Research Article On the System of Diophantine Equations

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1 e Scientific World Journal, Article ID 63617, 4 pages Research Article On the System of Diophantine Equations x 6y = 5and x=az b Silan Zhang, 1, Jianhua Chen, 1 and Hao Hu 1 1 School of Mathematics and Statistics, Wuhan University, Hubei 437, China College of Science, Huazhong Agricultural University, Hubei 437, China Correspondence should be addressed to Silan Zhang; @163.com Received 3 March 14; Accepted 6 June 14; Published 17 June 14 Academic Editor: Gustaaf Schoukens Copyright 14 Silan Zhang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Mignotte and Pethö used the Siegel-Baker method to find all the integral solutions(x,y,z)of the systemof Diophantine equations x 6y = 5and x=z 1. In this paper, we extend this result and put forward a generalized method which can completely solve the family of systems of Diophantine equations x 6y = 5and x=az bfor each pair of integral parameters a, b.the proof utilizes algebraic number theory andp-adic analysis which successfully avoid discussing the class number and factoring the ideals. 1. Introduction Let Z, N, andq be the sets of all integers, positive integers, and rational numbers, and let a, b be the integers. The system of Diophantine equations x 6y = 5, x=az b (1) isaquarticmodelofanellipticcurvethathasbeeninvesti- gatedinmanypapers.mignotteandpethő[1]usedthesiegel- Baker method to solve (1) fora = and b = 1;however, their method was complicated as a combination of algebraic and transcendental number theory. In 1998, Cohn [] gave an elementary proof of the above system of equations for a= and b=1.in4,le[3]usedasimilarelementarymethod to extend the result of Cohn s work and proposed an effective method solving the system of equations x Dy =1 D, x=z 1, () where D 1is the power of an odd prime. As an example, solutions of the equations for D=6and D=8are given in the paper so as to show the effectiveness of the method. In this paper, we use algebraic number theory and Skolem s p-adic method [4] to solve(1), and the method is relatively simple. In the proposed method, both the consideration of the class number in the field and the factorization of ideals of integral ring are avoided. Moreover, a faster algorithm proposed in [5] to compute the fundamental unit and the set of nonassociated factors is used. In order to well interpret the main result, the symbol notation used in this paper is defined as below. Here, we assume that m is an integer and ε =5+ 6 is the fundamental unit in the field Q( 6), andleta denote (βε m + βε m )/(β + β), whereβ =±ηε or β =±η,with η=1+ 6; β denotes the conjugate of β in Q( 6). Themainresultofthispaperisasfollows. Theorem 1. Let (x,y,z)be an integral solution of (1).Then z exists only when it satisfies one of the following four equations for k Z: (N1) az + θa = ±αε k with θ =a(1 5), (N) az + θa = ±αε k with θ =4a(1+ 5), (N3) az + θa = ±αε k with θ =a( 1 5), (N4) az + θa = ±αε k with θ =4a( 1+ 5), where ε is the fundamental unit of the totally complex quartic field of Q(θ), α is the nonassociated factor such that αα = 4a(b 5), α denotes the relative conjugate of α, anda is referred to above.

2 The Scientific World Journal As the application to the theorem, we give the following corollary. Corollary. The system of (1) for a = and b = 1 has exactly six integral solutions: (x, y, z) = (16561, ±6761, ±91), (71, ±9, ±6), (17, ±7, ±3), (7, ±3, ±), (1, ±1, ±1), and ( 1, ±1, ).. Proof of the Theorem Before the proof of the theorem, Lemma 3 is needed. Lemma 3. If m, ε, β, andβ are defined as before, then A= (βε m + βε m )/(β + β) is an algebraic number in the field Q( 5). Proof. Rewriting A,wehave A= βεm + βε m = (βεm + βε m )() (β + β) = β ε m + β ε m +ββ(ε m + ε m ) β + β. +ββ Since β + β, β ε m + β ε m,andε m + ε m are all rational integers and ββ = 5,then clearly A is an algebraic number. Thus, the lemma is proven. Proof of Theorem. There are four separate cases in consideration during the proof. Since the process is very similar in each case, some details will be omitted for simplicity. Now we prove the theorem. After rewriting the first equation of (1), factorization in the field Q( 6) yields Then we have (3) (x + 6y) (x 6y) = (1 + 6) (1 6). (4) (x + 6y) = ± ηε n, (x 6y) = ± ηε n, Adding (5), we get with n Z. (5) x=±(ηε n + ηε n ). (6) The solution of (6)issplitintofourcases. Case 1. Assume that n=m+1is odd and x = ηε n + ηε n. Then x=ηε m+1 + ηε m+1 =() ( βεm + βε m ) ββ, with β =ηε. (7) Since (β + β) =34+ 5 = (1 5) (3 + 5), ββ = 5, by inserting (7) andthesecondequationof(1) into(6), we get (8) az θ A =(b 5), (9) where θ =1 5 and A = (3+ 5)((βε m +βε m )/()). We multiply both sides of the equation by a to obtain (az) θ A =4a(b 5) = αα. (1) Without loss of generality, we also denote θ =a(1 5) and A = (3 + 5)((βε m + βε m )/(β + β)). Factoring (9) in the field Q(θ),wehave (N1) az + θa = ±αε k, with θ =a(1 5), k Z. (11) Case. In another case, when n=mis even and x = ηε n + ηε n,then x=ηε m + ηε m =() ( βεm + βε m ) ββ, (1) with β =η. Similarly, we have (az) θ A =4a(b 5) = αα, (13) where θ = 4a(1 + 5) and A = (βε m + βε m )/(β + β). Furthermore, we know that z satisfies (N) az + θa = ±αε k with θ =4a(1+ 5), k Z. (14) Case 3. Consider n=m+1is odd and x = (ηε n + ηε n ). By the same consideration, we deduce that z satisfies (N3) az + θa = ±αε k with θ =a( 1 5), k Z, (15) where A = (3 5)((βε m + βε m )/(β + β)). Case 4. Consider n=mis even and x = (ηε n + ηε n ). Similarly, we know that z satisfies (N4) az + θa = ±αε k with θ =4a( 1 + 5), k Z, (16) where A = (βε m + βε m )/(β + β). Thus, we complete the proof of the theorem.

3 The Scientific World Journal 3 3. Proof of the Corollary Remark 4. The method of the proof of the corollary is a special instance of general procedure for the computation of integral points on some quartic model of elliptic curves. The method is relatively simple, because it avoids the use of class number and ideal factorization in imaginary quartic fields. Before the proof, we give the following lemma. Lemma 5. For any integer m = and i > 1, one has V p (p i ( m i )) >V p (p ( m 1 )),wherev p ( ) denotes the standard padic valuation and p is an odd prime. Proof. We know field Q( 5).Ifweexpandz+θA in the basis 1, θ, θ,andθ 3 of Q(θ),weobtainz+θA = a+bθ+cθ +dθ 3 (a,b,c,d Q); then, from Lemma 3, wealsoknowthatc=.inshort,we denote this fact by (z + θa) =. To use the p-adic analysis, a suitable prime p is needed. Here, we take p=7. A straightforward computation shows that ε 8 1(mod 7). () Since (αε ) (mod 7), (αε 7 ) (mod 7), and(αε i ) (mod 7)(i=,...,6),wegetk, 7(mod 8). (i) Let k (mod8); sinceε 8 1(mod7), weobtain ε 8 =1+7ξ;then V p (p i ( m i )) V p (p( m 1 )) =i 1+V p ( m i ) V p ( m 1 ) ± (z + θa) =α(1+7ξ) m =α+7( m 1 ) (αξ) +7 ( m )(αξ )+ +7 m ( m m )(αξm ), (1) (m 1) (m i+1) =i 1+V p ( ) i! =i 1+V p ((m 1) (m i+1)) V p (i!) i 1 V p (i!) =i 1 ([ i i ]+[ p p ]+ ) i 1 i p 1 ( p p 1 )i 1>, with i>1. (17) Therefore, for i>1,thep-adic valuation of the term of p i ( m i ) exceeds the p-adic valuation of the term of p ( m 1 ). Thus,we complete the proof of Lemma 5. To complete the proof of the corollary, the fundamental unit in the totally complex quartic field Q(θ) is computed. Furthermore, nonassociated factor of 5 in the ring of integers of Q(θ) is also calculated. The idea of computation of fundamental unit and nonassociated factorization stems from Zhu and Chen [5, 6] andbuchmann swork[7], which offered a fast implementation scheme. Results are obtained viamathematica7., whicharelistedin Table 1. Proof of Corollary. Case 1. Substituting a=and b=1into (9), we know that z satisfies 4z θ A = 5 =αα. (18) Equation (N1) is reduced to (N5) z + θa = ±αε k with θ =1 5, k Z. (19) From Table 1,wegetε = 11 + θ 6θ 4θ 3 and α= 6+4θ θ 3.FromLemma 3,weknowthatA is an element in the with k=8m.sowehave ±(z + θa) = (α+7( m 1 ) (αξ) + +7m ( m m )(αξm )). () From Lemma 5 and (αξ) = ( 3 + 3θ + 6θ + 3θ 3 ) (mod 7),wegetm=and z=±3by working modulo 7 r+ on formula (13), where 7 r m. (ii) When k 7(mod 8),wehave ± (z + θa) =αε 1 (1+7ξ) m =αε 1 +7( m 1 )(αε 1 ξ) + +7 m ( m m )(αε 1 ξ m ), with k= 1+8m.Sowehave ±(z + θa) =(αε 1 +7( m 1 )(αε 1 ξ) + +7 m ( m m )(αε 1 ξ m )). Similar deduction shows that m=and z=±68. (3) (4) Case. Secondly, we similarly consider the following equation: (N6) z + θa = ±αε k, with θ =+ 5, k Z. (5) From Table 1,wegetε = 49 θ + 3θ + (7/)θ 3 and α= +θ. By the same argument we choose p=3and ε 11 1(mod 3).Similarlywehavek,1(mod 11);thenwecan deduce z=±1and z=±91,respectively. Case 3. Consider the following equation: (N7) z + θa = ±αε k with θ = 1 5, k Z. (6)

4 4 The Scientific World Journal Table 1: Associated number field. θ Integral basis Fundamental unit Nonassociated factor of , θ, θ,θ θ 6θ 4θ 3 6+4θ θ , θ, 1 θ, 1 4 θ3 49 θ + 3θ + 7 θ3 +θ 1 5 1, θ, θ,θ θ + 3θ + 38θ 3 4+θ , θ, 1 θ, 1 4 θ3 1 θ θ 1 θ3 1 θ Table : Positive z-value of the solution. a=1 a= a=3 a=4 a=5 a=6 a=7 a=8 a=9 a=1 b = 1 z = z =, 1,, 3, 6, 91 z = z = z = z = z = z =, 1, 3 z = z = b= z=1,3 NS z=1 NS NS NS NS NS z=1 NS b = 3 z = z = 1, 59 NS z=1 z= NS NS NS NS z = 1, 83 b=4 NS NS z=1,5 NS z=1 NS NS NS NS NS b=5 z= NS z= z=1 NS z=1 NS NS NS NS b=6 NS NS NS NS z=1 NS z=1,5 NS NS NS b=7 z= z=, z= z= z= z=,1, z= z=,1 z= z= b=8 z=1,3,5,41 NS NS NS NS NS z=1 NS z=1 NS b = 9 z = 4 z = 1,, 9 NS z= z=4 NS NS z=1 NS z=1 b = 1 z = 3, 9 NS z=1,3 NS NS NS NS NS z=1,3 NS NS refers to no solution. We also have ε = θ + 3θ + 38θ 3, α=4+θ 3, p=5, and ε 4 1(mod 5). Direct deductions show that z=±and z = ± Case 4. The final equation is (N8) z + θa = ±αε k with θ = + 5, k Z. (7) It corresponds to ε=1 θ θ (1/)θ 3, α=1 θ, p=13, and ε 5 1(mod 13). Asimilardeductionyieldsz=and z=±6. All in all, if (x,y,z)is an integral solution of (1), ±z = 3, 68, 1, 91,, 11798,, 6. (8) Substituting (4) into the system of Diophantine equations (1), we get all integral solutions, namely, (x,y,z) = (16561, ±6761, ±91), (71, ±9, ±6), (17, ±7, ±3), (7, ±3, ±), (1, ±1, ±1),and( 1, ±1, ). This completes the proof of the corollary. Remark 6. Like before, we can solve a family of systems of Diophantine equations (1). As a direct application of this theorem, systems of equations with parameters 1 a 1 and 1 b 1are solved and results are listed in Table. For simplicity, we only list the positive z-value of solutions (x,y,z). Acknowledgments ThisworkispartlysupportedbytheFundamentalResearch Funds for the Central Universities (no. 6614QC1) and National Natural Science Foundation of China (6135). References [1] M. Mignotte and A. Pethő, On the system of Diophantine equations x 6y = 5 and x = z 1, Mathematica Scandinavica,vol.76,no.1,pp.5 6,1995. [] J. H. E. Cohn, The Diophantine system x 6y = 5,x= z 1, Mathematica Scandinavica, vol.8,no.,pp , [3] M. H. Le, On the Diophantine system x Dy =1 Dand x=z 1, Mathematica Scandinavica,vol.95,no.,pp , 4. [4] H. Brandt and Th. Skolem, Diophantische Gleichungen. (Erg. d. Math. u. ihrer Grenzgebiete, 5. H. 4), Verlag Julius Springer, Berlin, Germany, [5]H.L.ZhuandJ.H.Chen, Integralpointsoncertainelliptic curves, Rendiconti del Seminario Matematico della Università di Padova,vol.119,pp.1,8. [6] H. Zhu and J. Chen, Integral points on y =x 3 + 7x 6, Mathematical Study,vol.4,no.,pp ,9. [7] J. Buchmann, The computation of the fundamental unit of totally complex quartic orders, Mathematics of Computation, vol.48,no.177,pp.39 54,1987. Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.

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