Uniqueness of Solutions to Simultaneous Pell Equations

Abstract

Let m, n be integers with $1<m<n$, and let $d=\gcd (m,n)$. In this paper, using the diophantine approximation method, we prove that if $d \ge 2(n/d)^4$, then the simultaneous Pell equations $x^2-(m^2-1)y^2=1$ and $z^2-(n^2-1)y^2=1$ have only the positive integer solution $(x,y,z)=(m,1,n)$.

1 Introduction

Let $\mathbb {Z}$, $\mathbb {N}$ be the sets of all integers and positive integers, respectively. Let a, b be distinct positive integers. The simultaneous Pell equations

$$\begin{aligned} X^2-aY^2=1,\quad Z^2-bY^2=1 \end{aligned}$$

(1.1)

arise in connection with a variety of classical problems on number theory and arithmetic algebraic geometry (see [13]).

Let N(a, b) denote the number of solutions (X, Y, Z) of (1.1). As early as the 1920s, using the diophantine approximation method of Thue [18], Siegel [17] proved that N(a, b) is always finite. However, his result is ineffective. An effective upper bound for N(a, b) was given by Schlickewei [15]. Using the subspace theorem of Schmidt [16], he proved that $N(a,b)<4\cdot 8^{2^{78}}$. In 1996, using the Padé approximation method (see [14]), Masser and Rickert [12] improved considerably the above mentioned upper bound to prove that $N(a,b) \le 16$. Two years later, Bennett [3] further proved that $N(a,b) \le 3$. Simultaneously, since there is no known pair (a, b) which attains $N(a,b)=3$, he proposed the following conjecture:

Conjecture 1.1

$N(a,b) \le 2$.

Around 2001, using the Baker method, Le [9, 10] and Yuan [20] independently proved that if $\max \{a,b\}<C$, where C is an effectively computable absolute constant, then $N(a,b)\le 2$. In 2006, Bennett et al. [4] completely verified Conjecture 1.1, i.e., they unconditionally proved that $N(a,b)\le 2$.

As explained in [3], any positive solution of equations (1.1) gives rise to a positive solution of the simultaneous Pell equations

$$\begin{aligned} x^2-(m^2-1)y^2=1,\quad z^2-(n^2-1)y^2=1 \end{aligned}$$

(1.2)

for some integers m, n with $1<m<n$. Obviously, (1.2) has a solution $(x,y,z)=(m,1,n)$. In this respect, Bennett [2] showed that if

$$\begin{aligned} n=\frac{\alpha ^{2l}-\bar{\alpha }^{2l}}{4\sqrt{m^2-1}},\quad l \in \mathbb {N}, \end{aligned}$$

(1.3)

where

$$\begin{aligned} \alpha =m+\sqrt{m^2-1},\quad \bar{\alpha }=m-\sqrt{m^2-1}, \end{aligned}$$

(1.4)

then (1.2) has another solution $(x,y,z)=((\alpha ^{2l}+\bar{\alpha }^{2l})/2,2n,2n^2-1)$. Thus, Yuan [20] proposed a stronger conjecture as follows:

Conjecture 1.2

If $N(m^2-1,n^2-1) \ge 2$, then m, n must satisfy (1.3) with (1.4).

The above conjecture has not been solved yet. Let $d=\gcd (m,n)$. In 2012, Le [11] showed that if $d>n^{\delta }$ and $n>C(\delta )$, where $\delta$ is a real number with $1/2<\delta <1$ and $C(\delta )$ is an effectively computable constant depending only on $\delta$, then (1.2) has only the positive solution $(x,y,z)=(m,1,n)$. Afterwards, He et al. [8] proved the same result for $m<n \le m+m^{1/5}$. However, it should be pointed out that the proof of Lemma 2.4 of [11] is incorrect. Therefore, the result in [11] has not been confirmed.

In this paper, using the diophantine approximation method and along the same approach as in [11], we prove the following result:

Theorem 1.3

If $d \ge 2(n/d)^4$, then (1.2) have only the positive integer solution $(x,y,z)=(m,1,n)$.

Obviously, the above theorem shows that the result in [11] is true for $4/5<\delta <1$.

2 Lower Bounds for Solutions

Any positive solution to simultaneous Pell equations (1.2) can be expressed as

$$\begin{aligned} y=v_r=v_s' \end{aligned}$$

for some positive integers r and s, where $\{v_k\}_{k=1}^{\infty }$ and $\{v_l'\}_{l=1}^{\infty }$ are sequences defined by

$$\begin{aligned} v_k=\frac{\alpha ^k-\alpha ^{-k}}{2\sqrt{m^2-1}},\quad v_l'=\frac{\beta ^l-\beta ^{-l}}{2\sqrt{n^2-1}} \end{aligned}$$

(2.1)

with

$$\begin{aligned} \alpha =m+\sqrt{m^2-1},\quad \beta =n+\sqrt{n^2-1}. \end{aligned}$$

We quote the following lemma on the indices of sequences.

Lemma 2.1

[11, Lemma 2.3] Let r and s be positive integers with $\min \{r,s\}>1$. If $v_r=v_s'$ holds, then we have

• (1) $r>s$.

• (2) $r \equiv s \pmod 2$.

• (3) If r is odd, then $r \equiv s \pmod 4$.

The goal of this section is to prove the following.

Lemma 2.2

Let r and s be positive integers with $\min \{r,s\}>1$. Assume that $v_r=v_s'$ holds.

• (1) If r is even, then either

  $$\begin{aligned} r>\frac{d^2}{2n_1} \end{aligned}$$

  or

  $$\begin{aligned} r>\frac{3^{1/3}d^{2/3}}{m_1} \quad \text {and}\quad rm_1=sn_1. \end{aligned}$$

• (2) If r is odd, then either

  $$\begin{aligned} r>\frac{\sqrt{2}\,d}{n_1} \end{aligned}$$

  or

  $$\begin{aligned} r>\frac{24^{1/4}d^{1/2}}{m_1}\quad \text {and}\quad (r^2-1)m_1^2=(s^2-1)n_1^2. \end{aligned}$$

Proof

Note that the proof proceeds along the same lines as the one of [11, Lemma 2.4] until halfway.

(1) Since r is even, we have

$$\begin{aligned} v_r=\sum _{i=0}^{r/2-1}\left( \begin{array}{c}r\\ 2i+1 \end{array}\right) m^{r-2i-1}(m^2-1)^i, \end{aligned}$$

(2.2)

from which we get

$$\begin{aligned} v_r \equiv rm(m^2-1)^{r/2-1}\equiv (-1)^{r/2-1}rm \pmod {m^3}. \end{aligned}$$

Similarly, since s is also even by Lemma 2.1, we have

$$\begin{aligned} v_s' \equiv (-1)^{s/2-1}sn \pmod {n^3}. \end{aligned}$$

It follows that $rm_1 \equiv \lambda sn_1 \pmod {d^2}$ with $\lambda \in \{\pm 1\}$, which implies that either

$$\begin{aligned} rm_1+sn_1 \ge d^2 \end{aligned}$$

(2.3)

or

$$\begin{aligned} rm_1=sn_1 \end{aligned}$$

(2.4)

holds. If (2.3) holds, then

$$\begin{aligned} r>\frac{rm_1+sn_1}{2n_1} \ge \frac{d^2}{2n_1}. \end{aligned}$$

Assume that (2.4) holds. Considering (2.2) modulo $m^5$ yields

$$\begin{aligned} v_r \equiv (-1)^{r/2-1}rm\left( \frac{(r-2)(r-4)}{6}m^2+1\right) \pmod {m^5}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} v_s' \equiv (-1)^{s/2-1}sn \left( \frac{(s-2)(s-4)}{6}n^2+1\right) \pmod {n^5}. \end{aligned}$$

From $v_r=v_s'$, $m/m_1=n/n_1=d$ and (2.4) we deduce either

$$\begin{aligned} rm_1 \left( (r-2)(r-4)m^2+(s-2)(s-4)n^2+12 \right) \ge 6d^4 \end{aligned}$$

(2.5)

or

$$\begin{aligned} (r-2)(r-4)m^2=(s-2)(s-4)n^2. \end{aligned}$$

(2.6)

Note that (2.4) together with $\gcd (m_1,n_1)=1$ implies that $r \equiv 0 \pmod {n_1}$.

Suppose first that (2.5) holds. Then,

$$\begin{aligned} rm_1\left( 2r^2m_1^2-6rm_1(m_1+n_1)+8(m_1^2+n_1^2)+\frac{2}{d^2}\right) \ge 6d^2. \end{aligned}$$

If $m_1=1$ and $r=n_1$, then (2.4) gives $s=1$, which contradicts the assumption. Hence we have $rm_1 \ge 2n_1$, which implies that

$$\begin{aligned} -6rm_1(m_1+n_1)+8(m_1^2+n_1^2)+\frac{2}{d^2}&\le -12n_1(m_1+n_1)+8(m_1^2+n_1^2)+\frac{2}{d^2}\\&<0. \end{aligned}$$

It follows that $r^3m_1^3>3d^2$, which immediately gives $r>3^{1/3}d^{2/3}/m_1$.

Suppose second that (2.6) holds. Then, (2.4) shows that

$$\begin{aligned} 3rm_1=3sn_1=4(m_1+n_1). \end{aligned}$$

Since $\gcd (m_1,n_1)=1$, $r \equiv s \equiv 0 \pmod 2$ and $r>s$, we obtain $(m_1,n_1)=(1,2)$ and $(r,s)=(4,2)$. In view of $d \ge 2$, we have

$$\begin{aligned} v_r&=\frac{1}{2\sqrt{d^2-1}}\left( (d+\sqrt{d^2-1})^{4} -(d-\sqrt{d^2-1})^{4}\right) \nonumber \\&>\frac{0.9}{2\sqrt{d^2-1}}(2d^2-1+2d\sqrt{d^2-1})^2. \end{aligned}$$

(2.7)

On the other hand, we see that

$$\begin{aligned} v_s'&=\frac{1}{2\sqrt{4d^2-1}}\left( (2d+\sqrt{4d^2-1})^2-(2d-\sqrt{4d^2-1})^2\right) \nonumber \\&<\frac{1}{2\sqrt{4d^2-1}}(2d+\sqrt{4d^2-1})^2. \end{aligned}$$

(2.8)

Since $0.9/\sqrt{d^2-1}>1/\sqrt{4d^2-1}$ and $2d^2-1+2d\sqrt{d^2-1}>2d+\sqrt{4k^2-1}$, inequalities (2.7) and (2.8) contradict $v_r=v_s'$. Thus, (2.6) does not hold. Therefore we obtain the assertion of (1).

(2) Since r is odd, we have

$$\begin{aligned} v_r=\sum _{i=0}^{(r-1)/2}\left( \begin{array}{c}r\\ 2i+1 \end{array}\right) m^{r-2i-1}(m^2-1)^i, \end{aligned}$$

(2.9)

from which we get

$$\begin{aligned} v_r \equiv (-1)^{(r-3)/2}\left( -1+\frac{r^2-1}{2}m^2\right) \pmod {m^4}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} v_s' \equiv (-1)^{(s-3)/2}\left( -1+\frac{s^2-1}{2}n^2\right) \pmod {n^4}. \end{aligned}$$

It follows from $v_r=v_s'$ and Lemma 2.1(iii) that

$$\begin{aligned} (r^2-1)m_1^2 \equiv (s^2-1)n_1^2 \pmod {2d^2}, \end{aligned}$$

which implies that either

$$\begin{aligned} \max \left\{ (r^2-1)m_1^2,(s^2-1)n_1^2 \right\} >2d^2 \end{aligned}$$

(2.10)

or

$$\begin{aligned} (r^2-1)m_1^2=(s^2-1)n_1^2 \end{aligned}$$

(2.11)

holds. If (2.10) holds, then

$$\begin{aligned} r>\frac{1}{n_1}\cdot \max \left\{ m_1\sqrt{r^2-1},n_1\sqrt{s^2-1}\right\} >\frac{\sqrt{2}\,d}{n_1}. \end{aligned}$$

Assume that (2.11) holds. Considering (2.9) modulo $m^6$ yields

$$\begin{aligned} v_r \equiv (-1)^{(r-1)/2}\left( \frac{(r^2-1)(r^2-9)}{24}m^4-\frac{r^2-1}{2}m^2+1 \right) \pmod {m^6}. \end{aligned}$$

Similarly we have

$$\begin{aligned} v_s' \equiv (-1)^{(s-1)/2}\left( \frac{(s^2-1)(s^2-9)}{24}n^4-\frac{s^2-1}{2}n^2+1 \right) \pmod {n^6}. \end{aligned}$$

It follows from $v_r=v_s'$ and Lemma 2.1(iii) that

$$\begin{aligned} \frac{(r^2-1)(r^2-9)}{24}m^4-\frac{r^2-1}{2}m^2 \equiv \frac{(s^2-1)(s^2-9)}{24}n^4-\frac{s^2-1}{2}n^2 \pmod {d^6}, \end{aligned}$$

which implies that either

$$\begin{aligned} \max \left\{ (r^2-1)m^2\left( (r^2-9)m^2-1\right) ,(s^2-1)n^2\left( (s^2-9)n^2-1\right) \right\} \ge 24d^6 \end{aligned}$$

(2.12)

or

$$\begin{aligned} (r^2-1)m^2\left( (r^2-9)m^2-12\right) =(s^2-1)n^2\left( (s^2-9)n^2-12\right) . \end{aligned}$$

(2.13)

If (2.12) holds, then by (2.11) we have $r^4m_1^4>24d^2$, that is,

$$\begin{aligned} r>\frac{24^{1/4}d^{1/2}}{m_1}. \end{aligned}$$

If (2.13) holds, then (2.11) shows that $(r^2-9)m^2=(s^2-9)n^2$ and thus $m^2=n^2$, which contradicts $m<n$. This completes the proof of Lemma 2.2.

3 Upper Bounds for Solutions

Let

$$\begin{aligned} \theta _1=\frac{\sqrt{n^2-1}}{n} \quad \text {and}\quad \theta _2=\frac{\sqrt{m^2-1}}{m}. \end{aligned}$$

Then, the following inequality can be easily deduced from Eq. (1.2).

Lemma 3.1

Let (x, y, z) be a positive solution to Eq. (1.2). Then,

$$\begin{aligned} \max \left\{ \left| \theta _1-\frac{z}{ny}\right| ,\left| \theta _2-\frac{x}{my}\right| \right\} <\frac{1}{2m\sqrt{m^2-1}}\cdot y^{-2}. \end{aligned}$$

The following result is a version of [14, Theorem] or [3, Theorem 3.2].

Proposition 3.2

Let $m_0$, $n_0$ and N be positive integers with $n_0 \ge 4$, $n_0-m_0 \ge 3$ and $N \ge 3.724m_0'n_0^2(n_0-m_0)^2$, where $m_0'=\max \{n_0-m_0,m_0\}$. Assume that N is divisible by $m_0n_0$. Then, the numbers $\theta _1$ and $\theta _2$ satisfy

$$\begin{aligned} \max \left\{ \left| \theta _1-\frac{p_1}{q}\right| ,\left| \theta _2-\frac{p_2}{q}\right| \right\} >\left( \frac{1.417\cdot 10^{28}m_0'n_0N}{m_0}\right) ^{-1}q^{-\lambda } \end{aligned}$$

(3.1)

for all integers $p_1$, $p_2$, q with $q>0$, where

$$\begin{aligned} \lambda =1+\frac{\log (10m_0^{-1}m_0'n_0N)}{\log (2.686m_0^{-1}n_0^{-1}(n_0-m_0)^{-2}N^2)}<2. \end{aligned}$$

Proof

For $0 \le i,\,j \le 2$ and integers $a_0$, $a_1$, $a_2$, let

$$\begin{aligned} p_{ij}(1/N)=\sum _{ij}\left( \begin{array}{c}k+1/2\\ h_j\end{array}\right) C_{ij}^{-1}\prod _{l \ne j}\left( \begin{array}{c}-k_{ij}\\ h_l\end{array}\right) , \end{aligned}$$

with

$$\begin{aligned} C_{ij}=\frac{N^k}{(N+a_j)^{k-h_j}}\prod _{l \ne j}(a_j-a_l)^{k_{il}+h_l}, \end{aligned}$$

where $k_{il}=k+\delta _{il}$ with $\delta _{il}$ the Kronecker delta, $\sum _{ij}$ denotes the sum over all non-negative integers $h_0$, $h_1$, $h_2$ satisfying $h_0+h_1+h_2=k_{ij}-1$, and $\prod _{l\ne j}$ denotes the product from $l=0$ to $l=2$ omitting $l=j$. We take $a_0=-n_0$, $a_1=-m_0$, $a_2=0$ and $N=m_0n_0N_0$ for some integer $N_0$. From the proof of [5, Theorem 2.2], we see that

$$\begin{aligned} p_{ijk}:=2^{-1}\left( 4m_0n_0(n_0-m_0)^2\right) ^k\frac{4.09\cdot 10^{13}}{1.6^k}p_{ij}(1/N) \in \mathbb {Z}. \end{aligned}$$

It follows from the proof of [7, Theorem 21] (see also the proof of [3, Theorem 3.2]) that

$$\begin{aligned} |P_{ijk}|<pP^k \quad \text {and}\quad \left| \sum _{j=0}^2 p_{ijk}\theta _j \right| < l L^{-k}, \end{aligned}$$

where

$$\begin{aligned} p&=\frac{4.09\cdot 10^{13}}{2}\left( 1+\frac{m_0'}{2N}\right)<2.047\cdot 10^{13},\\ P&=\frac{8(1+m_0/(2N))\cdot 4m_0n_0(n_0-m_0)^2N}{1.6\zeta }<\frac{10m_0'n_0N}{m_0},\\&\quad \left( \text {where}~\zeta ={\left\{ \begin{array}{ll}(n_0-m_0)^2(m_0+n_0) &{} \text {if}\quad 2m_0 \ge n_0,\\ m_0^2(2n_0-m_0) &{} \text {if}\quad 2m_0<n_0, \end{array}\right. }\right) \\ l&=\frac{4.09\cdot 10^{13}}{2}\cdot \frac{27}{64}\left( 1-\frac{n_0}{N}\right) ^{-1}<8.649\cdot 10^{12},\\ L&=\frac{1.6}{4m_0n_0(n_0-m_0)^2N}\cdot \frac{27}{4}\left( 1-\frac{n_0}{N}\right) ^2N^3>\frac{2.686N^2}{m_0n_0(n_0-m_0)^2}. \end{aligned}$$

Now the assertion follows from [3, Lemma 3.1] by noting that

$$\begin{aligned} \lambda =\frac{\log (10m_0^{-1}m_0'n_0N)}{\log (2.686m_0^{-1}n_0^{-1}(n_0-m_0)^{-2}N^2)}<2 \end{aligned}$$

and

$$\begin{aligned} c^{-1}:=4pP(2l)^{\lambda -1}<1.417\cdot 10^{28}, \end{aligned}$$

the latter of which is the coefficient of $q^{-\lambda }$ in the right-hand side of inequality (3.1). $\square$

Lemma 3.3

If $y=v_r$ with $r \ge 2$ and $m \ge 32$, then

$$\begin{aligned} \log y > (r-1) \log (1.999m). \end{aligned}$$

Proof

By (2.1), the asserted inequality is equivalent to

$$\begin{aligned} f(r):=\log (\alpha ^r-\alpha ^{-r})-\log (2\sqrt{m^2-1})-(r-1)\log (1.999m)>0. \end{aligned}$$

Since it is easy to see that f(r) is an increasing function of r, we have $f(r) \ge f(2)$ for $r \ge 2$. Noting that $m \ge 32$, we obtain $f(2)>0$ from

$$\begin{aligned} \frac{\alpha ^2-\alpha ^{-2}}{2\sqrt{m^2-1}}&>\alpha -\alpha ^{-3}>2\sqrt{m^2-1}-(2m)^{-3}\\&=2m\left( \sqrt{1-\frac{1}{m^2}}-\frac{1}{(2m)^4}\right) > 1.999m. \end{aligned}$$

Now we are ready to get upper bounds for solutions.

Proposition 3.4

Let m and n be integers with $1<m<n$. Assume that $d \ge 2n_1^4$. Suppose that Eq. (1.2) have a positive solution (x, y, z) other than $(x,y,z)=(m,1,n)$.

• (1) If $n_1^2<2m_1^2$, then $m_1 \le 6$, $n_1 \le 7$, $d \le 5184$ and $r \le 72$.

• (2) If $n_1^2>2m_1^2$, then $m_1 \le 5$, $n_1 \le 20$, $d \le 349791$ and $r \le 7160$.

Proof

Note that the assumption implies that $d \ge 2n_1^4 \ge 32$. We apply Proposition 3.2 with $m_0=m_1^2$, $n_0=n_1^2$, $N=m_1^2n_1^2d^2$, $p_1=mz$, $p_2=nx$, $q=mny$. Then, we see from Lemma 3.1 that

$$\begin{aligned} y^{2-\lambda }<7.089\cdot 10^{27}m_1^2n_1^6d^4M_1, \end{aligned}$$

where $M_1=\max \{m_1^2,n_1^2-m_1^2\}$ and we used $\lambda <2$ and $\sqrt{m^2-1}>0.9995m_1d$. Taking the logarithm of both sides, one can deduce from Lemma 3.3 that

$$\begin{aligned} r-1.001<\frac{\log (7.089\cdot 10^{27}m_1^2n_1^6d^4M_1)\log (2.686m_1^2n_1^2(n_1^2-m_1^2)^{-2}d^4)}{\log (1.999m_1d)\log (0.2686m_1^2n_1^{-2}(n_1^2-m_1^2)^{-2}M_1^{-1}d^2)}. \end{aligned}$$

Since it is easy to see that the right-hand side is a decreasing function of d, the assumption $d \ge 2n_1^4$ shows that

$$\begin{aligned} r-1.001<\frac{\log (1.135\cdot 10^{29}m_1^{2}n_1^{22}M_1)\log (42.976m_1^{2}n_1^{16} (n_1^2-m_1^2)^{-2})}{\log (3.998m_1n_1^4)\log (1.0744m_1^2n_1^6 (n_1^2-m_1^2)^{-2}M_1^{-1})}. \end{aligned}$$

(1) In the case where $n_1^2<2m_1^2$, we have $M_1=m_1^2$. Noting that

$$\begin{aligned} (n_1^2-m_1^2)^{-2}&\le (4^2-3^2)^{-2} =\frac{1}{7},\\ n_1^4(n_1^2-m_1^2)^{-2}&=\left( 1+\frac{m_1^2}{n_1^2-m_1^2}\right) ^2>4, \end{aligned}$$

we have

$$\begin{aligned} r-1.001<\frac{\log (1.135\cdot 10^{29}m_1^4n_1^{22})\log (0.8771m_1^2n_1^{16})}{\log (3.998m_1n_1^4)\log (4.2976n_1^2)}. \end{aligned}$$

Since $n_1/\sqrt{2}<m_1<n_1$, we obtain

$$\begin{aligned} r-1.001<\frac{\log (1.135\cdot 10^{29}n_1^{26})\log (0.8771\cdot n_1^{18})}{\log (2.827n_1^5)\log (4.2976n_1^2)}=:g(n_1). \end{aligned}$$

Suppose first that r is even.

If $r>d^2/(2n_1)$, then $d^2<2n_1 \left( g(n_1)+1.001\right)$, which together with $d \ge 2n_1^4$ implies that

$$\begin{aligned} n_1^7<0.5 \left( g(n_1)+1.001\right) . \end{aligned}$$

Thus, we obtain $n_1<1$, which is a contradiction.

If $r>3^{1/3}d^{2/3}/m_1$ and $rm_1=sn_1$, then $d^{2/3}<3^{-1/3}m_1\left( g(n_1)+1.001\right)$ and $d \ge 2n_1^4$ together show that

$$\begin{aligned} n_1^{5/3}<12^{-1/3}\left( g(n_1)+1.001\right) , \end{aligned}$$

which yields $n_1 \le 7$, $m_1 \le 6$, $r \le 72$ and $d \le 5184$.

Suppose second that r is odd.

If $r> \sqrt{2}\,d/n_1$, then $d<n_1 \left( g(n_1)+1.001\right) /\sqrt{2}$ and $d \ge 2n_1^4$ together imply that

$$\begin{aligned} n_1^3<2^{-3/2}\left( g(n_1)+1.001\right) , \end{aligned}$$

which means $n_1 \le 3$, contradicting $n_1^2<2m_1^2$.

If $r>24^{1/4}d^{1/2}/m_1$ and $(r^2-1)m_1^2=(s^2-1)n_1^2$, then by $24^{1/4}d^{1/2}<m_1 \left( g(n_1)+1.001\right)$ and $d \ge 2n_1^4$ we have

$$\begin{aligned} n_1<24^{-1/4}\cdot 2^{-1/2}\left( g(n_1)+1.001\right) , \end{aligned}$$

which yields $n_1 \le 21$, $m_1 \le 20$ and $r \le 65$. However, there are no integers $m_1$, $n_1$, r in the ranges above satisfying $m_1^2<n_1^2<2m_1^2$ and

$$\begin{aligned} \gcd (m_1,n_1)=1,\quad r^2-1 \equiv 0 \pmod {n_1^2},\quad (r^2-1)m_1^2+n_1^2=\square . \end{aligned}$$

(3.2)

(2) In the case where $n_1^2>2m_1^2$, we have $M_1=n_1^2-m_1^2$ and $m_1^2<n_1^2-m_1^2<n_1^2$, which together show that

$$\begin{aligned} r-1.001 < \frac{\log (1.135\cdot 10^{29}m_1^2n_1^{24})\log (4.776m_1^2n_1^{16})}{\log (3.998m_1n_1^4)\log (1.0744m_1^2)}=:g_2(m_1,n_1). \end{aligned}$$

(3.3)

Note that $g_2(m_1,n_1)$ is a decreasing function of $m_1$.

Suppose first that r is even.

If $r>d^2/(2n_1)$, then $d^2<2n_1(g_2(1,n_1)+1.001)$, which together with $d \ge 2n_1^4$ implies that

$$\begin{aligned} n_1^7<0.5(g_2(1,n_1)+1.001). \end{aligned}$$

Hence we have $n_1=2$, $m_1=1$, $r \le 3542$ and $d \le 119$.

If $r>3^{1/3}d^{2/3}/m_1$ and $rm_1=sn_1$, then $d^{2/3}<3^{-1/3}m_1(g_2(n_1)+1.001)$, which again together with $d \ge 2n_1^4$ shows that

$$\begin{aligned} n_1^{8/3}<12^{-1/3}m_1(g_2(m_1,n_1)+1.001). \end{aligned}$$

(3.4)

It follows from $\sqrt{2}\,m_1<n_1$ that

$$\begin{aligned} n_1^{5/3}<12^{-1/3}\cdot 2^{-1/2}(g_2(m_1,n_1)+1.001). \end{aligned}$$

(3.5)

Assume now that $m_1 \ge 6$. Then, we see from (3.3) and (3.5) that $n_1 \le 8$, which contradicts $n_1>2m_1^2$. Thus we have $m_1 \le 5$. As for the remaining upper bounds, inequalities (3.3) and (3.4) with $m_1=1$ give the worst bounds $n_1 \le 20$, $r \le 7160$ and $d \le 349791$.

Suppose second that r is odd.

If $r>\sqrt{2}\,d/n_1$, then $d<2^{-1/2}n_1(g_2(m_1,n_1)+1.001)$ and $d \ge 2n_1^4$ together imply that

$$\begin{aligned} n_1^3<2^{-3/2}(g_2(m_1,n_1)+1.001). \end{aligned}$$

(3.6)

Assume that $m_1 \ge 3$. Then, (3.6) shows that $n_1 \le 3$, which contradicts $n_1^2>2m_1^2$. Hence we have $m_1 \le 2$, $n_1 \le 13$, $r \le 6549$ and $d \le 60200$.

If $r>24^{1/4}d^{1/2}/m_1$ and $(r^2-1)m_1^2=(s^2-1)n_1^2$, then from $24^{1/4}d^{1/2}<m_1(g_2(m_1,n_1)+1.001)$ and $d \ge 2n_1^4$ we deduce that

$$\begin{aligned} n_1^2<24^{-1/4}m_1(g_2(m_1,n_1)+1.001). \end{aligned}$$

(3.7)

Since $n_1^2>2m_1^2$, we obtain

$$\begin{aligned} n_1<96^{-1/4}(g_2(m_1,n_1)+1.001). \end{aligned}$$

(3.8)

Assume now that $m_1 \ge 20$. Then by (3.8) we have $n_1 \le 28$, which contradicts $n_1>\sqrt{2}\,m_1 \ge 20\sqrt{2}>28$. Thus we have $m_1 \le 19$. Finding an upper bound for $n_1$ using (3.7) for each $m_1$ with $1 \le m_1 \le 19$, we obtain $n_1 \le 62$ and $r \le 8736$ (which corresponds to the case where $(m_1,n_1)=(1,62)$). However, it is easy to check that there are no integers $m_1$, $n_1$, r in the ranges above satisfying (3.2), $n_1^2>2m_1^2$ and $2n_1^4\,(\le d)<24^{-1/2}m_1^2r^2$. This completes the proof of Proposition 3.4. $\square$

4 Proof of Theorem 1.3

Proof of Theorem 1.3

The equality $v_r=v_s'$ implies that

$$\begin{aligned} 0<\varLambda := r\log \alpha - s \log \beta + \frac{1}{2} \log \frac{n^2-1}{m^2-1} < \frac{\alpha ^2}{\alpha ^2-1} \alpha ^{-2r} \end{aligned}$$

(4.1)

(see [19, (61)]). Since we know by Proposition 3.4 upper bounds for $m=m_1d$, $n=n_1d$ and r, we may use (4.1) and apply [6, Lemma 5], which is based on the Baker–Davenport reduction lemma [1, Lemma]. In each of the cases $n_1^2<2m_1^2$ and $n_1^2>2m_1^2$, we obtained the new bound $r \le 4$ around in two seconds and in 40 minutes, respectively, where in fact the reduction started from the larger bound $r \le 10^{10}$, as it was faster than using the actual bounds. On the other hand, since the case where $r>24^{1/4}d^{1/2}/m_1$ and $(r^2-1)m_1^2=(s^2-1)n_1^2$ does not occur from the proof of Proposition 3.4, Lemma 2.2 together with the assumption $d \ge 2n_1^4$ and $n_1 \ge 2$ implies $r>3^{1/3}\cdot 2^{2/3}n_1^{5/3}>7$ even in the worst case, which is incompatible with $r \le 4$. $\square$