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International Journal of Modern Engineering Research (IJMER) Vol.3, Issue.3, May-June. 2013 pp-1424-1427 ISSN: 2249-6645

Integral solution of the non-homogeneous heptic equation in terms of the generalised Fibonacci and Lucas sequences x5  y5  ( x3  y3 ) xy  4z 2 w  3( p2  T 2 )2 w3

S.Vidhyalakshmi, 1 K.Lakshmi, 2 M.A.Gopalan3 123

Department of Mathematics, Shrimati Indira Gandhi College, Trichy-620002

Abstract: We obtain infinitely many non-zero integer sextuples ( x, y, z, w, p, T ) satisfying the Non-homogeneous 5 5 3 3 2 2 2 2 3 equation of degree seven with six unknowns given by x  y  ( x  y ) xy  4 z w  3( p  T ) w . The solutions are obtained in terms of the generalised Fibonacci and Lucas sequences. Recurrence relations for the variables are given. Various interesting relations between the solutions and special numbers, namely polygonal numbers, Pyramidal numbers, centered hexagonal pyramidal numbers and Four Dimensional Figurative numbers are presented.

Keywords: Fibonacci and Lucas sequences, heptic equation, integral solutions, Non-homogeneous equation, special numbers. MSC 2000 Mathematics subject classification: 11D41. NOTATIONS:

GFn (k , s ) 

 n   n  k  k 2  4s k  k 2  4s   ,     2 2 

-Generalised Fibonacci sequence

   k  k 2  4s k  k 2  4s  -Generalised Lucas sequence GLn (k , s)   n   n    ,  2 2     Tm, n -Polygonal number of rank n with size m

Pnm - Pyramidal number of rank n with size m CPn,6 - Centered hexagonal pyramidal number of rank n F4, n,3 -Four Dimensional Figurative number of rank n whose generating polygon is a triangle I. Introduction The theory of diophantine equations offers a rich variety of fascinating problems. In particular, homogeneous and non-homogeneous equations of higher degree have aroused the interest of numerous Mathematicians since antiquity [13].Particularly in [4, 5] special equations of sixth degree with four and five unknowns are studied. In [6-8] heptic equations with three and five unknowns are analysed. This paper concerns with the problem of determining non-trivial integral solution of the nonhomogeneous equation of seventh degree with six unknowns given 5 5 3 3 2 2 2 2 3 by x  y  ( x  y ) xy  4 z w  3( p  T ) w , in terms of the generalised fibonacci and lucas sequences. Recurrence relations for the variables are also given.Various interesting properties between the solutions and special numbers are presented.

II. Method of Analysis The Diophantine equation representing the non- homogeneous equation of degree seven is given by (1) x5  y5  ( x3  y3 ) xy  4z 2 w  3( p2  T 2 )2 w3 Introduction of the transformations x  u  v, y  u  v, z  2v, w  u, p  v  1, T  v  1 , v  1 (2) In (1) leads to v  2u  1 The above equation (3) is a pellian equation, whose general solution is given by 2

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(3)

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International Journal of Modern Engineering Research (IJMER) Vol.3, Issue.3, May-June. 2013 pp-1424-1427 ISSN: 2249-6645

n 1 n 1   1 3 2 2  3 2 2    2   (4)  n 1 n 1  1   un  3 2 2  3 2 2  n  0,1, 2,...... 2 2    The values of un and v n can be written in terms of the generalised Fibonacci and Lucas sequences.



vn 











1  vn  GLn 1(6, 1)  2  un  2GFn 1(6, 1)  

(5)

In view of (2) and (5) the non-zero distinct integral solutions of (1) in terms of the generalised Fibonacci and Lucas sequences are obtained as

        wn  2GFn 1 (6, 1)   1 pn  GLn 1 (6, 1)  1  2   1 Tn  GLn 1 (6, 1)  1, n  0,1, 2,3.... 2  1 xn  2GFn 1 (6, 1)  GLn 1 (6, 1) 2 1 yn  2GFn 1 (6, 1)  GLn 1 (6, 1) 2 zn  GLn 1 (6, 1)

(6)

A few numerical examples are tabulated below:

0 1 2 3 4 5 6

5 29 169 985 5741 33461 195025

-1 -5 -29 -169 -985 -5741 -33461

6 34 198 1154 6726 39202 228486

2 12 70 408 2378 13860 80782

4 18 100 578 3364 19602 114244

2 16 98 576 3362 19600 114242

The above values of xn , yn , z n , wn , pn , Tn , satisfy the following recurrence relations respectively.

xn  2  6 xn 1  xn  0 yn  2  6 yn 1  yn  0 zn  2  6 zn 1  zn  0 wn  2  6wn 1  wn  0 pn  2  6 pn 1  pn  4 Tn  2  6Tn 1  Tn  4 2.1 Properties:

1 1 GLn 1(6, 1)  2GFn  2 (6, 1)  GLn  2 (6, 1)  0 2 2 2. xn  yn  2wn  zn  pn  Tn  8GFn 1 (6, 1)  2GLn 1(6, 1) 1. 2GFn 1 (6, 1) 

3. xn  yn  pn  Tn 4. T2n 1 

1 GL2n  2 (6, 1)  1  0 2 www.ijmer.com

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International Journal of Modern Engineering Research (IJMER) Vol.3, Issue.3, May-June. 2013 pp-1424-1427 ISSN: 2249-6645

5. x3n  2  y3n  2  GL3n  3 (6, 1) 6. w2n 1  2GLn 1 (6, 1).GFn 1 (6, 1) 7. GL3n  3 (6, 1)  2T3n  2  0(mod 2) 8. xn  yn  2 wn 9. Each of the following is a nasty number: a) 6( z2 n 1  4) b) 3(2 p2n 1  GLn 1 (6, 1)) c)

6( x2n 1  y2n 1)  2 p25

48wn2  24 F4,1,3

10. Each of the following is a cubical integer: 2 a) 4( zn  2 zn wn  2 x2n 1) b) 2 p3n  2  3 zn  2 c) x3n  2  y3n  2  3 zn d) x3n  2  y3n  2  6 pn  6 e) 2T3n  2  3( xn  yn )  2 f) 2T3n  2  3 zn  2 g) 2T3n  2  6 pn  4 11. Each of the following is a biquadratic integer: 5 a) 2 p4n 3  8T2n 1  2 p2 b) x4n  3  y4n  3  4 z2n 1  2T3,2 c) 2T4n  3  4( x2 n 1  y2 n 1 )  CP2,6 d) 2T4 n  3  8T2 n 1  T4,4 2 e) 8( z2n 1  8wn ) g) 2 p4n  3  4( x2n 1  y2n 1  2) h) x4n  3  y4n  3  4( x2n 1  y2n 1 )  6 12. 2 zn wn  z2 n 1  2 y2n 1  0 13. w2n 1  zn wn  0 14. x2n 1  y2 n 1  2 zn wn  0 15. x2n 1  y2 n 1  4 pn wn  4wn  0 2 16. zn  x2n 1  y2n 1  0(mod 2) 17. x3n  2  y3n  2  2wn ( z2n 1  1) 18. w2n 1  2 wn pn  2wn  0 19. Define:

X  zn , Y=2(pn  1), Z  z2n 1  2 It is to be noted that the triple ( X , Y , Z ) satisfies the Elliptic Paraboloid X 20. Define:

 Y 2  2Z

X  zn , Y=2Tn  2, w  x 2n 1  y2n 1  2 It is to be noted that the triple ( X , Y ,W ) satisfies the Hyperbolic Paraboloid 2X www.ijmer.com

Y 2  W 1426 | Page

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International Journal of Modern Engineering Research (IJMER) Vol.3, Issue.3, May-June. 2013 pp-1424-1427 ISSN: 2249-6645

21. Define:

X  z2n 1  2, Y=2T2n+1  4, Z  x 2n 1  y2n 1  2 It is to be noted that the triple ( X , Y , Z ) satisfies the Cone X 22. Define:

 Y 2  2Z 2

Y

X  2Tn  2, Y= x 2n 1  y2n 1  2 It is to be noted that the pair ( X , Y ) satisfies the parabola X

III. Conclusion One may be able to get the solutions to (1) in terms of other choices of number sequences. For example, the solution to (1) is also written in terms of Pell and Pell-Lucas sequence as follows:

        wn  2 P2n  2 (2,1)   1 pn  PL2n  2 (2,1)  1  2   1 Tn  PL2n  2 (2,1)  1, n  0,1, 2,3.... 2  1 xn  2 P2n  2 (2,1)  PL2n  2 (2,1) 2 1 yn  2 P2n  2 (2,1)  PL2n  2 (2,1) 2 zn  PL2n  2 (2,1)

(7)

The corresponding properties can also be obtained in terms of number sequences.

References [1] [2] [3]

L.E.Dickson, History of Theory of Numbers Vol.11, Chelsea Publishing company,New York 1952. L.J.Mordell, Diophantine equation Academic Press, London1969 Carmichael, R.D.,The theory of numbers and Diophantine Analysis Dover Publications, New York 1959

[4]

M.A.Gopalan, S.Vidhyalakshmi and K.Lakshmi, on the non- homogeneous sextic equation x International Journal of Applied Mathematics and Applications, 4(2), 2012, 171-173 M.A.Gopalan, S.Vidhyalakshmi and K.Lakshmi, Integral Solutions of the sextic

[5]

x  y  z  w  3( x  y)T [6]

564 M.A.Gopalan

and

sangeetha.G,

 2( x2  w) x2 y 2  y 4  z 4 , equation

with

five

Unknowns

International Journal of Engineering Sciences & Research Technology, 1(10,) 2012, 562parametric

integral

solutions

the

heptic

equation

with

5unknown

x4  y4  2( x3  y3 )( x  y)  2( X 2  Y 2 ) z5 , Bessel Journal of Mathematics 1(1), 2011, 17-22 [7] [8]

M.A.Gopalan and sangeetha.G, on the heptic diophantine equations with 5 unknowns x  y  ( X Journal of Mathematics, 9(5), 2012, 371-375 Manjusomnath, G.sangeetha and M.A.Gopalan, on the non-homogeneous heptic equations 7 x3  (2 p  1) y5  z , Diophantine journal of Mathematics, 1 (2), 2012, 17-121

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 Y 2 )z with

, Antarctica unknowns

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