On the System of Diophantine Equations 𝑥²−6𝑦²=−5 and 𝑥=𝑎𝑧²−𝑏

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.

1. Introduction

Let ℤ, 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 𝑥²−6𝑦²=−5,𝑥=𝑎𝑧²−𝑏 (1) 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 𝑥²−𝐷𝑦²=1−𝐷,𝑥=2𝑧²−1, (2) 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’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≥0 is an integer and 𝜀₀=5+2√6 is the fundamental unit in the field ℚ(√6), and let 𝐴 denote (𝛽𝜀^𝑚₀+𝛽𝜀₀^𝑚)/(𝛽+𝛽), where 𝛽²=±𝜂𝜀₀ or 𝛽²=±𝜂, with 𝜂=1+√6; 𝛽² 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=±𝛼𝜀^𝑘 with 𝜃²=2𝑎(1−√−5),(N2)2𝑎𝑧+𝜃𝐴=±𝛼𝜀^𝑘 with 𝜃²=4𝑎(1+√−5),(N3)2𝑎𝑧+𝜃𝐴=±𝛼𝜀^𝑘 with 𝜃²=2𝑎(−1−√−5),(N4)2𝑎𝑧+𝜃𝐴=±𝛼𝜀^𝑘 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, Lemma 3 is needed.

Lemma 3. If 𝑚, 𝜀₀, 𝛽, and 𝛽 are defined as before, then 𝐴=(𝛽𝜀^𝑚₀+𝛽𝜀₀^𝑚)/(𝛽+𝛽) is an algebraic number in the field ℚ(√−5).

Proof. Rewriting 𝐴, we have 𝐴=𝛽𝜀^𝑚₀+𝛽𝜀₀^𝑚𝛽+𝛽=𝛽𝜀^𝑚₀+𝛽𝜀₀^𝑚𝛽+𝛽𝛽+𝛽²=𝛽²𝜀^𝑚₀+𝛽²𝜀₀^𝑚+𝛽𝛽𝜀^𝑚₀+𝜀₀^𝑚𝛽²+𝛽²+2𝛽𝛽. (3)
Since 𝛽²+𝛽², 𝛽²𝜀^𝑚₀+𝛽²𝜀₀^𝑚, and 𝜀^𝑚₀+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 𝑥+√6𝑦𝑥−√6𝑦=1+√61−√6. (4) Then we have 𝑥+√6𝑦=±𝜂𝜀^𝑛₀,𝑥−√6𝑦=±𝜂𝜀₀^𝑛,with𝑛∈ℤ. (5)
Adding (5), we get 2𝑥=±𝜂𝜀^𝑛₀+𝜂𝜀₀^𝑛. (6) The solution of (6) is split into four cases.
Case 1. Assume that 𝑛=2𝑚+1 is odd and 2𝑥=𝜂𝜀^𝑛₀+𝜂𝜀₀^𝑛. Then 2𝑥=𝜂𝜀₀^2𝑚+1+𝜂𝜀₀^2𝑚+1=𝛽+𝛽²𝛽𝜀^𝑚₀+𝛽𝜀₀^𝑚𝛽+𝛽²−2𝛽𝛽,with𝛽²=𝜂𝜀₀. (7) Since 𝛽+𝛽²=34+2√−5=1−√−53+√−5²,𝛽𝛽=√−5, (8) by inserting (7) and the second equation of (1) into (6), we get 2𝑎𝑧²−𝜃²𝐴²=2𝑏−√−5, (9) where 𝜃²=1−√−5 and 𝐴=(3+√−5)((𝛽𝜀^𝑚₀+𝛽𝜀₀^𝑚)/(𝛽+𝛽)). We multiply both sides of the equation by 2𝑎 to obtain (2𝑎𝑧)²−𝜃²𝐴²=4𝑎𝑏−√−5=𝛼𝛼₋. (10) Without loss of generality, we also denote 𝜃²=2𝑎(1−√−5) and 𝐴=(3+√−5)((𝛽𝜀^𝑚₀+𝛽𝜀₀^𝑚)/(𝛽+𝛽)). Factoring (9) in the field ℚ(𝜃), we have (𝑁1)2𝑎𝑧+𝜃𝐴=±𝛼𝜀^𝑘,with𝜃²=2𝑎1−√−5,𝑘∈ℤ. (11)Case 2. In another case, when  𝑛=2𝑚 is even and 2𝑥=𝜂𝜀^𝑛₀+𝜂𝜀₀^𝑛, then 2𝑥=𝜂𝜀₀^2𝑚+𝜂𝜀₀^2𝑚=𝛽+𝛽²𝛽𝜀^𝑚₀+𝛽𝜀₀^𝑚𝛽+𝛽²−2𝛽𝛽,with𝛽²=𝜂. (12)
Similarly, we have (2𝑎𝑧)²−𝜃²𝐴²=4𝑎𝑏−√−5=𝛼𝛼₋, (13) where 𝜃²=4𝑎(1+√−5) and 𝐴=(𝛽𝜀^𝑚₀+𝛽𝜀₀^𝑚)/(𝛽+𝛽). Furthermore, we know that 𝑧 satisfies (𝑁2)2𝑎𝑧+𝜃𝐴=±𝛼𝜀^𝑘with𝜃²=4𝑎1+√−5,𝑘∈ℤ. (14)Case 3. Consider 𝑛=2𝑚+1 is odd and 2𝑥=−(𝜂𝜀^𝑛₀+𝜂𝜀₀^𝑛). By the same consideration, we deduce that 𝑧 satisfies (𝑁3)2𝑎𝑧+𝜃𝐴=±𝛼𝜀^𝑘with𝜃²=2𝑎−1−√−5,𝑘∈ℤ, (15) where 𝐴=(3−√−5)((𝛽𝜀^𝑚₀+𝛽𝜀₀^𝑚)/(𝛽+𝛽)).
Case 4. Consider 𝑛=2𝑚 is even and 2𝑥=−(𝜂𝜀^𝑛₀+𝜂𝜀₀^𝑛). Similarly, we know that 𝑧 satisfies (𝑁4)2𝑎𝑧+𝜃𝐴=±𝛼𝜀^𝑘with𝜃²=4𝑎−1+√−5,𝑘∈ℤ, (16) where 𝐴=(𝛽𝜀^𝑚₀+𝛽𝜀₀^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 integer 𝑚≠0 and 𝑖>1, one has 𝑉_𝑝(𝑝^𝑖(^𝑚_𝑖))>𝑉_𝑝(𝑝(^𝑚₁)), where 𝑉_𝑝(⋅) denotes the standard 𝑝-adic valuation and 𝑝 is an odd prime.

Proof. We know 𝑉_𝑝𝑝^𝑖𝑚𝑖−𝑉_𝑝𝑝𝑚1=𝑖−1+𝑉_𝑝𝑚𝑖−𝑉_𝑝𝑚1=𝑖−1+𝑉_𝑝(𝑚−1)⋯(𝑚−𝑖+1)𝑖!=𝑖−1+𝑉_𝑝((𝑚−1)⋯(𝑚−𝑖+1))−𝑉_𝑝(𝑖!)≥𝑖−1−𝑉_𝑝()𝑖𝑖!=𝑖−1−𝑝+𝑖𝑝²𝑖+⋯≥𝑖−1−≥𝑝−1𝑝−2𝑝−1𝑖−1>0,with𝑖>1. (17) Therefore, for 𝑖>1, the 𝑝-adic valuation of the term of 𝑝^𝑖(^𝑚_𝑖) 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 field ℚ(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.

Proof of Corollary. Case 1. Substituting 𝑎=2 and 𝑏=1 into (9), we know that 𝑧 satisfies 4𝑧²−𝜃²𝐴²=2−2√−5=𝛼𝛼₋. (18)
Equation (N1) is reduced to (𝑁5)2𝑧+𝜃𝐴=±𝛼𝜀^𝑘with𝜃²=1−√−5,𝑘∈ℤ. (19)
From Table 1, 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 the 𝑝-adic analysis, a suitable prime 𝑝 is needed. Here, we take 𝑝=7. A straightforward computation shows that 𝜀⁸≡1(mod7). (20) Since (𝛼𝜀⁰)₂≡0(mod7), (𝛼𝜀⁷)₂≡0(mod7), and (𝛼𝜀^𝑖)₂≢0(mod7)(𝑖=2,…,6), we get 𝑘≡0,7(mod8).
(i) Let 𝑘≡0(mod8); since 𝜀⁸≡1(mod7), we obtain 𝜀⁸=1+7𝜉; then ±(2𝑧+𝜃𝐴)=𝛼(1+7𝜉)^𝑚𝑚1=𝛼+7(𝛼𝜉)+7²𝑚2𝛼𝜉²+⋯+7^𝑚𝑚𝑚𝛼𝜉^𝑚, (21) with 𝑘=8𝑚. So we have ±(2𝑧+𝜃𝐴)₂=𝑚1𝛼+7(𝛼𝜉)+⋯+7^𝑚𝑚𝑚𝛼𝜉^𝑚₂. (22) From Lemma 5 and (𝛼𝜉)₂=(−30+30𝜃+60𝜃²+30𝜃³)₂≢0(mod7), we get 𝑚=0 and 𝑧=±3 by working modulo 7^𝑟+2 on formula (13), where 7^𝑟‖𝑚.
(ii) When 𝑘≡7(mod8), we have ±(2𝑧+𝜃𝐴)=𝛼𝜀⁻¹(1+7𝜉)^𝑚=𝛼𝜀⁻¹𝑚1+7𝛼𝜀⁻¹𝜉+⋯+7^𝑚𝑚𝑚𝛼𝜀⁻¹𝜉^𝑚, (23) with 𝑘=−1+8𝑚. So we have ±(2𝑧+𝜃𝐴)₂=𝛼𝜀⁻¹𝑚1+7𝛼𝜀⁻¹𝜉+⋯+7^𝑚𝑚𝑚(𝛼𝜀⁻¹𝜉^𝑚)₂. (24)
Similar deduction shows that 𝑚=0 and 𝑧=±68.
Case 2. Secondly, we similarly consider the following equation: (𝑁6)2𝑧+𝜃𝐴=±𝛼𝜀^𝑘,with𝜃²=2+2√−5,𝑘∈ℤ. (25) 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(mod23). Similarly we have 𝑘≡0,1(mod11); then we can deduce 𝑧=±1 and 𝑧=±91, respectively.
Case 3. Consider the following equation: (𝑁7)2𝑧+𝜃𝐴=±𝛼𝜀^𝑘with𝜃²=−1−√−5,𝑘∈ℤ. (26) We also have 𝜀=−85+34𝜃+30𝜃²+38𝜃³, 𝛼=4+𝜃³, 𝑝=5, and 𝜀⁴≡1(mod5). Direct deductions show that 𝑧=±2 and 𝑧=±11798.
Case 4. The final equation is (𝑁8)2𝑧+𝜃𝐴=±𝛼𝜀^𝑘with𝜃²=−2+2√−5,𝑘∈ℤ. (27) It corresponds to 𝜀=1−2𝜃−𝜃²−(1/2)𝜃³, 𝛼=12−𝜃, 𝑝=13, and 𝜀⁵≡1(mod13). A similar deduction yields 𝑧=0 and 𝑧=±6.
All in all, if (𝑥,𝑦,𝑧) is an integral solution of (1), ±𝑧=3,68,1,91,2,11798,0,6. (28) 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 

Positive 𝑧-value of the solution.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work is partly supported by the Fundamental Research Funds for the Central Universities (no. 2662014QC010) and National Natural Science Foundation of China (61202305).