DOI 10.4010/2016.1727 ISSN 2321 3361 © 2016 IJESC `

Research Article Volume 6 Issue No. 6

On the Solutions of the Homogeneous Biquadratic Diophantine Equation x4  y 4  82z 2  w2 p2 G. Janaki1, S. Vidhya2 Assistant Professor1, 2 Department of Mathematics Cauvery College for Women, Trichy, Tamil Nadu, India

Abstract: Different methods of the non-zero distinct integer solutions of the homogeneous biquadratic Diophantine equation with five unknowns x4  y4  82z 2  w2 p2 are obtained. A few interesting relations among the special and the solutions are observed. The recurrence relations among the solutions are also given.

Keywords: Homogeneous biquadratic equation with five unknowns, integer solutions, Polygonal numbers.

Notations

Tm,n = Polygonal of rank n with sides m . 4 Pn = Square of rank n. 5 Pn = Pentagonal pyramidal number of rank n.

Gnon = Gnomonic number of rank n.

SOn = Stella Octangula number of rank n.

Starn = of rank n.

PTn = Pentatope number of rank n.

CH n = Centered of rank n.

4DFn = Four dimensional

Pn = .

I. INTRODUCTION Method 1 Biquadratic equation is sometimes used as quartic equation. Pattern 1 Such equations are easy to solve, since they reduce to a Assume quadratic equation and hence can be solved for using the (4) 82  9  i9  i quadratic formula [1-5]. Biquadratic Diophantine equations, 2 2 homogeneous and non-homogeneous have stimulated the and p  a b  a iba ib (5) interest of numerous mathematicians. In the context one may Using (4) and (5) in (3) and employing the method of refer [6-11] for various problems on the Diophantine factorization, we get equations with two, three and four variables. u ivu iv 9i9ia ib2 a ib2 In this communication, the biquadratic equation with five Equating the like factors, we get 4 4 2 2 2 2 unknowns given by x  y  82z  w p is considered u  iv  9  ia  ib for its non-zero distinct integer solutions. The recurrence u iv  9 ia ib2 relations among the solutions are also obtained. Equating real and imaginary parts, we get II. METHOD OF ANALYSIS u  9a 2  9b2  2ab The homogeneous biquadratic Diophantine equation with five unknowns to be solved for its non-zero distinct integral v  a 2  b 2 18ab solution is Substituting u and v in equation (2), the non-zero distinct x4  y4  82z 2  w2 p2 (1) integer solutions are x  x(a,b) 10a 2 10b2 16ab Introducing the linear transformations y  y(a,b)  8a 2 8b2  20ab x  u  v, y  u  v, z  2u  v, w  2u  v (2) z  z(a,b) 19a 2 19b2 14ab Equation (1) becomes w  w(a,b) 17a 2 17b2  22ab u 2  v2  82 p2 (3) p  p(a,b)  a 2  b2

International Journal of Engineering Science and Computing, June 2016 7275 http://ijesc.org/ Properties Properties

1. 8T8,a  x(a,1) 10 p(a,1)  2 is a nasty number. 216 1. x(a,1)  y(a,1)  4DF T  0 mod5 . 2 a 12,a   2. . a y(a,1)  z(a,1)  w(a,1) 164T6,a  0 mod 2

1 2. 2T28,a  z(a,1)  w(a,1)  30Gnoa  32 is a nasty 3. is a x(1,b)  y(1,b)  z(1,b)  41T12,b 1066 number. 210 quadratic integer. 3. 1 is 4324DFa  w(a,1) p(a,1) 11SOa  3T26,a 17 Pattern 4 19 Rewriting equation (3) as a biquadratic integer. 2 2 2 Pattern 2 1v  82 p u (10) 82 can also be written in the form as Write 1 as 82  1 9i1 9i (6) 82 1 82 1 1    (11) Using (5) and (6) in (3) and employing the method of 81 factorization, we get Assume 2 2 u  ivu  iv  1 9i1 9ia  ib a  ib v  82a2  b2   82a  b 82a  b (12) Proceeding as in pattern 1, the non-zero distinct integer Using (11) and (12) in (10) and employing the method of solutions of (1) are factorization, we get 2 2 x  x(a,b) 10a 10b 16ab  82 1 82 1 2 2  82a  b  82a b   82 p  u 82 p u 2 2 81 y  y(a,b)  8a  8b  20ab Equating the like factors, we get 2 2 z  z(a,b) 11a 11b  34ab  82 1 2  82a  b   82 p  u w  w(a,b)  7a 2  7b2  38ab 9 2 2  82 1 2 p  p(a,b)  a  b  82a b   82 p u 9 Properties Equating rational and irrational parts, we get 1. p(a,1)  z(a,1)  x(a,1)  2  0 mod 9. 1 2 2  2. 78y(1,b)  w(1,b)  p(1,b)  27Gno CH  42 is a p  82a  b  2ab  b b 9  (13)  nasty number. 1 u  82a 2  b2 164ab    3. x(a,1)  y(a,1)  4T3,a  20Gnoa  22  0. 9  As our interest is to find only integer solutions, so replacing Pattern 3 a  9A and b  9B in (12) and (13), we get Rewriting equation (3) as p  738A2  9B2 18AB 2 2 2 (7) 1u  82 p v u  738A2  9B 2 1476AB Assume v  6642A2 81B 2 u  82a2  b2  82a  b 82a  b (8)    Substituting and in equation (2), The non-zero distinct Write 1 as, integer solutions are (9) 1  82 9 82 9 x  x(A, B)  7380A2  72B 2 1476AB Using (8) and (9) in (7) and employing the method of y  y(A, B)  5904A2  90B 2 1476AB factorization, we get 2 2 2 2 z  z(A, B)  8118A  63B  2952AB 82 9 82 9 82a b 82a b  82 p  v 82 p v          w  w(A, B)  5166A2  99B 2  2952AB Equating the like factors, we get 2 2 2 p  p(A, B)  738A  9B 18AB  82  9 82a  b   82 p  v Properties 2  82  9 82a  b   82 p  v 1. 10 p(1,B)  y(1,B)  648Gno 12636  0. Equating rational and irrational parts, we get B 2. w(1,B) 10 p(1,B) 1386Gno 11160 is a p  82a2  b2 18ab B quadratic integer. v  738a2  9b2 164ab 3. x(1,B) 8p(1,B)  z(1,B)  666Gno  0 mod 9. Substituting u and v in equation (2), the non-zero distinct B integer solutions are Pattern 5 x  x(a,b)  820a 2  8b 2 164ab Equation (3) can be written as 2 2 y  y(a,b)  656a 10b 164ab z  z(a,b)  902a 2  7b 2 164ab u 2  p2  81p2 v2 w  w(a,b)  574a 2 11b 2 164ab p  p(a,b)  82a 2  b 2 18ab u  pu  p 9p  v9p v (14)

International Journal of Engineering Science and Computing, June 2016 7276 http://ijesc.org/ which is represented in the form of ratio as Solutions of Choice 2: x  x(A,B)  8A2  8B 2  20AB u  p 9p  v A y  y(A,B) 10A2 10B 2 16AB   , B  0 9p  v u  p B z  z(A, B)  7A2  7B 2  38AB w  w(A, B) 11A2 11B 2  34AB This is equivalent to the following two equations p  p(A, B)  A2  B 2 Properties Bu  B 9Ap  Av  0 1. is  Au  A 9Bp  Bv  0 7y(A,1) 10z(A,1)  StarA  249GnoA  248 a nasty number. Solving the above equations by cross ratio method, we get 2. 4 x(A,1) p(A,1) 192PTA 84PA  6T24,A  0 mod 2. 3. w(A,1) 11p(A,1)  2T  7Gno  7  0 u  A2  B 2 18AB 24,A A 2 2 v  9A  9B  2AB Method 2 p  A2  B 2 Here the integer solutions of (1) are obtained by employing a Substituting the values of u and v in (2), the non-zero distinct different procedure to solve (3), integer solutions are Let u0  9v, p0  v be the least positive integer solution of x  x(A,B)  8A2  8B 2  20AB (3). To obtain the other solutions of (3), Consider the Pellian 2 2 y  y(A,B) 10A 10B 16AB equation 2 2 z  z(A, B)  7A2  7B 2  38AB u  82 p 1 w  w(A, B) 11A2 11B 2  34AB ~ ~ whose initial solution un , pn is given by p  p(A, B)  A2  B 2 ~ 1 Properties un  f n 2 1. x(1, B)  y(1, B)  z(1, B)  w(1, B)  6ProB  57GnoB  51  0. ~ 1 2. 3w(A,1) 3p(A,1) 12T 30 is a nasty pn  gn 21,A 2 82 number. n1 n1 where f n  163 18 82  163 18 82 and 1 3. is a bi- n1 n1 y(A,1) p(A,1) 8SOA 12GnoA  2 10 gn  163 18 82  163 18 82 quadratic integer. Applying Brahmagupta lemma between u0 , p0  and Note ~ ~ un , pn  , the sequence of non-zero distinct integer solutions There are other two different choices of ratio methods which of (1) are obtained as 1  n1 n1   are given by xn1  v 163 18 82 9  82 163 18 82 9  82 1   Choice 1: 2  1  n1 n1   yn1  v 163 18 82 9  82 163 18 82 9  82 1 2    n1 n1 u  p 9p  v A z  v 163 18 82 9  82  163 18 82 9  82  1   , B  0 n1          9p  v u  p B     n1 n1   Choice 2: wn1  v 163 18 82 9  82 163 18 82 9  82 1    u  p 9p  v A 1  n1 n1  pn1  v 16318 82  82  9 16318 82  82 9   , B  0   9p  v u  p B 2 82 The recurrence relations satisfied by the solutions of equation Solutions of Choice 1: 2 2 (1) are x  x(A,B)  10A 10B 16AB x  326x  x  324 n1 n2 n3 y  y(A,B)  8A2 8B 2  20AB yn1  326yn2  yn3  324 2 2 z  z(A,B)  11A 11B  34AB zn1  326zn2  zn3  324 2 2 w  w(A,B)  7A  7B  38AB wn1  326wn2  wn3  324 2 2 p 326 p  p  0 p  p(A,B)  A  B n1 n2 n3 Properties 1. x(A,2)  z(A,2)  p(A,2)  0 mod 4. III. CONCLUSION One may search for other non-zero distinct integer solutions of 2. 1 5 is a cubic the considered biquadratic equations with multivariables. y(A,1) 8p(A,1)  40PA 16 20 integer. IV. REFERENCES 3. 4x(1, B)  5y(1, B)  82Gno  82  0. [1] Carmicheal R.D, The Theory of Numbers and Diophantine B Analysis, Dover Publications, New York, 1959.

International Journal of Engineering Science and Computing, June 2016 7277 http://ijesc.org/ [2] Dickson L.E, History of Theory of Numbers, Vol 11, Chelsea publishing company, New York, 1952.

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