. Since by construction f(p) is a multiple of PIP2 ... PHI, it follows that f is not a solution of the problem. The result follows. 0 Proof. Note that all polynomials of the form f(X) = axm work. Suppose f(X) = xmg(X) is such a polynomial which is not of this form. Then g(X) is also such a polynomial. Thus we may assume that f is nonconstant and f(O) =1= O. Let P be a prime not dividing f(O), hence P does not divide f(p)Âˇ By Dirichlet's Theorem there are infinitely many primes q of the form q = P + kf(p)2. Choose such a q with If(q)1 > If(p)l. For such a q we have

490

Chapter 10. Arithmetic Properties of Polynomials

f(q) == f(p) (mod f(p)2). Thus f(q) is divisible by every prime factor of f(p), with exactly the same multiplicities, and at least one additional prime factor. Iterating this construction, we can find a sequence of primes (Pn) for which f(Pn) has at least n prime factors. Thus the only solutions are the ones already ~~. 0 9. Find all polynomials f E Z[X] with the property that for any relatively prime integers m, n, the numbers f(m), f(n) are also relatively prime. Iranian TST

Proof. The solutions are the polynomials ±Xd for d 2 o. Trivially, these are solutions of the problem. Consider now any nonconstant solution f. Suppose that p is a large prime for which p does not divide f (p). Then p and p+ f (p) are relatively prime and thus f (p) and f (p + f (p)) are relatively prime. However, f (p) divides f (p + f (p)). Thus necessarily If (p) I = 1 or f (p) = O. Since p was large, none of this happens (f is nonconstant). Thus, for p large enough we have plf(p), forcing plf(O). This implies that f(O) = 0 and so we can write f(X) = Xg(X) for some polynomial g with integral coefficients. Of course, g satisfies the same property as f and has smaller degree. By an easy induction on the degree of f, we deduce that f is indeed of the form ±Xd for some d. 0 Proof. We prove first the following easy lemma: Lemma 10.2. If f E Z[X] is not of the form ±xn, then there exist different primes p,q such that qlf(p).

Proof. Assuming the contrary, we may assume that the leading coefficient of f is positive. So, for all large enough p, f(p) = pkp for some integer kp . Since kp = lO~~~) converges to deg(f), it follows that kp = deg(f) for all sufficiently large p and so f(p) = pdeg(j) for all sufficiently large p. The result follows. 0 Coming back to the proof, choose a solution f of the problem and suppose that f is not of the form ±xn. Pick primes p, q as in the lemma. Then q divides f(p) and f(p + q) (since q divides f(p + q) - f(p)). But this is impossible, since p,p+q are relatively prime and so f(p) and f(p+q) are relatively prime. The result follows. 0

491

10.1. The a - blf(a) - f(b) trick

It is a classical fact that for any integer a and any positive integers m and n we have

gcd(a m - 1, an

1) = agcd(m,n) - 1.

A sequence (an)n with the property that gcd(am,a n ) = agcd(m,n) is called a Mersenne sequence. Thus (an - 1)n is a Mersenne sequence. In chapter 3, problem 3 we proved that for any Mersenne sequence an there exists a sequence of integers bn such that an = TIdln bd. The next problem shows how to construct a Mersenne sequence by iterating a polynomial with integer coefficients.

tf 10.

Let

f be a polynomial with integer coefficients and let ao = 0 and an = f(an-I)

for all n 2 1. Prove that (a n )n2:o is a Mersenne sequence. Romanian TST

Proof. Let g = fd, the composite of f with itself d times. If m = du and n = dv, with gcd(u, v) = 1, are any positive integers, note that an = gV(O), am = gU(O) and ad = g(O). So, it is enough to prove that for any g with integer coefficients and any gcd(u,v) = 1 we have gcd(gU(O),gV(O)) = g(O). Let us remark that for any polynomial h with integer coefficients and any k we have h(O)lhk(O). This is obvious. Applying this to g already shows that g(O) divides gcd(gU(O),gV(O)). Conversely, let x = gcd(gU(O),gV(O)). Applying the previous remark to h = gU and h = gV, we obtain that for all A, B 2 1 we have xlgAu(O) and xlgBv(O). Taking A, B such that Bv = Au + 1, we obtain that xlg(gAu(O)) and xlgAu(O). It is then clear that x divides g(O) and the conclusion follows. 0 11. Find all positive integers k such that if a polynomial with integer coefficients f satisfies

0::; f(O), f(1), ... , f(k then f(O)

+ 1) ::; k,

= f(1) = ... = f(k) = f(k + 1). IMO 1997 Shortlist

Chapter 10. Arithmetic Properties of Polynomials

492

Proof. If k ::; 2 we can easily find counterexamples, for instance f(X) = X(2 X) for k = 1 and f(X) = X(3 - X) for k = 2. So, assume that k 2:: 3 and let f be a polynomial with integral coefficients such that

0::; f(O), f(l), ... , f(k + 1) ::; k. As f(k + 1) - f(O) is between -k and k and since it is a multiple of k + 1, we must have f(O) = f(k + 1). Thus, we can write

f(X) - f(O)

X(X - (k

+ l))g(X)

for some polynomial g, which clearly has integer coefficients (note that X divides f(X) - f(O) and X is relatively prime to X - (k + 1)). Thus, f(O) + i(i - k - l)g(i) is between 0 and k for all 0 ::; i ::; k + 1. In particular, we have k 2:: i(k + 1 - i)lg(i)1 for i in this range. However, i(k + 1 - i) > k for 2 ::; i ::; k - 1, so that g(i) = 0 for 2 ::; i ::; k - 1. In particular, we can write

g(X) = (X - 2)··· (X - k + l)h(X) for some polynomial h, again with integer coefficients. But then

f(k) - f(O)

-k(k - 2)!h(k),

f(l) - f(O)

(_l)k-lk(k - 2)!h(1).

This implies that k(k - 2)!lh(x)1 ::; k for x E {l, k}, which clearly implies that h(l) = h(k) = 0, unless k = 3. Therefore, unless k = 3 we can definitely conclude that f(O) = f(l) =:= ... = f(k + 1) and so all k 2:: 4 are solutions of the problem. It remains to study the case k = 3. But we note that we can actually have equality in all previous inequalities, yielding without much difficulty the polynomial f(X) = X(X - 2)2(4 - X). This proves that k = 3 0 is not a solution and the problem is solved. Fermat's little theorem implies that p divides 2P 2 for any prime p. The following question is then rather natural to ask, but not easy to answer.

12. Find all polynomials f with integer coefficients such that f(p) 12P any odd prime number p.

2 for

Gabriel Dospinescu, Peter Scholze

10.1. The a - blf(a) - f(b) trick

493

Proof. We will use the fact that f(p) divides f(p+kf(p)) for any k and choose k such that p + kf(p) is again a prime. Of course, this is not possible unless p is relatively prime to f(p). Since f(X) = X is a solution (by Fermat's little theorem), a reasonable first step will be to prove that any solution satisfies f(O) = O. This is the hardest step. Assume for a moment that we proved that any nonconstant solution satisfies f(O) = 0, then we can write f(X) = Xg(X) and of course 9 satisfies the same conditions. So either 9 is constant or it is a multiple of X. Continuing like this we obtain that f(X) = ±2a Xb for some a, b 2:: O. Clearly a ::; 1 and by taking p = 3 we obtain that b ::; 1. Thus, all solutions are of the form f(X) = ±2a X b with a, b ::; 1 (these are clearly solutions of the problem). Now, let us prove that if f is nonconstant, then f(O) = O. Assume that f(O) -I- 0 and, by changing f with - f, that the leading coefficient of f is positive. Then for large primes p, p is relatively prime to f(p) (as if p divides f(p), then p divides f(O)). By Dirichlet's theorem for such a prime p we can choose infinitely many k 2:: 1 (depending on p, of course) such that q = p+kf(p) is again a prime. Then f(p) divides 2P -2 and also 2P+k!(p) -2. But then f(p) also divides 2(2 gcd (p-l,p-l+k!(p)) - 1). As f(p) == f(l) (mod p - 1), we deduce that f(p) divides 2(2 gcd (p-l,k!(1)) - 1). Now, if f(l) = 0, all this work will accomplish nothing, but the good news is that f(l) cannot be 0: otherwise, f(p) would be a multiple of p - 1 and so p - 1 would divide 2P - 2 for any prime p. This is of course false (take p = 5). Thus f(l) -I- O. However, there is a big obstacle in the previous approach: k depends on p and we have absolutely no control on it. The crucial remark is that we can however control gcd(p - 1, kf(l)) by choosing more carefully k. This is the object of the next paragraph. We will choose k of the form k = 2 + r(p - 1) for suitable integers r. In order to do that, we need to ensure that p+ f(p)(2+r(p-1)) is a prime. This is realized for infinitely many r if p + 2f(p) is relatively prime to (p - l)f(p). Now, if l is a prime factor of p + 2 f (p) that divides (p - 1) f (p), then we cannot have llf(p) (as otherwise l divides p and so I = p divides f(p), a contradiction), so l divides p - 1 and also 1 + 2f(1) (since f(p) == f(l) (mod p -1)). But we can now do the following: choose from the beginning large primes p such that p-1 is relatively prime to 1+2f(1), for instance primes p == 2 (mod 1+2f(1)).

494

Chapter 10. Arithmetic Properties of Polynomials

There are infinitely many such primes by Dirichlet's theorem. For such p, we saw that p + 2J(p) is relatively prime to (p - l)f(p), so we can indeed choose k of the form 2 + rep - 1) such that q = p + kJ(p) is a prime. The previous paragraph shows that J(p) divides 2(2 gcd (p-1,k f (1Âť -1). But gcd(k,p-1) = 2, so that f(p) actually divides 22f(1) - 1. But since f(p) can take arbitrarily large values and since J(l) -=I 0, this is a (so desired) contradiction. This shows that f(O) = 0 and then the problem is solved by the first paragraph. 0

10.2

Derivatives and p-adic Taylor expansions

Most of the time when working with congruences mod p it is enough to know that f(a) == J(b) (mod p) whenever a == b (mod p). However, when dealing with congruences with prime power moduli, one is sometimes obliged to use more precise estimates. Assume that f E Z[X] has degree nand consider its Taylor expansion n

J(x

+ h) = ~ J

(k)()

k! x hk,

k=O

which is valid for any complex numbers x, h. Note that f(~ix) is an integer. Indeed, by linearity it suffices to prove this when J(X) = xn and in this case J(k)(X) =

(n)

k x

495

erivatives and p-adic Taylor expansions 10.2. D

. d 1 t f E Z[X] be a polynomial. If f(O), f(l), ... , 13. Let p be a prIme an e .. d b 2 rove that the f(p2 - 1) give distinct remainders when dlVlde ~ p ,p h d' 'ded numbers f(O),/(l), ... ,/(p3 - 1) give distinct remamders w en IVI by p3.

Putnam 2008

, _ f( ') (mod 3) for some i, j. Since f(i) == f(j) Proof. Assume that f,(z~ ~ . J d.f we deduce that i == j (mod p2), say (mod p2) and since f IS m)ectlVe mo p 'k = 0 (mod p). Assume that this is j = i + p2 k. It is enough to prove that not the case. We have f(i) == f(j) == f(i

+ kp2) ==

f(i)

+ kp2 !,(i)

(mod p3),

so p divides kf'(i), hence p divides !,(i). But then

f(i

+ kp) ==

f(i)

+ kp!'(i) ==

f(i)

(mod p2),

. ld '+ k - i (mod p2) a contradic0 which, combined with the) hYPdO~~Si'S'(~~d ~~) ;h:result follO\~S. tion. Thus k == 0 (mod p an Z = J I â&#x20AC;˘ . 1 with integer coefficients such that P(O) = 0 and 14. Let P be a po1ynomia gcd(P(O), P(l), P(2), ... ) = 1. Show that there are infinitely many n such that

n-k

gcd(P(n) _ P(O), pen + 1) - P(l), pen + 2) - P(2), ... )

= n.

USA TST 2010

It follows easily from this that we have an equality in Z[X] (thus in R[X] for any commutative ring R)

Proof. Let us try to study first d = gcd(P(n) - P(O), pen + 1) - P(l), ... ) n

In particular, we have the very useful congruence f(a + b) == f(a) + bf'(a) (mod b2 ) whenever J E Z[X] and a, bE Z. We give a few applications of these ideas in this section.

. nts Let q be a prime factor of dn , 1 ial P with integer coeffi Cie . B t for any po ynom d )f 11 k . e Pis n-periodic modulo q. u ) = 1 then P is I-periodic so that Pen + k) == P(k) (mo q or a 'f ' Id(' P is also q-periodic modulo q, Thus, 1 gc q, n ,

496

Chapter 10. Arithmetic Properties of Polynomials

modulo q (by Bezout's lemma) and so q divides P(n + 1) - P(n) for all n. Then q divides P(n) - P(O) for all n, so if P(O) = 0, then q must divide gcd(P(O), P(1), .. .). In particular, for our polynomial we must have qln for any prime factor q of dn . The previous paragraph suggests taking for n a power of a prime, say n = pN. Then we saw that d n is also a power of p. Note that d n is a multiple of n, since n divides P( n + k) - P( k) for all k. It remains to see if we can have pN+IIP(k + pN) - P(k) for all k. Since

P(k + pN) == P(k)

+ pNP'(k)

(mod pN+l) ,

10.3. Hilbert polynomials and Mahler expansions

monic) to find q, r E Z[X] such that f(X) = (XP - X)q(X) + r(X) and deg r < p. Then p divides r( x) for all integers x (by Fermat's little theorem and the hypothesis) and the result follows from Lagrange's theorem. Assume that the result holds for k and that k+ 1 ::; p. If pH I divides f (x) for all x, then by the inductive hypothesis we can write f(X) = L~=Opk-i(XP - X)i . gi(X) for some gi E Z[X]. Pick any integers x and z and write x P - x = py for some integer y. Then (x + pz)P - (x + pz) == p(y - z) (mod p2), thus k

f(x

+ pz) == Lpk(y -

z)igi(x + pz) == pk L(y - z)igi(x)

i=O

this would imply that p divides P'(k) for all k. Now we see how to choose our numbers n: pick and fix once and for all a value k such that P' (k) i- O. For all sufficiently large p, p does not divide P' (k). For any such p, the previous arguments show that dn = n for all n = pN. The conclusion follows. 0 Before tackling the next problem, we need to recall a very standard result, known as Lagrange's theorem. Consider f E Z[X] and a prime p. Let] E lFp[X] be the reduction of f mod p and assume that] i- O. As lFp is a field, it .!:.ollows that] has at most deg] ::; deg f distinct roots in lFp . Therefore, if f i- 0 (which is equivalent to the fact that at least one coefficient of f is not a multiple of p), then the congruence f (x) == 0 (mod p) has at most deg f pairwise distinct solutions. The next result is a generalization of this classical theorem; it will also be used in the solution of problem 20. This was one of the problems proposed in the Iranian TST 2011, but the result is much older, see for instance theorem 27, chapter II of the beautiful book [21]. 15. Let p be a prime, k a positive integer and f E Z[X] such that pk divides f (x) for all x E Z. If k ::; p, prove that there are polynomials go, gl,·· ., gk E Z[X] such that

497

(mod pk+l).

i=O

Therefore the hypothesis on f implies that p divides L~=o zigi(x) for all integers z. Using the fact that k + 1 ::; p and Lagrange's theorem, it follows that p divides gi (x) for all i and all x. By the case k = 1 we can write gi(X) = (XP - X)hi(X) + pri(X) for some hi, ri E Z[X]. Replacing these expressions in f(X) = L~=opk-i(XP - X)i. gi(X) yields the desired result. 0

10.3

Hilbert polynomials and Mahler expansions

Consider the polynomial

(~) = X(X -1)(X

2)···(X-n+1) n!

known as the nth Hilbert polynomial. It is clear that this polynomial has rational (not necessarily integer) coefficients and it is not difficult to check that it takes integer values at all integers (note that x( x - 1) ... (x n + 1) = n! (~) if x ;:::: 1, while

f(X) = Lpk-i(XP - X)i . gi(X). i=O

Proof. The proof is by induction on k. If k = 1, perform the division algorithm in Z[X] for the polynomials f and XP - X (which we can do, as XP - X is

if x < 0). These polynomials playa fundamental role in the theory of integervalued polynomials because of the following beautiful and classical result of Polya:

498

I I

Chapter 10. Arithmetic Properties of Polynomials

Theorem 10.3. Let f be a polynomial of degree n, with real coefficients. Then f (Z) c Z if and only if there exist integers ao, a I, a2, ... ,an such that

Proof. One implication follows from the previous discussion, so assume that f(Z) c Z. The polynomials (~),(~), ... ,(~) have degrees O,l, ... ,n, thus they form a basis of the JR.-vector space of real polynomials with degree at most n. Thus there exist unique real numbers ao, aI, ... ,an such that

10.3.

499

Hilbert polynomials and Mahler expansions

We call aj the Mahler coefficients of f and we refer the reader to the addendum 3.B for more details on these coefficients, which play an absolutely fundamental role in p-adic analysis. Another consequence of the proof of theorem 10.3 is the following very useful result. Proposition 10.4. If f is a polynomial of degree n with coefficients in an arbitrary commutative ring, then

I) -It+

n+l

() 1 f(X

+ k) = O.

k=O

Consider the operator b,.f(X) = f(X

+ 1) -

f(X) and observe that

Proof. The left-hand side is b,.n+lf(X). However, we have degb,.g < degg for any nonzero polynomial g. We deduce that b,.n+l f is the zero polynomial and the result follows. 0

The next problem is an immediate application of theorem 10.3. The reader might try to solve it in a different way and he will realize that the problem is actually quite tricky. Applying b,. successively to the relation 'fl6. Let n be a positive integer. What is the least degree of a monic polynomial f with integer coefficients such that nlf(k) for any integer k?

we deduce that aj = b,.j f(O) for all j. On the other hand, an immediate induction shows that

[;.kf(X)

~ ~(-l)k-jG)f(X + j).

Thus, if f(O), f(l), .. . ,f(n) are integers, then so are ao, al," ., an路 The result 0 follows. The proof of the previous theorem shows that ak =

2) _l)k-J"(k)" f(j)路 k

j=O

fC:)

Proof. Let d = deg f. By theorem 10.3 we can write = 2:::%=0 ad~) for some integers ai. Considering the leading coefficients in this equality shows that d! is a multiple of n. On the other hand, if d! is a multiple of n, then we can take f(X) = X(X + 1)路路. (X + d - 1). Thus the answer is: d is the smallest integer such that d! is a multiple of n. 0

'..f 17. Let f be a polynomial such that f(n) E Z for all n E Z. Prove that for any integers m, n the number lcm[l, 2, ... ,deg(f)] . f(~=~(n) is an integer. MOSP 2001 Proof. Fix m, n distinct integers and let d = m-n and g(X) = f(n+X). Then g has rational coefficients, sends integers to integers and deg(f) = deg(g). So,

500

Chapter 10. Arithmetic Properties of Polynomials

10.3. Hilbert polynomials and Mahler expansions

a) m - n divides f(m) - f(n) for all O:S m

we need to prove that lcm[l, 2, ... , deg(g)) .

g(d) - g(O) d

c) m - n divides f(m) - f(n) for all m

f(i) - f(j) =

'1(~) E z.

(tD, which is clearly an integer. D

Holden Lee Proof. Since f(Z) C Z, theorem 10.3 yields the existence of integers ao, al, ... , such that

~ ~akG)

The result follows now from the previous problem and the following general result. D Lemma 10.5. Let ai E Z and let

f(X)

~ ~ak (~)

and Lk

k=O

18. Let f be a polynomial of degree d such that f(Z) c Z and for which f(m) - f(n) is a multiple of m - n for all 0 :S m, n :S d. Prove that j(m) - f(n) is a multiple of m - n for all integers m, n with m #- n.

f(X)

d = degf路

~ lern(l, 2",., k)

(by convention Lo = 1). Then the following assertions are equivalent:

#- n

Proof. Suppose that a) holds. We will prove by induction on i that Li divides ai. This is clear for i = 0, so assume that ao, . .. ,ai-1 are multiples of L o , L 1, ... , Li-1 and fix 0 :S j < i. Then j - i divides

It is thus enough to prove that for any 1 :S i :S D we have

But the left-hand side is equal to lcm[1'7,路路路,Dj The result follows.

#- n:S

b) Lk divides ak for all 0 :S k :S d.

is an integer. Let D = degg, so, using theorem 10.3, there exist integers ao, aI, ... ,aD such that

lcm[1,2, ... ,D)

501

ak (

(~) - (~) )

= ai

L. ak ( (~) - (~) ) . O~k<~

By the previous problem and the inductive hypothesis, each of the numbers

(G) - G))

ak with 0 :S k < i is a multiple of i - j. We deduce that i - j divides ai and since j < i was arbitrary, it follows that Li divides ai. Hence a) implies b). The previous problem shows that b) implies c) and since it is D trivial that c) implies a), the result follows. We are done. Remark 10.6. The fact that the polynomial f(X) = L~=o ak . (~) satisfies f(at:.J(b) E Z for all a #- bE Z if and only if lcm(l, 2, ... , k)lak for all k seems to have been first noticed by R.R.Hall and LZ.Ruzsa.

The following problem strongly suggests using Lagrange's interpolation theorem, but there are some difficulties in making the argument work, since the polynomials appearing in Lagrange's formula don't have integer coefficients. The problem is quite tricky and makes again use of Hilbert's polynomials. <V 19.

a) Prove that for all positive integers n there is a polynomial f E Z[X) such that all numbers f(l) < f(2) < ., . < f(n) are powers of 2. b) Let a > 1 be an integer and let n be a positive integer. Prove that there exists a polynomial f of degree n, having integer coefficients, such that f(O), f(l), ... , f(n) are pairwise distinct positive integers, all of the form 2a k + 3 for some integer k. Chinese TST 2004

502

,, ,

Chapter 10. Arithmetic Properties of Polynomials

503

10.3. Hilbert polynomials and Mahler expansions

for some suitable integers ai. The reason is that f (i) only depends on ai, so that it will be rather easy to adjust ai in order to make f(i) prime. Indeed,

Proof. a) We will choose f of the form

note that

f(i) for some suitable integers A, B, to be chosen later. By the binomial formula, for any 0 :S i :S n we have f( i) = A(1 + B)i. Now, of course we want f to have integral coefficients and unfortunately (~) does not have integral coefficients. However, n!(~) has integral coefficients for alII :S i :S n. Note however that we cannot simply take for B a multiple of n!, since then A(1 +B)i has no chance of being a power of 2. However, we can profit from the presence of A: take A to be 2v2 (n!) and B an odd multiple of the greatest odd divisor of nL Then the previous remarks show that f has integral coefficients. Finally, we want 1 + B to be a power of 2. Thus, we can choose (for example) 1 + B = 2'P(d), where d is the largest odd divisor of nL With these choices, f(l) < f(2) < ... < f(n) are powers of 2, as needed. b) This uses the same idea as a). Write n! = m· q, where all prime factors of m are among those of a and where gcd(q, a) = 1. Let b = a'P(q) - 1, so q divides b. Finally, define

f(X) = 2a

~ (~)bi + 3.

(_1)n-i(n

It has integer coefficients because i!ln!la 1 :S k :S n we have

P(i) = 2a m . (b + l)i

+ 1.

Now we will choose the ai's inductively so that f(l) < ... < f(n) are primes. We ~ill heavily use theorem 9.6, according to which for all n there are infinitely many primes of the form 1 + kn. Thus, there exists an integer al such. that 1 + (_1)n-l(n 1)!al = f(l) is a prime. Fix such al and c~~~se (agalI~ by the cited result) an integer a2 such that 1 + a2(n - 2)!( -1) = f(2) IS a prime greater than f (1). Continuing like this, we find aI, a2, ... ,an such that f(l) < f(2) < ... < f(n) are primes and the problem is solved.

20. Suppose that n is a positive integer not divisible by the cube of a prime number. Consider all sequences (Xl, X2, ... ,xn) with Xi E ZjnZ. For how many of these can we find a polynomial f with integer coefficients such that f(i) (mod n) = Xi for all i? USA TST 2008

Proof. Let An be the additive group of those sequences (Xl, X2, .... ' Xn) E (ZjnZt associated to integer polynomials, as in the problem. We wIll show that the map

<])(ao, al,···, an-I) m

i)!(i -1)!ai

(1(1), f(2), ... , f(n)),

b for all 0 :S i :S n. Moreover, for

+3 =

where

2a m +'P(q)i + 3.

Remark 10.7. It is also true that for all positive integers n there is a polynomial f E Z[X] such that all numbers f(l) < f(2) < ... < f(n) are prime numbers. We will try to find such a polynomial of the form

f(X) = al(X - 2)(X - 3)··· (X - n) + a2(X -1)(X - 3).·· (X - n) + ... + an(X - 1).·· (X - n + 1) + 1,

n-l

f(X) =

+ I:ai I1(X i=l

j),

j=l

is an isomorphism of abelian groups <]) : TI~:~ g Cd ~ An. First, note that <]) is well-defined: indeed, since a product of d consecutive integers is a multiple of d!, it is clear that the sequence (1(i)hSiSn does not depend on the choice of representatives ai for ai· Let us prove that <]) is surjective. Repeated division algorithm sho:vs th~t any polynomial with integer coefficients of degree at most d can be wntten m

(Z/ 0,k!)Z)

504

Chapter 10. Arithmetic Properties of Polynomials

the form

f(X) = ao

i=l

j=l

Proof. We will exhibit a bijection between Amn and Am

+ I:ai II(X - j)

505

An. Consider the

map

<p : Amn

for some integers ai. We may restrict to d < n since all ak with k 2:: n do not matter when considering f(i) (mod n). This yields the surjectivity. It is clear that <I> is a group homomorphism. It remains to prove that <I> is injective, so suppose f satisfies f(i) == 0 (mod n) for 1 ::S i ::S n. We want to show that ak is a multiple of gcd(~,k!) for 0 ::S k < n. Assuming the contrary, there is some least k for which this does not hold. Then we may assume aj = 0 for j < k (since replacing them by 0 does not change the values of f mod n). But then plugging in X = k + 1 gives f(k + 1) = k!ak == 0 (mod n). and ak is a multiple of gc n, kl)' . contrary to our assumption. Thus the number of polynomial sequences (Xl, . .. ,xn ) is

n-l

10.3. Hilbert polynomials and Mahler expansions

II gcd(n,n k!)'

k=O

If a prime p has vp(n) = 1, then there are p factors in this product (those with k = 0,1, ... ,p - 1) which are multiples of p. If a prime p has vp(n) = 2, then

there are p factors which are multiples of p2 (k = 0, ... ,p 1) and p more factors which are multiples of just p (k = p, ... , 2p-1). Therefore for n which are not divisible by the cube of a prime, we have

o p:vp(n)=l

--t

An,

<p(Smn(J)) = (Sm(J), Sn(J)).

Note that it is well-defined, for if Smn(J) = Smn(g), we have f(x) == g(x) (mod mn) for all 1 ::S X ::S mn and so Sm(J) = Sm(g) and Sn(J) = Sn(g). We claim that <p is bijective. Injectivity is very easy, for if <p( Smn (J)) = <p( Smn (g)), then m divides (J - g) (i) for all 1 ::S i ::S m and n divides (J - g) (i) for all 1 ::S i ::S n. By the division algorithm, it follows that m divides (J g) (x) for all integers x and the same for n. Since gcd(m, n) = 1, we have mnlf(x) - g(x) for all x and we are done. For surjectivity, we need to prove that if f, g E Z[X] are arbitrary, then there exists h E Z[X] such that h(i) == f(i) (mod n) for all 1 ::S i ::S nand h(i) == g(i) (mod m) for all 1 ::S i ::S m. Simply choose h = Amf + Bng, where A, B are integers such that Am == 1 (mod n) and Bn == 1 (mod m). The lemma is proved. 0 Thanks to this lemma and the hypothesis that n is cube free, it remains to find IApl and IAp21. The first task is trivial: for any p-tuple (al, ... ,ap) of remainders mod p there exists a polynomial f such that f(i) = ai (mod p) for all 1 ::S i ::S p, by Lagrange's interpolation formula. Thus IApl = pp. Finding IAp21 is more delicate, since Zjp2Z is not a field and so Lagrange's interpolation formula is useless. Moreover, there are nontrivial relationships between the numbers f(i) (mod p2), which implies that IAp21 is definitely smaller than p 2p2 . The first point is to note that

Proof. For a polynomial f E Z[X] and n 2:: 1, let Sn(J) = (J(1)

(mod n), f(2)

(mod n), ... , f(n)

(mod n)) E (ZjnZt

and let An = {Sn(J)lf E Z[X]}. We want to find IAnl, at least when n is cube-free. First, we will prove, the following Lemma 10.8. IAnl is multiplicative, i.e.

for all gcd( m, n)

= l.

as for any f E Z[X] one can find g E Z[X] of degree smaller than 2p and such that f(x) == g(x) (mod p2) for all x. Indeed, simply take for g the remainder of f when divided by (XP - X)2. Let G = {J E Zjp2Z[X] I deg(J) < 2p}, an 2

abelian group of order (p2)2p = p4p and let S : G --t (Zjp2Z)P be defined by S(J) = (J(1), f(2), ... ,f(p2)). The previous observation shows that Ap2 is the image of the map S : G --t (Zjp2z)p2. As IIm(S)1 = 'Kl~'s),' it remains to find IKer(S)I. We can actually describe Ker(S) rather explicitly, thanks

Chapter 10. Arithmetic Properties of Polynomials

506

I, I

to problem 15. Indeed, this problem (for k = 2) implies that Ker(S) is in bijection with the set of polynomials u E lFp[X] of degree smaller than p. As this set has pP elements, it follows that IAp21 = p3P . The problem is finally solved. D

507

10.4. p-adic estimates

which is clear (simply order the y/s). Coming back to the problem, we infer that (n-d-1)d+l lcm(n-1,n-2, ... ,n-(d+1))2':n

IDA

p-adic estimates

In this section we consider a few problems dealing with p-adic properties of polynomials. .... 21. Let (a n )n2l be an increasing sequence of positive integers such that for some polynomial f E Z[X] we have an :::; f(n) for all n. Suppose also that m - nlam - an for all distinct positive integers m, n. Prove that there exists a polynomial 9 E Q[X] such that an = g(n) for all n. USAMO 1995 Proof Let d be the degree of f and choose a polynomial P of degree at most = ai for 1 :::; i :::; d + 1. This is possible by Lagrange's interpolation formula. Choose (and fix) N 2': 1 such that h = N P has integral coefficients. Then h( i) = N ai and h has degree d. Fix any integer n > d + 1 and observe that m - n divides Na m - Nan and m-n divides h(m) - h(n). Thus, if m :::; d+ 1, then m-n divides Nan - h(n). Consequently, Nan - h(n) is a multiple of lcm(n - 1, ... , n - (d + 1)). Note that INan h( n) I :::; Cn d for some constant C, because an is bounded by f and because h has degree at most d. On the other hand, we have the following result: d with rational coefficients and such that P( i)

Lemma 10.9. For any positive integers Xl, X2," ., Xn ) lcm(xl, X2," ., xn) a multiple of IT ,XIX2'~~~(x x ) â&#x20AC;˘ lSt<JSn

Proof It is enough to prove that for any prime p, the p-adic valuation of XIX2"'X n ) . at I Icm ( Xl,X2"",Xn ) IS east th a t 0 f IT lSi<jSn d(' ,)' W n't'mg Yi = Vp ( Xi, gc XZ'X J this comes down to the inequality

max(yd 2':

LYi - L min(Yi' Yj), i<j

l::;i<j::;d+l gc

d(' ')' n - z, n - J

which is greater than Cln d+l for some constant C l > 0 and all large n (this is because gcd(n - i, n - j) divides j - i). Thus, if n is large enough, then necessarily lcm( n - 1, n - 2, ... , n - (d

+ 1)) > IN an -

h( n) IÂˇ

Combining this with the result of the previous paragraph, we infer that there is no such that for all n 2': no we have Nan = h (n ). Finally, pick any m 2': 1 and observe that for all n 2': no we have m nlNam - Nan and m - nlh(m) - h(n). Since Nan = h(n), we deduce that m - n divides Na m - h(m) for all'n 2': no, forcing Na m = h(m). Thus, we proved the existence of a polynomial 9 = -fith, with rational coefficients, such that an = g(n) for all n. D Remark 10.10. There are a lot of non-polynomial maps f such that m n divides f(m) - f(n) for all m -=I=- n. For instance, pick any sequence of integers an, infinitely many of which are nonzero and define f(n)

I::aj .

+ j)(n + j

- 1) ..... (n - j).

j20

It is an easy exercise to check that f satisfies the desired congruences and that f is not polynomial. Remark 10.11. It is not true that if an = g(n) for all n, then 9 E Z[X]. For . . () n4 _n 2 h g(m)-g(n). . t instance, one can eaSIly check that If 9 n = - 2 - ' t en m-n IS an m eger for all different integers m and n. 22. Consider all sequences (f(1) (mod n), f(3) (mod n), ... , f(1023) (mod n)), where n = 1024 and f is an arbitrary polynomial with integer

508

Chapter 10. Arithmetic Properties of Polynomials coefficients. Prove that at most 235 of these sequences are permutations of 1,3,5, ... ,1023 (mod n). USA TST 2007

Proof. Define Po(X) = 1 and Pi(X) = r1i=l(X -(2k-1)) for i 2: 1. Repeated euclidean division (taking into account that ~ is monic) shows that each polynomial f E Z[X] can be written f = coPo + ClPl + ... + enPn for integers Ci, where n = degf. Note that Pi (2n - 1) is a multiple of 2i . if, so by Legendre's formula we have V2 (~( x)) 2: ai for all odd numbers x, where ai = 2:::%':0 ~ Note that ao = 0, al = 1, a2 = 3, a3 = 4, a4 = 7, a5 = 8, and ai 2: 10 for i 2: 6. Say a polynomial f is good if its associated sequence (as in the statement of the problem) is a permutation of 1,3,5, ... ,1023. Let

l J.

f(X)

10.4. p-adic estimates

509

p-adic analogue of a classical theorem of Weierstrass, stating that continuous functions on a compact interval can be uniformly approximated by polynomials). It is not difficult to see that the characteristic function of 2Z2 + 1 is continuous, so it can be uniformly approximated with polynomials. The next problem gives a more precise estimate. bJ723. Prove that for all n there exists a polynomial f with integer coefficients and degree not exceeding n such that 2n divides f(x) for all even integers x and 2n divides f (x) - 1 for all odd integers x.

P. Hajnal, KaMaL

Proof. Define O(x) = 1 if x is odd and 0 otherwise. Then we compute that

CiPi(X)

0:::; i:::; n

be a good polynomial. If we delete the terms with Pi for i 2: 6 (where ai 2: 10), we get a polynomial with the same associated sequence as f, so we are only interested in Co, Cl,·· . ,C5' Note that CO is odd and that f(l) ¢. f(3) (mod 4), as f(l) == f(4k + 1) (mod 4) and f(3) == f(4k + 3) (mod 4) for all k. But since f(3) - f(l) == Cl(Pl (3) - H(l)) == 2Cl (mod 4), it follows that Cl is odd. Finally, note that if we mod out Ci by 2 1O - a;, we get a polynomial with the same associated sequence as f, so we are only interested in the values of Ci (mod 2 1O - a;). But since Co is odd, there are at most 2 9 choices for it; for the same reason there are at most 28 choices for Cl (mod 29 ) and for 2 ~ i ~ 5 there are at most 21O - a ; choices for Ci (mod 21O - ai ). Hence the number of complete remainder sequences is at most

Since the binomial coefficients are all integers it follows that truncating after the nth term gives a polynomial

g(X)

(~) (_2)k-l E Q[X]

k=l

of degree n such that for all integers m, g(m) is an integer with g(m) == O(m) (mod 2n). Recall that v2(k!) = k - s2(k) ~ k - 1. Therefore every coefficient of 9 has an odd denominator. Hence we can choose a polynomial f (X) E Z[X] of degree at most n such that every coefficient of f is congruent to the corresponding coefficient of 9 mod 2n. Then f(m) == O(m) (mod 2n) for all 0 m and we are done.

29 .2 8

II 2

10 -

;

= 29 . 2 8 . 27 . 26 • 23 . 22 = 235.

i=2

It is a classical result of Mahler that continuous functions on the ring of p-adic integers Zp can be uniformly approximated by polynomials (this is the

Proof. Define O(x) = 1 if x is odd and 0 otherwise. We want to find a polynomial f of degree at most n and such that f(x) == O(x) (mod 2n) for all integers x. We will prove the existence by induction, but in order to prove the inductive step, we will need the following key result:

510

Chapter 10. Arithmetic Properties Lemma 10.12.

Let f be a polynomial of degree at most n

. olynomzals

F(x) = f(x) - O(x).

at most n such that f(x) == O(x) (mod 2n- 1 ) for all integers x. We will choose our polynomial of the form n

Then for all integers x we have

g(X) = f(X)

F(x Proof Since

+ n + 1) ==

t;(

511

10.4. p-adic estimates

_1)n-k (n; I)F(x

+ k)

+ Lak'

(X - j).

(mod 2n). It clearly has degree at most n and for all 0 :S k :S n we have

f has degree a t most n, proposition 10.4 shows that

g(k) = f(k)

+ (-1)n- kak . k!(n -

k)!.

n+l

t;( (n; 1) _1)k

Thus, to prove the lem

f(x

+ k)

By hypothesis, there are integers bk such that f(k) = O(k) O:S k :S n. Now, since

+ bk . 2n- 1 for

all

., . rna IS remallls to prove that 2 n divides

n+l

t;( (n; _1)k

I)O(x

+ k).

However, if x is even this is equal to - 2n , b ecause

;;=

c::11)

= 2n.

Similarly, if x is odd, then it is equal to 2 n b , ecause

;;=

(n2:

1) =

2n.

This ends the proof of the lemma. The crucial point that we deduce from t h ' . 0 ((x) == O(x) (mod 2n) for all 0 < x < th IS le~ma IS the following: if llltegers x. This follows triviaU; b in~' /n f(x} = O(x) (mod 2n) for all lemma. y uc lon, USlllg the congruence in the We. ca~ now finally prove by induction the existen ce of fÂˇ For n = 1 everythlllg IS clear, so assume that we constructed a polynomial f f ' o degree

it is clear that we can choose integers ak such that 2n divides

By construction, 2n will divide g(k) - O(k) for all 0 :S k :S n and thus for all x, by the previous results. The inductive step is thus proved and the problem is solved. 0 We end this section with a challenging problem having a very elementary but delicate solution. The main ideas are the pigeonhole principle and p-adic estimates, but the way in which they are combined is rather subtle. A special case of this problem (whose proof is very similar) is discussed in chapter 3, problem 13.

\f24. Let J E Z[X] be a polynomial and let a be an integer. Consider the sequence ao = a, an+l = J(a n ). If an -+ 00 and the set of prime divisors of (an)n is finite, prove that J(X) = AX d for some A, d. Tuymaada Olympiad, 2003

Froof Let

be the composition of

with itself d times. Assume that ao, al, .... First, we will

pI,P2,'" ,PN comprise all prime factors of all numbers

Chapter 10. Arithmetic Properties of Polynomials

512

prove that one of the numbers f(O), f 2(0), ... ,f N (0) is O. To do this, fix any positive integer r and no such that for all n 2: no we have an> (PIP2'" PN y. Such no exists by assumption. Thus, for all n 2: no there exists i such that vPi (an) > r. Considering N + 1 consecutive values n = no, no + 1, ... , no + N, the i's that we find take only N possible values, so two of them will be equal. Thus, we can find U E {I, 2, ... , N}, i 2: no and k E {I, 2, ... ,N} such that vPk (ai) > rand vPk (aHu) > r. Since aHu = r(ai) == r(O) (mod ai), it follows that Pk divides r(O). Thus, we proved that for any r we can find U E {I, 2, ... , N} and k such that Pk divides r(O). It follows that one of f(O), P(O), ... , fN (0) has infinitely many divisors and the claim is proved. If fi(O) = 0, then working with fi instead of f (the iterates of a under fi form a subsequence of the sequence (an)n, and so will still have only finitely many prime divisors), we may assume that f(O) = 0 (it is easy to see that if fi(X) is of the form AXd for some i 2: 1, then f itself is of this form). Write then f(X) = Xd(b o + blX + ... + bexe) for some integers bo, .. . , be and some d 2: 1 such that bobe =I- o. Assume that the conclusion of the problem does not hold, so that e 2: 1. Note that an divides an+l. Since for each i there is ni such that ani is a multiple of Pi, the number an1 +.+ nN is a multiple of Pl'" PN and so there is no such that for all n 2: no we have PI ... pNlan . It follows easily from this that the prime factors of bo are all amongst the Pi. Consider now some i such that pilbo. Then

vpi(an+J) = dVpJa n ) + vpi(b o + blan + ... 2: vpi(an ) +min(l,vpi (a n )),

+ bea~)

from which is trivially follows that vPi (an) -+ 00 for n -+ 00 (we know that for large enough n we have vpi(a n ) > 0 by the previous paragraph). Now, we are finally ready to conclude: let Pill'" ,Pik be those primes among PI,路路路 ,PN that divide boo By the previous paragraph, eventually by increasing no, we may assume that that v pi . (an) > VPi . (b o) for all 1 ::; j ::; k and all n 2: no. If q is a prime factor of 60 + ... + 6ea~ with n .2: no, then q divides an+l, so q is one of the Pi's. But all Pi's divide an, so q must divide bo and so q E {Pill'" ,Pik}' Moreover, since vPi (an) > vPi (b o) for all 1 ::; j ::; k, we J J have VPi . (b o+" 路+bea~) = v pi . (bo). Putting together these observations shows J

10.5. Miscellaneous problems

513

that bo + bla n + ... + bea~ ::; bo for all sufficiently large n, contradicting the fact that an -+ 00. The result finally follows. 0

10.5

Miscellaneous problems

The following problem is a version of a very classical topic, known as the Prouhet-Tarry-Escott problem. It concerns disjoint sets of n integers having the same sum of dth powers for all 1 ::; d ::; k (k depending on n).

\f 25. Let

d, r be positive integers with d 2: 2. Prove that for any nonconstant polynomial f with real coefficients and of degree smaller than r, the numbers f(O), !(1), ... ,!(dr - 1) can be divided into d disjoint groups such that the sum of the elements of each group is the same.

J. O. Shallit, AMM E 3032 Proof. Define the sets

where Sd(X) is the sum of digits of x when written in base d. We will prove that LXEAi(r) f(x) is independent of i, for any polynomial f of degree smaller than r. The proof will be by induction on r, the case r = 1 being clear (since then each Ai (r) has exactly one element and f must be constant). Assuming that the result holds for r, observe that (by linearity) it is enough to prove that LXEAi(r+l) xr is independent of i in order to prove the inductive step. But, by definition (and considering the last digit in base d) we have

Ai(r + 1) = U1:6{dx

+ jlx E Ai_j(r)},

the union being disjoint (and the sets Ai being numbered mod d). Thus

d-I xr=L L (dx+jY j=O XEAi_j(r) xEAi(r+l) L

514

Chapter 10. Arithmetic Properties of Polynomials

Separate in the last sum the term corresponding to xr and observe that it does not depend on i, since it is just

dr L xr. xEAo(r)U···UAd_l (r)

515

26. Let p ~ 5 be a prime and let a, b, c be integers such that p does not divide any of the numbers a - b, b - c, c - a. Let i, j, k be nonnegative integers such that i + j + k is divisible by p - 1 and such that for all integers x, the number

(x - a)(x - b)(x - c)[(x - a)i(x - b)j(x - c)k - 1]

The other terms are independent of i by the induction hypothesis. Thus I:xEAi(r+l) xr is also independent of i, finishing the proof of the inductive ~~.

10.5. Miscellaneous problems

is divisible by p. Prove that each of i, j, k is divisible by p - 1.

Kiran Kedlaya and Peter Shor, USA TST 2009

Proof. This will use the same sets Ai, but the proof that they satisfy the desired conditions is different. Let z be any dth root of unity different from 1 and let

Proof. We start with some formal reductions. First, note that we may assume that i, j, k < p - 1, as we can replace i, j, k with their remainders mod p - 1, without affecting the hypothesis or the conclusion (this uses Fermat's little theorem). We want to prove that i = j = k = 0, so assume the contrary. By hypothesis, i + j + k = p - 1 or 2(p - 1). In the second case, replace each x E {i,j, k} with p - 1 - x. As this does not change the hypothesis or the conclusion, we can assume from now on that i + j + k = p - 1. Finally, we can clearly assume that i is the largest among i, j, k. MultiI?lying the congruence

i::l.af(X) = f(X)

+ zf(X + a) + z2 f(X + 2a) + ... + zd-l f(X + (d -

l)a).

Since 1 + z + ... + zd-l = 0, we have deg(i::l.af) < deg(f). Let our disjoint sets be A o, AI, . .. A d- l as in the previous solution and consider

A(z) = d-l Lzk k=O

(

f(X) ) .

XEA k

°: ;

b)j(x - c)k - 1] ==

(x - a)(x - b)(x - c)[(x - a)i(x

(mod p)

by (x - a)Hk and using Fermat's little theorem, we deduce that

By an immediate induction, we obtain that

f(x) = (x - a)(x - b)(x - c)[(x - b)j(x for all integers x. On the other hand,

c)k - (x - a)j+k] ==

(mod p).

has degree at most

But deg(f) < r, so that deg(i::l. dr-li::l. dr-2 . . . i::l.di::l.d(X)) < 0 and so A(z) = 0. Since this holds for all such z, it follows that the polynomial A(t) is a multiple of 1 + t + ... + t d - 1 and hence I:xEAk f(x) is independent of k. 0 The following problem is rather technical, but the idea is very simple: after some algebraic manipulations, everything comes down to the fact that a nonzero polynomial of degree at most d cannot vanish at more than d distinct points.

(for p ~ 5) and p different roots mod p. Thus f vanishes in lFp[X] and we deduce the equality (X - b)j(X - c)k = (X - a)Hk in lFp[X]. Note that j + k =I- 0, as i < p -1 and i + j + k = p 1. Thus (X - b)j(X - c)k vanishes at b or c. But this is impossible, as by hypothesis (X - a)Hk does not vanish at either b or c. 0

Chapter 10. Arithmetic Properties of Polynomials

516

The next problem is very exotic and tricky. 27. Prove the existence of a number c > 0 with the following property: for any prime p, there are at most cp2/3 positive integers n such that p divides n! + 1. Chinese TST 2009

Proof. Of course, if pin! + 1, then n :S p - 1. Let p > 2 and let 1 < nl < n2 < ... < nm < p be all solutions of the equation n! : : : : -1 (mod p). Assume that m > 1 (otherwise everything is clear). The congruences ni! : : : : -1 (mod p) and ni+l! ::::::: -1 (mod p) imply that

(ni

+ l)(ni + 2)· .. (ni + ni+l -

nd : : : : 1

(mod p).

10.5. Miscellaneous problems

In particular, p > follows.

+ l)(x + 2)··· (x + k)

:::: 1

(j + 1)(j + 2) > m > j(j + 1) . 2

Since m 2: j(jtl) = 'L{=l j, when the differences ni+l - ni are written in ascending order, the first is at least 1, the next two are at least 2, and so on, each time the next i differences are at least i (this is because for a fixed k, 1 :S k < p, ni+l - ni = k has at most k solutions i). Thus

~(ni+l -

ni) 2: 12 + 22

+ ... + i =

j(j

+ 1)~2j + 1).

< (3p)1/3. Since m < (j

if' 28.

Let n be an even positive integer. Find the least positive integer k for which one can find polynomials with integer coefficients f, g such that

f(X)(X

+ It + g(X)(Xn + 1) = k. IMO Shortlist 1996

Proof. Let us write n = 2r . m for some odd integer m and assume that we have

f(X)(X

+ l)n + g(x)(xn + 1) =

I1;:1

The idea is to find

p> nm - nl

j(j + 1)(2j + 1) 6

for some f, g E Z[X] and some positive integer k. Taking for X a root Zi of the polynomial X 2r + 1, we deduce that f(Zi)(Zi + l)n = k. Multiplying all these relations and taking into account to (1 + Zi) = 2, it follows that n f(Zi) . 2 = k2r. Since f(Zi) is an integer (by theorem 9.10), 2n 2r divides k and so k must be a multiple of 2m . In particular, k 2: 2m . We will prove now that k = 2m works. Let us see what happens when m = 1 first. We need to find polynomials f, g with integer coefficients such that f(X)(X + 1)2r + g(X)(X2r + 1) = 2.

I1;:1

i=l

We deduce that

I?, the result

(mod p).

Since the polynomial (x + 1)(x + 2)··· (x + k) - 1 E (ZjpZ)[x] has at most k distinct roots modulo p, it follows that for each 1 < k < p there are at most k indices i such that niH - ni = k. We will prove that this is enough to force m < cp2/3. Choose a positive integer j such that

and so j

Before passing to the next problem, let us recall that Bezout's lemma does not hold in Z[X]. However, if f and g are polynomials with integer coefficients and with no common complex root, then they are relatively prime and by Bezout's lemma in Q[X] there is a nonzero integer c and polynomials A, B E Z[X] such that Af + Bg = c. Usually lei > 1 and it is not at all clear what is the smallest possible value of Icl, given f and g. The following problem discusses the case f = (X + l)n and g = xn + 1. It is a challenging problem, combining several ideas in a very tricky way.

Letting k = ni+l - ni, we see that x = ni is a solution to

517

such that

f(z)(z

+ 1)2r

= 2

518

Chapter 10. Arithmetic Properties of Polynomials 2r

for some root Z of X + 1. Indeed, since X + 1 is irreducible over the rational numbers (because (X + 1)2r + 1 is Eisenstein for the prime 2), this would imply 2r that X + 1 divides f(X)(X + 1)2r - 2, which would give us g. The key point 2r is to take Z = e ¥, because all the other roots Zi of X + 1 are of the form zj, with odd j. Thus, if Zl = Z, .. . , Z2r are the roots of X 2r + 1, then we can write Zi + 1 =r(z + l)Qi(z) for some polynomials Qi with integer coefficients. And since (1 + Zi) = 2, it follows that (1 + z)2r Qi(Z) = 2 which gives us the polynomial f and finishes the proof in the case m = 1. Finally, it is rather formal to deduce the general case from the case m = 1. Namely, pick polynomials with integer coefficients f, 9 such that

IT;=l

with ad >

+ 1?r + g(X)(X2T + 1) =

and d 2': 2. Note that we may assume that ao =I- 0, otherwise the

problem is trivial. . . First let us consider the equation fen!) === 0 (mod p). Unless p dIVIdes ao, this f~rces n < p. So, let us look for n = p - k with k > O. We have to compute first (p k)! (mod p), which is very easy by Wilson's theorem: -1=== (p-1)! === (p _ k)!(p - k

IT;:l

f(X)(X

519

10.5. Miscellaneous problems

+ 1)··' (p -

=== (p _ k)!( _l)k-l(k - 1)!

1) (mod p).

Thus, we have fen!) === 0 (mod p) if and only if p\Xk, where

Xk = ao(k _1)!d

+ al(k

1)!d-l(_1)k

+ ... + ad( _1)kd.

Then

f(x)m(x

+ It =

(2 - g(X)(X2r

+ 1»m =

2m + (X2r

+ l)h(X)

for some h E Z[X]. The last equality follows from the binomial formula. Now, replace X by xm in the previous equality, to get

f(xm)m(xm

+ 1)n = 2m + (xn + l)h(xm)

and observe that (xm + 1)n = (X + 1)nA(X) for some A E Z[X] (because m 0 is odd). The conclusion is now clear. The idea of the following problem is very natural, but there are a lot of technical details one has to deal with, which makes the proof rather long. ~

29. Suppose that f is a polynomial of degree at least 2, with positive leading coefficient and integer coefficients. Show that there are infinitely many n such that f (n!) is composite. IMO Shortlist 2005

Proof. We will try first to find prime numbers P and positive integers n such that plf(n!). Then, we will ensure that n is large enough and finally we will have to get rid of the cases fen!) = O,p, -p. Write f(X) = adXd + ad_l Xd - 1 + ... + ao,

We will prove first the existence of large prime factors of Xk, more precisely such that p 2': k. This is the content of the following

Lemma 10.13. There exists ko such that for all k > ko, there exists a prime

factor Pk of Xk such that Pk 2': k. es

Proof This is easy: choose kl such that Vp«kl - 1)!) > vp(ad) for all pr.in: < \~ \ If all prime factors p of Xk are less than k for some k 2': kl' they dIVIde p _ d· . . «k -1)') > (k -1)! and Xk, so they divide ad' But for such a pnme p, smce vp . a) we must have v (Xk) = vp(ad)' We deduce that IXkl S ladl· Now, Vp ( d, P k h' h . sible as 0 choose ko > kl such that IXkl > ladl for all k 2': 0, W IC IS pos ao =I- O. Fix now ko and Pk as in the lemma. Fix also a positi:-e inte~er N ~nd assume that none ofthe numbers fen!) with n 2': N is composIte. By mcrea~mg N we may assume that x ---t f(x!)-x is increasing on [N, 00). By constructIOn, f«p k -k)') Pk 'd"d IVI es ., Thus , ifPk-k> - N, then we must have f«Pk-k)!) = 'tPk and this will happen if we ensure that k, k + 1, ... , k + N - 1 are compos~ e. To have this, we can choose k = ka = a(N + 1)! + 2 for a 2': 1. De~otmg _ k we deduce that f(x a !) = Xa + a(N + I)! + 2 for all suffiCIently xa - Pk a a, ' h th a t Xa ---t 00 , k) In particular the last relatIOn sows Iarge a (so tha t ka >

°.

520

Chapter 10. Arithmetic Properties of Polynomials

because the map a t-+ Xa is injective. In particular, for infinitely many a we have Xa+l 2: Xa + 1 and so

f(xa!) - Xa

+ (N + I)! =

f(Xa+I!) - Xa+l 2: f((xa

+ I)!) -

(xa

+ 1).

This implies that

f((xa

+ I)!) -

f(xa!) ::; 1 + (N + I)!,

which is certainly impossible because f(j(~:!~)!) -+ 00 for a -+ 00. Thus our assumption was wrong and at least one of the numbers f(n!) with n 2: N is composite. Since N was arbitrary, the conclusion follows. D

Remark 10.14. The hypothesis that d 2: 2 was not used in the proof, so the result still holds for linear polynomials with positive leading coefficient. For instance, this also solves the following Chinese TST 2011 problem: prove that for all positive integers d there are infinitely many n such that d . n! - 1 is composite.

10.6

Notes

The following people helped us with solutions and we would like to thank them: Alexandru Chirvasitu (problem 24), Xiangyi Huang (problems 6, 12, 27), Holden Lee (problem 22), Mitchell Lee (problem 2), Fedja Nazarov (problem 16), Hunter Spink (problem 10), Richard Stong (problems 7, 8, 20, 23, 25), Qiaochu Yuan (problem 1), Victor Wang (problem 14), Gjergji Zaimi (problems 3, 5), Alex Zhu (problems 14).

Chapter 11

Lagrange Interpolation Formula It is a standard fact that a nonzero polynomial f with coefficients in a field K has at most deg f roots in K. Lagrange's interpolation formula is in some sense a refinement of this standard result: it makes precise the way in which deg f + 1 points of K determine the polynomial f.

Theorem 11.1. Let K be a field and let xo, Xl, . .. ,Xn be distinct elements of K. Then for any polynomial f E K[X] of degree at most n we have

f(X) =

L k=O

f(Xk) .

X-X. _ xJ .

II X j#

Proof. Simply note that the difference between the polynomials in both sides of the desired equality is a polynomial of degree at most n which vanishes at Xo, Xl, ... ,Xn , thus it is the zero polynomial. D The following corollary is an immediate consequence of the previous theorem, but it is rather useful in practice, especially when proving complicated identities.

522

523

Chapter 11. Lagrange Interpolation Formula

Corollary 11.2. Let f E K[X] be a polynomial of degree at most n and let Xo, Xl, ... ,Xn E K be distinct elements. Then

2. A polynomial

of degree n satisfies 1 f(k)

f(Xk) k=O TIjtk(Xk - Xj)

for all 0 ::; k ::; n. Find f(n

+ 1). Titu Andreescu, IMO Shortlist 1981

is 0 if deg f < n and equals the leading coefficient of f otherwise. Proof. Simply identify the coefficients of xn in Lagrange's identity.

= (ntl)

Proof. This is also immediate using theorem 11.1:

The first two problems are direct applications of Lagrange's interpolation formula.

if 1.

A polynomial p of degree n satisfies p( k) its value at n + 1.

= 2k for all 0 ::; k ::; n. Find n

Murray Klamkin

= 2)-lt- k k=O 1- (_l)n+l

Proof. Using theorem 11.1, we obtain

o 3. Prove that for any real number a we have the following identity

Tepper's identity

Remark 11.3. There is also a neat solution without use of the interpolation formula: consider the polynomial

Proof. Use corollary 11.2 for the polynomial P(X) = (a - x)n and the points Xk = k. 0

f(X) =

(~) + (~) + ... + (~).

It has degree n and satisfies f (k) = 2k for all 0 ::; k ::; n by the binomial formula. Thus we must have f = p and then clearly p(n+ 1) = 2n+l_1. Even though very neat, this solution is not so conceptual.

4. Let

Prove that 8 1

= Xo + xl + ... + Xn and compute 82.

Chapter 11. Lagrange Interpolation Formula

525

Proof. Actually, the following method shows how to compute Sp in general. The idea is to consider the remainder f(X) of Xn+p when divided by

for all x E JR. It is not really obvious that Tn exists, but the reader can easily check the existence inductively, by establishing the recurrence relation Tn+l (X) + T n- 1 (X) = 2XTn(X). This also implies that the leading coefficient of Tn is 2n-l. A fundamental theorem of Chebyshev states that for any monic polynomial f E JR[X] of degree n we have maxXE[-l,l]lf(x)1 2: 2nl_l' with equality if and only if f = Âą 2nl_l Tn. This explains why Chebyshev polynomials are so important. For a proof of Chebyshev's theorem, see problem 14 in this chapter.

524

and to apply Lagrange's interpolation formula to it. Identifying leading coefficients and using the fact that f(xj) = x'j+P, we deduce that Sp is precisely the leading coefficient of f(X). Now, for p = 1 we clearly have

f(X) = X n+1

5. Let a, b, c be real numbers and let f (x) = ax 2 + bx + c be such that

IT (X -

Xj), max{lf(Âą1)1, If(O)I} ::; 1.

j=O so that its leading coefficient is must have

L.:7=0 Xi.

Xn+2 = Q(X)

On the other hand, for p = 2 we

IT (X -

Prove that if Ixl ::; 1 then If(x) I ::;

Xj)

~ and

IX2 f

(~) I ::; 2.

+ f(X)

Spain 1996

j=O for some polynomial Q. Comparing degrees and leading coefficients shows that = X + c for some constant c. To determine c, we impose the condition that deg(J) ::; n. This implies that c = L.:'j=o x j. Then, it is easy to find

Proof. Using Lagrange interpolation, we can write

Q(X)

the coefficient of xn in f(X) and the answer is (L.:'j=o Xj L.:OSi<jsn XiXj. Actually, we leave as a nice exercise for the reader to prove that for all p we have

f(X)

f(0)(1 - X2)

The following problems deal with extremal properties of polynomials. The underlying philosophy is that imposing conditions on the values of a polynomial at sufficiently many points of an interval automatically imposes conditions at all other values. Lagrange's interpolation formula is a very handy tool in such situations, but there. is an extra ingredient which appears all the time, namely the Chebyshev polynomials. Recall that the nth Chebyshev polynomial Tn is the unique polynomial of degree n such that cos( nx) = Tn (cos x)

X2+X 2

+ f( -1)

X2_X 2 .

We deduce that for all Ixi ::; 1 we have

If (x ) I ::; 1 - x2 where in the sum above the ai are nonnegative integers.

+ f(1)

+ xl

Ix2

= 1 -:-

+ Ix I ::; "4'

the last inequality being equivalent to (ixi - ~)2 2: O. Similarly, we find that

Ix 2f (1/ x ) I ::; 1 - x 2 +

1+x 2

1-x

+ -2-

2 - x ::; 2.

6. Find the maximal value of the expression a 2 + b2 + c2 if Iax 2 + bx + cl ::; 1 for all x E [-1,1]. Laurentiu Panaitopol

526 Chapter 11. Lagrange Interpolation Formula

527

Proof Define A

= f(I),

= f(O),

= f(-I).

Remark 11.4. For third degree polynomials, Chebyshev's theorem is very easy to prove using Lagrange's interpolation formula: if f is a monic polynomial of third degree, identifying leading coefficients in Lagrange's formula yields the equality

Then we easily obtain

a=~-B A-C 2 ,b=_

e=B.

Therefore, an immediate computation gives

+ +e

A2 + C2 2

+ 2B2 -

B(A + C).

Since IAI, IBI, ICI ::; 1, the last expression' 1 the obvious estimate for each term f 't) T~s c early bounded above by 5 (use To see that it is optimal . 1 ~ 1 . us, we always have a 2+b2+e2 < 5 , sImp y ta e Chebyshev's polynomial 2X2 _ 1 -

7. Define pea, b, e) = max Ix 3 - ax2 _ bx _

xE[O,3] value of this function over 1R3?

e. What is the least possible

Proof Let Pa,b,c(X) = X3 - aX 2 _ bX _ e . . [0,3] to [-1,1] via an affine map and th t' The Idea IS to map the interval theorem in order to bound f b I en 0 Use Chebyshev's least deviation rom eowmaxXE[O,3] IF.a,b,c ()I x . Note that

iel

. 27/8, Chebyshev's theorem . h th~rd degree wIth leading coefficient gIves us t e estImate max XE[-l,l] Thus P(a b e) "

> -

27 32

IF.a,b,c (3(X 2+ 1))

27 2': 32'

d h' . an t IS IS optimal, since equality holds for

F.a,b,c -_32T3(2X/3 27 -

Proof Choose distinct numbers xo, Xl, X2, X3 and identify the coefficients of X in Lagrange's formula

We deduce that

max IPa b (3(X + 1)) I XE[-l,l] , ,c 2 .

(3(X+l)). a,b,c ~ IS a polynomial of

Since F.

8. Let a, b, e, d E 1R such that lax 3 + bx 2 + ex + dl ::; 1 for all x E [-1,1]. What is the maximal value of lei? For which polynomials is the maximum attained? Gabriel Dospinescu

Chinese TST 2001

xr;}[~~]IPa,b,c(x)1 = '

1 = 2f(1) _ 4f(1/2) + 4f( -1/2) _ 2f( -1). 3 3 3 3 Of course, this can also be triviall;y checked by hand. This equality and the triangle inequality imply that maXXE[-l,l]lf(x)1 2': ~ for any such f, with equality when f(X) = X 3 - ~X.

1),

where T3(X) = 4X3 - 3X.

I'"

X X XIX2 + X2 3 + X3 l ~ (xo - xd(xo - X2)(XO - X3) XIX2 + X2 X3 + X3 Xl ::; (xo - XI)(XO - X2)(XO - X3) . =

f(xo)

The problem is to find a 4-tuple (XO,XI,X2,X3) which minimizes the last expression. As usual in this kind of problems, a good idea is to choose the points where IT3(x)1 takes its maximal value 1 on the interval [-1,1]. These are the points Xo = -1, Xl = -~, X2 = ~ and X3 = 1. It is easy to compute the last sum in this case and we find that lei ::; 3. Since this value is attained for the polynomial T3(X) = 4X3 - 3X, this is the maximal value. Also, it is not difficult to check that equality appears in the above chain of inequalities only for T3 and -T3. 0

Chapter 11. Lagrange Interpolation Formula

528

9. If a polynomial f E ~[X] of degree n satisfies If(x)1 ::::; 1 for all x E [0,1]' then

529

Remark 11.5. The identity ) ~(_1)n-k ( n~l p(k)

n+l

Proof. We will use Lagrange's interpolation formula at the points kin for We have

o : : ; k ::::; n.

which shows that

(-~) I::::;

tIT Ik ~ )1~ t (n ++~) j

k=Oj#

= 2n+l_1.

for polynomials of degree at most n is extremely useful. We proved it here as a consequence of Lagrange's interpolation formula, but it also follows from the theory of finite differences, for a glimpse of which we refer the reader to section 10.3. 11. Let a, b, c, d E ~ such that lax 3 + bx 2 + ex Prove that

z=O,l, ... ,n+l

Proof. The crucial observation is that we have

(a - l)n+l =

I:(

-It-

k=O

Proof. Let us look at the values of P(X) = aX3 + bX2 + eX + d at -1, -1/2, 1/2, 1, which are the classical interpolation points in Chebyshev's theorem (for n = 3). Writing A = f(l),

-ll n +l <

B = f(1/2),

= f( -1/2),

D = f( -1),

we can easily express a, b, c, d in terms of A, B, C, D. An easy computation gives 2 4 a = 3(A - D) - 3(B - C), and

(p(k) _

ak ).

Thus, if Ip(k) - akl < 1 for all 0 ::::; k ::::; n

(n ~ 1)

::::; 1 for all x E [-1,1].

IMO Shortlist 1996

~(-l)n-t:l)p(k) ~O. This is simply Lagrange's interpolation formula written in a slightly changed form (though a more conceptual way of seeing this is using the finite differences theory). We deduce from this and the binomial formula that

+ dl

lal + Ibl + lei + Idl : : ; 7.

k=O k

10. Let a 2': 3 be a real number and let p be a real polynomial of degree n. Prove that . max la i - p(i)1 2': 1.

+ 1, we must have

I: (n ~ 1)

= 2n +1,

k=O contradicting the fact that a 2': 3. This finishes the solution.

This shows that f(a, b, c, d) = lal + Ibl + lei + Idl is actually a convex function of A, B, C, D E [-1, 1]. Thus it attains its maximum when all A, B, C, Dare equal to 1 or -1. Now, it is a tedious matter to check that in all cases the expression is at most 7. We have equality for the Chebyshev polynomial (or its opposite) of degree 3. 0

531

Chapter 11. Lagrange Interpolation Formula

530

12. Let A = {p E lR[X] I degp::; 3, Ip(±1)1 ::; 1, Ip (

±~) I::; 1}.

Find sup max Ip" (x) I. pEA Ixl:S1 IMC 1998

Proof. Write p(X) = aX3 + bX 2 + eX + d. Since p"(x) is an affine function on [-1,1]' the maximum of its absolute value is obtained for x = 1 or x = -1. Thus, we need to bound in an optimal way the number max(16a + 2bl, 1- 6a + 2bl)· Taking the Chebyshev polynomial shows that this can be already as large as 24._ But it is always at most 24 (under the given hypothesis on p), since for instance using the formulas in the solution of problem 11 yields

6a + 2b =

Dividing both sides by z and using the binomial formula yields (1- z)n-1 = 0, a contradiction. 0

Remark 11.6. There are many other approaches. The most efficient is the method of finite differences, which yields

~n-1 P(1) =

n-1

2) -1)j (n -:- 1)p(n -

j) = z(z - 1)n-1 =I-

j=O

Another very neat proof considers the polynomial P(X + 1) - zP(X). It has degree at most n - 2, it is clearly nonzero and vanishes at 1,2, ... , n - 1, a contradiction. We leave the land of routine problems in the Lagrange interpolation formula and tackle some more delicate results. The following is taken from [66].

3 P (1) - 3P( -1) - 3 P (1/2) + 3 P ( -1/2).

Thus 16a + 2bl ::; 24 by the triangle inequality. We proceed in a similar way to bound I - 6a + 2bl or we can observe that if p satisfies the conditions in the problem, so does p( -x). 0 1//13. Let n 2': 3 and let f, 9 E lR[X] be polynomials such that the points (f(1),g(1)), (f(2),g(2)), ... , (f(n),g(n)) are the vertices of a regular ngon in counterclockwise order. Prove that max (deg f, deg g) 2': n - 1. Putnam 2008

Proof. It is enough to prove that P(X) = f(X) + ig(X) has degree at least n -1. Clearly, we may assume that the regular n-gon has vertices z, z2, ... ,zn 2i1r for z = en (simply apply a translation and a hQmothety to reduce the general case to this one). So, it remains to prove that if a polynomial P satisfies P( i) = zi for all 1 ::; i ::; n, then deg P 2': n - 1. Assume that this is not the case. Using Lagrange's interpolation formula, we obtain

14. Let f be a complex polynomial of degree n such that If(x)1 ::; 1 for all x E [-1,1]. Prove that If(k)(x)1 ::; IT~k)(x)1 for all k and all real numbers x such that Ixl 2': 1. Deduce Chebyshev's theorem: if f is monic of degree n, then 1

max If(x)l2': -1· xE[-l,l] 2n W.W. Rogosinski

Proof. Pick for the moment n + 1 points Xo, .. . , Xn in [-1,1] and apply Lagrange's interpolation formula: f(X) = tf(Xi) i=O

II X -

Xj.

#i Xi - Xj

Next, differentiate this equality k times, so if we set Pi(X) = TI#i(X - Xj), then

f(k)(x)

f(Xi) Pi(k)(x)

i=O Pi(Xi)

532

533

Chapter 11. Lagrange Interpolation Formula

for all x. Choose any Ixi ?: 1 and apply the triangle inequality to the previous identity, together with the hypothesis that If(Xi)1 :::; 1. We infer that

If(k)(x)1 :::;

6;'

"15. The polynomials

E JR[XJ and a E JR[X, YJ satisfy

f(x) - f(y) = a(x, y)(g(x) - g(y))

IPi(k)(x)l. IPi(Xi)l

for all x, y. Prove that we can find a polynomial h such that

f(x) = h(g(x)).

But we also have

T(k) (x) = n

~To

L... i=O

(xo)Pi(k)(x) ~ P( 0) ,

Russia 2004

~ x~

Proof. Fix an integer N > deg f. If 9 is constant, so is f and everything is clear. So, assume that 9 is not constant, thus we can choose N + 1 distinct numbers xo, Xl, ... ,X N at which 9 takes different values. By Lagrange's interpolation formula, there exists a polynomial h of degree at most N such that h(g(Xi)) = f(Xi) for all 0 :::; i :::; N. Note that by hypothesis

so it would be really nice if we could ensure that

n (k) ( ) k I (k) ( ) I LTn(Xi)Pi x = L ~ x . i=O Pi(Xi) i=O IPi(Xi)1 In order to have this, we already need ITn(xdl = 1 for all i, which suggests taking the old friends Xi = cos for 0 :::; i :::; n. Then Tn(Xi) = (_1)i and

e;)

p(k)(x) (-1)~. A(Xi) o

we would like to prove that the numbers have the same sign. If this were true, we would be done by the previous arguments. Note that Xo > Xl > ... > Xi, so that

Pi(Xi) = (Xi - XO)(Xi - Xl)'" (Xi - Xi-l)(Xi - Xi+1)'" (Xi - xn) has the sign (_1)i. So, it remains to see if the signs of the numbers Pi(k)(x) are independent of i as long as Ixi ?: 1. The crucial remark is that p?) have all their roots in [-1,1J. Indeed, this follows from Rolle's theorem and the fact that Pi have all their roots in [-1, 1J. But then it is clear that the sign of p?)(x) only depends on the degree of p?) (which is independent of i) and on the position of X with respect to 1 (ie whether X ?: 1 or X :::; -1). So, it is independent of i and we are done. Finally, let us prove Chebyshev's theorem using this result. Suppose that If(x)1 < 2n~1 for all X E [-1,1J. By compactness there exists c > 2n- l such that /cf(x)1 :::; 1 for all X E [-1,1J. Using the result established above, we

Ixi

f(X) - f(Xi) = a(X, Xi)(g(X) - g(Xi)), so that g(X) - g(xd divides f(X) - f(xd. On the other hand, g(X) - g(Xi) also divides h(g(X)) - h(g(Xi)) = h(g(X)) - f(Xi) (this follows by linearity and the fact that g(X) - g(Xi) divides g(X)k - g(Xi)k for all k ?: 1). Thus h(g(X)) - f(X) is a multiple of g(X) - g(Xi)' Now, the polynomials g(X) g(Xi) are clearly pairwise relatively prime, since the numbers g(Xi) are pairwise distinct .. Thus h(g(X)) - f(X) is a multiple of Il~o(g(X) - g(Xi)), but has degree smaller than that of Il~o(g(X) - g(xd). Therefore f(X) = h(g(X)) and we are done. 0 The next problem presents P.J. O'Hara's beautiful proof [63J of a famous inequality of Bernstein. 16.

a) Prove that for all complex polynomials and for all X E <C

having degree at most n

( ) +;.1" L...f XZk (1- Zk)2' n

'() nO) xf X = 2f(x

?: 1 we have Icf(n)(x)1 :::; ITJn) (x) As f is monic of degree n and since Tn has degree n and leading coefficient 2n- l , this yields c :::; 2n - l , a contradiction. 0

deduce that for

f, 9

2Zk

k=l

where

are the roots of the polynomial

xn + 1.

534

Chapter 11. Lagrange Interpolation Formula

b) Deduce Bernstein's inequality: for all complex polynomials. have II!'II ~ nllfll, where Ilfll = max If(z)l.

f we

Izl=l

P.J. O'Hara

535

Simply choose f(X) =

xn in the equality

x1'(x) =

~n ~(f(XZi) ~

f(x)) ( 2Zi)2' 1- z·

i=l

Ilfll

b) By scaling f, we may assume that use the identity in (a) to deduce that

Proof. a) The polynomial

gx (

X) = f(xX) - f(x) X-1

11' (z) I ~

has degree at most n - 1 and it is easy to see that gx(1) = Xf'(X). Lagrange's interpolation formula for the points Zi yields

The miracle is that all numbers 2Zk ' (1 - Zk)2 1 (Zk 2

~2 + ~n

k=l

= 1. Choose any

2Zk (1 - Zk)2

Izl

= 1 and

I·

+.1...) - 1 Zk

are actually negative, so the previous inequality becomes

which combined with the equalityl

II( Zi - Zj )= nZi #i

n-l

11'(z)1 ~

n = --

n _ ~~ 2Zk . 2 n ~ (1- Zk)2 k=l

Zi Since we've already seen that

gives

Taking X = 1 and remembering that gx (1) =

f' (x),

we can now safely conclude that 1!,(z)1 problem.

we obtain

n, finishing the proof of this hard 0

We end this chapter with a beautiful inequality due to Gelfond, who used it in his solution of Hilbert's seventh problem. This famous problem asked to prove that a b is transcendental whenever b is an algebraic irrational number and a is an algebraic number different from 0 and 1.

so we only need to prove that

17. Let f be a complex polynomial of degree at most n and let Zo, Zl,' .. ,Zd be the zeros of the polynomial Xd+l - 1, where d > n. Define n

10btained by differentiating the identity

xn + 1 =

II(X i=l

Zi)

and taking X =

Zi.

Ilfll =

max If(z)l· Izl=l

536

Chapter 11. Lagrange Interpolation Formula a) Prove that if there exist n + 1 pairwise distinct numbers XO,Xl, .. ·,Xn among ZO,Zl, ... ,Zd such that I/(xdl ~ 2ld then IIIII < 1.

11.1. Notes All in all, each term Il#i l~i-=-:i)1 can be bounded from above by

b) Deduce that

d+l

11/11·lIgll ~ 4

deg

(f)+deg(g)

2d - n

·lIlglI· A.O. Gelfond

Prool· a) The first step is rather clear: Lagrange's interpolation formula for the points Xo, Xl, ... ,Xn yields n

L I(Xi) II Z - Xj . , ° '-/.' Xi - X

I(z) =

I/(z)1

tIl

Iz-xjl. 2d i=O #i IXi - xjl

We brutally bound each Iz - Xjl by 2, so that Il#i Iz part is to deal with Il#i IXi - xjl. Write {Yl,"" Yd-n}

Xjl ~ 2n. The subtle

= {zo, ... , Zd} - {xo, ... , xn}

and observe that

j#i

d-n xjl·

II IXi - Ykl = II (Xi -

Zj) = d + 1,

Zj#Xi

k=l

because Ilz j #Xi (Xi - Zj) is exactly the derivative of Xd+ 1 absolute value d + 1. Since IXi - Ykl ~ 2, we infer that

d+1

x'-x'i > 2 -. II #i I

d- n

2d d

+ l'

Since there are n + 1 terms and n + 1 < d + 1, we deduce that for all Izi = 1 we have I/(z)1 < 1 and so 11/11 < 1. b) This part is an easy consequence of the first one. Scaling I and g allows us to assume that 11111 = Ilgll = 1. Assuming that Illgll < 4~' we With deduce that for any 0 ~ i ~ d we have I/(Zi)1 < 2ld or Ig(zi)1 < d = deg(f) + deg(g), it follows that either we can find at least deg(f) + 1 Zi'S such that I/(zi)1 < 2ld or we can find at least deg(g) + 1 Zi'S for which Ig(Zi)1 < Both cases are excluded by part a), and the result follows. 0

:b.

id'

Jr~

Combining this with the triangle inequality and the hypothesis, we obtain for Izi = 1

II IXi -

537

1 at Xi, which has

11.1

Notes

We would like to thank the following people for providing solutions to the following problems: Xiangyi Huang (problems 6, 7, 8, 9, 12), Qiaochu Yuan (problems 1, 3, 10).

Chapter 12

Higher Algebra Combinatorics

â&#x20AC;˘

This last chapter deals with applications of linear algebra in combinatorics. There is a huge literature on this subject, which is still very active, so all we can do in this chapter is to barely scratch the surface and present the reader with some interesting elementary examples. Of course, this does not mean that there aren't a lot of difficult problems in this chapter. Before passing to problems, let us recall some basic results of linear algebra. Let K be a field. A K-vector space V is an abelian group endowed with an external multiplication K x V --+ V, denoted (a, v) --+ aÂˇ v, which is compatible with the additive operation of V and with the operations in K. A subspace of V is an additive subgroup which is stable under multiplication by elements of K. A family (Vi)iEI of elements of V is called linearly independent if for all finite subsets SCI and all (ai)iEs E Klsl, the relation I:sEs as 'V s = 0 forces as = 0 for all s E S. A family (VdiEI is called generating if any vector v E V can be written v = I:sEs asÂˇ Vs for some finite subset SCI and some as E K. Finally, a basis of V is a linearly independent family which is simultaneously generating. Using Zorn's lemma, one can prove that any vector space has a basis and that any linearly independent family of vectors can be extended to a basis. In most of the following problems we will only consider finite dimensional spaces, i.e. vector spaces having a finite generating family.

Chapter 12. Higher Algebra in Combinatorics

540

The dimension of such a space is by definition the number of elements of a basis (it is not obvious, but it can be proved that this does not depend on the choice of the basis). If V and Ware vector spaces over K, a map f : V -+ W is called linear if f (Vi + V2) = f (vt) + f (V2) for all Vi, V2 E V and f (a . v) = a . f (v) for all v E V and a E K. The kernel of f is then the subspace Ker(J) = {v E Vlf(v) = O} of V. A fundamental result in linear algebra is the following formula dim V = dim Ker(J)

+ dim Im(J).

Traditionally, dim Im(J) is called the rank of f. Let f : V -+ V be a linear map, where V is a K-vector space of dimension n. If ei, e2, ... ,en is a basis of V, the matrix of f in this basis is defined by the equalities f(ei) = 'Lj=l ajiej' It is easy to check that the matrix associated to the composition of f and g is simply the product of the matrices associated to f and g. Conversely, any matrix can be naturally seen as the matrix of an endomorphism of V in the given basis. If A is an n x n matrix with entries in some commutative ring R, its determinant is defined by det A =

10(0' )alo-(1)a2o-(2) ... ano-(n) ,

o-ESn Sn being the group of permutations of {1,2, ... ,n} and 10(0') being the signature of a. The most important (and rather surprising!) property of the determinant is its multiplicativity, i.e. det(AB) = det A . det B. Another important property is that A is invertible in Mn(R) if and only if det A is a unit of R. Thus, if R is a field, we can test if a matrix is invertible by testing whether its determinant is nonzero and this happens if and only if its associated endomorphism is bijective. Let Mn(K) be the space of n x n matrices with entries in K. If A E Mn(K), its characteristic polynomial is det(X . In - A), a monic polynomial of degree n with coefficients in K. Its roots in an algebraic closure K of K are called the eigenvalues of A. Hence, if A is an eigenvalue of A, then we can find v E K" such that A . v = A . v, i.e. 'Lj=l aijvj = A . Vi for all i. It is not difficult to check that the sum of the eigenvalues of A (counted with multiplicities) is equal to the trace of A, which is by definition the sum

12.1.

541

The determinant trick

of the diagonal elements of A. Also, det A is the product of the eigenvalues of A (again counted with multiplicities). It is a little bit trickier to prove that the eigenvalues of g(A) (where g E K[X]) are g(At), g(A2), ... ,g(An)'

12.1

The determinant trick

We present in this section a few tricky combinatorial problems in which the determinant of a matrix plays an important role. 1. Let n 2: 2. Find the greatest p such that for all k E {1, 2, ... ,p} we have

(tiO'(i))k

o-EAn

L o-EBn

~=l

(tiO'(i))k ~=l

where An, Bn are the sets of all even and odd permutations of the set {1, 2, ... , n} respectively. Gabriel Dospinescu

Proof. The first ingredient is the following: Lemma 12.1. Let ai, a2,' .. ,am and bl , b2, ... ,bm be positive integers and let N be a positive integer. Then

+ a~ + ... + a~ = bT + b~ + ... + b~

for all 1 ::::; k ::::; N if and only if (X - l)N+l divides Xal

+ X a2 + ... + Xam

_ Xbl _ X b2

••• _

Xb m.

Proof. This is quite easy: if f(X)

= xal + X a2 + ... + Xam _ Xbl _ X b2

••• _

xbm,

then (X - l)N+1 divides f if and only if f(j)(l) = 0 for all 0 ::::; j ::::; N. This happens if and only if m

L ai (ai i=l

1) ... (ai - j

+ 1) = L i=l

bi (b i - 1) ... (b i - j

+ 1)

542

Chapter 12. Higher Algebra in Combinatorics

12.1.

for all 0 ~ j ~ N. As the polynomials X(X - 1) ... (X - j + 1) for 1 ~ j ~ N span the same vector space as the polynomials X, X2, ... , XN (this is

The determinant trick

543

where the sum is taken over all permutations a of {1, 2, ... , n}. Putnam 2005

immediate), the previous equality holds if and only if

Proof. We start with the exotic proof: observe that

a~ + a~ + ... + a~ = b~ + b~ + ... + b~ for all 1 ~ k ~ N. The conclusion follows.

Using the previous result, we deduce that p + 1 is the multiplicity of 1 as a root of the polynomial

f(X) =

XS(cr) -

crEAn where S(a) = a(1)

f(X)

XS(cr) ,

crEEn

+ 2a(2) + ... + na(n).

The key point is that we can write

c(a)xa(l) . X2cr(2) ... xncr(n)

c(a)

~ cr 1+f(a)

n(n+l) f(X) = X-2.

c(a)

1 1 r xf(cr)dx r (Lcr c(a)xf(cr)) dx. Jo Jo =

detA(x) = (x

crESn where aij = Xij, Sn is the set of all permutations of {1, 2, ... , n} and c(a) is the signature of a (i.e. 1 if a E An, -1 if a E Bn). Finally, using properties of Vandermonde determinants, we can write

We recognize the determinant of the matrix A(X)ij = x 1i=j under the integral. But det(A(x)) can also be computed using row-column operations: add all 1 and then columns to the first one, take out the common factor x + n subtract the first column (which consists now only of ones) from all the others, to make the elements in the first row equal to 0, except for the first one. Next, expand the determinant using the first row. We obtain in this way that

det(A),

+ n -1)(x _l)n-1.

Thus

" c(a) = (x _ 1 + n)(x - 1t- 1dx ~ 1 + f(a) Jo cr

1 0

(xj - Xi)

-1

l:Si<jSn and since the multiplicity of 1 in xj - Xi is 1 for all i p + 1 = G)Âˇ Thus the answer to the problem is G) - 1.

(x+n)xn- 1dx

= _ (_l)n+1 _ (_1)n =1= j,

we deduce that 0

n+1

= (_1)n+1_n_

n+ l'

The next problem is rather strange, both because of the statement and because of one of its proofs. We also present a more natural proof, based on the inclusion-exclusion principle. 2. For a permutation a of {1,2, ... ,n} let c(a) = 1 if a is even and -1 otherwise. Let f(a) be the number of fixed points of a. Prove that " c(a) _ (1)n+1 n ~ 1 + f(a) - . n + l'

which is precisely the desired result.

Proof. Here is a more natural proof: we will partition the permutations a according to their fixed points. For a permutation a, let Fix( a) be the set of its fixed points and let f(a) = IFix(a)l. As

c(a)

cr 1 + f(a)

L ( L

k=O k + 1 1A1 =k

Fix(cr)=A

c(a)) ,

Chapter 12. Higher Algebra in Combinatorics

544

we naturally try to understand F(A) = LFix(a)=A c(o} We will compute F(A) using the inclusion-exclusion principle, according to which

F(A) =

(_l)IBI-IAI

AeB

On the other hand,

F(C) =

c(o}

alB=l

The notation criB = 1 means that cr(b) = b for all b E B. Now, permutations such that criB = 1 correspond to permutations of the complement of B, so if n - IBI 2 2 we have LalB=l c(cr) = O. We deduce that LBeG F(G) = 0 unless IBI 2 n - 1, when it is equal to 1. Combining this with the previous formula (inclusion-exclusion principle) yields F(A) = (_1)n- 1AI (IAI + 1 - n). We deduce that

L a

f) + f (cr ) = f) c(cr)

_1)n-k

k=O

(n) k + + n (n) _nt + (n) k

k=O

The determinant trick

545

Proof. The answer is negative. First, we will use the standard trick of counting pairs (P,{C l ,G2 }), where C l ,G2 are distinct circles among the 22 ones and P is a point among the 22 ones that belongs to G l n G2 . Each point belongs to at least 7 circles, so for each P there are at least sets {Gl , G2} living On the other in a triple with P. So the number of triples is at least 22 . hand, for each {Gl , G 2 }, G l n G2 has at most 2 elements, so there are at most 2Âˇ (222) triples. The miracle is that we actually have 2Âˇ (222) = 22 . G). Thus, all previous inequalities are actually equalities. Let H, P2, ... , P 22 be the points and let Gl, G2, ... , G 22 be the circles. Consider now the matrix A whose (i, j)-entry is 1 if Pj E Gi and 0 otherwise. The previous paragraph shows that P . pt is the matrix whose elements on the main diagonal are equal to 7, all other elements being equal to 2. An easy computation shows that the determinant of this matrix is 49 . 521 . But this determinant is also equal to (det A)2, which is a perfect square. We deduce that 5 is a perfect square, which is a bit difficult to make happen. . . 0

F(G).

BeG

12.1.

G).

We end this section with another rather challenging problem on permutations.

1k 1

4. A permutation cr of {1,2, ... ,n} is called k-limited if Icr(i) il:S k for all 1 :S i :S n. Prove that the number of k-limited permutations of {1,2, ... ,n} is odd if and only if n == 0,1 (mod 2k + 1).

_1 (_l)n-k k=O k 1 k

-n ~_1_(n+1)(_lt-k L...- n +1 k+1

Putnam 2008

k=O

and the result follows immediately.

The next problem crucially uses the multiplicativity of the determinant. It is (rather amusingly) a consequence of the irrationality of V5. 3. Is there in the plane a configuration of 22 circles and 22 points on their union (the union of their circumferences) such that any circle contains at least 7 points and any point belongs to at least 7 circles? Gabriel Dospinescu, Moldavian TST 2004

Proof. Let M be the n x n matrix defined by Mij = 1Ii - j l:S k ' It is clear that the number of k-limited permutations has the same parity as det M. So the question is when M is invertible in M n (IF2). From now on, we always work in lF2. Let us prove that if n == 2, ... ,2k (mod 2k + 1), then M is not invertible. Pick integers 0 :S a < b :S k such that n + a + b == 0 (mod 2k + 1) (it is clear that they exist under the assumption that n == 2, ... , 2k (mod 2k + 1)) and set j = (n + a + b)/(2k + 1). If ri is the i-th row of M, then the vector all .of whose components are 1 can be written as rk+l-a+(2k+l)i and as Li~t rk+l-b+(2k+1)i' Hence there is a nontrivial combination of the rows of M which yields the zero vector. Thus M is not invertible in this case.

Lt::t

Chapter 12. Higher Algebra in Combinatorics

546

Assume now that n == 0,1 (mod 2k+ 1) and that aI, a2, ... , an are scalars such that alrl +a2r2+" ·+anrn is the zero vector. Put ai = if i tJ. {I, ... , n}. Then am-k + am-k+l + ... + am+k = for all m. By comparing the relations for m and m + 1, we obtain am-k = am+k+l for 1 :::; m < n, so ai repeat with period 2k + 1. Taking m = 1, ... ,k further yields the equalities ak+2 = ... = a2k+1 = 0. Taking m = n - k, ... , n - 1 gives another chain of equalities a n-2k = ... = an-I-k = 0. If n == (mod 2k + 1), this can be rewritten as al = ... = ak = 0, whereas for n == 1 (mod 2k + 1), it can be rewritten as a2 = ... = ak+1 = 0. In either case, since we also have al + ... + a2k+1 = 0, we deduce that all of the ai must be zero. The conclusion follows. 0

12.2

Matrices over lF2

The reduction map Z -+ Zj2Z induces a natural map Mn(Z) -+ Mn (Zj2Z), which is easily seen to be a ring homomorphism. This map is very useful when dealing with applications of linear algebra in combinatorics, since for a lot of problems the parity gives already enough information. Also, the incidence matrices of families of sets are binary matrices and so are the adjacency matrices of graphs. To simplify notation, we let IF2 = Zj2Z.

~ 5. 2n + 1 real numbers have the property that no matter how we eliminate one of them, the rest can be divided into two groups of n numbers, the sum of the numbers in the two groups being the same. Prove that the numbers are equal. Proof. Let Xl, X2, ... ,X2n+1 be real numbers as in the problem and let X be the column vector whose coordinates are the Xi'S. We can write the hypothesis in the form AX = for some matrix A = (aij) with aij E {-1,0, I}, aii = and such that the sum of the numbers in each row of the matrix (aij) is zero. Of course, this linear system of equations has the trivial solution Xl = X2 = . .. = X2n+1 and the problem asks to prove that this is the only solution. Now, the dimension of the vector space of solutions is dim KerA, which is also 2n + 1 - rkA. Thus, it is enough to prove that A has rank 2n. Of course, the rank is at most 2n, since the sum of the elements in each line is

12. 2.

Matrices over IF2

547

0. But if we see A as a matrix over IF2, A becomes the matrix with on the diagonal and 1 elsewhere. Since 2n + 1 is odd, it is easy to check that this matrix has rank 2n (the associated system of equations over IF2 is simply X2 + ... + X2n+1 = 0, ... ,Xl + ... + X2n = 0, which clearly has the only solution Xl = X2 = ... = X2n+1)' But then A must have rank at least 2n, since the rank can only decrease after reduction mod 2 (this is simply saying that a nonzero element of IF2 lifts to an odd, thus nonzero, integer). The result follows. 0

Proof. The classical proof is done in three steps. Let Xl,"" X2n+1 be the numbers in the problem. In the first step we assume that all Xi are integers. The hypothesis clearly implies that all Xi have the same parity as Xl + x2 + ... + X2n+l. Thus all Xi have the same parity. Writing them either as 2Yi or as 2Yi + 1, it is clear that Yl, ... ,Y2n+l satisfy the same property as the Xi'S. Since the sum of the absolute values of the Yi'S is smaller than that of the Xi'S, after finitely many steps deduce that all Xi are equal (because if the process continues forever then all the Xi must have been 0). The second step treats the case when the Xi'S are rational numbers. Multiplying them by a suitable positive integer N to make them integers, we reduce trivially this case to the one considered above. Finally, we consider the general case when the Xi'S are real numbers. Let el, e2," ., ek be a basis of the (Q-vector space generated by the Xi'S. Write each Xi = ailel + ai2e2 + ... + aikek. Since the e/s are linearly independent, it follows that for each fixed j, the numbers alj, a2j, ... ,a2n+1,j satisfy the same properties as the Xi'S. Since these numbers are rational, it follows that they are equal by the second step. But since this holds for every j, we have o Xl = X2 = ... = X2n+ 1· Binary matrices are very useful when dealing with iterations of some process on a combinatorial structure. The fact that 2 kills M n (IF2) implies that for any commuting matrices A, B E Mn(IF2) and for any k > 1 we have (A + B)2k = A2k + B2k. This is very useful in computations.

"f 6.

The edges of a regular 2n-gon are colored red and blue. A step consists in recoloring e;1ch edge whose neighbors have the same color in red and recoloring each edge whose neighbors have different colors in blue. Prove

548

Chapter 12. Higher Algebra in Combinatorics that after 2n - l steps all of the edges will be red and that this need not hold after fewer steps. Iranian Olympiad 1998

Proof. The easy part is the second question: if there is exactly one blue edge, then clearly after k < 2n - l steps the k-th edge after this blue edge is blue, so it is impossible to make all edges red after less than 2n - l steps. Let us now prove that after 2n - l steps all edges are red. Let Xj be the vector giving the state of the edges after the j-th step, so Xj is the column vector with 2n coordinates, the k-th coordinate being 1 if the k-th edge is blue after j steps and 0 otherwise. By definition we have Xj+l = AXj , where everything is taken modulo 2, A = B + B- 1 and Bij = 1 if j == i + 1 (mod 2n) and 0 otherwise. Since B and B- 1 commute, the binomial formula and the fact n 1 1 1 1 that k- ) is even for alII S k S 2n-l_l show that A 2n - == B 2n - +B- 2n (mod 2). However, it is very easy to compute the successive powers of B: a trivial induction shows that Bk is simply the matrix whose entry is 1 if 2n = I and so by the j - i == k (mod 2n) and 0 otherwise. In particular, B 1 2n previous formula A - == 0 (mod 2). But since Xj = Aj X o, it is clear that n l after 2 - steps all edges are red. D

The following problem is a classical old result, which has a very beautiful linear-algebraic proof. lot 7. The map s : ]Rr

---t ]Rr

is defined by

Prove the equivalence of the following statements: i) for all nonnegative integers aI, a2, . .. , ar, there exists n such that the n-th iterate of s evaluated at (aI, a2, ... , ar ) is (0,0, ... ,0); ii) r is a power of 2. Ducci's problem

12.2. Matrices over lF2

549

Proof. Suppose first that r is a power of 2. Observe that the maximal coordinate of s(al, a2, ... , ar) is clearly smaller than or equal to the maximal coordinate of (aI, a2, ... , ar ). Start with some nonnegative integers (aI, a2, ... , ar ) and define Xo = (aI, a2, .. . , ar ) and X n+l = S(Xn). By the previous remark, all Xn's live in the hypercube [0, max air. We will prove that the Xn's become arbitrarily divisible by 2, thus one of them will have to be equal to O. To prove this, it is enough to prove that one of the Xn's has all coordinates even numbers, since then we can repeat the procedure to make the coordinates of another Xn multiples of 4 and so on. Now, note that if Xn = (al(n), a2(n), ... , ar(n)) then Xn+l == (1 + T)(Xn) (mod 2), where T(Xl,X2, ... ,Xr ) = (X2,X3, ... ,xd. Thus Xn == (1 +T)nxo (mod 2). Now, since G) is even for alII S k S r - 1 (because r is a power of 2), we have (1

+ Ty Xo ==

(~)TkXO == Xo + T rXo

(mod 2).

k=O

But trivially T r = 1, the identity map, so Xo + T r Xo == 2Xo == 0 (mod 2). Thus all coordinates of Xr are even and we are done. Assume now that i) holds. We deduce that for any vector x E lF2 there exists n such that (1 + T)nx = 0 .. Indeed, any such x is the reduction mod 2 of an r-tuple of nonnegative integers (aI, a2, ... , ar ) and we know that we can find n such that sn((al, a2, ... , ar)) = (0, ... ,0). But we also saw that

Sn((al,a2, ... ,ar )) == (I+T)n(Xl,X2, ... ,Xr )

(mod 2).

Now, since there are finitely many x E lF 2, it follows that there exists n 2: 1 such that (1 + T)n;r = 0 for all x E lF2 (simply choose nx for each x and put n = maxn x ). We claim that this forces r to be a power of 2. Notice that the minimal polynomial of T as an endomorphism of lF2 is xr + 1 (T is simply the shift, so we can easily compute P(T) for any polynomial P). Thus, we must have xr + II(X + l)n in lF2[X], In particular, we must have xr + 1 = (X + l)j for some j S n and so is even for alII Sus j -1, forcing j to be a power of 2. But then (1 + X)j = 1 + xj and so r = j is also a power of 2. D

Before discussing the next problems, we need some preliminaries. First, an easy but fundamental remark: suppose that f E Z[X l , X 2, .. . , Xn] satisfies

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Chapter 12. Higher Algebra in Combinatorics

f(xl, X2,·· ., x n ) = 0 for all integers Xi. Then an easy induction on n shows that all coefficients of f are zero. Thus, for all fields K and all Xl, X2, . .. , Xn E K we have f(xl, X2, . .. ,xn ) = 0 E K. Let us apply this observation to establish the following

Theorem 12.2. Let K be a field and let A E Mn(K) be such that aij +aji for all i, j. If n is odd and if aii = 0 for all i, then det(A) = o.

Proof· Consider the polynomial

0 al2 -al2 0 f(al2, aI3,· .. ,an,n-l) = -a13 -a23

al3 a23 0

-al n -a2n -a3n

al n a2n a3n E Z[al2' aI3,·· ., an,n-l].

By the previous discussion, it suffices to prove that f vanishes when evaluated at any aij E Z. But if X E Mn (Z) is antisymmetric, then det(X)

= det(X ) = det( -X) = (_l)n det(X) = -

det(X),

the first equality being standard, the second one the hypothesis and the last one a consequence of the fact that n is odd. Thus det(X) = 0 for all antisymmetric matrices X E Mn(Z). But this is precisely what we wanted to prove. 0 We consider next two applications of the previous theorem.

III 8. Suppose that n teams compete in a tournament wherein each team plays against any other team exactly once. In each game, 2 points are given to the winner, 1 point for a draw, and 0 points for the loser. It is known that for any subset S of teams, one can find a team (possibly in S) whose total score in the games with teams in S is odd. Prove that n is even. D. Karpov, Russian Olympiad 1972

551

12.2. Matrices over lF2

Proof. Define the matrix A = (aij) byaij = 1 if there is a draw between i, j and 0 otherwise. Clearly, aii = 0 and aij = aji for all i, j. See A as a matrix with coefficients in lF 2 . We claim that A is invertible. Indeed, otherwise we can find X E lF~ nonzero such that Ax = O. If I = {I :::; i :::; nlxi = I}, then the condition Ax = 0 can be written I:.jEI aij = 0 for all 1 :::; i :::; n. Since each match which is not a draw yields an even number of points, the above condition implies that for all i, the total score in the games between i and the teams in I is even, contradicting the hypothesis. Now, it is enough to use theorem 12.2 to conclude that n is even. 0

The following problem is essentially the same as the previous one, but written in different language.

~ 9. A simple graph has the property: given any nonempty set H of its vertices, there is a vertex x of the graph such that the number of edges connecting x with the points in H is odd. Prove that the graph has an even number of vertices. Proof. If A is the adjacency matrix of the graph, seen as matrix over lF2, then the hypothesis of the problem translates into: for all nonzero x E lF~ we have Ax #- O. Thus A is invertible over lF2. Since A is clearly antisymmetric (because it is symmetric and we are in characteristic two) with zero diagonal, the result follows from theorem 12.2. 0

The following problem is rather easy. 10. Let AI, A 2, ... ,An be subsets of A = {I, 2, ... ,n} such that for any nonempty subset T of A, there is i E A such that IAi n TI is odd. Suppose that BI, B2 are subsets of A such that IAi n BII = IAi n B21 = 1 for all i. Prove that BI = B2. Gabriel Dospinescu, Mathematical Reflections Proof. Let's consider the subsets of A as vectors in ~, where the entry in the ith position is 1 if the subset contains the element, 0 otherwise. Let the vectors corresponding to Ai'S form the rows of the matrix M. The first condition can

552

Chapter 12. Higher Algebra in Combinatorics

553

12.3. Applications of bilinear algebra

be expressed as M v =1= 0 for all v =1= O. Let v be the vector all of whose components are 1. The second condition says that Mb j = v for all j, where bj are the vectors representing the Bj's. So we have M(b l - b2 ) = 0, whence bl = b2 and BI = B2. 0

This implies that VI + V2 + ... + Vn+l = o. If n is odd, then we are quite stuck, but if n is even, then last equality certainly cannot hold, as each component of VI + V2 + ... + Vn+l is an odd number. Fortunately, n = 50 is even in our problem and so we are done. 0

12.3

aI, a2, ... ,a51 and call the attributes PI, P 2, ... ,PlOo . We will count triples

Proof. Let us assume that the book has 51 pairwise dissimilar plants, call them

Applications of bilinear algebra

In problems concerning incidences, it is sometimes useful to consider properties of inner products and bilinear algebra. If K is any field, we define the inner product of two vectors x = (Xl, X2, ... , x n ) E K n and n Y = (YI, Y2, ... , Yn) E K as

(x, y)

= XIYI

+ X2Y2 + ... + XnYn·

If K = lR, we let Ilxll = ~ and call it the norm of x. The Cauchy-Schwarz inequality is equivalent to the triangle inequality Ilx + yll ::; Ilxll + IIYII. Let us start with a rather standard problem, which is nearly optimal. As the remark following the proof of it shows, this problem is actually rather subtle. We will see some variants of it later in the chapter.

11. A handbook classifies plants by 100 attributes (each plant either has a given attribute or does not have it). Two plants are dissimilar if they differ in at least 51 attributes. Show that the handbook cannot give 51 plants all dissimilar from each other. Tournament of the Towns 1993 Proof. More generally, assume that the handbook considers 2n attributes, that two plants are dissimilar if they differ in at least n + 1 of these, and that we have n+ 1 plants all dissimilar from each other. For each i, let Vi be the vector whose j-th coordinate is 1 if plant i has attribute) and -1 otherwise. Let (,) be the standard inner product and observe that the hypothesis implies that (Vi,Vj) < 0 for all i =1= j. It is however clear that (Vi,Vj) is an even integer, thus we actually have (Vi, Vj) ::; -2 for all i =1= j. But then n+l

II ~ vil1

n+l

(ai, aj, Pk) with i < j so that ai and aj differ on attribute Pk . On the one hand, for each (ai, aj) there are at least 51 triples containing this pair, so the total number of triples is at least 51 (5D. On the other hand, if Ak is the set of plants having attribute Pk, then there are IAkl(51 - IAkl) ::; 25·26 pairs (ai, aj) living in triples with attribute Pk . We deduce that

~(2n) + 2 i;(Vi' Vj)

(

::; 2n(n + 1) - 4 n;

= O.

100· 25 . 26 2': 51 . a contradiction.

c;)

-¢=}

2600 2': 2601,

Remark 12.3. A Hadamard matrix is a square matrix with entries ±1 such that the rOws are pairwise orthogonal vectors. It is a famous conjecture that for all n E 4N there exists an n x n Hadamard matrix. Assume that n + 1 is even and suppose that the previous conjecture holds. Thus we can find a Hadamard matrix of order 2n + 2. By multiplying some rows by -1, we may assume that all entries of the first column are 1. Since the columns are orthogonal, each column except the first now has n + 1 positive and n + 1 negative terms. Now consider all the rows in which the first two terms are + 1. There are n + 1 such rows. Take these rows and chop away the first two columns. The result is n + 1 vectors of length 2n such that the inner product of every pair of vectors is - 2. Interpreting the rows as plants and the columns as attributes, we get the required set of n + 1 plants which pairwise differ in strictly more than half of their 2n attributes (in fact, every pair differs in exactly n + 1 attributes). One can actually exhibit Hadamard matrices of order 104, yielding 52 plants which pairwise differ in 52 out of 102 attributes.

The next two problems are variations on the following nice and rather classical result. We will also present purely combinatorial proofs and we will leave the reader to decide which method is more efficient.

Chapter 12. Higher Algebra in Combinatorics

554

Theorem 12.4. Let V1, V2, .. . , Vk be nonzero vectors in

for all i

-# j.

i -# j, the inner product (Vi, Vj) is simply the difference between the number of positions where Vi and Vj coincide and the number of positions where the two vectors differ. Taking into account the hypothesis, we deduce that (Vi, Vj) :S for all i -# j. Of course, Vi -# for all i. The result follows from theorem 12.4. 0

such that

Proof. We will induct on n. For n = 1, it is an obvious application of the pigeonhole principle. Now, suppose that the result holds for n - 1 and consider V1, V2,.·., Vk -# in ~n such that (Vi, Vj) :S for all i -# j. Dividing each Vi by its norm and working with the new set of vectors, we may assume that Vi have norm 1. Choose an isometry A sending V1 to the vector e1, whose first coordinate is 1 and all the others are 0. Working with the vectors AV1, AV2, ... ,AVk instead of VI, V2, ... ,Vk (they satisfy the same hypothesis, as an isometry is invertible and preserves the inner product), we may assume that A is the identity and so V1 = e1. Then the first coordinate of each of the vectors V2, V3, . .. ,Vk is non-positive by hypothesis. Consider the vectors W2, W3, ... , Wk E ~n-1 obtained by deleting the first coordinate of V2, V3, ... ,Vk. They are the projections of V2,V3, ... ,Vk on the orthogonal complement to e1. Since the first coordinate of each of V2, V3, ... ,Vk is non-positive and since (Vi, Vj) :S for 2 :S i -#j :S k, we also have (Wi, Wj) :S for all i -# j. If W2, w3, ... , Wk are nonzero we obtain k :S 2n - 1 < 2n by the inductive hypothesis. So, assume that some wi's are zero. We claim that at most one of them is zero. If we manage to prove this, it will follow that k - 2 :S 2(n - 1) and the inductive step will be proved. So, assume that W2 = W3 = 0, then V1, V2, V3 are multiples of e1 and since they are norm 1, each of them is equal to e1 or -e1. But then we can find 1 :S i < j :S 3 such that Vi = Vj. As (Vi, Vj) :S 0, it follows that Vi = 0, a contradiction. The result follows. 0

t • .c ~

Proof. The crucial ingredient in the proof is the following Lemma 12.5. If in an m x n binary matrix every two rows differ in at least d positions and if 2d > n, then m :S 2J~n· Proof. Call a pair (x, y) of matrix entries good if x and yare in the same column and if (x, y) = (0,1). If di is the number of ones in the i-th colum~, then dearly there are 2:7=1 di(m - di ) good pairs. Since di(m di):S ~ , there are at most n. ~2 good pairs. However, if we fix two rows, we find at least d good pairs, by hypothesis. Thus we find at least d(r;:) good pairs and so d(r;:) :S n· ~2, which immediately yields the lemma.

Iranian Olympiad

Proof. It is more convenient to replace all zeros in the table by -1. Of course, this does not change anything to the hypothesis or conclusion of the problem. See each row as a vector in ~n and let V1, V2, .. . , Vm be these vectors. For any

Passing now to the proof, let A1 be the number of rows whose first element is 1 and let A2 be the number of rows whose first element is 0. If we consider only the rows whose first element is 1 and then delete the first element in each such row, we obtain a set of rows such that any two differ in at least ~ positions. By the previous lemma, we deduce that A1 :S nand A2 :S n. Since m = A1 + A 2, the result follows. 0

I\' 13. Let n be a positive integer. If u = (Ui);:l and V = (Vi);:l are binary sequences of length 2n , let the distance between them be

and 1 such that any two rows differ in at least n/2 positions. Prove that m :S 2n.

S*4..

Then k :S 2n.

t:I 12. An m x n matrix is filled with

555

12.3. Applications of bilinear algebra

d(u,v) =

lUi - Vii·

i=l

Find the greatest number of binary sequences of length 2n whose pairwise distances are at least 2n-1. China TST 2005

556

Chapter 12. Higher Algebra in Combinatorics

Proof. Replacing each coordinate 0 by -1 in each vector, we must find the maximal number of vectors of length 2n with coordinates ±1 and whose pairwise inner products are nonpositive. Theorem 12.4 shows that this maximal number is at most 2n+1. The problem is thus solved if we can exhibit 2n+1 vectors with coordinates ±1 in ]R2n and with nonpositive inner products. We will consider the vectors ±el, ±e2, ... , ±e2n, where el, e2, ... , e2n is an orthogonal basis of ]R2n and all ei have coordinates ±l. Such a basis can be constructed by induction: for n = 1 choose el = (1,1) and e2 = (1, 1). Assuming that we have such a basis for n, consider the vectors fi = (ei' ei) for 1 ~ i ~ 2 n (where (x, y) is the vector obtained by considering the coordinates of x followed by the coordinates of y) and fi = (-ei-2n, ei-2n) for 2n + 1 ~ i ~ 2n+1. We can immediately check that these vectors are linearly independent and by construction they are orthogonal. The inductive step is proved. D Remark 12.6. A more general question is the following: given nand k, what is the maximal number of binary sequences of length n such that any two differ in at least k places? For some nontrivial cases of this problem, we refer the reader to chapter 10 of [7]. Here is another very classical and nontrivial result. There are purely combinatorial proofs, but they are really not very illuminating. On the other hand, the algebraic proof is very neat. 14. Let AI, A 2 , ... ,Am be distinct subsets of a set A with n 2: 2 elements. Suppose that any two of these subsets have exactly one common element. Prove that m ~ n. Fisher's inequality

Proof. We may assume that A = {1, 2, ... , n}. FQllowing the usual technique for this kind of problem, consider the incidence matrix M of the sets, that is the m x n matrix defined by mi,j = 1 if j E Ai and 0 otherwise. By hypothesis, M . Mt is the matrix with i-th row (1,1, ... , IAil, 1, ... ,1). For technical reasons, we have to discuss two cases. The first case is when IAil = 1 for some i. But then by assumption all Aj have to contain Ai and so Aj \ Ai for j i- i are pairwise disjoint nonempty subsets of a set with n - 1

12.3. Applications of bilinear algebra

557

elements, thus certainly m - 1 ~ n - 1 and we are done. The hard case is when all IAil 2: 2 and the key point (to be proved below) is that in this case M . Mt is invertible, thus of rank m. Since the rank of M . Mt is at most that of M, which is at most n, this will be enough to prove the result. In order to prove the key point, we will argue by contradiction: if M . Mt is not invertible, there exists a nonzero vector with real coordinates Xl, X2, ... , Xm such that M . Mtx = O. But then txM . Mtx = 0 and so m

IAil x ; + 2

i=l

XiXj =

l:S,i<j:S,m

This can be also written as

Since IAil 2: 2 for all i, the last relation obviously implies that all Xi are zero, a contradiction. Thus, the claim is proved and the problem is solved. D

Remark 12.7. One can also prove the invertibility of M· Mt in a different way, if IAil 2: 2 for all i. Namely, if M . Mtx = 0, then (IAil- l)Xi + 5 = 0, where 5 = Xl + X2 + ... + X n · But then Xi = IA and so

i1-1 1

5 = -5·

i=l

IAI- l' t

which obviously implies that 5 = 0 and so all

are zero.

Remark 12.8. We leave as an exercise to the reader to adapt the previous proof and deduce the following more general result: let AI, A 2 , ... , Am be different subsets of a set with n elements X. If there is a E {1, 2, ... , n 1} such that IAi n Aj I = a for all i i- j, then m ~ n. It is also very hard to solve the following problem without linear algebra. On the other hand, even with the help of (bi)linear algebra, this problem is quite tricky.

Chapter 12.

558

Higher Algebra in Combinatorics

15. Let AI, A 2, ... , Am and B 1 , B2, ... , Bp be families of distinct subsets of {1, 2, ... , n} such that Ai n Bj is an odd number for all i and j. Then

12.3.

559

Applications of bilinear algebra

for all j. But then we also have

mp:::: 2n-1.

Benny Sudakov ProofÂˇ Let Vi be the incidence vector of Ai and let Wi be the incidence vector of B i , seen as vectors in IF!2. The hypothesis becomes: (Vi,Wj) = 1 for all i,j,

which can be written (taking into account that aij

where (,) is the standard inner product on IF!2. Let V be the vector space spanned by the vectors Wi. Since the vectors Vi + VI are m distinct vectors each orthogonal to every Wj, thus to V, we have As

m:::: 2dimSpan(vi+vIl :::: 2n-dimSpan(wj). + WI) = 0 for all j, k, {Wi + wIll :::: i :::: p}

Moreover, since (VI, Wj) = 1 and (VI, Wk { Wi 11

:::: i

:::: p} and

are disjoint subsets of Span( wd. Thus 2p :::: two inequalities yields the desired result.

2dimSpan(wi).

the sets

Multiplying these 0

The next three problems also require a preliminary discussion. Namely, we will recall the proof of the following classical, but nontrivial result:

Theorem 12.9. Let A E Mn(lF2) be a symmetric matrix and let d be the column vector whose coordinates are the entries on the main diagonal of A. Then there exists x E lF2 such that Ax = d.

The easiest proof uses bilinear algebra: consider the standard inner product on IF!2 and observe that it suffices to prove that any vector which is orthogonal to Im(A) is also orthogonal to d; But if V is such a vector, we have

vi =

+ aji = 0)

Vi, we are done.

16. Light bulbs LI, L 2 , . .. ,Ln are controlled by switches Sl,S2, ... ,Sn' Switch Si changes the onloff status of light Lf and possibly the status of some other lights. Suppose that if Si changes the status of Lj then Sj changes the status of Li. Initially all lights are off. Is it possible to operate the switches in such a way that all the lights are on? Uri Peled, AMM 10197 Proof. This is a very easy application of the previous result: define aij = 1 if Sj changes the status of Li and 0 otherwise. By hypothesis, we have aij = aji and aii = 1. Thus the matrix A = (aij), seen as matrix over IF 2, is symmetric

with diagonal elements 1. By the previous theorem, we can find a nonzero vector x E IF!2 such that Ax is the diagonal vector of A. Thus, 'L,"]=l aijXj = 1 for all i. If J is the set of those j with Xj = 1, the previous equality tells us that if we operate the switches Sj with j E J, then the state of each Li will change an odd number of times and so all lights Li will be on. 0 The following problem is a bit trickier. We also give a purely combinatorial proof, to see the difference ...

for all Wj E lF2 and by exchanging the two sums we deduce that n

LaijVi i=l

17. Let G be a graph. Prove that the set of its vertices can be partitioned in two groups (possibly empty) such that each group induces a subgraph in which all vertices have even degree. Gallai's Cycle-Cocycle partition theorem

560

Chapter 12. Higher Algebra in Combinatorics

Proof. Consider the adjacency matrix A of the graph and perform the following operations on it: for each vertex i of odd degree, add a 1 in position (i, i) of A. For all the other vertices i, leave position (i, i) as in the original matrix (Le. equal to 0). We thus get a new symmetric binary matrix B = (b ij ). By the previous theorem, there exists x E lF2' such that n

12.4. Matrix equations

G has n vertices 1,2, ... , n and consider the matrix B as in the solution of the previous problem. We claim that there is a natural bijection between admissible partitions and solutions of the system Bx = d, where d is the main diagonal of B. Indeed, let us identify admissible partitions Co, C 1 with vectors x E lF2', where Xi = 0 if i E Co and Xi = 1 if i E Cl. The admissibility condition can be written: for all i we have

bijxj = bii

i=1

== 0 (mod 2),

where C( i) is the class of the partition containing i. Just as in the proof of the previous problem, this condition can also be written n

L jEV1U{i}

bijxj = bii .

j=l

bij = Lbij j=1

and the last quantity is 0 by construction. Thus i has even degree in V2 and we are done. D

Thus, the number of admissible partitions is the number of vectors x E lF2' such that Bx = d. We know that there is at least one such vector Xo (by the discussion preceding problem 10 or by the previous problem). But then {xlAx

We continue with a nice variation on the previous result.

'.f 18.

bij=I,j#-i,jEC(i)

for all i. Let VI = {I :S i :S nlxi = I} and let V2 be its complement. We claim that this partition of the vertices satisfies the desired conditions. Indeed, for i E VI we have I:jEV1-{i} bij = 0, so i has an even degree in the subgraph induced by VI. For i E V2, we have L bij = Lbij jE V2-{i} j=l

561

and {xlAx At a certain mathematical conference, every pair of mathematicians are either friends or strangers. At mealtime, every participant eats in one of two large dining rooms. Each mathematician insists upon eating in a room which contains an even number of his or her friends. Prove that the number of ways that the mathematicians may be split between the two rooms is a power of two . USAMO 2008

Proof. Note that this problem implies the existence of at least one admissible partition (i.e. one that satisfies the conditions of the problem), which is precisely the content of the previous problem. Fortunately, the hard work was actually done. Let G be the graph whose vertices are the mathematicians, two vertices being connected if the mathematicians are friends. Suppose that

2dimKer(A).

12.4

= d} = Xo + {xlAx = O}

= O} =

Ker(A) is a linear subspace of lF2', so it has cardinality Since this is a power of 2, the conclusion follows. D

Matrix equations

To fully appreciate the power of linear algebra in combinatorics, the reader should try to find a purely combinatorial solution to the following problem.

'f 19. Let Gl, G2, .. . , Gk be complete bipartite subgraphs of the complete graph K 2n with 2n vertices. Assume that every edge of K2n is contained in an odd number of subgraphs G l , G2, ... , Gk. Prove that k 2: n. Proof. Let 1,2, ... , 2n be the vertices of K 2n and let L i , Ri be disjoint subsets of {I, 2, ... , 2n} such that the set of vertices of Gi is Li U ~ and two vertices

Chapter 12. Higher Algebra in Combinatorics

562

are connected if and only if one is in Li and the other is in ~. If li E JF§n is the vector whose nonzero coordinates are precisely those on positions belonging to Li and if ri is similarly defined, then the entry in position (u, v) of the matrix li . rf is precisely 1uEL i,vE R i. We deduce that the entry in position (u, v) of k

X = L(li . rf

+ ri . mE

M2n( JF 2)

i=l

is if u = v and it is the number of subgraphs containing the edge uv, taken mod 2, otherwise. Therefore, the hypothesis can be reformulated as an equality of matrices X = J - 12n, where J is the matrix all of whose entries are 1. Now, matrices of the form VI . v~ are precisely the matrices of rank at most 1. So, we expressed X as a sum of 2k matrices of rank at most 1. Since the rank function is sub-additive (thought of in terms of endomorphisms, this is obvious: the image of u + v is contained in the sum of the images of u and v for any endomorphisms u, v of a vector space), in order to solve the problem it is enough to prove that J - hn has rank 2n, i.e. that it is invertible. This is however trivial: if Jx = x for some vector x, then Xi = Xl + X2 + ... + X2n for all i, so first all Xi'S are equal and then Xi = 2nxi = 0, since we are working in characteristic 2. The result follows. 0 We also discussed the following beautiful problem in example 7, chapter 23 of [3]. We give here a very natural solution using the following classical result:

Theorem 12.10. Let X E Mn(lR) and Y E Mm(lR). There exists a nonzero matrix A E Mnxm(lR) such that XA = AY if and only if X, Y have a common eigenvalue.

Proof. Assume first that there exists a nonzero matrix A E Mnxm(lR) such that XA = AY, but X, Y have no common eigenvalue. Thus the characteristic polynomials P, Q of X, Yare relatively prime and so there exist R, S E qT] such that PR + QS = 1. Now, since XA = AY, we have by the CayleyHamilton theorem = P(X)A = AP(Y). But we also have AQ(Y) = 0, as Q(Y) = (again by Cayley-Hamilton). Thus

= A(PR)(Y) + A(QS)(Y) = AP(Y)R(Y) + AQ(Y)S(Y) = 0,

12.4. Matrix equations

563

a contradiction. This settles the first part. Next, if X, Y have a common eigenvalue A, then X and y t also have the common eigenvalue A and so one can find nonzero vectors v, w such that X v = AV, ytw = AW. It is then easy to check that A = vw t is a nonzero complex matrix such that X A = AY. Then one of Re(A) and Im(A) is a nonzero real matrix satisfying again XA = AY and we are done. 0

20. A grid divides an n x m sheet of paper into unit squares. The two sides of length n are taped together to form a cylinder. Prove that it is possible to write a real number in each square, not all zero, so that each number is the sum of the numbers in the neighboring squares, if and only if there exist integers k, 1 such that n + 1 does not divide k and

2l7r k7r 1 cos- +cos-- =-. m n+1 2 Ciprian Manolescu, Romanian TST 1998

Proof. See the grid as an n x m matrix A and define aO,i = an+l,i = i. The problem becomes: there exists a nonzero matrix A such that

for all

(for all 1 ~ i ~ n, j being taken mod m) if and only if there exist integers k, 1 as in the statement. The previous equality can be written very simply by introducing the matrices B E Mn(Z), C E Mm(Z) defined by bij = 1 if Ii - jl = 1 and otherwise, Cij = 1 if Ii - jl = 1 or (i,j) E {(I,m), (m, I)}. Indeed, the previous relation is equivalent to AC+BA = A, or BA = A(I -C). By theorem 12.10, the existence of the matrix A is equivalent to the existence of a common eigenvalue of B and I-C. This is equivalent to the existence of eigenvalues AI, A2 of B, respectively C such that Al + A2 = 1. Let us compute the eigenvalues of C. If Cx = AX for a nonzero vector x, we can write

... , and so if we extend Xi by m-periodicity, we can simply write this as

564

Chapter 12. Higher Algebra in Combinatorics

If rl, r2 are the roots of the equation t 2 - )..t + 1 = 0, then Xi = Glri + G2r~ (Gld +G2id if rl = r2). Imposing the condition of m-periodicity, we see that we should have r 1 = r2 = 1 and (because rl and r2 are reciprocal roots of unity) 2l7r ).. = rl +r2 = rl +rl = 2Re(rl) = 2cosm

for some ~ 1 < m. This suggests the form of the eigenvalues of G. To make this argument completely rigorous, it suffices to reverse the computations: . 1 h 1 211r 2i7r1 . SImp y c oose any ~ < m, set).. = 2 cos m' rl = e--;:;;:- and choose Xi = r i . Then X is an eigenvector with eigenvalue)... Since we have found the correct number of eigenvectors, we are done: the eigenvalues of G are 2 cos (the argument was a bit long, but we wanted to present it in this form as it is rather useful). The same kind of reasoning shows that the eigenvalues of B are 2 cos nk';l and the result follows from these computations and the previous discussion. 0

We give two proofs for the following beautiful and classical result. For the first proof we follow [29], while the second proof appears in [42].

21. In a society, acquaintance is mutual and any two persons have exactly one common friend. Then there is a person who knows all the others. The friendship theorem, Erdos, Renyi, Sos

Proof. First, we will prove that either there is a universal friend (i.e. a person knowing all the others) or any two persons have the same number of friends. If A and B are not friends, let aI, a2,"" ak be A's friends and let bl, b2 , .. . ,bl be B's friends. By assumption, ai and B have a unique common friend bf(i) and bj , A have a unique common friend ag(j). Since f and g are clearly inverses to each other we have k = l and we are done. Consider the obvious graph of enemy relationships in this society and suppose that it is disconnected. Then the society can be partitioned into two sets X and Y such that everyone in set X is friends with everyone in set Y. If both X and Y have two people or more, then two people in X have at least two common friends in Y, contradiction. On the other hand, if either set contains only one person then that person is a universal friend. Thus, if the enemy graph is disconnected, there must be

12.4. Matrix equations

565

a universal friend. On the other hand, if the enemy graph is connected, then every person has the same number of friends, because we saw that two enemies have the same number of friends. This proves the claim and also shows that we may assume that the enemy graph is connected in what follows. Let d be the number of friends of each person and let n be the number of people. Then, the number of pairs of people with a specific person as their common friend is (g), so by counting the number of pairs of people in two ways we get = n(g) and so n = d2 - d + 1. Finally, let A be the adjacency matrix of the graph. By our previous work, we can write A2 = (d - 1)1 + J, where J is the matrix having 1 at each entry. Trivially, (d - 1)1 + J has eigenvalues d 2 and d - 1 (with multiplicities 1 and n - 1 respectively). Thus the eigenvalues of A are d and ±v;r=l. Since the trace of A is zero, we deduce that d + (a - b)v;r=l = 0, where a, b are the multiplicities of the eigenvalues ±v;r=l. Squaring the last relation, we deduce that d - 1 divides d 2, so that d = 1 or d = 2. But both such cases trivially satisfy the theorem, which gives the contradiction we needed. 0

Proof. As in the previous solution, we prove that either there is a universal friend or the associated graph G is d-regular and connected, for some d such that n = d2 - d + 1. Now comes the beautiful idea: for a positive integer p, let c(p) be the number of cycles of people of length p such that each person in the cycle is friends with the next person (people can appear multiple times in such a cycle). Any such cycle can be constructed in the following way: pick a starting person, pick one of his friends, pick a friend of that friend, etc., until you are at the second to last person in the cycle. If the second to last person is different from the first person, then the last person must be their common friend, but if the second to last person is the same as the first then the last person can be any of the first person's d friends. Thus c(p)

ndP- 2 + (d - l)c(p - 2).

If d =1= and d =1= 2, pick a prime divisor p of d - 1. Then, since n = d 2 - d + 1, we have n == 1 (mod p) and so c(p) == 1 (mod p). On the other hand, c(p) is a multiple of p: there are no cycles that are shifts of themselves (since no person is a friend of himself or herself... ), so the number of cycles of length

566

Chapter 12. Higher Algebra in Combinatorics

p must be a multiple of p. The last two results are not possible at the same time, so that we must have d = 0 or d = 2. But in this case it is clear that

there is a universal friend, which finishes the proof.

Remark 12.11. For infinitely many persons, this result does not hold anymore. Remark 12.12. The theorem actually classifies all graphs as in the statement: all of them consist of a collection of triangles sharing a vertex. Remark 12.13. In [42] one can also find a combination of the first two solutions, which is very elegant. Namely, first we prove that the graph is d-regular for some d such that n = d 2 - d + 1 and that the adjacency matrix A of the graph satisfies A2 = (d - 1)1 + J, where J is the matrix all of whose entries are 1. Suppose that d 2: 3 and pick a prime p dividing d - 1. Working in Mn(lFp), we obtain A2 = J and since AJ = dJ, we have AJ = J. Then AP = J and so tr(AP) == n (mod p). On the other hand, an extension of Fermat's little "theorem yields tr(AP) == (trA)P == trA (mod p) (working with the eigenvalues of A this reduces easily to theorem 9.15). But clearly trA = o. We deduce that p divides n, which is clearly impossible. Thus d :::; 2. We end this section with another classical result in algebraic combinatorics. 22. In a graph G with n 2 + 1 vertices suppose that every vertex has degree n and every cycle has length at least 5. Then n E {I, 2, 3", 7, 57}. Hoffman-Singleton's theorem

Proof. The combinatorial part of the theorem is contained in

Lemma 12.14. If x, yare vertices of G, the number of common neighbors of x, y is 0 if x, yare connected and 1 otherwise. Proof. We will count triples (x, (y, z)) in which x, y, z are vertices and x is connected to both y and z. Counting them according to x, we obtain (n 2 + 1)(~) triples (since there are n 2 + 1 vertices, each of degree n). On the other hand, two connected vertices cannot belong to a triple (otherwise we would obtain a 3-cycle in G) and two non-connected vertices y, z can belong to at most one triple (otherwise G would have a 4-cycle). So the number

567

12.4. Matrix equations

(n2il)

of triples is at most

minus the number of edges of the graph, so at

most (n2il) - (n2~l)n = (n 2 + I)G). Since we have already established that there are precisely (n 2 + 1) G) triples, it follows that every two non-connected vertices belong to exactly one triple and the lemma is proved. 0 Let A be the incidence matrix of G and let J be the matrix all of whose entries are 1. The previous lemma yields

= nI + J

- I - A

(n - 1)1 - A

+ J.

Next, A is a symmetric matrix with real entries and zero diagonal, so it has zero trace and it is diagonalizable in an orthonormal basis (by the spectral theorem). In particular, all eigenvalues of A are real and the eigenvectors corresponding to different eigenvalues are orthogonal. Since the sum of entries in each row is n (because G is n-regular), we have AVl = nVl, where Vl is the vector all of whose coordinates are 1. So, if Av = ),v for some v =1= 0 and ), =1= n, then (Vl'V) = 0 and so Jv = O. But this implies that ),2 = n - 1 - ),. So, if rl, r2 are the roots of the equation x 2 + x = n - 1, any eigenvalue of A different from n is rl or r2. Next, we claim that n has multiplicity 1 as an eigenvalue of A. It suffices to prove that if Av = nv, then v is collinear to Vl. This is easy, since Av = nv implies n 2v = A 2v = (n - l)v - Av + Jv and so Jv = (n 2 + l)v, hence all coordinates of v are equal and v is collinear to Vl. Let a, b be the multiplicities of rl, r2 as eigenvalues of A. So a+b = n 2 by the result we have just established and arl + br2 + n = 0, as A has trace O. Since rl

-1+~ 2

the previous relation becomes ~a-b -a+b -2- + v4n-3-+n=O. 2

Now, we have two cases: if v4n - 3 is irrational, the last relation implies that a = band n = a. Since a + b = n 2, this gives n = 2 and we are done. If

568

Chapter 12. Higher Algebra in Combinatorics

k = vi4n - 3 is rational, write a + b = n 2 previous relation. We obtain 16k(a - b) - (k 2

(k2t3)

and replace this in the

+ 3)2 + 8(k 2 + 3) =

Reducing this mod k, we finally obtain kl15 and since n

k2t3, the result

~~.

Remark 12.15. The proof of the lemma actually yields the following result: if G is a graph in which any cycle has length at least 5 and any vertex has degree at least n, then G has at least n 2 + 1 vertices. So the theorem proved above actually classifies the extremal case. At the time of this writing, it is unknown whether or not there exists a graph as in the theorem for n = 57 (for all the other values of n, such graphs exist).

12.5

The linear independence trick

23. In an m x n table, real numbers are written such that for any two rows and any two columns, the sum of the numbers situated in opposite corners of the rectangle formed by them is equal to the sum of the numbers situated in the other two opposite corners. Some of the numbers are erased, but the remaining ones allow us to find the erased numbers using the above property. Prove that at least n + m - 1 numbers remained on the table. Russian Olympiad 1971 ProofÂˇ Consider the set X of all m x n matrices with real entries such that for any two rows and columns, the sum of the numbers situated in the opposite corners of the rectangle formed by them is equal to the sum of the numbers situated in the other two opposite corners. This set is obviously a linear subspace of Mmxn(IR). If 5 is the set of pairs (i,j) such that the number situated at the entry (i, j) has not been erased, then the hypothesis can be translated into: the map f : X -+ IRlsl obtained by sending a matrix A E X to the collection (ai,j)(i,j)ES is injective. But then the dimension of X has to

12.5.

569

The linear independence trick

be at most the dimension of IRlsl. Thus, 151 is at least equal to the dimension of X as IR-vector space. Therefore it is enough to exhibit m + n - 1 linearly independent matrices in X. This is easy: simply consider the matrices Aj having the j-th row equal to the vector (1,1, ... ,1) and the other rows zero (and this for all 1 ::; j ::; m), then the matrices Bi having the i-th column consisting of ones and the remaining entries 0 (and this for all 1 ::; i ::; n - 1) It is immediate to check that they are in X and linearly independent. D Once the necessary mathematical translations are done, the proof of the following problem is shorter than its statement ... 24. In a contest consisting of n problems, the jury defines the difficulty of each problem by assigning it a positive integral number of points. 1 Any participant who answers the problem correctly receives that number of points for the problem; any other participant receives 0 points. After the participants submitted their answers in such a contest, the jury realized that given any ordering of the participants, it could have defined the problems' difficulty levels to make that ordering coincide with the participants' ranking according to their total scores. 2 Determine, in terms of n, the maximum number of participants for which such a scenario could occur. Russian Olympiad 2001 Proof. It is clear that we may have n participants: impose that participant j solves problem j and nothing else. We will prove that we cannot have more than n participants. By associating to participant j the vector Xj consisting of the number of points he received for each problem, the hypothesis implies that Xj E IR+. and for every ordering of the x/s, there is a vector y E IR+. which gives that ordering of (Xj, y). On the other hand, if the number of participants m is greater than n, there is a nontrivial linear combination of the Xj'S that vanishes. By separating the coefficients of such a combination according to their sign, we find an equality LI G:iXi = LJ {3jXj for some disjoint subsets IThe same number of points may be assigned to different problems. 2Ties are not permitted.

570

Chapter 12. Higher Algebra in Combinatorics

I, J of indices and positive numbers ai, /3j such that clearly impossible to rank all Xi above all Xj'

I.:i ai 2: I.:j /3j.

But it is 0

There is no known combinatorial proof of the following result. On the other hand, the algebraic proof is rather simple. 25. Let m > n + 1 and let AI, A 2, ... , Am be subsets of {I, 2, ... , n}. Then there are disjoint sets I, J such that Ai = Aj and Ai = Aj.

n n

iEI

jEJ

iEI

jEJ

Lindstrom's theorem

12.5.

571

The linear independence trick

cardinality of a set shattered by F. For instance, let X = IR n and let F be the collection of all affine half-spaces. Then vc(F) = n + 1, though this is not obvious. Note that F shatters the set of vertices of an n + I-simplex, so vc( F) 2: n + 1. The converse inequality follows from the following classical theorem of Radon: if S c IR n is a set with more than n + 1 elements, then one can find disjoint subsets A, B of S such that the convex hulls of A and B have common points. The following beautiful result shows that large collections shatter big sets. 26. Let F be a family of subsets of {I, 2, ... ,n} such that

Proof. Associate to each set Ai a vector Vi E 1R 2n , say

where Xj = 1 if j E Ai and 0 otherwise. We obtain at least n + 2 vectors that lie in the (n + 1)-dimensional vector subspace of 1R2n defined by the equations Xl + YI = X2 + Y2 = ... = Xn + Yn' Thus, these vectors are linearly dependent and we can find aI, a2,'" ,am E IR not all zero and such that I.:~l aiVi = O. Define now I = {ilai > O} and J = {ilai < O}. Clearly I, J are disjoint and we prove that Ai = Aj and Ai = Aj. This is very easy: if j E

UAi but j

U U n n jEJ iEI jEJ t/. U A j , then by definition the jth component of all Vi with iEI

iEI jEJ i E J is 0, whereas there exists an i E I with the jth component of Vi nonzero. But then, looking at the jth component in the equality I.:~l aivi = 0 yields

a contradiction. By symmetry, we deduce that

UAi = UA

j .

Similarly,

iEI

jEJ looking at the n + 1, ... ,2nth coordinate in the equality I.:~l aivi = 0 we can

show that

n n Ai =

iEI

A j , which finishes the proof of the theorem.

jEJ

In order to motivate the following result, we need some preliminaries. Suppose that F is a collection of subsets of some set X. We say that F shatters a subset S of X if {A n SIA E F} contains all subsets of S. The Vapnik-Chervonenkis dimension of F (also denoted vc(F)) is the maximal

Then vc(F) 2: k. Sauer-Shelah's lemma

Proof. Let us assume that vc(F) ::; k - 1. We will prove that

Let Pk-l be the family of all subsets A of {I, 2, ... , n} such that IAI ::; k-1. To any X E F associate the map fx : Pk-l --+ IR defined by fx(A) = lACX (this is equal to 1 if A c X and to 0 otherwise). We will prove that the maps fx are linearly independent when X runs over the elements of F. This will clearly imply the desired bound on the size of F. Assume that I.:XEFaX fx = 0 for some real numbers ax, not all equal to zero. Evaluating at elements A E Pk-l, we deduce that for any such A we have I.:XEF,AcX ax = O. The key step is to choose a minimal subset A such that I.:XEF,AcX ax =f=. O. We will prove that for any B C A we can find X E F such that B = A n X. This will contradict the property of F and the result will follow. It is of course enough to prove that I.:xEF,B=xnA ax =f=. O. But by the inclusion-exclusion principle we have

L XE~xnA=B

L BcCcA

(_I)lc-BI.

(

Lax) . XEF,CCX

572

Chapter 12. Higher Algebra in Combinatorics

However, by minimality of A all sums I:XEF,ccX ax are zero, except for C = A. Thus the right-hand side is nothing more than (_I)IA-BI I:XEF,AcX ax, which is nonzero by the choice of A. This finishes the proof. 0 Proof. We will prove by induction on IFI the following statement: for any n and any family F of subsets of {I, 2, ... , n}, F shatters at least IFI subsets of {I, 2, ... ,n}. This trivially implies the desired result. Now, the result clearly holds for IFI = 1, so assume that it holds for all IFI < N and take a family F with N elements, all of them subsets of {I, 2, ... ,n}. Let Fo consist of the elements of F not containing 1 (we assume, without loss of generality, that 1 occurs in some, but not all elements of F). By induction, Fo shatters at least IFol subsets of {2, 3, ... , n}. Also by induction, Fl = {X\{I} : X E F, 1 EX} shatters at least IFll subsets of {2,3, ... ,n}. But if A ~ {2,3, ... ,n} is shattered by Fl then A is also shattered by {X E Fll E X}, and if A is shattered by both Fo and Fl then both A and A U {I} are shattered by F. We conclude that at least IFol+IFll = IFI subsets of {I, 2, ... , n} are shattered by F, and the result follows. 0 Remark 12.16. Here is a beautiful geometrical interpretation of the previous result: let S be a subset of {-I, l}n with more than I:t=o (~) elements. Then there exists a subset I C {I, 2, ... , n} such that

That is, a large subset of the vertices of the unit hypercube has a coordinate projection covering a unit hypercube in a smaller dimension! We present two proofs of the following classical theorem of Bollobas. One of the proofs needs some preliminaries on resultants. Let K be a field and let f = ao + alX + ... + apXP and 9 = bo + blX + ... + bqXq be two polynomials with coefficients in K. Let K[X]n be the n+ I-dimensional K-vector space of polynomials of degree at most n in K[X]. It is easy to see that gcd(f, g) = 1 if and only if the map cp : K[X]q-l x K[X]p_l -+ K[X]p+q-l, cp(U, V) = U f + V 9 is injective. Since the source and the target are K-vector spaces of the same finite dimension, it follows that gcd(f, g) = 1 if and only if cp is an invertible linear map, i.e. if and only if the matrix of cp in the natural

12.5.

573

The linear independence trick

bases of K[X]q-l x K[X]p-l and K[X]p+q-l is invertible. This matrix has the first q columns the coordinates of f, X f, ... ,Xq-l f with respect to the basis (1, X, . .. ,XP+q-l) and the last p columns the coordinates of g, Xg, .. . , Xp-lg with respect to the same basis. The determinant of this matrix is called the resultant of f and 9 and is denoted Res(f, g). The previous discussion shows that Res(f, g) = 0 if and only if f and 9 have a common nontrivial divisor, which is the same as saying that they have a common root in some algebraic closure of K. 27. Let a, b be positive integers and let AI, A 2 , ... ,Am and Bl, B2, ... , Bm be sets such that a)

IAll = IA21 =

... =

IAml =

a and

IBll =

IB21 = ... =

IBml =

b) Ai n Bj is nonempty if and only if i =I- j. Then m ::; (at b ).

Bollobas'theorem

Proof. We will only assume that Ai n Bj =I- for i < j. We may assume that Ai, Bj are subsets of {I, 2, ... ,n} for some n. Choose VI, V2, ... ,Vn E lR a + l such that any a + 1 vectors among them are linearly independent (this is easily achievable). Let V[ E 1Ra+1 be a vector orthogonal to all Vu with U E A k . Such a vector exists and is unique up to a scalar multiple as by construction the vectors (VU)uEAk span a hyperplane of IRa+!. Consider the polynomials fj(X) =

II (vx,X), XEBj

where (,) is the standard inner product and X = (Xl, X 2 ,·· ., Xa+l)' These are homogeneous polynomials of degree b in a + 1 variables, living thus in a space of dimension (at b ). We claim that fj(vf) = 0 for i < j and fi(vf) =I- 0, which implies that the fi are linearly independent and so m ::; (at b). To prove the claim, let i < j. Since AinBj =I- 0, there is x E BjnAi and so (vx,vf) = o. Thus fj(vf) = 0 for i < j. Also, since Ai n Bi = 0, we have (vx, vf) =I- 0 for 0 all x E B i , thus fi(vf) =I- O.

574

Chapter 12. Higher Algebra in Combinatorics

Proof. Of course, we may assume that all Ai, B j are subsets of R Consider the polynomials fi(X)

II (X -

a),

gj(X)

II (X -

12.5.

that VI, V2, ... , Vm, is a basis of V'. As can write

= 0, we have Xi E V for all i, so we

Xi = LX;l)Vl

b).

1=1

aEAi

By assumption, fi and gj have a common root if and only if i =f. j. Thus, if R(j, g) is the resultant of two polynomials f, g, we have R(fi, gj) = 0 if and only if i =f. j. Now, write

575

The linear independence trick

with X1 1) E Q. Note that there exists r such that x~m) =f. 0, as otherwise V would be included in the span of the vectors VI, V2,' .. , Vm-l, contradicting that dim V = m. . Let now Xi(t) = Xi + tx1m). The crucial fact is the following Lemma 12.17. There exists t E IR such that

and observe that the expression

1) If Xi(t) = Xj(t), then i = j.

is a homogeneous polynomial of degree a in the variables go, ... ,gb' Moreover, the previous results show that Ai(gjo, ... ,9jb) = 0 if and only if i =f. j. Thus, the polynomials Ai are linearly independent and homogeneous of degree a in b + 1 variables. Since the vector space of such polynomials has dimension (a~b), the result follows. 0

Proof. For all i =f. j, the equation Xi(t) = Xj(t) has at most one solution (recall that Xi =f. Xj), so 1) fails for at most finitely many t. On the other hand, Xl(t) = 0, so if r is chosen such that x~m) =f. 0, then the left-hand side of 2) is at least Xr + tx~m) and this becomes arbitrarily large for t Âť 0 or t < < O. The conclusion follows. 0

The following problem is very challenging. We present the author's proof. 28. Let Xl, X2, ... ,Xn be distinct real numbers and suppose that the vector space spanned by Xi - Xj over the rationals has dimension m. Then the vector space spanned only by those Xi - Xj for which Xi - Xj =f. Xk - Xl whenever (i,j) =f. (k,l) also has dimension m. Straus's theorem

Proof. By working with the numbers Xi - Xl, we may assume that Xl = O. Let V be the span of all differences Xi - Xj, so dim V = m. Say a difference Xi - Xj is unique if Xi - Xj = Xk - Xl implies that i = k and j = l. Let V' be the span of the unique differences, let m' be its dimension and suppose that m' < m. As V' is a subspace of V, there is a basis VI, V2,"" Vm of V such

Coming back to the proof, fix such a t and choose j, k such that Xj(t) = maxxi(t) and Xk(t) = minxi(t). We claim that the difference Xj - Xk is unique and X)m) =f. xim). This will yield a contradiction, as Xj - Xk E V' in this case, so its coordinate with respect to Vm has to be zero. To prove the claim, assume that Xj - Xk = Xu - Xv' Looking at the mth coordinate, we obtain X)m) - xim) = xLm) - x~m), so that Xj(t) - Xk(t) = xu(t) - xv(t). But then clearly Xj(t) = xu(t) and Xk(t) = xv(t), so that by the choice of t we m must have j = u and k = v. Finally, if X)m) = xi ), then

a contradiction.

Chapter 12. Higher Algebra in Combinatorics

576

12.6

Applications to geometry

In this section we discuss some rather hard problems with geometric flavor. Before passing to the first problem, we need to recall a few basic facts from convex geometry. Suppose that K is a closed convex subset of IRn. The dual of K is K* = {x E IRnl(x,v) 2: O,\iv E K}, where (.) is the standard inner product on IRn. It is easy to see that K* is again a closed convex set. Standard but nontrivial separation theorems in convex geometry imply the fundamental equality K** = K for any closed and convex subset K of IRn. Classical examples of closed convex sets are the convex hull of finitely many points and the cone generated by finitely many vectors. 3 Applying the equality K** = K to the cone generated by VI, V2, ... ,Vk, we obtain the following beautiful result, which is one of the many versions of Farkas'lemma: Theorem 12.18. (Farkas' lemma) Let v, VI, V2, .. " Vk E IR n and suppose that (w, v) 2: 0 whenever min(w,vi) > O. Then V is in the cone generated by VI, V2,"" Vk· The following hard problem is a very nice way to re-state Farkas' lemma. 29. A figure composed of! by 1 squares has the property that ifthe squares of a fixed m by n rectangle are filled with numbers the sum of all of which is positive, the figure can be placed on the rectangle 4 so that the numbers it covers also have positive sum. 5 Prove that a number of such figures can be placed on the rectangle such that each square is covered by the same number of figures. Russia 1998 E ]Rn, the cone generated by these vectors ~s the set of linear comb~nat~ons where ai are nonnegative real numbers. It is clear that this cone is a convex set, but it is not really obvious that it is a closed subset of]Rn. 4possibly after being rotated by a multiple of ~. 5However, the figure may not have any of its squares outside the rectangle.

12.6.

Applications to geometry

Proof. The first step (and the trickiest ... ) is to restate the problem in a more algebraic way. Suppose that we have N ways to put the figure in the rectangle, possibly after being rotated and call these ways 1,2, ... , N. To each such way i associate its incidence vector Vi E IRmn, where the jth coordinate of Vi is 1 if the square j is covered and 0 otherwise (we number the squares of the table in some way from 1 to mn). Now, the hypothesis becomes: for any V E IRmn such that (v,1) > 0, we also have max(v, Vi) > O. Here 1 E IRmn is the vector all of whose coordinates are 1. The result then follows from the following

Lemma 12.19. Let v, VI, V2, ... , VN E Qn be vectors, with v nonzero. Suppose that for all w E IR n , if (w, v) > 0, then max (w, Vi) > O. Then there are rational nonnegative numbers aI, a2, ... ,aN such that v = al VI + a2V2 + ... + aNVN. If we accept this for a moment, we deduce the existence of nonnegative integers a I, a2, ... , aN and of a positive integer a such that a· 1 = al VI + a2v2 + ... + aNVN. So, if we put ai copies of the ith placement of the figure, each square is covered a times and the problem is solved. Now, let us turn to the proof of the lemma. By Farkas' lemma, we can find nonnegative real numbers Xl, X2, ... , XN such that V = XlVI +X2V2+" ·+XNVN. Forgetting about those indices i such that Xi = 0, we find a linear system in the Xi'S (obtained by writing the previous equality coordinate by coordinate), whose associated matrix has rational entries and which has positive solutions. We claim that such a system also has positive rational solutions. This follows from the following standard:

Lemma 12.20. If a linear system of equations with rational coefficients has real solutions, then the rational solutions are dense in the set of real solutions. Proof. As the associated matrix has rational entries, if we put it in reduced row-echelon form (i.e. perform Gaussian elimination), we immediately deduce that the space of solutions has a basis with rational coordinates. The result follows from the density of Q in R 0

3If VI, V2, ... ,Vk

al VI

+ a2V2 + ... + akVk,

577

We're done.

To motivate the following problems, we need to recall a famous result of Dehn, also known as Hilbert's third problem: given two polyhedra in IR3 with

578

Chapter 12. Higher Algebra in Combinatorics

equal volume, can we always cut the first one into finitely many polyhedrons which can be reassembled to yield the second polyhedron? Dehn showed that the answer is negative in a rather complicated way, but nowadays there is a very beautiful and short proof due to Hadwiger, which we will sketch now. We claim that the regular simplex and the cube yield a negative answer to Hilbert's problem. Let a = arccos the measure of the dihedral angles of the regular simplex. It is easy to see that ~ is irrational (using the fact that if r and cosr-rr E Q, then cosr-rr E {±!, ±1, O}, which was explained in chapter 9, section 9.2). Let 0:1, ••. ,O:k be the dihedral angles appearing at the edges of the pieces of the dissection and let V be the Q-vector space generated by them. As ~ is irrational, there is a linear form f : V -+ Q such that f (1r) = 0 and f(a) = 1. If P is a polyhedron appearing in the dissection, define its Dehn invariant by H(P) = L:e lelf(o:), the summation being taken over the edges of P, lei being the length of e and 0: the dihedral angle at e. Simple geometry shows that this invariant is additive. Since the invariant of the cube is 0 and that of the regular simplex is not, it follows that we cannot cut the regular simplex into pieces that can be reassembled to form the unit cube. The following problem uses a similar argument, which is extremely useful in tiling problems. We need one more preliminary result: any vector space has a basis and any linearly independent set of vectors extends to a basis.

30. An a x b rectangle is divided into squares with sides parallel to that of . . a b the rectangle and wIth SIde lengths Xl, X2, ... , Xn . Prove that - and Xi Xi are rational numbers. Dehn's theorem Proof. The first ingredient is the following

Lemma 12.21. There exists an additive map f : lR -+ lR such that f(x) if and only if X is a rational multiple of a.

Proof. Consider a basis B of lR as a Q-vector space and such that a E B. Choose a bijection g : B - {a} -+ B (g exists, because B is an infinite set) and define f(a) = 0 and f(x) = g(x) for all X E B - {a}. To define f for any

12.6. Applications to geometry

579

real number r, express r as a finite combination r = Xl bl + X2b2 of elements bi E B with rational coefficients Xi and define

We obtain in this wayan additive map f and we have f(r) is a rational multiple of a. Indeed, if f (r) = 0 and r = Xl bl is as above, then

+ ... + xnbn

= 0 if and only if r

+ X2b2 + ... + xnbn

Now the elements f(b i ) are either equal to 0 (if bi = x) or equal to g(bd. Moreover, the g(bi ) are distinct (because g is injective and the bi'S are distinct) and are linearly independent (since they are elements of a basis B). The only possibility to have the previous relation is Xi = 0 whenever bi =F x, which means precisely that r is a rational multiple of a. 0 Define the invariant of a rectangle R with side-lengths x, y by H(R) = f(x)f(y).

The following lemma establishes the additivity of this invariant.

Lemma 12.22. If a rectangle R is partitioned into finitely many rectangles Rl, R2, ... , Rk, then H(R) = H(Rl) + H(R 2) + ... + H(Rk). Proof This is essentially obvious by additivity of f: any finite partition with rectangles can be refined to a partition which forms a grid (just extend all sides of all rectangles in the partition until they meet the sides of R), so it is enough to check the claim for a grid. This reduces then to proving that if R is partitioned into two rectangles R l , R2 by a line parallel to one of its sides, then H(R) = H(R l ) + H(R2). But this is clear, by definition of Hand because f is additive. 0

Coming back to the proof and using the lemma, we obtain n

0= f(a)f(b) = Lf(xd 2 , i=l

580

which yields f(xi)

Chapter 12. Higher Algebra in Combinatorics

= 0 for all i and so

:i are rational numbers. Since we could

have worked with b instead of a from the very beginning, it follows that b Xi are also rational numbers and the theorem is proved. o Remark 12.23. In [51], the following similar result is proved using Dehn's original method: a rectangle R has at least one rational sidelength and is tiled by smaller rectangles, each having a rational perimeter. Then all sidelengths of R and of the tiling rectangles are rational.

The following stronger result will be used in the next problems. The proof is very similar to that of problem 30. Theorem 12.24. (Dehn) Let the rectangle Ro be tiled with rectangles RI, R2,"" Rn. Let ri be the side-ratio of R i . Then ro E Q(rI, r2,.'Âˇ' rn). Proof. Suppose that ro tJ- L = Q(rI, r2, ... ,rn ), so we may pick a basis B of lR as L-vector space, such that the base and height bo, ho of Ro are in B. Define the invariant of a rectangle bx h to be Cl (b )c2(h)-Cl (h)c2(b), where Cl (x), C2(X) are the coordinates of x with respect to bo, ho E B. As Ci are L-linear maps, it is easy td check that Ri has invariant O. An argument similar to the one used in lemma 12.22 shows that the invariant is additive, so the invariant of Ro is the sum of the invariants of the Ri's, that is O. This contradicts the fact that Ro has invariant 1. 0

We end this chapter with a truly amazing result, whose proof is taken from the beautiful paper [34]. The proof requires the following rather technical theorem, which we will take for granted, since the proof would take us quite far afield. Theorem 12.25. (Wall) If r is an algebraic number all of whose conjugates have positive real part, then there exist n ~ 1 and positive rational numbers Cl,C2,.Âˇ .,en such that

581

12.6. Applications to geometry

31. Let r be a positive real number. Prove the equivalence of the following statements: 1) the unit square can be tiled with finitely many rectangles similar to the 1 x r rectangle and having sides parallel to the sides of the square.

2) r is algebraic and all of its conjugates have positive real part. Laczkovich-Szekeres theorem Proof. Prepare for a long but exciting battle. Recall that the side-ratio of a rectangle is the quotient between its horizontal side and its vertical side. Step O. Let us suppose that r is algebraic and all of its conjugates have positive real part. Choose CI, C2, .. . as in Wall's theorem and cut off a rectangle of side-ratio Clr from a unit square, by a vertical cut. From the remaining rectangle, cut off a rectangle of ratio _1_ by a horizontal cut. The C2 r

remaining rectangle has side-ratio C3r + C4r~ ... ' Repeating the process, we tile the unit square by rectangles with side-ratios Clr, _1_,C3r, .... We conclude C2T using the fact that the Ci'S are rational. This proves one implication. The other implication is harder and requires the following steps. Step 1. First, we prove that if there is a partition of a square with rectangles whose side-ratio is r or l/r, then r is algebraic. Let x be any real number which is transcendental over Q(r) and scale the vertical axis by a factor of x. Applying theorem 12.24 to this new situation, we deduce that x E Q(rx,x/r), so we have an equality xP(rx,x/r) = Q(rx,x/r) for P, Q E Q[X, Y]. As x is transcendental over Q(r), the previous equality yields XP(rX,X/r) = Q(rX,X/r) in Q(r)[X]. Looking at the leading coefficient, we easily obtain a nontrivial polynomial with rational coefficients killing r, thus r is algebraic. Step 2. We claim that if r > 0 is algebraic and has a conjugate with real part smaller than or equal to 0, then it has a conjugate with negative real part. Indeed, otherwise r has a purely imaginary conjugate and so there is a root x of the minimal polynomial p of r such that p( x) = p( - x) = O. As p is irreducible, p must divide p(X) + p(-X) and so p(r) + p(-r) = O. Thus -r < 0 is a root of p and we are done.

Chapter 12. Higher Algebra in Combinatorics

582

Step 3. Assume now that r has a conjugate with negative real part. Let p be the minimal polynomial of r and let Q be the companion matrix6 of p. Let B be a basis of ~ as Q(r)-vector space, chosen such that 1 E B. If x E ~, its coordinate with respect to 1 E B is of the form aO+al + .. ·+an_Irn - 1 with ai E Q and we let Vx be the column vector with coordinates ao, al,· .. , an-I. Thus Vrx = Qvx . The key point is the following: Lemma 12.26. There exists a symmetric matrix ME MnClR) such that

2) There exists s

E Q(r) such that v~Mvs

(and corresponds to complex

conjugate eigenvalues o:±i(3 of Q). Let T be the diagonal matrix whose entries are the real parts of the eigenvalues ofQ. We choose M = (P-I)tTP- I . Then for all v we have vtMQv = (P-Iv)tTD(P-Iv) =

L W;O:i 2: 0, i

6Recall that this means that Q is the matrix of the multiplication by r, seen as an endomorphism of the Q-vector space Q(r).

Notes

583

where Wi are the coordinates of p-Iv and O:i are the squares of the real parts of the eigenvalues of Q. SO the first property is satisfied. On the other hand, T has a negative entry (as r has a conjugate of negative real part by step 2 and the conjugates of r are exactly the eigenvalues of Q), which we may assume to be the first one. If el is the column vector whose first coordinate is 1 and the other are 0, we deduce that (Pel)t M Pel < O. Simply choose v E sufficiently close to Pel to ensure that vtMv < 0 and then choose s E Q(r) such that Vs = v. The lemma is proved. D

«:r

12.7

< O.

The proof of this lemma will be given in the next step. Let us see why this finishes the proof of the theorem. Define the invariant of a rectangle b x h to be vbM Vh. In the usual way, we obtain that this invariant is additive. The invariant of a rectangle b x br is vbMVbr = vbM QVb 2: 0 and the invariant of a rectangle br x b is equal to that of a rectangle b x br, as M is symmetric. So rectangles of side-ratio r have nonnegative invariant. On the other hand, 2) ensures the existence of an s x s square with negative invariant. This square cannot be tiled by rectangles with side-ratio r and we are done. Step 4. We prove the lemma. As p has no multiple root (because it is irreducible), Q is diagonalizable over C. As it has real entries, it is immediate to deduce the existence of a block-diagonal matrix D and of P E GLn(~) such that Q = P D p-l. Moreover, each block of D has side 1 (and corresponds to a real eigenvalue of Q) or is of the form (_0:(3

12.7.

Notes

The solutions to some of the problems in this chapter are due to the following people: Alon Amit (problem 11), Tom Belulovich (problem 4), Aart Blokhuis (problem 27), Iurie Boreico (problem 15), Zeb Brady (problem 21), Alexandru Chirvasitu (problem 13), Darij Grinberg (problems 12, 15, 19), Omid Hatami (problem 12), Xiangyi Huang (problems 15, 20), Laszlo Lovasz (problem 27), James Merryfield (problem 2), Jorge Miranda (problem 14), Fedja Nazarov (problems 7, 24, 29), Hunter Spink (problem 8), J. Steinhardt (problem 18), Gjergji Zaimi (problems 5, 7, 9, 10, 11, 20).

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