A parametric family of quartic Thue equations

Abstract

In this paper,we give all primitive solutions of a parameterized family of quartic Thue equations:

　　　　　 x^4-4c{x}^{3}y+(6c+2)x^2{y}^2+4cx{y}^3+y^4=96c+169,c>0.

Keywords

Extension of classical Legendre's theorem / Baker-Wüstholz's theorem / Thue equation

1 Introduction and main conclusions

Let the integers n⩾3,A_i\in \mathbb{Z}(0⩽i⩽n), and M be a given non-zero integer. If

f(x,y)=A_n{x}^n+A_{n-1}{x}^{n-1}y+A_{n-2}{x}^{n-2}{y}^2+\cdots +A_0{y}^n

is a homogeneous irreducible polynomial of degree n in two variables, then Thue equation is a Diophantus equation of one class which has the form

(1)\begin{array}{l}f(x,y)=M.\end{array}

In 1909, Thue [17] proved that equation (1) has a finite number of solutions, hence the name of equation (1). Thue's proof is based on the approximation theorem, i.e., let \alpha be the algebraic number of degrees n and n⩾2,\epsilon >0, then there exists a constant c_1(\alpha ,\epsilon ) such that for all p\in \mathbb{Z},q\in \mathbb{N}, the following inequality holds:

\left|\alpha -\frac{p}{q}\right|⩾\frac{c_1(\alpha ,\epsilon )}{q^{\frac{n}{2}+1+\epsilon }}.

However, there is no effective algorithm for solving the upper bound problem. In 1968, Baker [1] gave a valid bound based on the log-linear theory of algebraic numbers. In recent years, general and powerful methods inspired by Baker's approach have been developed to find the explicit solutions of the Thue equation [5, 15, 18]. Until 1989, Tzanakis and de Weger [18] gave an upper bound of |x| and |y|, but this upper bound was astronomical. In 1990, Thomas [16] was the first to study a parametric family of cubic Thue equations with positive discriminant. Since then, many types of Thue equations with parameters have been studied, as summarized in detail in the literature [12].

Interestingly, by the Tzanakis method [19], it is possible to transform a quartic Thue equation into the corresponding simultaneous quadratic Diophantus equation. Dujella [7, 8] used this method to study a quartic Thue equation with parameters like

(2)x^4-4c{x}^{3}y+(6c+2)x^2{y}^2+4cx{y}^3+y^4=\mu ,c⩾0.

It actually gives the number of positive integer solutions (x,y) to this equation for \mu =1,-2c,2c+1,-6c+1,6c+4.

In this paper, we delve into quadratic Thue equations of this class, using the extended classical Legendre theorem to try to find solutions of equation (2) when \mu takes larger values. In our discussion, we follow the techniques used in [6-8], such as the “congruence method” for solving the simultaneous Pell equation, a theorem on the simultaneous approximation of algebraic numbers in [4], and the Baker−Davenport reduction method, among others.

Since f(x,y)=x^4-4c{x}^{3}y+(6c+2)x^2{y}^2+4cx{y}^3+y^4 is a homogeneous polynomial. Therefore, when the equation f(x,y)=\mu is considered, if the integer solution (x,y) is nontrivial solution and \gcd (x,y)=d>1, then d^4\mid \mu. Conversely, if \mu does not contain the factor like d^4(d>1), we know that the solution (x,y) of the equation satisfies: \gcd (x,y)=1; if \mu contains the factor like d^4(d>1), then the equation degenerates to the solution of f(x,y)=\frac{\mu }{d^4}. For convenience, we only consider solutions that satisfied \gcd (x,y)=1 and call these solutions the primitive solutions of the equation. Since f(x,y)=f(-x,-y)=f(y,-x)=f(-y,x), only positive integer solutions need to be considered thereafter.

Theorem 1　 Let c be a non-zero integer, for the equation

(3)\begin{array}{l}f(x,y)=x^4-4c{x}^{3}y+(6c+2)x^2{y}^2+4cx{y}^3+y^4=96c+169,c⩾0,\end{array}

(a) when c\ne 0,2, the primitive positive integer solution of the equation has (x,y)=(3,2);

(b) when c=0, the primitive positive integer solutions of the equation have (x,y)=(2,3),(3,2);

(c) when c=2, the primitive positive integer solution of the equation has (x,y)=\left(2y_n+\frac{1}{2}{x}_n,\frac{1}{2\sqrt{5}}{y}_n\right), where

\begin{array}{rl}{x}_n& =(8+3\lambda \sqrt{5})(9+4\sqrt{5}{)}^n+(8-3\lambda \sqrt{5})(9-4\sqrt{5}{)}^n,\\ y_n& =(8+3\lambda \sqrt{5})(9+4\sqrt{5}{)}^n-(8-3\lambda \sqrt{5})(9-4\sqrt{5}{)}^n,\end{array}

or

\begin{array}{rl}{x}_n& =(\lambda +2\sqrt{5})(9+4\sqrt{5}{)}^n+(-\lambda +2\sqrt{5})(9-4\sqrt{5}{)}^n,\\ y_n& =(\lambda +2\sqrt{5})(9+4\sqrt{5}{)}^n-(-\lambda +2\sqrt{5})(9-4\sqrt{5}{)}^n.\end{array}

and \lambda \in \{1,-1\},n\in \mathbb{N}, when \lambda =1,n⩾0, when \lambda =-1,n⩾1.

2 The case of 0⩽c⩽4 and c\ne 3

We will give the primitive positive integer solution of the equation when 0⩽c⩽4 and c\ne 3.

When c=0, f(x,y)=x^4+2x^2{y}^2+y^4={\left(x^2+y^2\right)}^2=169, so x^2+y^2={13}^2, and it is easy to see that the solution of the equation is (x,y)=(2,3),(3,2).

When c=1, f(x,y)=x^4-4x^{3}y+8x^2{y}^2-4x{y}^3+y^4={\left(x^2-2xy+y^2\right)}^2+6x^2{y}^2=265, we observe that 6x^2{y}^2⩽265, so (x,y)=(3,2).

When c=2, f(x,y)={\left(x^2-4xy-y^2\right)}^2={19}^2, so we have x^2-4xy-y^2=\pm 19, i.e., (x-2y{)}^2-5y^2=\pm 19. There are two cases:

(1) when (x-2y{)}^2-5y^2=19, (x-2y)+y\sqrt{5}=(8\pm 3\sqrt{5})(9+4\sqrt{5}{)}^n;

(2) when (x-2y{)}^2-5y^2=-19, (x-2y)+y\sqrt{5}=(\pm 1+2\sqrt{5})(9+4\sqrt{5}{)}^n.

The solution of the above case is easy to represent, and we know that there are an infinite number of solutions.

When c=4, f(x,y)={\left(x^2-8xy-y^2\right)}^2-(6xy{)}^2=553, and \left(x^2-14xy-y^2\right)\left(x^2-2xy-y^2\right)= 7\times 79, from the decomposition of 553, we have 12xy=72, or 12xy=554, and we know that only (3,2) is the only solution obtained.

From now on, we can set c⩾3 and c\ne 4.

3 Application of the Tzanakis method

Tzanakis considers the quadratic Thue equation

(4)\begin{array}{c}f(x,y)=\mu ,\\ f(x,y)=a_0{x}^4+4a_1{x}^{3}y+6a_2{x}^2{y}^2+4a_{3}x{y}^3+a_4{y}^4\in \mathbb{Z}[x,y],a_0>0.\end{array}

The corresponding cubic resolvent is

(5)4{\rho }^3-g_2\rho -g_3=0,

with solutions {\rho }_1,{\rho }_2,{\rho }_3, and invariants g_2,g_3 defined as:

g_2=a_0{a}_4-4a_1{a}_3+3a_2^2\in \frac{1}{12}\mathbb{Z},g_3=\left|\begin{array}{lll}{a}_0& a_1& a_2\\ a_1& a_2& a_3\\ a_2& a_3& a_4\end{array}\right|\in \frac{1}{432}\mathbb{Z}.

The Tzanakis method is applied on the assumption that its corresponding quartic domain K is composed of two real quadratic domains, if and only if

(6)\frac{a_1^2}{a_0}-a_2⩾max\left\{{\rho }_1,{\rho }_2,{\rho }_3\right\}.

For f(x,y)=x^4-4c{x}^{3}y+(6c+2)x^2{y}^2+4cx{y}^3+y^4, we can get

\begin{array}{c}{g}_2=\frac{1}{3}\left(21c^2+6c+4\right),g_3=-\frac{1}{27}\left(81c^3+99c^2-18c-8\right),\\ {\rho }_1=\frac{1}{2}c+\frac{2}{3},{\rho }_2=c-\frac{1}{3},{\rho }_3=-\frac{3}{2}c-\frac{1}{3}.\end{array}

Therefore, condition (6) is satisfied, so that the Tzanakis method can be applied to obtain

(7)\begin{array}{l}\{\begin{array}{l}(2c+1)U^2-2c{V}^2=\mu ,\\ (c-2)U^2-c{Z}^2=-2\mu ,\end{array}\end{array}

where

U=x^2+y^2,V=x^2+xy-y^2,Z=-x^2+4xy+y^2.

Thus, by the Tzanakis method, equation (3) in the theorem has been transformed into a problem of solving a system of simultaneous Pell equations.

4 Application of the extended classical Legendre theorem

We will apply the extended classical Legendre theorem to give the primitive positive integer solution (U,V) of the equation

(8)\begin{array}{l}(2c+1)U^2-2c{V}^2=96c+169,\end{array}

where c⩾3.

The expansion of the simple continued fraction [13] of a quadratic irrational number \alpha =\frac{A+\sqrt{d}}{B} is periodic. So we suppose that \alpha has the following representation:

\alpha =\left[a_0,a_1,a_2,\dots ,a_m,\dots \right]=\left[a_0,\overline{a_1,a_2,\dots ,a_k}\right],

where the integers m>0,k>0, and k are the smallest positive periods of this continued fractional expansion. Let \frac{p_m}{q_m} be the mth asymptotic fraction of \alpha, and p_{-1}=1,q_{-1}=0. Then let \frac{P_m+\sqrt{d}}{Q_m} be the mth perfect quotient of \alpha, and P_0=A,Q_0=B. We have

a_m=\lfloor \frac{P_m+\sqrt{d}}{Q_m}\rfloor ,\frac{P_m+\sqrt{d}}{Q_m}=a_m+{\left(\frac{P_{m+1}+\sqrt{d}}{Q_{m+1}}\right)}^{-1},m⩾0.

(\lfloor X\rfloor denotes the largest positive integer that does not exceeding X.)

From the above discussion, we have \sqrt{\frac{2c+1}{2c}}=[1,\overline{4c,2}], and accordingly, the following relationship holds:

(9)\begin{array}{rl}& P_0=0,Q_0=2c,d=2c(2c+1),P_m=2c,Q_{2m-1}=1,Q_{2m}=2c,\\ & p_{-1}=1,p_0=1,p_{2m-1}=4c{p}_{2m-2}+p_{2m-3},p_{2m}=2p_{2m-1}+p_{2m-2},\\ & q_{-1}=0,q_0=1,q_{2m-1}=4c{q}_{2m-2}+q_{2m-3},q_{2m}=2q_{2m-1}+q_{2m-2}.\end{array}

The classical Legendre theorem [13] is: if , then \frac{e}{f} is an asymptotic fraction of \alpha. Fatou and Worley both generalize the classical Legendre theorem. In 1904, Fatou [10] proved that if , then we have \frac{e}{f}=\frac{p_m}{q_m}, or \frac{p_{m+1}\pm p_m}{q_{m+1}\pm q_m}. In 1981, Worley [20] proved that if , then we have

\frac{e}{f}=\frac{p_m}{q_m},\frac{p_{m+1}\pm p_m}{q_{m+1}\pm q_m},\frac{2p_{m+1}\pm p_m}{2q_{m+1}\pm q_m},\frac{3p_{m+1}+p_m}{3q_{m+1}+q_m},\frac{p_{m+1}\pm 2p_m}{q_{m+1}\pm 2q_m}\text{, or}\frac{p_{m+1}-3p_m}{q_{m+1}-3q_m}\text{.}

In 2004, Dujella [9] gave a better conclusion of a generalization of the classical Legendre theorem.

Lemma 1　 Let \alpha be an irrational number and \frac{p_m}{q_m} be the mth asymptotic fraction of \alpha. If e and f are non-zero integers that are relatively prime and satisfy the following inequality

(10)

where C is a positive real number. Then there exist non-negative integers m,r,s which satisfying such that

(e,f)=\left(r{p}_{m+1}\pm s{p}_m,r{q}_{m+1}\pm s{q}_m\right).

Lemma 2　 Suppose that (U,V) is any primitive positive integer solution of the equation

(11)\begin{array}{l}(2c+1)U^2-2c{V}^2=\mu ,\mu \in \mathbb{Z},\end{array}

and \frac{p_m}{q_m} is the mth asymptotic fraction of \sqrt{\frac{2c+1}{2c}}. Then V=r{p}_{m+1}\pm s{p}_m,U=r{q}_{m+1}\pm s{q}_m, where r,s,m are non-negative integers, C is a positive real number, and the relation is satisfied.

Proof　It is easy to see that \sqrt{\frac{2c+1}{2c}}>1,\frac{V}{U}>0, thus we have

(12)

Taking C=\frac{\mu }{2c}, from Lemma 1 we know that V=r{p}_{m+1}\pm s{p}_m,U=r{q}_{m+1}\pm s{q}_m, which gives us the proof of the Lemma.

Lemma 3　 Let non-negative integers r,s, satisfy , r and s are relatively prime, and \frac{p_m}{q_m} is the mth asymptotic fraction of \sqrt{\frac{2c+1}{2c}}, P_m,Q_m as defined in equation (9). If (-1{)}^m\left(s^2{Q}_{m+1}\pm 2rs{P}_{m+2}-r^2{Q}_{m+2}\right)=\mu, then (U,V)=\left(r{q}_{m+1}\pm s{q}_m,r{p}_{m+1}\pm s{p}_m\right) is the primitive solution of equation (11).

Proof　By

\frac{\sqrt{2c+1}}{\sqrt{2c}}=\frac{\left(P_{m+1}+\sqrt{2c(2c+1)}\right)p_m+Q_{m+1}{p}_{m-1}}{\left(P_{m+1}+\sqrt{2c(2c+1)}\right)q_m+Q_{m+1}{q}_{m-1}},

simplifying and comparing the coefficients on both sides of the equation, we have (2c+1)q_m=P_{m+1}{p}_m+Q_{m+1}{p}_{m-1},2c{p}_m=P_{m+1}{q}_m+Q_{m+1}{q}_{m-1}. So we have (2c+1)q_m^2-2c{p}_m^2=Q_{m+1}\left(p_{m-1}{q}_m-p_m{q}_{m-1}\right)=(-1{)}^m{Q}_{m+1}, (2c+1)q_{m-1}{q}_m-2c{p}_{m-1}{p}_m=(-1{)}^{m+1}{P}_{m+1}. (According chapter 10 of the literature [13], p_{m-1}{q}_m-p_m{q}_{m-1}=(-1{)}^m).

Hence

\begin{array}{rl}(2c+1)U^2-2c{V}^2& =(2c+1){\left(r{q}_{m+1}\pm s{q}_m\right)}^2-2c{\left(r{p}_{m+1}\pm s{p}_m\right)}^2\\ & =r^2\left((2c+1)q_{m+1}^2-2c{p}_{m+1}^2\right)\pm 2rs\left((2c+1)q_{m+1}{q}_m-2c{p}_{m+1}{p}_m\right)\\ & +s^2\left((2c+1)q_m^2-2c{p}_m^2\right)\\ & =r^2(-1{)}^{m+1}{Q}_{m+2}\pm 2rs(-1{)}^{m+2}{P}_{m+2}+s^2(-1{)}^m{Q}_{m+1}\\ & =(-1{)}^m\left(s^2{Q}_{m+1}\pm 2rs{P}_{m+2}-r^2{Q}_{m+2}\right).\end{array}

Thus, when (-1{)}^m\left(s^2{Q}_{m+1}\pm 2rs{P}_{m+2}-r^2{Q}_{m+2}\right)=\mu, (U,V) is the primitive solution of equation (11).

For equation (8), in fact, if (U,V)\ne (13,11), it is easy to prove \frac{V}{U}⩾1 by the converse method, and by Lemma 2, \left|\sqrt{\frac{2c+1}{2c}}-\frac{V}{U}\right|⩽ \frac{96c+169}{4c{U}^2}, so take C=38. From Lemma 3 and equation (9), the following lemma is obtained by direct calculation.

Lemma 4　 The primitive positive integer solution (U,V) of equation (8) is

\left(2q_{2m+1}+13q_{2m},2p_{2m+1}+13p_{2m}\right),\left(13q_{2m}-2q_{2m-1},13p_{2m}-2p_{2m-1}\right),

where m⩾0,(13,11) is contained in the associative class of the latter solution.

From (9) we have

\begin{array}{rl}& p_{2m+1}=(8c+2)p_{2m-1}-p_{2m-3},p_{2m}=(8c+2)p_{2m-2}-p_{2m-4},\\ & q_{2m+1}=(8c+2)q_{2m-1}-q_{2m-3},q_{2m}=(8c+2)q_{2m-2}-q_{2m-4},\end{array}

that is, we have p_{m+4}=(8c+2)p_{m+2}-p_m,q_{m+4}=(8c+2)q_{m+2}-q_m. By Lemma 4, U=2q_{2m+1}+13q_{2m} or 13q_{2m}-2q_{2m-1}, i.e., there exists a non-negative integer m such that U=v_m, or U=v_m^{\mathrm{\prime }}, the sequences \left\{v_m\right\} and \left\{v_m^{\mathrm{\prime }}\right\} are defined as follows:

(13)\begin{array}{rl}& v_0=13,v_1=96c+13,v_{m+2}=(8c+2)v_{m+1}-v_m,\\ & v_0^{\mathrm{\prime }}=13,v_1^{\mathrm{\prime }}=8c+13,v_{m+2}^{\mathrm{\prime }}=(8c+2)v_{m+1}^{\mathrm{\prime }}-v_m^{\mathrm{\prime }}.\end{array}

Let {\epsilon }_1=4c+1+2\sqrt{2c(2c+1)} and {\overline{\epsilon }}_1=4c+1-2\sqrt{2c(2c+1)}. {\epsilon }_1,{\overline{\epsilon }}_1 are all the units in the ring \mathbb{Z}\sqrt{2c(2c+1)}. Solving the recurrence relation (13) we get

(14)v_m=\frac{1}{2\sqrt{2c+1}}\left[(13\sqrt{2c+1}+11\sqrt{2c}){\epsilon }_1^m+(13\sqrt{2c+1}-11\sqrt{2c}){\overline{\epsilon }}_1^m\right],

(15)v_m^{\mathrm{\prime }}=\frac{1}{2\sqrt{2c+1}}\left[(13\sqrt{2c+1}-11\sqrt{2c}){\epsilon }_1^m+(13\sqrt{2c+1}+11\sqrt{2c}){\overline{\epsilon }}_1^m\right].

Consider the equation set

(16)\begin{array}{l}\{\begin{array}{l}(2c+1)U^2-2c{V}^2=96c+169,\\ (c-2)U^2-c{Z}^2=-2(96c+169).\end{array}\end{array}

If (U,V,Z) is a positive integer solution of (16). By Lemma 4, the solution (U,V) of equation (8) has the following form:

U\sqrt{2c+1}+V\sqrt{2c}=(13\sqrt{2c+1}\pm 11\sqrt{2c})(4c+1+2\sqrt{2c(2c+1)}{)}^n,

and (U,Z) is the solution of the equation (c-2)U^2-c{Z}^2=-2(96c+169), which is given by the following form:

U\sqrt{c-2}+Z\sqrt{c}=\left(U^{\ast }\sqrt{c-2}+Z^{\ast }\sqrt{c}\right)(c+1+\sqrt{c(c-2)}{)}^n,

\left(U^{\ast },Z^{\ast }\right) is a fundamental solution in a certain associative class of solutions.

It is easy to verify that when k is even, q_k\equiv 1(\text{mod}c); when k is odd, q_k\equiv 0(\text{mod}c). By Lemma 4, U=2q_{2m+1}+13q_{2m} or 13q_{2m}-2q_{2m-1}, so |U|\equiv 13(\text{mod}c). It is easy to see that |U|\equiv \left|U^{\ast }\right|(\text{mod}c), with \left|U^{\ast }\right|\equiv 13(\text{mod}c). Also according to Theorem 1 on page 29 of Ref. [14], {\left|U^{\ast }\right|}^2⩽96c+169. When c⩾3, , we get \left|U^{\ast }\right|=13, or \left|U^{\ast }\right|=c+13. So Z^{\ast }=19. Then there exists a non-negative integer m such that U=w_m, or U=w_m^{\mathrm{\prime }}, where the sequences \left\{w_m\right\} and \left\{w_m^{\mathrm{\prime }}\right\} are defined as follows:

(17)\begin{array}{rl}& w_0=13,w_1=32c-13,w_{m+2}=(2c-2)w_{m+1}-w_m,\\ & w_0^{\mathrm{\prime }}=-13,w_1^{\mathrm{\prime }}=6c+13,w_{m+2}^{\mathrm{\prime }}=(2c-2)w_{m+1}^{\mathrm{\prime }}-w_m^{\mathrm{\prime }}.\end{array}

Let {\epsilon }_2=c-1+\sqrt{c(c-2)} and {\overline{\epsilon }}_2=c-1-\sqrt{c(c-2)}.{\epsilon }_2,{\overline{\epsilon }}_2 are units in the ring \mathbb{Z}\sqrt{c(c-2)}. Solving the recurrence relation (17), we get

(18)w_m=\frac{1}{2\sqrt{c-2}}\left[(13\sqrt{c-2}+19\sqrt{c}){\epsilon }_2^m-(-13\sqrt{c-2}+19\sqrt{c}){\overline{\epsilon }}_2^m\right],

(19)w_m^{\mathrm{\prime }}=\frac{1}{2\sqrt{c-2}}\left[(-13\sqrt{c-2}+19\sqrt{c}){\epsilon }_2^m-(13\sqrt{c-2}+19\sqrt{c}){\overline{\epsilon }}_2^m\right].

Thus the following conclusion holds.

Lemma 5　 Let (U,V,Z) be a positive integer solution of the equation set (16). Then there exist non-negative integers m and n such that

U=v_m=w_n,\mathrm{o}\mathrm{r}\mathrm{U}={\mathrm{v}}_{\mathrm{m}}={\mathrm{w}}_{\mathrm{n}}^{\mathrm{\prime }},\mathrm{o}\mathrm{r}\mathrm{U}={\mathrm{v}}_{\mathrm{m}}^{\mathrm{\prime }}={\mathrm{w}}_{\mathrm{n}},\mathrm{o}\mathrm{r}\mathrm{U}={\mathrm{v}}_{\mathrm{m}}^{\mathrm{\prime }}={\mathrm{w}}_{\mathrm{n}}^{\mathrm{\prime }}.

5 Congruence relation

In this section, some congruence relations are obtained for the sequences \left\{v_m\right\},\left\{v_m^{\mathrm{\prime }}\right\},\left\{w_n\right\},\left\{w_n^{\mathrm{\prime }}\right\}.

Lemma 6　 Let \left\{v_m\right\},\left\{v_m^{\mathrm{\prime }}\right\},\left\{w_n\right\},\left\{w_n^{\mathrm{\prime }}\right\} as shown in (13), (17). Then for all non-negative integers m,n, we have

(20)\begin{array}{l}{v}_m\equiv 13+4m(13m+11)c\left(\text{mod}64c^2\right),\end{array}

(21)\begin{array}{l}{v}_m^{\mathrm{\prime }}\equiv 13+4m(13m-11)c\left(\text{mod}64c^2\right),\end{array}

(22)\begin{array}{l}{w}_n\equiv (-1{)}^n(13-n(13n+19)c)\left(\text{mod}4c^2\right),\end{array}

(23)\begin{array}{l}{w}_n^{\mathrm{\prime }}\equiv (-1{)}^{n-1}(13-n(13n-19)c)\left(\text{mod}4c^2\right).\end{array}

Proof　Prove by mathematical induction. Only prove equation (20), the rest of the formulae can be proved similarly. Equation (20) clearly holds when m=0, 1,v_0=13,v_1=13+4(13\times 1+11)c. Suppose that equation (20) holds for all 0⩽m⩽k+1. For v_{k+2}, we have

\begin{array}{rl}{v}_{k+2}& =(8c+2)v_{k+1}-v_k\\ & =(8c+2)[13+4(k+1)(13(k+1)+11)c]-(13+4k(13k+11)c)\\ & =13+32c^2\left(24+37k+13k^2\right)+4c\left(74+63k+13k^2\right)\\ & =13+32c^2[24+(37+13k)k]+4c(k+2)(13k+37),\end{array}

and 24+(37+13k)k\equiv 0(\text{mod}2), so v_{k+2}\equiv 13+4(k+2)(13(k+2)+11)c(\text{mod}64c^2). Thus, for m=k+2, equation (20) holds. In summary, (20) holds.

Let us first discuss the case v_m=w_n. Assume that the integers m⩾1,n⩾1. In particular, v_m=w_n\left(\text{mod}4c^2\right). By Lemma 6, 1\equiv (-1{)}^n(1-n(n+1)c)\equiv (-1{)}^n(\text{mod}2), so n is even.

Suppose n⩽\sqrt{\frac{4c}{65}}-\frac{63}{130}, from (14),(18) and v_m=w_n, we have m⩽n. From v_m=w_n\left(\text{mod}4c^2\right), Lemma 6, we have 4m(13m+1)\equiv -n(13n+19)(\text{mod}4c). Let the integers

\mathcal{A}=4m(13m+1)+n(13n+19).

We have . So when m⩾1 and n⩾1, . Combining this with the discussion above, we have \mathcal{A}\equiv 0(\text{mod}4c), and we get the contradiction. So n>\sqrt{\frac{4c}{65}}-\frac{63}{130}.

Similarly, discussing the equations v_m=w_n^{\mathrm{\prime }},v_m^{\mathrm{\prime }}=w_n,v_m^{\mathrm{\prime }}=w_n^{\mathrm{\prime }}, we come to a similar conclusion.

Lemma 7　 Let \left\{v_m\right\},\left\{v_m^{\mathrm{\prime }}\right\},\left\{w_n\right\},\left\{w_n^{\mathrm{\prime }}\right\} be as shown in (13), (17). Then for all positive integers m,n:

(i) if v_m=w_n,n>\sqrt{\frac{4c}{65}}-\frac{63}{130}, and n is an even number;

(ii) if v_m=w_n^{\mathrm{\prime }},n>\sqrt{\frac{4c}{65}}-\frac{25}{130}, and n is an odd number;

(iii) if v_m^{\mathrm{\prime }}=w_n,n>\sqrt{\frac{4c}{65}}+\frac{25}{130}, and n is an even number;

(iv) if v_m^{\mathrm{\prime }}=w_n^{\mathrm{\prime }},n>\sqrt{\frac{4c}{65}}+\frac{63}{130}, and n is an odd number.

6 Simultaneous Diophantus Approximation and the Baker-Davenport reduction method

Let

{\theta }_1=\sqrt{\frac{2c+1}{2c}},{\theta }_2=\sqrt{\frac{c-2}{c}},

{\theta }_1 and {\theta }_2 are two quadratic irrational numbers that are both very close to 1.

Lemma 8　 Let (U,V,Z) be a positive integer solution of the equation set (16) with c⩾5, which satisfies the following inequality:

Proof　It has been proved before that . From , we have \frac{Z}{U}⩾\sqrt{\frac{c-2}{c}}>0. Thus we have . When c⩾5, .

Lemma 9 (see [4], Theorem 3.2)　 Let a_i,p_i,q be integers, 0⩽i⩽2, and be a non-zero integer, and there exists some j,0⩽j⩽2,a_j=0. N>M^9, where M=\underset{0⩽i⩽2}{max}\left\{\left|a_i\right|\right\}. Then we have

\underset{0⩽i⩽2}{max}\left\{\left|\sqrt{1+\frac{a_i}{N}}-\frac{p_i}{q}\right|\right\}>(130N\gamma {)}^{-1}{q}^{-\lambda },

where

and (\log (X) denotes the natural logarithm of X),

Lemma 10　 Let c be a positive integer. Then (i) when the only solution of c⩾244461,v_m=w_n is (m,n)=(0,0);(\mathrm{i}\mathrm{i}) when c⩾244054, the only solution of v_m^{\mathrm{\prime }}=w_n is (m,n)=(0,0); (iii) when c⩾244285,v_m=w_n^{\mathrm{\prime }} has no solution; (iv) when c⩾243880,v_m^{\mathrm{\prime }}=w_n^{\mathrm{\prime }} has no solution.

Proof　Take a_0=-4,a_1=0,a_2=1,p_0=Z,p_1=U,p_2=V,q=U, N=2c,M= \underset{0⩽i⩽2}{max}\left\{\left|a_i\right|\right\}=4, and N>M^9. Then we have c>131072. By Lemma 9, we have

It is easy to see that , so we have \gamma =\frac{400}{9}. From , we have \lambda =1+\frac{\log \left(33\cdot (2c)\cdot \frac{400}{9}\right)}{\log \left(1.7\cdot \frac{4c^2}{400}\right)}. So . Also c⩾172550,2-\lambda >0. Therefore,

(24)

Further,

(25)

From (18) and (19), and using mathematical induction (for n) can prove the inequality

(26)w_n>w_n^{\mathrm{\prime }}>(c-1+\sqrt{c(c-2)}{)}^n>(2c-3{)}^n.

First consider the case where v_m=w_n. By Lemma 7, n>\sqrt{\frac{4c}{65}}-\frac{63}{130}, so when n⩾1 and m⩾1,

(27)\log U>n\log (2c-3)>\left(\sqrt{\frac{4c}{65}}-\frac{63}{130}\right)\log (2c-3).

From (25) and (27) we have

(28)

Thus, when c⩾243880, equation (28) does not hold. Similarly, we obtain results for the remaining cases.

We will use the Baker-Davenport reduction method to solve (16), let \mathcal{C}=\{244460, 244053,244284,243879\}.

Lemma 11　 Let c be an integer, and \left\{v_m\right\},\left\{v_m^{\mathrm{\prime }}\right\},\left\{w_n\right\},\left\{w_n^{\mathrm{\prime }}\right\} be as shown in (13), (17). If v_m=w_n, v_m^{\mathrm{\prime }}=w_n,v_m=w_n^{\mathrm{\prime }}, or v_m^{\mathrm{\prime }}=w_n^{\mathrm{\prime }},m⩾2,n⩾2, then when c=3, we have

(29)

when 5⩽c⩽M_c,M_c\in \mathcal{C},

(30)

where {\epsilon }_1,{\epsilon }_2\in \{1,-1\}.

Proof　 Define

\begin{array}{c}P=\frac{1}{\sqrt{2c+1}}\left[\left(13\sqrt{2c+1}+11{\epsilon }_2\sqrt{2c}\right)(4c+1+2\sqrt{2c(2c+1)}{)}^m\right],\\ Q=\frac{1}{\sqrt{c-2}}\left[\left(19\sqrt{c}+11{\epsilon }_1\sqrt{c-2}\right)(c-1+\sqrt{c(c-2)}{)}^n\right].\end{array}

From this, we have P+P^{-1}\cdot \frac{96c+169}{2c+1}=Q-Q^{-1}\cdot \frac{2(96c+169)}{c-2}, obviously Q>P, and

Consider c\ne 3 at first. By c⩾5, we know

so \frac{5c(96c+169)}{(2c+1)(c-2)} is a decreasing function, thus , i.e., .

Also by m,n⩾1,c⩾5, we have P>8c+1⩾41, i.e., P⩾42, thus

We start by proving an important estimator:

(31)

From Taylor’s expansion, when , we have

Therefore, the estimate (31) holds. It can also be seen that is crucial for our derivation, in practice, we take k=3.

So, using (31), we can derive

When c=3, by the assumption that m⩾2, we get

Using (31), we get

End of proof.

Baker−Wüstholz’s Theorem [3, 11]　 Let {\alpha }_1,{\alpha }_2,\dots ,{\alpha }_l be l algebraic numbers and b_1,b_2,\dots ,b_l be l rational integers that are not all 0. Let

\mathrm{\Lambda }=b_1\log {\alpha }_1+b_2\log {\alpha }_2+\cdots +b_l\log {\alpha }_l.

Then

\log \mathrm{\Lambda }⩾-18(l+1)!l^{l+1}(32d{)}^{l+2}{h}^{\mathrm{\prime }}({\alpha }_1)h^{\mathrm{\prime }}({\alpha }_2)\cdots h^{\mathrm{\prime }}({\alpha }_l)\log (2ld)\log B,

where

\begin{array}{c}B=max\left\{\left|b_1\right|,\dots ,\left|b_l\right|\right\},d=\left[\mathbb{Q}\left({\alpha }_1,{\alpha }_2,{\alpha }_3\right):\mathbb{Q}\right],\\ h^{\mathrm{\prime }}(\alpha )=\frac{1}{d}max\{h(\alpha ),|\log \alpha |,1\}.\end{array}

h(\alpha ) is called the Weil height of the standard logarithm of \alpha,

h(\alpha )=\log a_0+\sum _{i=1}^{\mathrm{deg}\alpha }\log max\left\{1,\left|{\alpha }^{(i)}\right|\right\},

a_0\in \mathbb{Z} is the first term coefficient of the minimum polynomial of \alpha, {\alpha }^{(i)} is the conjugate of \alpha, 1⩽i⩽\mathrm{deg}\alpha.

We apply the Baker−Wüstholz’s Theorem to (30) of Lemma 11.

Take l=3,B=n,{\alpha }_1=c-1+\sqrt{c(c-2)},{\alpha }_2=4c+1+2\sqrt{2c(2c+1))},

{\alpha }_3=\frac{\sqrt{2c+1}\left(19\sqrt{c}+13{\epsilon }_1\sqrt{c-2}\right)}{\sqrt{c-2}\left(13\sqrt{2c+1}+11{\epsilon }_2\sqrt{2c}\right)}.

Calculating the corresponding h^{\mathrm{\prime }}(\alpha ), we can get , h^{\mathrm{\prime }}\left({\alpha }_2\right)=\frac{1}{2}\log \left(4c+1+2\sqrt{2c(2c+1)}\right).

Firstly consider 5⩽c⩽M_c,M_c\in \mathcal{C}. Let {\alpha }_3^{(1)},{\alpha }_3^{(2)},{\alpha }_3^{(3)},{\alpha }_3^{(4)} be the four conjugates of {\alpha }_3. Then we have

Since a_0⩽(c-2{)}^2(169+96c), such that

\begin{array}{rl}{h}^{\mathrm{\prime }}\left({\alpha }_3\right)& =\frac{1}{4}max\left\{h\left({\alpha }_3\right),\left|\log {\alpha }_3\right|,1\right\}\\ & =\frac{1}{4}h\left({\alpha }_3\right)=\frac{1}{4}\left(|{\alpha }_3^{(1)}|\cdot |{\alpha }_3^{(3)}|\cdot |{\alpha }_3^{(4)}|\right)\\ & ⩽\log \left[2(c-2)(2c+1)(169+96c)\frac{19\sqrt{c}+13\sqrt{c-2}}{13\sqrt{2c+1}-11\sqrt{2c}}\right].\end{array}

On the one hand, assuming m⩾2, from Lemma 11 we know that

On the other hand, by the Baker−Wüstholz’s Theorem, we have

\log \mathrm{\Lambda }⩾-3.821024857\times {10}^{15}\cdot \log n\cdot h^{\mathrm{\prime }}\left({\alpha }_1\right)\cdot h^{\mathrm{\prime }}\left({\alpha }_2\right)\cdot h^{\mathrm{\prime }}\left({\alpha }_3\right).

Thus, .

(32)

Known by m⩾2,c⩾5, .

\log \frac{\sqrt{2c+1}\left(19\sqrt{c}+13{\epsilon }_1\sqrt{c-2}\right)}{\sqrt{c-2}\left(13\sqrt{2c+1}+11{\epsilon }_2\sqrt{2c}\right)}>\log \frac{\sqrt{2c+1}(19\sqrt{c}-13\sqrt{c-2})}{\sqrt{c-2}(13\sqrt{2c+1}+11\sqrt{2c})},

and .

Thus, \log \frac{\sqrt{2c+1}\left(19\sqrt{c}+13{\epsilon }_1\sqrt{c-2}\right)}{\sqrt{c-2}\left(13\sqrt{2c+1}+11{\epsilon }_2\sqrt{2c}\right)}>\log 0.25⩾-1.3862943611199.

From Lemma 11, we have

Thus,

(33)

Combining this we get . Solving this inequality, we get .

Discussing c=3 in the same way. We have , and solving this inequality, we have .

A variant of the Baker−Davenport reduction method [2, 11] is used below to reduce this upper bound on n.

Lemma 12 [6-8]　 Let M be a positive integer. \frac{p}{q} is an asymptotic fraction of the expansion of \kappa continued fraction, satisfying q>10M, and \epsilon =\Vert \mu q\Vert -M\cdot \Vert \kappa q\Vert, where \Vert \vartheta \Vert denotes the distance from \vartheta to its nearest integer. If \epsilon >0, then inequality satisfying

\frac{\log \left(\frac{Aq}{\epsilon }\right)}{\log B}⩽m⩽M

has no integer solution (m,n).

Using Lemma 12, let

\kappa =\frac{\log {\alpha }_2}{\log {\alpha }_1},\mu =\frac{\log {\alpha }_3}{\log {\alpha }_1},A=\frac{\overline{C}}{\log {\alpha }_1},B=(4c+1+2\sqrt{2c(2c+1)}{)}^2.

When c=3, \overline{C}=123.5128,M=7.08626\times {10}^{18}; when c⩾5, \overline{C}=128.8385,M= 6.663903617\times {10}^{18}.

We use the Pari/Gp software (version 2.4.1) to implement Lemma 12. Write the relevant programs and set the precision. “\mathrm{\setminus }p\mathrm{\setminus }4000”; the function used for continued fraction expansion: contfrac, and setting the parameter of number of items to lmax=88; this ensures that, for four cases: v_m=w_n,v_m^{\mathrm{\prime }}=w_n,v_m=w_n^{\mathrm{\prime }}, or v_m^{\mathrm{\prime }}=w_n^{\mathrm{\prime }}, and the corresponding c:c=3, or 5⩽ c⩽M_c,M_c\in \mathcal{C}, all with q>10M, and \epsilon >0. We use a computer with a Central Processing Unit (CPU): AMD Athlon(TM) 64 Bit Dual Core Processor 4200+, 2.20GHz+.

For c=3, we get m⩽15, with a running time of about 78 ms. For the case 5⩽c⩽M_c,v_m=w_n, the runtime is about 17270.921 s (about 4 h 47 min 50 s), the time of the remaining cases is approximately the same. Combining (33) and the parity of n in Lemma 7, the rest of the calculation can be done using the Pari/GP software. This gives the following lemma.

Lemma 13　 Let c be an integer and c=3 or 5⩽c⩽M_c,M_c\in \mathcal{C}. Then the only solution to both the equation v_m=w_n and the equation v_m^{\mathrm{\prime }}=w_n is (m,n)=(0,0); and the equations v_m=w_n^{\mathrm{\prime }} and v_m^{\mathrm{\prime }}=w_n^{\mathrm{\prime }} have no solution.

It follows directly from Lemmas 7 and Lemmas 13 that

Lemma 14　 Let c be an integer, c⩾3 and c\ne 4. The only primitive positive integer solutions of the equation set (16) are (13,11,19).

7 Proof of Theorem 1

Let (x,y) be the primitive positive integer solution of the equation (3), and let

U=x^2+y^2,V=x^2+xy-y^2,Z=-x^2+4xy+y^2.

By the Tzanakis method, we know that (U,V,Z) is a solution to the equation set (16). Let \gcd (U,V,Z)=d⩾1, so that we can write U=d{U}_1,V=\pm d{V}_1,Z=\pm d{Z}_1, where U_1,V_1,Z_1 are positive integers. By Lemma 14, all the primitive positive integer solutions of the equation set (16) can obtained, so \left(U_1,V_1,Z_1\right)=(13,11,19).

Assertion　If d>1, then d can only be 2 and 5. Use the Contradiction. Suppose d\ne 2,5, then we have

d\mid U,d\mid V,d\mid Z.

Therefore,

d\mid 2y^2-xy,d\mid 5xy.

By our hypothesis, d\mid xy, and then d\mid y^2. We know that U=x^2+y^2, so we have d\mid x^2, which gives us \gcd (x,y)>1, which contradicts the assumption that (x,y) is the primitive solution of the equation. This proves the assertion to be true.

Easy to know that for d=1, (3,2) is the only solution to equation (3). For d=2 and d=5, equation (3) has no corresponding solution. The theorem is proved.

References

[1] Baker A. Contributions to the theory of Diophantine equations I, on the representation of integers by binary forms. Philos Thans Roy Soc London Ser A 1968; 263: 273–291

[2] Baker A, Davenport H. The equation 3x2−2=y2 and 8x2−7=z2. Quart J Math Oxford 1969; 20(2): 129–137

[3] Baker A, Wüstholz G. Logarithmic forms and group varieties. J Reine Angew Math 1993; 442: 19–62

[4] Bennett M A. On the number of solutions of simultaneous Pell equations. J. Reine Angew Math 1998; 498: 173–199

[5] Bilu Y, Hanrot G. Solving Thue equations of high degree. J Number Theory 1996; 60: 373–392

[6] Dujella A, Pethö A. Generalization of a theorem of Baker and Davenport. Quart J Math Oxford 1998; 49: 291–306

[7] Dujella A, Jadrijević B. A parametric family of quartic Thue equations. Acta Arith 2002; 101: 159–170

[8] Dujella A. A family of quartic Thue inequalities. Acta Arith 2004; 111: 61–76

[9] Dujella A. Continued fractions and RSA with small secret exponent. Tatra Mt Math Publ 2004; 29: 101–112

[10] Fatou P. Sur l’approximation des incommenurables et les series trigonometriques. C R Acad Sci Paris 1904; 139: 1019–1021

[11] GaálI. Diophantine Equations and Power Integral Bases: New computational Methods. New York: Springer-Verlag, 2002

[12] Heuberger C, Pethö A, Tichy R F. Complete solution of parametrized Thue equations. Acta Math Inform Univ Ostraviensis 1998; 6: 93–113

[13] HuaL K. Introduction to Number Theory. New York: Springer, 1982

[14] KeZSunQ. Talking about Indeterminate Equations. New York: Shanghai Educational Publishing House, 1980

[15] Pethö A, Schulenberg R. Effectives Lösen von Thue Gleichungen. Publ Math Debrecen 1987; 34: 189–196

[16] Thomas E. Complete solutions to a family of cubic Diophantine equations. J Number Theory 1990; 34: 235–250

[17] Thue A. Über Annäherungswerte algebraischer Zahlen. J Reine Angew Math 1909; 135: 284–305

[18] Tzanakis N, de Weger B M M. On the Practical Solution of the Thue Equation. J Number Theory 1989; 31: 99–132

[19] Tzanakis N. Explicit solution of a class of quartic Thue equations. Acta Arith 1993; 64: 271–283

[20] Worley R T. Estimating |α−p/q|. J Austral Math Soc 1981; 31: 202–206

RIGHTS & PERMISSIONS

Higher Education Press 2023