# On the System of Diophantine Equationsx2-6y2=-5 and x=az2-b

Authors: Silan Zhang; Jianhua Chen; Hao Hu
Journal: The Scientific World Journal
 (2014)

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## Abstract

Mignotte and Pethö used the Siegel-Baker method to find all the integral solutions(x,y,z) of the system of Diophantine equations x2-6y2=-5 and x=2z2-1. In this paper, we extend this result and put forward a generalized method which can completely solve the family of systems of Diophantine equations x2-6y2=-5 and x=az2-b for each pair of integral parameters a,b. The proof utilizes algebraic number theory and p-adic analysis which successfully avoid discussing the class number and factoring the ideals.


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## Body

## 1. Introduction

LetZ, N, and Q be the sets of all integers, positive integers, and rational numbers, and let a, b be the integers. The system of Diophantine equations
(1)x2-6y2=-5,x=az2-b
is a quartic model of an elliptic curve that has been investigated in many papers. Mignotte and Pethő [1] used the Siegel-Baker method to solve (1) for a=2 and b=1; however, their method was complicated as a combination of algebraic and transcendental number theory. In 1998, Cohn [2] gave an elementary proof of the above system of equations for a=2 and b=1. In 2004, Le [3] used a similar elementary method to extend the result of Cohn’s work and proposed an effective method solving the system of equations
(2)x2-Dy2=1-D,x=2z2-1,
where D-1 is the power of an odd prime. As an example, solutions of the equations for D=6 and D=8 are given in the paper so as to show the effectiveness of the method.In this paper, we use algebraic number theory and Skolem’sp-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 thatm≥0 is an integer and ε0=5+26 is the fundamental unit in the field Q(6), and let A denote (βε0m+β¯ε0¯m)/(β+β¯), where β2=±ηε0 or β2=±η, with η=1+6; β¯2 denotes the conjugate of β2 in Q(6).The main result of this paper is as follows.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)
2az+θA=±αεk with θ2=2a(1--5),(N2)
2az+θA=±αεk with θ2=4a(1+-5),(N3)
2az+θA=±αεk with θ2=2a(-1--5),(N4)
2az+θA=±αεk with θ2=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, and A is referred to above.As the application to the theorem, we give the following corollary.Corollary 2.
The system of (1) for a=2 and b=1 has exactly six integral solutions: (x,y,z)=(16561,±6761,±91), (71,±29,±6), (17,±7,±3), (7,±3,±2), (1,±1,±1), and (-1,±1,0).


## 2. Proof of the Theorem

Before the proof of the theorem, Lemma3 is needed.Lemma 3.
Ifm, ε0, β, and β¯ are defined as before, then A=(βε0m+β¯ε0¯m)/(β+β¯) is an algebraic number in the field Q(-5).Proof.
RewritingA, we have
(3)A=βε0m+β¯ε0¯mβ+β¯=(βε0m+β¯ε0¯m)(β+β¯)(β+β¯)2=β2ε0m+β¯2ε0¯m+ββ¯(ε0m+ε0¯m)β2+β¯2+2ββ¯.
Sinceβ2+β¯2, β2ε0m+β¯2ε0¯m, and ε0m+ε0¯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
(4)(x+6y)(x-6y)=(1+6)(1-6).
Then we have
(5)(x+6y)=±ηε0n,(x-6y)=±η¯ε0¯n,withn∈Z.
Adding (5), we get
(6)2x=±(ηε0n+η¯ε0¯n).
The solution of (6) is split into four cases.
Case 1. Assume that n=2m+1 is odd and 2x=ηε0n+η¯ε0¯n. Then
(7)2x=ηε02m+1+η¯ε0¯2m+1=(β+β¯)2(βε0m+β¯ε0¯mβ+β¯)2-2ββ¯,kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkwithβ2=ηε0.
Since
(8)(β+β¯)2=34+2-5=(1--5)(3+-5)2,ββ¯=-5,
by inserting (7) and the second equation of (1) into (6), we get
(9)2az2-θ2A2=2(b--5),
where θ2=1--5 and A=(3+-5)((βε0m+β¯ε0¯m)/(β+β¯)). We multiply both sides of the equation by 2a to obtain
(10)(2az)2-θ2A2=4a(b--5)=αα-.
Without loss of generality, we also denote θ2=2a(1--5) and A=(3+-5)((βε0m+β¯ε0¯m)/(β+β¯)). Factoring (9) in the field Q(θ), we have
(11)(N1)2az+θA=±αεk,withθ2=2a(1--5),k∈Z.Case 2. In another case, when  n=2m is even and 2x=ηε0n+η¯ε0¯n, then
(12)2x=ηε02m+η¯ε0¯2m=(β+β¯)2(βε0m+β¯ε0¯mβ+β¯)2-2ββ¯,kkkkkkkkkkkkkkkkkkkkkkkkkkkkikkwithβ2=η.
Similarly, we have(13)(2az)2-θ2A2=4a(b--5)=αα-,
where θ2=4a(1+-5) and A=(βε0m+β¯ε0¯m)/(β+β¯). Furthermore, we know that z satisfies
(14)(N2)2az+θA=±αεkwithθ2=4a(1+-5),k∈Z.Case 3.Consider n=2m+1 is odd and 2x=-(ηε0n+η¯ε0¯n). By the same consideration, we deduce that z satisfies
(15)(N3)2az+θA=±αεkwithθ2=2a(-1--5),k∈Z,
where A=(3--5)((βε0m+β¯ε0¯m)/(β+β¯)).
Case 4.Consider n=2m is even and 2x=-(ηε0n+η¯ε0¯n). Similarly, we know that z satisfies
(16)(N4)2az+θA=±αεkwithθ2=4a(-1+-5),k∈Z,
where A=(βε0m+β¯ε0¯m)/(β+β¯).
Thus, we complete the proof of the theorem.


## 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 integerm≠0 and i>1, one has Vp(pi(mi))>Vp(p(m1)), where Vp(·) denotes the standard p-adic valuation and p is an odd prime.Proof.
We know(17)Vp(pi(mi))-Vp(p(m1))=i-1+Vp(mi)-Vp(m1)=i-1+Vp((m-1)⋯(m-i+1)i!)=i-1+Vp((m-1)⋯(m-i+1))-Vp(i!)≥i-1-Vp(i!)=i-1-([ip]+[ip2]+⋯)≥i-1-ip-1≥(p-2p-1)i-1>0,withi>1.
Therefore, for i>1, the p-adic valuation of the term of pi(mi) exceeds the p-adic valuation of the term of p(m1). Thus, we complete the proof of Lemma 5.To complete the proof of the corollary, the fundamental unit in the totally complex quartic fieldQ(θ) is computed. Furthermore, nonassociated factor of 2-2-5 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] and Buchmann’s work [7], which offered a fast implementation scheme. Results are obtained via MATHEMATICA 7.0, which are listed in Table 1.Table 1
Associated number field.
θ
2
Integral basis
Fundamental unit
Nonassociated factor of2-2-5
1
-
-
5
1
,
θ
,
θ
2
,
θ
3
11
+
2
θ
-
6
θ
2
-
4
θ
3
6
+
4
θ
-
θ
3
2
+
2
-
5
1
,
θ
,
1
2
θ
2
,
1
4
θ
3
-
49
-
20
θ
+
3
θ
2
+
7
2
θ
3
2
+
θ
-
1
-
-
5
1
,
θ
,
θ
2
,
θ
3
-
85
+
34
θ
+
30
θ
2
+
38
θ
3
4
+
θ
3
-
2
+
2
-
5
1
,
θ
,
1
2
θ
2
,
1
4
θ
3
1
-
2
θ
-
θ
2
-
1
2
θ
3
12
-
θProof of Corollary.
Case 1.Substituting a=2 and b=1 into (9), we know that z satisfies
(18)4z2-θ2A2=2-2-5=αα-.
Equation(N1) is reduced to
(19)(N5)2z+θA=±αεkwithθ2=1--5,k∈Z.
From Table1, we get ε=11+2θ-6θ2-4θ3 and α=6+4θ-θ3. From Lemma 3, we know that A is an element in the field Q(-5). If we expand 2z+θA in the basis 1,θ,θ2, and θ3 of Q(θ), we obtain 2z+θA=a+bθ+cθ2+dθ3(a,b,c,d∈Q); then, from Lemma 3, we also know that c=0. In short, we denote this fact by (2z+θA)2=0.
To use thep-adic analysis, a suitable prime p is needed. Here, we take p=7. A straightforward computation shows that
(20)ε8≡1(mod⁡7).
Since (αε0)2≡0(mod⁡7), (αε7)2≡0(mod⁡7), and (αεi)2≢0(mod⁡7)(i=2,…,6), we get k≡0,7(mod⁡8).
(i) Letk≡0(mod⁡8); since ε8≡1(mod⁡7), we obtain ε8=1+7ξ; then
(21)±(2z+θA)=α(1+7ξ)m=α+7(m1)(αξ)+72(m2)(αξ2)+⋯+7m(mm)(αξm),
with k=8m. So we have
(22)±(2z+θA)2=(α+7(m1)(αξ)+⋯+7m(mm)(αξm))2.
From Lemma 5 and (αξ)2=(-30+30θ+60θ2+30θ3)2≢0(mod⁡7), we get m=0 and z=±3 by working modulo 7r+2 on formula (13), where 7r∥m.
(ii) Whenk≡7(mod⁡8), we have
(23)±(2z+θA)=αε-1(1+7ξ)m=αε-1+7(m1)(αε-1ξ)+⋯+7m(mm)(αε-1ξm),
with k=-1+8m. So we have
(24)±(2z+θA)2=(αε-1+7(m1)(αε-1ξ)iii+⋯+7m(mm)(αε-1ξm))2.
Similar deduction shows thatm=0 and z=±68.
Case 2. Secondly, we similarly consider the following equation:
(25)(N6)2z+θA=±αεk,withθ2=2+2-5,k∈Z.
From Table 1, we get ε=-49-20θ+3θ2+(7/2)θ3 and α=2+θ. By the same argument we choose p=23 and ε11≡1(mod⁡23). Similarly we have k≡0,1(mod⁡11); then we can deduce z=±1 and z=±91, respectively.
Case 3. Consider the following equation:
(26)(N7)2z+θA=±αεkwithθ2=-1--5,k∈Z.
We also have ε=-85+34θ+30θ2+38θ3, α=4+θ3, p=5, and ε4≡1(mod⁡5). Direct deductions show that z=±2 and z=±11798.
Case 4. The final equation is
(27)(N8)2z+θA=±αεkwithθ2=-2+2-5,k∈Z.
It corresponds to ε=1-2θ-θ2-(1/2)θ3, α=12-θ, p=13, and ε5≡1(mod⁡13). A similar deduction yields z=0 and z=±6.
All in all, if(x,y,z) is an integral solution of (1),
(28)±z=3,68,1,91,2,11798,0,6.
Substituting (24) into the system of Diophantine equations (1), we get all integral solutions, namely, (x,y,z)=(16561,±6761,±91), (71,±29,±6), (17,±7,±3), (7,±3,±2), (1,±1,±1), and (-1,±1,0).
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≤10 and 1≤b≤10 are solved and results are listed in Table 2. For simplicity, we only list the positive z-value of solutions (x,y,z).Table 2
Positivez-value of the solution.
a
=
1
a
=
2
a
=
3
a
=
4
a
=
5
a
=
6
a
=
7
a
=
8
a
=
9
a
=
10
b
=
1
z
=
0
z
=
0,1
,
2
,
3,6
,
91
z
=
0
z
=
0
z
=
0
z
=
0
z
=
0
z
=
0,1
,
3
z
=
0
z
=
0
b
=
2
z
=
1,3
NS
z
=
1
NS
NS
NS
NS
NS
z
=
1
NS
b
=
3
z
=
2
z
=
1,59
NS
z
=
1
z
=
2
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
=
2
NS
z
=
2
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
=
0
z
=
0,2
z
=
0
z
=
0
z
=
0
z
=
0,1
,
2
z
=
0
z
=
0,1
z
=
0
z
=
0
b
=
8
z
=
1,3
,
5,41
NS
NS
NS
NS
NS
z
=
1
NS
z
=
1
NS
b
=
9
z
=
4
z
=
1,2
,
29
NS
z
=
2
z
=
4
NS
NS
z
=
1
NS
z
=
1
b
=
10
z
=
3,9
NS
z
=
1,3
NS
NS
NS
NS
NS
z
=
1,3
NS
“NS” refers to “no solution.”



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