ON THE TERNARY QUADRATIC EQUATION

Research Paper

Mathematics

E-ISSN No : 2454-9916 | Volume : 3 | Issue : 1 | Jan 2017

ON THE TERNARY QUADRATIC EQUATION x 2  y 2  xy  3z 2

Dr. R. Anbuselvi 1 | K. S. Araththi 2 1

Associate Professor, Department of Mathematics, A.D.M. College for women, Nagapattinam, India.

2

Research Scholar, Department of Mathematics, A.D.M. College for women, Nagapattinam, India.

ABSTRACT The ternary Quadratic Diophantine equation x  y  xy  3z is analyzed for its non – zero distinct integer solutions. Five different patterns of non-zero distinct integer solutions to the equation under consideration are obtained. A few interesting relations between the solutions and special numbers are exhibited 2

2

2

KEYWORDS: Integral solutions, Ternary quadratic.

I. INTRODUCTION The Ternary Quadratic Diophantine Equation offers an unlimited field for research because of their variety [1-2]. For an extensive review of various problems, one may refer [3-10].

This communication concerns with yet another interesting Also a few interesting relations among the solutions have been presented.

u  6ab; v  a 2  3b 2 ; …… (5) z  3b 2  a 2 Using (5) in (2), we obtain the integer solutions to (1) as given below:

x  x(a, b)  a 2  3b 2  6ab

II. NOTATIONS obl n- Oblong number of rank ‘n’

y  y (a, b)  a 2  3b 2  6ab

tm,n- Polygonal number of rank ‘n’ with sides’m’

z  z (a, b)  a 2  3b 2

III. METHOD OF ANALYSIS The ternary quadratic equation to be solved in integers is

PROPERTIES (i) x(1, n)  y (1, n)  0(mod 2) (ii) y (1, n)  z (1, n)  6obln  0(mod 6)

x 2  y 2  xy  3z 2 ……………………

(1)

(iv)

Now, introducing the linear transformations

x  u  v ; y  u  v …………………

(iii)

(2)

x(1, n)  z (1, n)  0(mod 6)

3x(1, n)  6 y (1, n)  9n(n  2)  0(mod 9) 4 y (1, n)  2 z (1, n)  0(mod 6)

(v) It is observed that (3) may also be written in the following three ways

In (1), it leads to

u 3( z  v) A   ………….……… zv u B

a.

x  x(a, b)  3b 2  a 2  6ab

(3)

y  y (a, b)  a 2  3b 2  6ab z  z (a, b)  a 2  3b 2

We solve equation (3) in five different methods and obtain five sets of solutions A. PATTERN:1 (3) is equivalent to the system of equations

WAY: 1

PROPERTIES x(1, n)  z (1, n)  6obln  0 (i) (ii)

bu  az  av  0  au  3bz  3bv  0

(iii)

from which we get

(v)

(iv)

x(1, n)  y (1, n)  0(mod12) 3 y (1, n)  z (1, n)  6obln  0 6 x(1, n)  3 y (1, n)  0(mod 3) x(1, n)  z (1, n) `12n  6obln

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Research Paper b.

E-ISSN No : 2454-9916 | Volume : 3 | Issue : 1 | Jan 2017

WAY: 2

REFERENCES 1. R. Anbuselvi K S Araththi on ternary quadratic Equations 2 y 2  xy  4 z 2 , Indian journal for research analysis, 2016, 9(6) 379-381.

x(a, b)  b 2  3a 2  6ab

y (a, b)  b 2  3a 2  6ab z (a, b)  3a 2  b 2

2.

R. Anbuselvi K S Araththi on ternary quadratic Equations x  4 y  40z , Global journal for research analysis, 2016, 8(5)256-258.

2

PROPERTIES x(1, n)  y (1, n)  0(mod 6) (i) (ii) (iii) (iv) (v) c.

x(n,1)  3 z (n,1)  6n  0(mod10) 6 x(n,1)  2 y (n,1)  0(mod 6) 3x(n,1)  3nz(n,1)  18obln y (n,1)  z (n,1)  6obln  0(mod 6)

Meena K, Vidhyalakshmi S,Gopalan M.A,Priya K,Integral points on the cone 3( x 2  y 2 )  5 xy  47 z 2 , Bulletin of Mathematics and Statistics and Research,2014,2(1),65-70.

4.

Gopalan M.A,Vidhyalakshmi S,Nivetha S,on Ternary Quadratic Equation 4( x 2  y 2 )  7 xy  31z 2 Diophantus J.Math,2014,3(1),1-7.

5.

Gopalan M. A, Vidhyalakshmi S, Kavitha A, Observation on the Ternary Cubic Equation x 2  y 2  xy  12 z 3 Antarctica J. Math, 2013; 10(5):453-460.

6.

Gopalan M. A,Vidhyalakshmi S, Lakshmi K, Lattice points on the Elliptic Paraboloid, 16 y 2  9 z 2  4 x 2 Bessel J. Math, 2013, 3(2), 137-145.

7.

Gopalan M. A, Vidhyalakshmi S, Umarani J, Integral points on the Homogenous Cone x 2  4 y 2  37 z 2 , Cayley J. Math, 2013, 2(2), 101-107.

8.

Gopalan M.A,Vidhyalakshmi S,Sumathi G,Lattice points on the Hyperboloid of one sheet

y  y (a, b)  3a 2  b 2  6ab z  z (a, b)  3a 2  b 2 PROPERTIES (i) (ii) (iii) (iv)

x(1, n)  y (1, n)  2n 2  0(mod 6) y(1, n)  z(1, n)  2t 4,n  0(mod 6)

x(1, n)  z(1, n)  2t 4,n  0(mod5) y (1, n)  z (1, n)  2n(n  3)  0

B. PATTERN:2 Assume u  3v  3 z 2

2

2

2

3.

WAY: 3

x  x(a, b)  3a 2  b 2  6ab

2

4 z 2  2 x 2  3 y 2  4 , The Diophantus J. Math, 2012,

(7)

1(2), 109-115.

Also, write 4 as

9.

3  (i 3) * (i 3)

(8)

z  a 2  3b 2

(9)

Substitute (8) and (9) in (7) and employing the method of factorization, define

u  i 3v  (i 3)[a  i 3b]2

Gopalan M. A, Vidhyalakshmi S, Lakshmi K, Integral points on the Hyperboloid of two sheets 3 y 2  7 x 2  z 2  21 , Diophantus J. Math, 2012, 1(2), 99-107.

10. Gopalan M. A, Vidhyalakshmi S, Mallika S, Observation on Hyperboloid of one sheet x 2  2 y 2  z 2  2 Bessel J. Math, 2012, 2(3),

(9)

Equating the real and imaginary parts, we get

u  6ab v  a 2  3b 2 z  a 2  3b 2

and hence x = x(a,b) = a  3b  6ab 2

y = y(a,b) =  a

2

 3b 2  6ab 2 2 z = z(a,b) = a  3b 2

IV. CONCLUSION To conclude, one may search for other patterns of solutions to the equation under consideration.

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