Advances in Pure Mathematics, 2011, 1, 155159 doi:10.4236/apm.2011.14028 Published Online July 2011 (http://www.SciRP.org/journal/apm) Copyright © 2011 SciRes. APM 155 On Polynomials Solutions of Quadratic Diophantine Equations Amara Chandoul Institut Supérieure d’Informatique et de Multimedia de Sfax, Sfax, Tunisia Email: amarachandoul@yaho o.fr Received April 11, 2011; revised May 14, 2011; accepted May 25, 2011 Abstract Let be a polynomial in :PPt \0,1.X In this paper, we consider the number of polynomial solutions of Diophantine equation 2 :4244=0EXP PYPXPPY 22 2 . We also obtain some formulas and recurrence relations on the polynomial solution , nn Y.E =1ax by 22 bxyydx ey 22 = of Keywords: Polynomial Solutions, Pell’s Equation, Diophantine Equation 1. Introduction A Diophantine equation is an indeterminate polynomial equation that allows the variables to be integers only. Diophantine problems have fewer equations than un known variables and involve finding integers that work correctly for all equations. In more technical language, they define an algebraic curve, algebraic surface or more general object, and ask about the lattice points on it. The word Diophantine refers to the Hellenistic mathematician of the 3rd century, Diophantus of Alexandria, who made a study of such equations and was one of the first mathematicians to introduce symbolism into algebra. The mathematical study of Diophantine problems Diophantus initiated is now called Diophantine analysis. A linear Diophantine equation is an equation between two sums of monomials of degree zero or one. While individual equations present a kind of puzzle and have been considered th rough ou t h istory, th e f ormulat ion of g en eral theories of Diophantine equations was an achievement of the twentieth century. For example, the equation is known the linear Diophantine equation. In general, the Diophantine equation is the equation given by =0fax c The equation DyND, with given integers and and unknowns N and , is called Pell’s equation If is negative, it can have only a finite number of solutions. If is a perfect square, say , the equation reduces to y = D 2 =Da D ayx ayN 1D and again there is only a finite number of solutions. The most interesting case of the equation arises when be a positive nonsquare. Although J. Pell contributed very little to the analysis of the equation, it bears his name because of a mistake by Euler. Pell’s equation was solved by Lagrange in terms of simple continued fractions. Lagrange was the first to prove that has infinitly many solutions in integers if is a fixed positive nonsquare integer. If the lenght of the periode of 22 =1xDy 22 =1xDy1DD l21 =vk xP is , all positive solutions are given by and 21vk =yQ if k is odd, and by 1vk =P=yQ and 1vk if is even, where and k=1,2,v nn denotes the nth convergent of the continued fraction expansion of PQ .D=xP (2 1)(1) =, vk yQ : =1,2,yv Incidentally, (2 1)(1)vk and v 22 = , are the positive solutions of xDy l22 =1xDy ,:=1,2, vv xyv provided that is odd. There is no solution of other than given by 11 = v vv Dyx Dy , , where 11 y is the least positive solution called the fundamental solution, which there are different method for finding it. The reader can find many references in the subject in th e book [1]. We recall that there are many papers in which are considered different types of Pell's equation. Many authors such as Tekcan [2], Kaplan and Williams [3], Matthews [4], Mollin, Poorten and Williams [5], Stevenhagen [6] and the others consider eome specific Pell equations and their integer solutions. In [2,7], It
156 22 =4xDy A. CHANDOUL considered the equation xD 22 and the equa tion and he obtained some formulas for its integer solutions. He mentioned two conjecture which was proved by A. S. Shabani [8]. In [9], we extend the work in [2,7] by considering the Pell equation 22 =9,y 2 = Dy kkD1 22 =1PDQ DP Q DD PQ 1, Dt 1 [ ,...,] when be a positive nonsquare and , we obtain some formulas for its integer solutions. 2 In [10], A. Dubickas and J. Steuding are interested in the polynomial solutions of the Pell equation given by where is a fixed polynomial, and are polynomials in the same variables as and with coefficients in the same field or ring as those of . The solutions and are called trivial. All other solutions are called nontrivial. The main difficulty in solving polynomial Pell equations is to determine whether nontrivial solutions exist or not. In case, if there is at least one nontrivial solution, all solutions are obtained as powers of the smallest nontrivial solution. They prove this for polynomials in one variable with coefficients in The proof is purely algebraic and extends without change to arbitrary polynomials in several variables ,=1, 0 0 Dt over every field of characteristic t 2 4 4 =0tty 2 16 0t :PPt[ ]\{0,1}.X 42 =0 Pt X , n 2. In [11,12], the number of integer solutions of Dio phantine equation and Diophantine equation over is considered, where 22 2 42xttyt x 22 2 16 4xttyt x 2.t16 =ty 2. Main Results Let be a polynomial in In this paper, we consider the number of polynomial solutions of Diophantine equation 2 22 2 : 44 E XPtPtY PtPt Y (1) We also obtain some formulas and recurrence relations on the polynomial solution n Y.E E E E = = of Note that the resolution of in its present form is difficult, that is, we can not determine how many solutions has and what they are. So, we have to transform into an appropriate Diophantine equation which can be easily solved. To get this let : UH YVK T (2) be a translation for some and By applying the transformation to we get T,E 22 2 2 :: 424 4 =0 TEE UHPtPtVK PtUHPtPtV K (3) In (3), we obtain 224UH Pt and 22 2244VKPt KPtPt Pt So we get =2 1HPt and Consequently for =2.K =2 1XU Pt = 2,YV and we have the Dio phantine equation 2 22 :1EUPtPt V . (4) which is a Pell equation. Now, we try to find all polynomial so lutions ,UV nn of TE and then we can retransfer all results from TE E.T E E to by using the inverse of Theorem 2.1: Let be the Diophantin e equation in (3), then 1) The fundamental solution of is 11 ,=2 1,2UV Pt 2) Define the sequence , nn UV 1 1 1 21 1 21 2 212 2,2 221 n n n UPt V UU PtPt Ptn VV Pt by (5) ,UV E Then nn 3) The solutions is a solution of , nn UV 2 11 11 =2 122 =22 1 nn n nn n UPtUPtPtV VUPtV satisfy the recurrence relations 2n (6) for 4) The solutions , nn UV satisfy the recurrence relations 12 3 12 3 =4 3 =4 3 nnnn nnnn UPtUUU VPt VVU 4nn (7) for , nn UV can be given by 5) The th solution Copyright © 2011 SciRes. APM
A. CHANDOUL Copyright © 2011 SciRes. APM 157 1 =1;2,22, ,2,2 n nn times UPt PtPt V 2,2 ,1 n 2 1,2Pt E 2 2=1Pt =1n 1,2 ,n 2 =1. n Pt V 1.n 1 1 1 1 1 2 1 1 2 1 2 1 2 1 n n n n nn nn U Pt V U Pt V U V Pt V V 2 2 2 1 1 n nn Pt V Pt V 11 , nn UV E (8) Proof. 1) It is easily seen that is the fundamental polynomial solution of since 11 ,=UV 22 21Pt Pt 2) We prove it using the method of mathematical induction. Let , by (5) we get ,=UV 2Pt 11 which is the fundamental solu tion and so is a so lution of . Now, we assume that the Diophantine equation (4) is satisfied for that is E 2 2 : n EU Pt We try to show that this equation is also satisfied for Applying (5), we find that 2 1 1 2 2 2 2 212 22 212 2 22 212 22 212 =22 212 =22 n n UPt Pt VPt PtPt Pt Pt Pt Pt Pt PtPt Pt Pt Pt UPt UPt (9) Hence, we conclude that 2 22 11 2 2 2 22 =21 2 22 == nn n nn UPtPtV PtU Pt Pt Pt U UPtPtV So is also solution of 3) Using (9), we find that 11 2n tPtV 2n =4,n 2 11 =2 12 =22 1 nn nn n UPtU P VUPtV for 4) We prove it using the method of mathematical induction. For we get 1 2 2 22 3 =2 1 =881 =324818 1 UPt UPt Pt UPtPt Pt 432 4=128256160321UPtPtPtPt and 432 4 32 22 2 32 1 128256160321 43 324010 21 =4 3 3248181 881 21 =4 3 UPtPt PtPt PtPtPt Pt Pt Pt PtPtPtPtPt Pt PtUUU Hence 12 3 =4 3 nnnn UPtUUU So =4.n ,n is satisfied for Let us assume that this relation is satisfied for that is, 12 3 =4 3 nnnn UPtUUU (10) Then using (9) and (10), we conclude that 112 =4 3 nnnn UPtUUU completing the proof. Similarly, we prove that 12 3 =4 3,4 nnnn VtVVVn =1n 5) We prove it using the method of mathematical induction. For , we have 1 1 21221 1 ==1 22 2 =1;2 Pt Pt UPt V Pt E which is the fundamental solution of . Let us assume that the nth solution , nn UV is given by 1 =1;2,2 2,,2,2 2,2 n nn times UPt PtPt V 11 ,. nn Uy and we show that it hold s for U sing (6) , we have
A. CHANDOUL C APM 158 2 1 1 212 221211 22 122 1 1 11 21 nn nn n n nnn nn n n PtUPtPt VPtU UPtPtV U VUPtV UPtV Pt U Pt V 1 n Pt V as opyright © 2011 SciRes. 11 1 1 211 22 2 2()2 1 1 22() 2 n Pt Pt Pt V Pt Pt Pt 1 111 2( ) 22 n U Pt we get 1 1 11 21 22 1 222 =1;2,2 2,,2,2 2 n nt imes Pt V Pt Pt Pt Pt Pt 1 111 22 2 ,2 n U Pt E E . could be transformed into the Diophantine equation via the transformation T Also, we showed that =2 1XU Pt =2. E E.T D E and YV So, we can retransfer all results from to by applying the inverse of Thus, we can give the following mai n theorem: Theorem 2.2: Let be the Diophantine equation in (1). Then 1) The fundamental (minimal) solution of is completing the proof. As we reported above, the Diophantine equation 11 ,=4 2,4XYPt 2) Define the sequence 1 ,=21,2 nn nn n XYU PtV where , nn Y , nn defined in (5). Then Y .E is a solution of So it has infinitely many integer solutions ,. nn XY 3) The solutions , nn Y satisfy the recurrence relations (see (11)) 4) The solutions , nn Y satisfy the recurrence relations (see (12)) =2 1,Pt t Then Example 2.3: Let 11 ,=41,2UV t is the fundamental solution of 22 2 :42=1EUt tV 2 2 2 2 2 4132161 418 4 =2 241 12864 5 Uttt ttt Vttt 232 2 3 2 3 4 125619236 1 418 4 2 241 12864 5 Utttt ttt Vttt 22 11 11 =2 1228102 =228 6 nn n nn n XPtXPtPtYPtPt YXPY Pt 2 2 12 3 123 =4 316248 =4316 16 nnnn nnnn XPtX XXPtPt YPYYY Pt 4 and some other solutions are 1t for n (11) t for n (12)
A. CHANDOUL Copyright © 2011 SciRes. APM 159 2 41 260 1 7 tt t 2 5292 1 9 tt t 3 2 4 4 43 32 41 8 4 2 241 20482048 63 1024768 156 Ut ttt Vt tt tt 3 2 5 5 543 43 2 41 418 4 2 241 16192204808896 15 8096 8192 2656 304 Ut ttt Vt ttt ttt Further 1 1 =2 U V 41 ;2= 2 t t 2 3216 1 16 4 tt Vt 2 2 =2 ;2,4 ,2= Utt 32 19236 1 64 5 ttt tt 32 3 256 =2;2,4 ,2,4 ,2=128 Uttt V 4 4 43 32 =2 ;2,4 ,2,4 ,2,4 ,2 2048 2048 =1024 768 tttt V tt tt 2 63260 1 1567 U tt t 5 2 2 55292 1 304 9 U tt tt 22 22 :42 42168=0EXttYtXt tY 2 232202 16 6 Xtt t 32 25619240 2 64 7 ttt tt 432 04863264 2 156 9 tttt t 32 2 8961552962 2656304 11 ttt tt 22 xD 22 xm 5 54 3 43 =2;2,4 ,2,4 ,2,4 ,2,4 ,2 16192204808896 1 =8096 8192 2656 ttttt U ttt tt It can be concluded now, that the fundamental solution of is Some other solutions are 82,4.t 2 Y 3 X 2 3128 Y 42048 2 X 32 41024 768 Ytt 54 5 43 1619220480 8 8096 8192 Xtt Xtt 5 3. Acknowledgements We would like to thank Saäd Chandoul and Massöuda Loörayed for helpful discussions and many remarks. 4. References [1] I. Niven, H. S. Zuckerman and H. L. Montgomery, “An Introduction to the Theory of Numbers,” 5th Edition, Oxford University Press, Oxford, 1991. [2] A. Tekcan, “The Pell Equation =4 y ,” Applied Mathematical Sciences, Vol. 1, No. 8, 2007, pp. 363369. [3] P. Kaplan and K. S Williams, “Pell’s Equation =1,4y and Continued Fractions,” Journal of Number Theory, Vol. 23, No. 2, 1986, pp. 169182. doi:10.1016/0022314X(86)900879 [4] K. Matthews, “The Diophantine Equation =,> 22 xDy 22 xD ND,” Expositiones Mathematicae, Vol. 18, 2000, pp. 323331. [5] R. A. Mollin, A. J Poorten and H. C. Williams, “Halfway to a Solution of =3 y ,” Journal de Theorie des Nombres Bordeaux, Vol. 6, No. 2, 1994, pp. 421457. [6] P. Stevenhagen, “A Density Conjecture for the Negative Pell Equation, Computational Algebra and Number The ory,” Math. Appl., Vol. 325, 1992, pp. 187200. [7] A. Chandoul, “The Pell Equation =9y ,” Re search Journal of Pure Algebra, Vol. 1, No. 2, 2011, pp. 1115. 22 xD 22 xD [8] A. S. Shabani, “The Proof of Two Conjectures Related to Pell’s Equation =4y ,” International Journal of Computational and Mathematical Sciences, Vol. 2, No. 1, 2008, pp. 2427. [9] A. Chandoul, “The Pell Equation 2 = 22 Dy k ,” Ad vances in Pure Mathematics, Vol. 1, No. 2, 2011, pp. 1622. doi:10.4236/apm.2011.12005 [10] A. Dubickas and J. Steuding, “The Polynomial Pell Equa tion,” Elemente der Mathematik, Vol. 59, No. 4, 2004, pp. 133143. doi:10.1007/s0001700402147 [11] A. Tekcan, “Quadratic Diophantine Equation 22 22 42 4xttyt xt 4 =0ty,” Bulletin of Malaysian Mathematical Society, Vol. 33, No. 2, 2010, pp. 273280. [12] A. Chandoul, “On Quadratic Diophantine Equation 22 22 16 4916xttyt xt 16= 0ty,” Interna tional Mathematical Forum, Vol. 6, No. 36, 2011, pp. 17771782.
