 Advances in Pure Mathematics, 2011, 1, 155-159 doi:10.4236/apm.2011.14028 Published Online July 2011 (http://www.SciRP.org/journal/apm) Copyright © 2011 SciRes. APM 155On Polynomials Solutions of Quadratic Diophantine Equations Amara Chandoul Institut Supérieure d’Informatique et de Multimedia de Sfax, Sfax, Tunisia E-mail: 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 ,nnXY.E=1ax by22bxyydx 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 =0fax c The equation xDyND, with given integers and and unknowns Nx 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=D2=Da Dxayx ayN1D and again there is only a finite number of solutions. The most interesting case of the equation arises when  be a positive non-square. 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 non-square integer. If the lenght of the periode of 22=1xDy22=1xDy1DDl21=vkxP is , all positive solutions are given by  and 21vk=yQ if k is odd, and by 1vkx=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)=,vkyQ: =1,2,yv Incidentally, (2 1)(1)vk and v 22=, are the positive solutions of xDy1l22=1xDy,:=1,2,vvxyv provided that is odd. There is no solution of other than  given by 11=vvvxDyx Dy,, where 11xy 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 . We recall that there are many papers in which are considered different types of Pell's equation. Many authors such as Tekcan , Kaplan and Williams , Matthews , Mollin, Poorten and Williams , Stevenhagen  and the others consider eome specific Pell equations and their integer solutions. In [2,7], It 156 22=4xDyA. CHANDOUL considered the equation xD22 and the equa- tion and he obtained some formulas for its integer solutions. He mentioned two conjecture which was proved by A. S. Shabani . In , we extend the work in [2,7] by considering the Pell equation 22=9,y2=xDykkD122=1PDQDP QDDPQ1,Dt1[ ,...,] when be a positive non-square and , we obtain some formulas for its integer solutions. 2In , 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 non-trivial. The main difficulty in solving polynomial Pell equations is to determine whether non-trivial solutions exist or not. In case, if there is at least one non-trivial solution, all solutions are obtained as powers of the smallest non-trivial 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sDt over every field of characteristic t24 4 =0tty216 0t:PPt[ ]\{0,1}.X42=0Pt X,n2.In [11,12], the number of integer solutions of Dio- phantine equation and Diophantine equation over is considered, where 22 242xttyt x 22 216 4xttyt x 2.t16 =ty 2. Main Results Let be a polynomial in In this paper, we consider the number of polynomial solutions of Diophantine equation   2222:44E XPtPtYPtPt Y (1) We also obtain some formulas and recurrence relations on the polynomial solution nXY.EEEE== 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 :XUHYVKT (2) be a translation for some H and By applying the transformation to we get T,E  22 22::424 4=0TEE UHPtPtVKPtUHPtPtV K   (3) In (3), we obtain 224UH Pt  and 222244VKPt KPtPt Pt So we get =2 1HPt and Consequently for =2.K=2 1XU Pt= 2,YV  and we have the Dio- phantine equation 222:1EUPtPt V.K  (4) which is a Pell equation. Now, we try to find all polynomial so lutions ,UVnn of TE and then we can retransfer all results from TE E.TE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 ,nnUV111211212212 2,2221nnnUPtVUUPtPt PtnVVPt  by  (5) ,UV EThen nn3) The solutions is a solution of ,nnUV21111=2 122=22 1nn nnn nUPtUPtPtVVUPtV satisfy the recurrence relations  2n (6) for 4) The solutions ,nnUV satisfy the recurrence relations 12 312 3=4 3=4 3nnnnnnnnUPtUUUVPt VVU  4nn (7) for ,nnUV can be given by 5) The -th solution Copyright © 2011 SciRes. APM A. CHANDOUL Copyright © 2011 SciRes. APM 157  1=1;2,22, ,2,2nnn timesUPt PtPtV 2,2 ,1n2 1,2PtE22=1Pt=1n1,2,n2=1.nPt V1.n11111211212121nnnnnnnnUPtVUPtVUVPt VV22211nnnPt VPt V11,nnUV E (8) Proof. 1) It is easily seen that is the fundamental polynomial solution of since 11,=UV2221Pt Pt 2) We prove it using the method of mathematical induction. Let , by (5) we get ,=UV2Pt11 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22:nEU Pt We try to show that this equation is also satisfied for Applying (5), we find that     211222221222212 222212 22212=22212=22nnUPt PtVPtPtPt PtPtPt PtPtPtPt PtPtPt UPtUPt   (9) Hence, we conclude that    2221122222=21 2 22==nnnnnUPtPtVPtU PtPt Pt UUPtPtV  So is also solution of 3) Using (9), we find that 112ntPtV2n=4,n211=2 12=22 1nnnn nUPtU PVUPtV for 4) We prove it using the method of mathematical induction. For we get   122223=2 1=881=324818 1UPtUPt PtUPtPt Pt 4324=128256160321UPtPtPtPt and    43243222 232 112825616032143 32401021=4 33248181 88121=4 3UPtPt PtPtPtPtPt PtPtPtPtPtPtPtPtPtPtUUU Hence  12 3=4 3nnnnUPtUUUSo =4.n,n is satisfied for Let us assume that this relation is satisfied for that is, 12 3=4 3nnnnUPtUUU  (10) Then using (9) and (10), we conclude that 112=4 3nnnnUPtUUU completing the proof. Similarly, we prove that 12 3=4 3,4nnnnVtVVVn =1n 5) We prove it using the method of mathematical induction. For , we have 1121221 1==122 2=1;2Pt PtUPtVPtE which is the fundamental solution of . Let us assume that the n-th solution ,nnUV is given by  1=1;2,2 2,,2,2 2,2nnn timesUPt PtPtV  11,.nnUyand we show that it hold s for U sing (6) , we have A. CHANDOUL C APM 158  211212 22121122 122 111121nnnn nnnnn nnnnPtUPtPt VPtU UPtPtVUVUPtV UPtVPtUPt V   1nPt V  as opyright © 2011 SciRes.  111 1211222 2()21 122() 2nPt Pt PtVPtPtPt 11112() 22nUPt  we get  11112122 1222=1;2,2 2,,2,2 2nntimesPtVPtPtPt Pt Pt 1111222,2nUPtEE. could be transformed into the Diophantine equation via the transformation T Also, we showed that =2 1XU Pt=2.EE.TDE 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,=21,2nn nnnXYU PtV where ,nnY,nn defined in (5). Then XXY.E is a solution of So it has infinitely many integer solutions ,.nnXY  3) The solutions ,nnXY satisfy the recurrence relations (see (11)) 4) The solutions ,nnXY 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222224132161418 4=2241 12864 5UttttttVttt  23223234 125619236 1418 42241 12864 5UtttttttVttt    221111=2 1228102=228 6nn nnn nXPtXPtPtYPtPtYXPY Pt  2 212 3123=4 316248=4316 16nnnnnnnnXPtX XXPtPtYPYYY Pt  4 and some other solutions are  1t for n (11)t for n (12) A. CHANDOUL Copyright © 2011 SciRes. APM 159241260 17ttt25292 19ttt 32444332418 4224120482048 631024768 156UttttVttttt   325554343 241418 4224116192204808896 158096 8192 2656 304UttttVttttttt   Further 11=2UV41;2= 2tt 23216 116 4ttVt22=2 ;2,4 ,2=Utt 3219236 164 5ttttt323256=2;2,4 ,2,4 ,2=128UtttV 444332=2 ;2,4 ,2,4 ,2,4 ,22048 2048 =1024 768ttttVtttt263260 11567Uttt 52255292 1304 9Utttt22 22:42 42168=0EXttYtXt tY 223220216 6Xttt 3225619240 264 7ttttt43204863264 2156 9ttttt32289615529622656304 11ttttt22xD22xm554343=2;2,4 ,2,4 ,2,4 ,2,4 ,216192204808896 1 =8096 8192 2656tttttUttttt It can be concluded now, that the fundamental solution of is Some other solutions are 82,4.t2Y3X 23128Y42048 2X 3241024 768Ytt545431619220480 88096 8192XttXtt   5 3. 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