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Use Style: Paper Title DOI 10.4010/2016.1727 ISSN 2321 3361 © 2016 IJESC ` Research Article Volume 6 Issue No. 6 On the Integer Solutions of the Homogeneous Biquadratic Diophantine Equation x4 y 4 82z 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 82z 2 w2 p2 are obtained. A few interesting relations among the special numbers 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 number of rank n with sides m . 4 Pn = Square pyramidal number 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 = Star number of rank n. PTn = Pentatope number of rank n. CH n = Centered Hexagonal number of rank n. 4DFn = Four dimensional figurate number Pn = Pronic number. 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 82 9 i 9 i (4) quadratic formula [1-5]. Biquadratic Diophantine equations, 2 2 homogeneous and non-homogeneous have stimulated the and p a b a iba 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 ivu iv 9i9ia ib2 a ib2 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 82z w p is considered u iv 9 ia ib for its non-zero distinct integer solutions. The recurrence u iv 9 ia ib2 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 82z 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 2 2 2 2 2 w w(a,b) 17a 17b 22ab u v 82 p (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 4324DFa 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 1v 82 p u (10) 82 can also be written in the form as Write 1 as 82 1 9i1 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 ivu iv 1 9i1 9ia 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. 78y(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) 1u 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 and in equation (2), the non-zero distinct B integer solutions are Pattern 5 x x(a,b) 820a 2 8b 2 164ab u v Equation (3) can be written as y y(a,b) 656a 2 10b 2 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 pu p 9p v9p 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 9Ap Av 0 1. is Au A 9Bp 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 equation y y(A,B) 10A 10B 16AB 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. n1 n1 where f n 163 18 82 163 18 82 and 1 3. is a bi- n1 n1 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 n1 n1 are given by xn1 v 163 18 82 9 82 163 18 82 9 82 1 2 Choice 1: 1 n1 n1 yn1 v 163 18 82 9 82 163 18 82 9 82 1 2 n1 n1 u p 9p v A z v 163 18 82 9 82 163 18 82 9 82 1 , B 0 n1 9p v u p B n1 n1 Choice 2: wn1 v 163 18 82 9 82 163 18 82 9 82 1 u p 9p v A 1 n1 n1 pn1 v 16318 82 82 9 16318 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 n1 n2 n3 y y(A,B) 8A2 8B 2 20AB yn1 326yn2 yn3 324 2 2 z z(A,B) 11A 11B 34AB zn1 326zn2 zn3 324 2 2 w w(A,B) 7A 7B 38AB wn1 326wn2 wn3 324 2 2 p 326 p p 0 p p(A,B) A B n1 n2 n3 Properties 1.
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