2 Resolution of The Diophantine Equation x 2 (t 2 t)y 2 (16t 4)x + (16t 2 16t)y =0

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1 International Mathematical Forum, Vol. 6, 20, no. 36, On Quadratic Diophantine Equation x 2 t 2 ty 2 6t 4x + 6t 2 6ty 0 A. Chandoul Institut Supérieure d Informatique et de Multimedia de Sfax, Route de Tunis km 0, B.P. 242, Sfax, Tunisia amarachandoul@yahoo.fr Abstract Let t 2 be a positive integer. Extending the work of A. Tekcan, here we consider the number of integer solutions of Diophantine equation E : x 2 t 2 ty 2 6t 4x + 6t 2 6ty 0. We also obtain some formulas and recurrence relations on the integer solution x n,y n of E. Keywords: Pell s equation, Diophantine equation Introduction Let t 2 be an integer. In [2], A. Tekcan consider the number of integer solutions of Diophantine equation D : x 2 t 2 ty 2 4t 2x+4t 2 4ty 0 over Z. He also derive some recurrence relations on the integer solutions x n,y n of D. In the present paper, we consider the integer of Diophantine equation E : x 2 t 2 ty 2 6t 4x + 6t 2 6ty 0 over Z, where t 2 be an integers. The reader can find many references in the subject in []. 2 Resolution of The Diophantine Equation x 2 t 2 ty 2 6t 4x + 6t 2 6ty 0 Note that the resolution of E in its present form is very difficult, that is, we can not determine how many solutions E has and what they are. So, we have

2 778 A. Chandoul to transform E into an appropriate Diophantine equation which can be easily solved. To get this let T : be a translation for some h and k. { x u + h y v + k 2 By applying the transformation T to E, we get T E : Ẽ : u + h2 t 2 tv + k 2 6t 4u + h +6t 2 6tv + k 0 3 In 3, we obtain u2h +4 6t and v 2kt 2 +2kt +6t 2 6t. So we get h 8t 2 and k 8. Consequently for x u +8t 2 and y v +8, we have the Diophantine equation which is a Pell equation. Ẽ : u 2 t 2 tv 2 32t Results Now, we try to find all integer solutions u n, oft E and then we can retransfer all results from T E toe by using the inverse of T. Theorem 3. Let Ẽ be the Diophantine equation in 3, then The fundamental solution of Ẽ is u,v 8t 2, 8. 2 Define the sequence u n, by u v un 8t 2 8 2t 2t 2 2t 2 2t n u v 2, n 2. 5 Then u n, is a solution of Ẽ.

3 Diop. Eq. x 2 t 2 ty 2 6t 4x + 6t 2 6ty The solutions u n, satisfy the recurrence relations u n 2t u n +2t 2 2t for n 2 2u n +2t 6 4 The solutions u n, u n 2t u n +2t 2 2t 2u n +2t 7 for n 4 5 The n-th solution u n, can be given by u n t ; 2, 2t 2,, 2, 2t 2,, 3, n. 8 }{{} n times Proof. It is easily seen that u,v 8t 2, 8 is the fundamental solution of Ẽ, since 8t t 2 t 32t We prove it using the method of mathematical induction. Let n,by 5 we get u,v 8t 2, 8 which is the fundamental solution and so is a solution of Ẽ. Now, we assume that the Diophantine equation 4 is satisfied for n, that is Ẽ : u2 n t2 tvn 2 32t +4. We try to show that this equation is also satisfied for n +. Applying 5, we find that un+ + 2t 2t 2 2t 2 2t n u v 2 2t 2t 2 2t 2 2t un 9 2t un +2t 2 2t 2u n +2t Hence, we conclude that u 2 n+ t 2 tv 2 n+ [2t u n +2t 2 2t ] 2 t 2 t[2u n +2t ] 2 u 2 n t2 tvn 2 32t +4.

4 780 A. Chandoul So u n+,+ is also solution of Ẽ. 3 Using 9, we find that u n 2t u n +2t 2 2t for n 2 2u n +2t 4 We prove it using the method of mathematical induction. For n 4, we get u 8t 2 u 2 32t 2 28t +2 u 3 48t 2 +72t 2 and u 4 52t 4 960t t 2 92t +2. Hence u 4 4t 3u 3 + u 2 u. So u n 4t 3u n + u n 2 u n 3. is satisfied for n 4. Let us assume that this relation is satisfied for n, that is, u n 4t 3u n + u n 2 u n 3. 0 Then using 9 and 0, we conclude that u n+ 4t 3u n + u n u n 2, completing the proof. Similarly, we prove that 4t , n 4. 5 We prove it using the method of mathematical induction. For n, we have u 8t 2 t + v 8 [t ;, 3] which is the fundamental solution of Ẽ. Let us assume that the n-th solution u n, is given by u n t ; 2, 2t 2,, 2, 2t 2,, 3. }{{} n times and we show that it holds for u n+,y n+.

5 Diop. Eq. x 2 t 2 ty 2 6t 4x + 6t 2 6ty 0 78 Using 6, we have u n+ + 2t u n +2t 2 2t 2u n +2t 2t u n + u n +2t t +t 2u n +2t as t + t + u n t + u n t +t + + 2t 2t 2t + 2t we get u n+ + t + 2t 2t + 2t completing the proof. t ; 2, 2t 2,, 2, 2t 2,, 3. }{{} n times

6 782 A. Chandoul As we reported above, the Diophantine equation E could be transformed into the Diophantine equation Ẽ via the transformation T. Also, we showed that x u +8t 2 and y v +8. So, we can retransfer all results from Ẽ to E by applying the inverse of T. Thus, we can give the following main theorem Theorem 3.2 Let D be the Diophantine equation in. Then The fundamental minimal solution of E is x,y 6t 4, 6 2 Define the sequence {x n,y n } n {u n +8t 2, +8}, where {x n,y n } defined in 5. Then x n,y n is a solution of E. So it has infinitely many integer solutions x n,y n Z Z. 3 The solutions x n,y n satisfy the recurrence relations x n 2t x n +2t 2 2ty n 32t 2 +36t 4 y n 2x n +2t y n 32t +20 for n 2. 4 The solutions u n, satisfy the recurrence relations x n 4t 3x n + x n 2 x n 3 64t 2 +80t 6 y n 4t 3y n + y n 2 y n 3 64t for n 4. Acknowledgements We would like to thank Saäd Chandoul and Massöuda Loörayed for helpful discussions and many remarks. References [] A. Chandoul, The Pell equation x 2 Dy 2 ±k 2, accepted in Advances of pure mathematics. [2] A. Tekcan, Quadratic Diophantine Equation x 2 t 2 ty 2 4t 2x + 4t 2 4ty 0, Bull. Malays. Math. Sci. Soc, , Received: March, 20