A NOTE ON THE GOORMAGHTIGH EQUATION CONCERNING DIFFERENCE SETS

Abstract

Let p be a prime and let r, s be positive integers. In this paper, we prove that the Goormaghtigh equation $(x^m-1)/(x-1)=(y^n-1)/(y-1)$, $x,y,m,n \in {\mathbb {N}}$, $\min \{x,y\}>1$, $\min \{m,n\}>2$ with $(x,y)=(p^r,p^s+1)$ has only one solution $(x,y,m,n)=(2,5,5,3)$. This result is related to the existence of some partial difference sets in combinatorics.

1 Introduction

Let ${\mathbb {N}}$ be the set of all positive integers. One hundred years ago, Ratat [27] and Rose and Goormaghtigh [28] conjectured that the equation

(1.1) $$ \begin{align} \frac{x^m-1}{x-1}=\frac{y^n-1}{y-1} \quad \mbox{for all } x,y,m,n \in {\mathbb{N}}, x \ne y, \min\{x,y\}>1, \min\{m,n\}>2, \end{align} $$

has only two solutions $(x,y,m,n)=(2,5,5,3)$ and $(2,90,13,3)$ with $x<y$ . Equation (1.1) is usually called the Goormaghtigh equation. The above conjecture is a very difficult problem in Diophantine equations. It was solved for some special cases (see [3, 5, 6, 9, 10, 12, 14–18, 22, 23, 26, 29–37]). But, in general, the problem is far from solved. The solution of (1.1) is closely related to some problems in number theory, combinatorics and algebra (see [1, 4, 13, 19, 21]). For example, while discussing the partial geometries admitting Singer groups in combinatorics, Leung et al. [19] found that the existence of partial difference sets in an elementary abelian $3$ -group is related to the solutions $(x,y,m,n)$ of (1.1) with

(1.2) $$ \begin{align} (x,y)=(2^r,3), \end{align} $$

where r is a positive integer. In [19], they proved that (1.1) has no solutions $(x,y,m,n)$ satisfying (1.2).

Let p be a prime and let r, s be positive integers. In this paper, we discuss the solutions $(x,y,m,n)$ of (1.1) with

(1.3) $$ \begin{align} (x,y)=(p^r,p^s+1). \end{align} $$

Thus, we generalise the above-mentioned result in [19] to prove the following theorem.

Theorem 1.1. Equation (1.1) has only one solution $(x,y,m,n)=(2,5,5,3)$ with (1.3).

Combining Theorem 1.1 and [19, Corollary 37] with $q=\alpha +1=2^s+1$ , we immediately obtain the following corollary which may be regarded as a generalisation of [19, Corollary 44].

Corollary 1.2. Suppose that a proper partial geometry $\varPi $ has at least two subgroup lines and that the parameters of the corresponding partial difference set have the form in [19, (34)]. Then, $\varPi $ cannot be expressed as

$$ \begin{align*} \varPi=\mathrm{pg}((2^s+1)^u,(2^r-1)(2^s+1)^u+2^s+1,2^s) \end{align*} $$

with $r,s,t \in {\mathbb {N}}$ .

The organisation of the paper is as follows. In Section 2, we prove Theorem 1.1 in the case where $r \le s$ using an upper bound for the number of solutions of the generalised Ramanujan–Nagell equations due to Bugeaud and Shorey [8]. In Section 3, using a lower bound for linear forms in three logarithms due to Matveev [24], we show that if $r>s$ and $p^r>3.436\times 10^{15}$ , then (1.1) has no solutions $(x,y,m,n)$ with (1.3). Thus, the remaining case to be checked is $r>s$ and $p^r<3.436\times 10^{15}$ . For this, we appeal to the reduction method due to Dujella and Pethő [11], based on [2, Lemma] by Baker and Davenport, to complete the proof of Theorem 1.1 in Section 4.

2 The case $r \le s$

Lemma 2.1 [20]

The equation

(2.1) $$ \begin{align} \frac{X^k-1}{X-1}=Y^l \quad \mbox{for all } X,Y,k,l \in {\mathbb{N}}, X>1, Y>1, k>2, l>1, \end{align} $$

has only two solutions, $(X,Y,k,l)=(3,11,5,2)$ and $(7,20,4,2)$ with $2 \mid l$ .

Let $D_1$ and $D_2$ be coprime positive integers and let p be a prime with $p \nmid D_1D_2$ . Further, let $N(D_1,D_2,p)$ denote the number of solutions $(X,Z)$ of the equation

(2.2) $$ \begin{align} D_1X^2+D_2=p^Z\quad \mbox{for all } X,Z \in {\mathbb{N}}. \end{align} $$

Combining the results in [7, 8], we immediately obtain the following two lemmas.

Lemma 2.2. We have $N(D_1,D_2,2) \le 1$ , except for the following cases:

1. (i) $N(1,7,2)=5$ , $(X,Z)=(1,3),(3,4),(5,5),(11,7)$ and $(181,15)$ ;

2. (ii) $N(3,5,2)=3$ , $(X,Z)=(1,3),(3,5)$ and $(13,9)$ ;

3. (iii) $N(7,1,2)=2$ , $(X,Z)=(1,3)$ and $(3,6)$ ;

4. (iv) $N(1,2^{k+2}-1,2)=2$ , $(X,Z)=(1,k+2)$ and $(2^{k+1}-1,2k+2)$ , where k is a positive integer with $k>1$ ;

5. (v) $N(3,29,2)=2$ , $(X,Z)=(1,5)$ and $(209,17)$ ;

6. (vi) $N(5,3,2)=2$ , $(X,Z)=(1,3)$ and $(5,7)$ ;

7. (vii) $N(13,3,2)=2$ , $(X,Z)=(1,4)$ and $(71,16)$ ;

8. (viii) $N(21,11,2)=2$ , $(X,Z)=(1,5)$ and $(79,17)$ ; and

9. (ix) if $D_1a^2{\kern-1pt}={\kern-1pt}2^k{\kern-1pt}-{\kern-1pt}\delta $ and $D_2=3\cdot 2^k{\kern-1pt}+{\kern-1pt}\delta $ , where a, k are positive integers with $k{\kern-1pt}>{\kern-1pt}1$ and $\delta \in \{1,-1\}$ , then $N(D_1,D_2,2)=2$ , $(X,Z)=(a,k+2)$ and $((2^{k+1}+\delta )a, 3k+2)$ .

Lemma 2.3. If $p \ne 2$ , then $N(D_1,D_2,p)\le 1$ , except for the following cases:

1. (i) $N(2,1,3)=3$ , $(X,Z)=(1,1), (2,2)$ and $(11,5)$ ; and

2. (ii) if $4D_1a^2=p^k-\delta $ and $4D_2=3p^k+\delta $ , where a, k are positive integers and $\delta \in \{1,-1\}$ , then $N(D_1,D_2,p)=2$ , $(X,Z)=(a,k)$ and $((2p^k+\delta )a,3k)$ .

Proposition 2.4. If $r \le s$ , then (1.1) has only one solution $(x,y,m,n)=(2,5,5,3)$ with (1.3).

Proof. We now assume that $(x,y,m,n)$ is a solution of (1.1) with (1.3). Then

(2.3) $$ \begin{align} \frac{p^{rm}-1}{p^r-1}=\frac{(p^s+1)^n-1}{p^s}. \end{align} $$

When $r=s$ , by (2.3),

(2.4) $$ \begin{align} \frac{p^{r(m+1)}-1}{p^r-1}=(p^r+1)^n. \end{align} $$

If $2 \mid n$ , by (2.4), the equation (2.1) has a solution $(X,Y,k,l)=(p^r,p^r+1,m+1,n)$ with $2 \mid l$ . However, since $m>2$ , by Lemma 2.1, this is impossible. So $2 \nmid n$ and $n \ge 3$ .

Since $p^r+1>2$ and $p^r \equiv -1 \pmod {(p^r+1)}$ , by (2.4),

$$ \begin{align*} 0 \equiv (p^r-1)(p^r+1)^n\equiv p^{r(m+1)}-1\equiv (-1)^{m+1}-1 \pmod{(p^r+1)}, \end{align*} $$

from which we get $2 \mid m+1$ . Hence, by (2.4),

(2.5) $$ \begin{align} \frac{(p^{2r})^{(m+1)/2}-1}{p^{2r}-1}=(p^r+1)^{n-1}. \end{align} $$

Recall that $2 \nmid n$ and $n \ge 3$ . We see from (2.5) that if $(m+1)/2>2$ , then (2.1) has a solution $(X,Y,k,l)=(p^{2r},p^r+1,(m+1)/2,n-1)$ with $2 \mid l$ . But, by Lemma 2.1 again, this is impossible. Therefore, since $2 \nmid m$ and $m \ge 3$ , we get $m=3$ , and by (2.5),

$$ \begin{align*} \frac{(p^{2r})^{(m+1)/2}-1}{p^{2r}-1}=\frac{p^{4r}-1}{p^{2r}-1}=p^{2r}+1=(p^r+1)^{n-1} \ge (p^r+1)^2>p^{2r}+1, \end{align*} $$

which is a contradiction. Thus, (1.1) has no solutions $(x,y,m,n)$ with (1.3) and $r=s$ .

When $r<s$ , by (2.3),

(2.6) $$ \begin{align} (p^r-1)(p^s+1)^n+(p^s-p^r+1)=p^{rm+s}. \end{align} $$

Since $r<s$ , $p^r-1$ , $p^s+1$ and $p^s-p^r+1$ are positive integers satisfying

(2.7) $$ \begin{align} \gcd((p^r-1)(p^s+1),p^s-p^r+1)=1,\quad p \nmid (p^r-1)(p^s+1)(p^s-p^r+1). \end{align} $$

If $2 \mid n$ , by (2.6), the equation (2.2) has a solution

$$ \begin{align*} (X,Z)=((p^s+1)^{n/2},rm+s) \end{align*} $$

for $(D_1,D_2)=(p^r-1,p^s-p^r+1)$ . Notice that (2.2) has another solution $(X,Z)=(1,s)$ for $(D_1,D_2)=(p^r-1,p^s-p^r+1)$ . So

(2.8) $$ \begin{align} N(p^r-1,p^s-p^r+1,p) \ge 2. \end{align} $$

However, by (2.7), using Lemmas 2.2 and 2.3, (2.8) is false.

Similarly, if $2 \nmid n$ , by (2.6), the equation (2.2) has a solution

$$ \begin{align*} (X,Z)=((p^s+1)^{(n-1)/2},rm+s) \end{align*} $$

for $(D_1,D_2)=((p^r-1)(p^s+1),p^s-p^r+1)$ . Moreover, (2.2) has another solution $(X,Z)=(1,r+s)$ for $(D_1,D_2)=((p^r-1)(p^s+1),p^s-p^r+1)$ . So

(2.9) $$ \begin{align} N((p^r-1)(p^s+1),p^s-p^r+1,p) \ge 2. \end{align} $$

Applying Lemmas 2.2 and 2.3 to (2.9), we can only obtain

(2.10) $$ \begin{align} (p,r,x)=(2,1,2). \end{align} $$

Therefore, by (1.3) and (2.10), we get $(D_1,D_2)=(5,3)$ and $(x,y,m,n)=(2,5,5,3)$ . Thus, the proposition is proved.

3 The case $r>s$

In this section, we assume that $r>s$ and $(x,y,m,n)$ is a solution of (1.1) with (1.3).

Lemma 3.1. If $(p,s)\ne (2,1)$ , then $n> p^r$ .

Proof. By (2.3),

$$ \begin{align*} \frac{p^{rm}-1}{p^r-1}=\sum_{i=0}^{m-1}p^{ri}=\sum_{j=1}^n \binom{n}{j}p^{s(j-1)}=\frac{(p^s+1)^n-1}{p^s}, \end{align*} $$

from which we get

(3.1) $$ \begin{align} p^r\bigg(\frac{p^{r(m-1)}-1}{p^r-1}\bigg)=(n-1)+\sum_{j=2}^n\binom{n}{j}p^{s(j-1)}. \end{align} $$

Since $n>2$ and $p \nmid (p^{r(m-1)}-1)/(p^r-1)$ , we see from (3.1) that $p \mid n-1$ and

(3.2) $$ \begin{align} p^r\, \|\, (n-1)+\sum_{j=2}^n\binom{n}{j}p^{s(j-1)}. \end{align} $$

Let

(3.3) $$ \begin{align} p^t\,\|\,n-1 \end{align} $$

and

(3.4) $$ \begin{align} p^{t_j}\,\|\,j \quad \mbox{for all } j=2,\ldots,n, \end{align} $$

where t is a positive integer and $t_j\ (j=2,\ldots ,n)$ are nonnegative integers. Then

(3.5) $$ \begin{align} t_j \le \frac{\log j}{\log p}\le j-1 \quad \mbox{for all } j=2,\ldots,n. \end{align} $$

Notice that both symbols ‘ $\le $ ’ in (3.5) can be taken by equal signs ‘ $=$ ’ if and only if $(p,t_j,j)=(2,1,2)$ . It follows from (3.5) that if $(p,t_j)\ne (2,1)$ , then

(3.6) $$ \begin{align} t_j<j-1 \quad \mbox{for all } j=2,\ldots,n. \end{align} $$

Hence, since $\gcd (j,j-1)=1$ and $(p,s)\ne (2,1)$ , by (3.3), (3.4) and (3.6),

(3.7) $$ \begin{align} \binom{n}{j}p^{s(j-1)}\equiv n(n-1)\binom{n-2}{j-2}\frac{p^{s(j-1)}}{(j-1)j}\equiv 0 \pmod{p^{t+1}} \quad \mbox{for all } j=2,\ldots,n. \end{align} $$

Therefore, by (3.3) and (3.7),

(3.8) $$ \begin{align} p^t\,\|\,(n-1)+\sum_{j=2}^n\binom{n}{j}p^{s(j-1)}. \end{align} $$

Comparing (3.2) and (3.8),

(3.9) $$ \begin{align} t=r. \end{align} $$

Further, since $n>1$ , by (3.3) and (3.9), we obtain $n-1 \ge p^r$ and $n>p^r$ . The lemma is proved.

Let ${\mathbb {Z}}$ , ${\mathbb {Q}}$ and ${\mathbb {R}}$ be the sets of all integers, rational numbers and real numbers, respectively. Let $\alpha $ be an algebraic number of degree d and let $\alpha ^{(1)},\ldots ,\alpha ^{(d)}$ denote all the conjugates of $\alpha $ . Further, let

$$ \begin{align*} f(X)=a\prod_{i=1}^d(X-\alpha^{(i)})\in {\mathbb{Z}}[X] \quad \mbox{for all } a \in {\mathbb{N}} \end{align*} $$

denote the minimal polynomial of $\alpha $ . Then

$$ \begin{align*} \mathop{\mathrm{h}}\nolimits(\alpha)=\frac{1}{d}\bigg(\log a +\sum_{i=1}^d\log \max\{1,|\alpha^{(i)}|\}\bigg) \end{align*} $$

is called the absolute logarithmic height of $\alpha $ .

Lemma 3.2 ([24, 25])

Let $\alpha _1$ , $\alpha _2$ , $\alpha _3$ be three distinct real algebraic numbers with $\min \{\alpha _1,\alpha _2,\alpha _3\}>1$ and let $b_1$ , $b_2$ , $b_3$ be three positive integers with $\gcd (b_1,b_2,b_3)=1$ . Further, let

$$ \begin{align*} \varLambda=b_1\log \alpha_1+b_2\log \alpha_2-b_3\log \alpha_3. \end{align*} $$

If $\varLambda \ne 0$ , then

$$ \begin{align*} \log|\varLambda|>-CD^2A_1A_2A_3\log(1.5e DB\log(e D)), \end{align*} $$

where

(3.10) $$ \begin{align} D&=[\,{\mathbb{Q}}(\alpha_1,\alpha_2,\alpha_3):{\mathbb{Q}}\,],\quad D'=[\,{\mathbb{R}}(\alpha_1,\alpha_2,\alpha_3):{\mathbb{R}}\,], \kern1pt\qquad\end{align} $$

(3.11) $$ \begin{align} A_j&\ge \max\{D\mathop{\mathrm{h}}\nolimits(\alpha_j),|\log \alpha_j|\} \quad \mbox{for } j=1,2,3, \qquad\qquad\qquad\, \end{align} $$

(3.12) $$ \begin{align} B&\ge \max\bigg\{b_j\frac{A_j}{A_1}\,\bigg|\,j=1,2,3\bigg\}, \qquad\qquad\qquad\qquad\qquad\qquad\kern1pt\end{align} $$

(3.13) $$ \begin{align} C&=\frac{5\times 16^5}{6D'}e^3(7+2D')\bigg(\frac{3e}{2}\bigg)^{D'}(26.25+\log(D^2\log(e D))).\ \ \end{align} $$

Proposition 3.3. If $r>s$ and $p^r>3.436\times 10^{15}$ , then (1.1) has no solutions $(x,y,m,n)$ with (1.3).

Proof. By [19], the proposition holds for $(p,s)=(2,1)$ . We can therefore assume that $(p,s)\ne (2,1)$ . By (2.3),

(3.14) $$ \begin{align} (p^r-1)(p^s+1)^n=p^{rm+s}+(p^r-p^s-1). \end{align} $$

Since $p^r-p^s-1>0$ , taking the logarithms of both sides of (3.14),

(3.15) $$ \begin{align} \log(p^r-1)+n\log(p^s+1)=(rm+s)\log p+\log \bigg(1+\frac{p^r-p^s-1}{p^{rm+s}}\bigg). \end{align} $$

Since $\log (1+\varepsilon )<\varepsilon $ for any $\varepsilon>0$ , by (3.15),

(3.16) $$ \begin{align} \begin{array}{ll} &0 <\log(p^r-1)+n\log(p^s+1)-(rm+s)\log p\\[6pt] &\quad =\log \bigg(1+\dfrac{p^r-p^s-1}{p^{rm+s}}\bigg)<\dfrac{p^r-p^s-1}{p^{rm+s}}. \end{array} \end{align} $$

Take

(3.17) $$ \begin{align} \alpha_1=p^r-1,\quad \alpha_2=p^s+1,\quad \alpha_3=p,\quad b_1=1,\quad b_2=n,\quad b_3=rm+s \end{align} $$

and

(3.18) $$ \begin{align} \varLambda=\log(p^r-1)+n\log(p^s+1)-(rm+s)\log p. \end{align} $$

By (3.16) and (3.18), we have $\varLambda>0$ and

(3.19) $$ \begin{align} (rm+s)\log p+\log \varLambda<\log(p^r-p^s-1)<\log(p^r-1). \end{align} $$

In order to apply Lemma 3.2, by (3.10), (3.11) and (3.17), we can choose the following parameters.

(3.20) $$ \begin{align} D& =D'=1, \end{align} $$

(3.21) $$ \begin{align} A_1=\log(p^r-1),\quad A_2& =\log(p^s+1),\quad A_3=\log p. \end{align} $$

Further, by (3.12), (3.13), (3.16), (3.20) and (3.21),

$$ \begin{align*} B=\frac{(rm+s)\log p}{\log(p^r-1)} \end{align*} $$

and

(3.22) $$ \begin{align} C<1.691\times 10^{10}. \end{align} $$

Applying Lemma 3.2 to (3.17) and (3.18), by (3.20)–(3.22),

(3.23) $$ \begin{align} \log \varLambda&>-1.691\times 10^{10}(\log(p^r-1))(\log(p^s+1))(\log p)\nonumber\\ &\quad\times \bigg(1.406+\log \bigg(\dfrac{(rm+s)\log p}{\log(p^r-1)}\bigg)\bigg). \end{align} $$

Substituting (3.23) into (3.19), we get

(3.24) $$ \begin{align} 1+1.691\times 10^{10}(\log(p^s+1))(\log p)\bigg(1.406+\log\bigg(\dfrac{(rm+s)\log p}{\log(p^r-1)}\bigg)\bigg)>\dfrac{(rm+s)\log p}{\log(p^r-1)}. \end{align} $$

Hence, since $(p,s)\ne (2,1)$ and $p^s+1 \ge 4$ , by (3.23), we can calculate that

(3.25) $$ \begin{align} \frac{(rm+s)\log p}{\log(p^r-1)}<1.501\times 10^{12}(\log(p^s+1))(\log p)(\log \log(p^s+1)). \end{align} $$

On the other hand, by (3.16),

(3.26) $$ \begin{align} \dfrac{(rm+s)\log p}{\log(p^r-1)}&>\bigg(1-\dfrac{p^r-p^s-1}{p^{rm+s}\log(p^r-1)}\bigg)+\dfrac{n\log(p^s+1)}{\log(p^r-1)}\nonumber\\ &>\dfrac{n\log(p^s+1)}{\log(p^r-1)}. \end{align} $$

Since $\log p \le (\log p^r)/2$ for $r \ge 2$ , the combination of (3.25) and (3.26) yields

(3.27) $$ \begin{align} n&<1.501\times 10^{12}(\log p)(\log(p^r-1))(\log \log(p^s+1))\nonumber\\ &<7.505\times 10^{11}( \log p^r)^2(\log \log p^r). \end{align} $$

Further, since $(p,s)\ne (2,1)$ , by Lemma 3.1, we have $n>p^r$ . Hence, by (3.27),

(3.28) $$ \begin{align} p^r<7.505\times 10^{11}(\log p^r)^2(\log \log p^r). \end{align} $$

Therefore, by (3.28), we obtain $p^r<3.436\times 10^{15}$ . Thus, if $r>s$ and $p^r>3.436\times 10^{15}$ , then (1.1) has no solutions $(x,y,m,n)$ with (1.3). The proposition is proved.

4 Proof of Theorem 1.1

We continue to assume that $r>s$ and that $(x,y,m,n)$ is a solution of (1.1) with (1.3). Put $m'=rm+s$ . By (3.25),

(4.1) $$ \begin{align} m'<1.501\times 10^{12}(\log(p^r-1))(\log(p^s+1))(\log \log(p^s+1)). \end{align} $$

Since Proposition 3.3 implies that $p^s \le p^{r-1}<1.718\times 10^{15}$ , we see from (4.1) that

(4.2) $$ \begin{align} m'<6.702\times 10^{15}. \end{align} $$

On the other hand, we deduce from Lemma 3.1 and (3.26) that

(4.3) $$ \begin{align} m'>\frac{n\log(p^s+1)}{\log p}>\frac{p^r \log(p^r+1)}{\log p}. \end{align} $$

Now, by (3.16),

(4.4) $$ \begin{align} 0<n-m'\kappa+\mu<AB^{-m'}, \end{align} $$

where

$$ \begin{align*} m'=rm+s,\quad \kappa=\frac{\log p}{\log(p^s+1)},\quad \mu=\frac{\log(p^r-1)}{\log(p^s+1)},\quad A=\frac{p^r-p^s-1}{\log(p^s+1)},\quad B=p. \end{align*} $$

Lemma 4.1. Let $\kappa $ , $\mu $ , $A>0$ and $B \ge 1$ be real numbers and let $M'$ be a positive integer. Let $p/q$ be a convergent of the continued fraction expansion of $\kappa $ such that $q>6M'$ , and put $\varepsilon =\|\mu q\|-M'\| \kappa q\|$ , where $\|\cdot \|$ denotes the distance from the nearest integer. If $\varepsilon>0$ , then inequality (4.4) has no integer solution $(n,m')$ satisfying

$$ \begin{align*} \frac{\log(Aq/\varepsilon)}{\log B}\le m' \le M'. \end{align*} $$

Proof. Since the assertion is identical with that of [11, Lemma 5a] if the middle term of inequalities (4.4) is multiplied by $-1$ , the lemma is proved in the same way as [11, Lemma 5a].

By Proposition 3.3, (4.2) and (4.4), we may apply Lemma 4.1 with $M'=6.702\times 10^{15}$ in the ranges

$$ \begin{align*} 2\le p <\sqrt{R},\quad 1 \le s<r<\log_p R \end{align*} $$

with $(p,s)\ne (2,1)$ , where $R=3.436\times 10^{15}$ . For $7 \le p <\sqrt {R}$ , the first step of reduction gives $m' \le 43$ , which contradicts (4.3) with $p \ge 7$ and $r \ge 2$ . For $p=5$ , the first and second steps of reduction give $m' \le 52$ and $m' \le 30$ , respectively. The latter contradicts (4.3) with $p=5$ and $r \ge 2$ . For $p=3$ , the first and second steps of reduction give $m' \le 75$ and $m'\le 45$ , respectively, which, together with (4.3), yields $r=2$ . For $p=2$ , the first and second steps of reduction give $m' \le 126$ and $m' \le 75$ , respectively, from which by (4.3) we obtain $r \in \{3,4\}$ .

Thus, it remains to consider the cases where

(4.5) $$ \begin{align} (p,r,s)\in \{(2,3,2),(2,4,2),(2,4,3),(3,2,1)\}. \end{align} $$

In view of the bounds for $m'=rm+s$ obtained above, it suffices to check that (3.14) with (4.5) has no solution $(m,n)$ in the ranges $m \le 24$ and $n \le 34$ , which can be easily done. Therefore, the theorem is proved.