1 Introduction
Let
${\mathbb {N}}$
be the set of all positive integers. One hundred years ago, Ratat [Reference Ratat27] and Rose and Goormaghtigh [Reference Rose and Goormaghtigh28] conjectured that the equation
$$ \begin{align} \frac{x^m-1}{x-1}=\frac{y^n-1}{y-1} \quad \mbox{for all } x,y,m,n \in {\mathbb{N}}, x \ne y, \min\{x,y\}>1, \min\{m,n\}>2, \end{align} $$
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 [Reference Balasubramanian and Shorey3, Reference Bennett, Garbuz and Marten5, Reference Bennett, Gherga and Kreso6, Reference Bugeaud and Shorey9, Reference Davenport, Lewis and Schinzel10, Reference He12, Reference Karanikolov14–Reference Le18, Reference Makowski and Schinzel22, Reference Makowski and Schinzel23, Reference Nesterenko and Shorey26, Reference Saradha29–Reference Yuan37]). 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 [Reference Abdollahi and Mohammadi Hassanabadi1, Reference Bateman and Stemmler4, Reference He and Togbé13, Reference Leung, Ma and Schmidt19, Reference Luca21]). For example, while discussing the partial geometries admitting Singer groups in combinatorics, Leung et al. [Reference Leung, Ma and Schmidt19] 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
$$ \begin{align} (x,y)=(2^r,3), \end{align} $$
where r is a positive integer. In [Reference Leung, Ma and Schmidt19], 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
$$ \begin{align} (x,y)=(p^r,p^s+1). \end{align} $$
Thus, we generalise the above-mentioned result in [Reference Leung, Ma and Schmidt19] to prove the following theorem.
Combining Theorem 1.1 and [Reference Leung, Ma and Schmidt19, Corollary 37] with
$q=\alpha +1=2^s+1$
, we immediately obtain the following corollary which may be regarded as a generalisation of [Reference Leung, Ma and Schmidt19, 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 [Reference Leung, Ma and Schmidt19, (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 [Reference Bugeaud and Shorey8]. In Section 3, using a lower bound for linear forms in three logarithms due to Matveev [Reference Matveev24], 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 [Reference Dujella and Pethő11], based on [Reference Baker and Davenport2, 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 [Reference Ljunggren20]
The equation
$$ \begin{align} \frac{X^k-1}{X-1}=Y^l \quad \mbox{for all } X,Y,k,l \in {\mathbb{N}}, X>1, Y>1, k>2, l>1, \end{align} $$
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
$$ \begin{align} D_1X^2+D_2=p^Z\quad \mbox{for all } X,Z \in {\mathbb{N}}. \end{align} $$
Combining the results in [Reference Bugeaud, Mignotte and Siksek7, Reference Bugeaud and Shorey8], 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:
-
(i)
$N(1,7,2)=5$
,
$(X,Z)=(1,3),(3,4),(5,5),(11,7)$
and
$(181,15)$
; -
(ii)
$N(3,5,2)=3$
,
$(X,Z)=(1,3),(3,5)$
and
$(13,9)$
; -
(iii)
$N(7,1,2)=2$
,
$(X,Z)=(1,3)$
and
$(3,6)$
; -
(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
$k>1$
; -
(v)
$N(3,29,2)=2$
,
$(X,Z)=(1,5)$
and
$(209,17)$
; -
(vi)
$N(5,3,2)=2$
,
$(X,Z)=(1,3)$
and
$(5,7)$
; -
(vii)
$N(13,3,2)=2$
,
$(X,Z)=(1,4)$
and
$(71,16)$
; -
(viii)
$N(21,11,2)=2$
,
$(X,Z)=(1,5)$
and
$(79,17)$
; and -
(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:
-
(i)
$N(2,1,3)=3$
,
$(X,Z)=(1,1), (2,2)$
and
$(11,5)$
; and -
(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
$((2p^k+\delta )a,3k)$
.
Proof. We now assume that
$(x,y,m,n)$
is a solution of (1.1) with (1.3). Then
$$ \begin{align} \frac{p^{rm}-1}{p^r-1}=\frac{(p^s+1)^n-1}{p^s}. \end{align} $$
When
$r=s$
, by (2.3),
$$ \begin{align} \frac{p^{r(m+1)}-1}{p^r-1}=(p^r+1)^n. \end{align} $$
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),
$$ \begin{align} \frac{(p^{2r})^{(m+1)/2}-1}{p^{2r}-1}=(p^r+1)^{n-1}. \end{align} $$
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),
$$ \begin{align} (p^r-1)(p^s+1)^n+(p^s-p^r+1)=p^{rm+s}. \end{align} $$
Since
$r<s$
,
$p^r-1$
,
$p^s+1$
and
$p^s-p^r+1$
are positive integers satisfying
$$ \begin{align} \gcd((p^r-1)(p^s+1),p^s-p^r+1)=1,\quad p \nmid (p^r-1)(p^s+1)(p^s-p^r+1). \end{align} $$
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
$$ \begin{align} N(p^r-1,p^s-p^r+1,p) \ge 2. \end{align} $$
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
$$ \begin{align} N((p^r-1)(p^s+1),p^s-p^r+1,p) \ge 2. \end{align} $$
Applying Lemmas 2.2 and 2.3 to (2.9), we can only obtain
$$ \begin{align} (p,r,x)=(2,1,2). \end{align} $$
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
$$ \begin{align} p^r\bigg(\frac{p^{r(m-1)}-1}{p^r-1}\bigg)=(n-1)+\sum_{j=2}^n\binom{n}{j}p^{s(j-1)}. \end{align} $$
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
$$ \begin{align} p^r\, \|\, (n-1)+\sum_{j=2}^n\binom{n}{j}p^{s(j-1)}. \end{align} $$
Let
$$ \begin{align} p^t\,\|\,n-1 \end{align} $$
and
$$ \begin{align} p^{t_j}\,\|\,j \quad \mbox{for all } j=2,\ldots,n, \end{align} $$
where t is a positive integer and
$t_j\ (j=2,\ldots ,n)$
are nonnegative integers. Then
$$ \begin{align} t_j \le \frac{\log j}{\log p}\le j-1 \quad \mbox{for all } j=2,\ldots,n. \end{align} $$
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
$$ \begin{align} t_j<j-1 \quad \mbox{for all } j=2,\ldots,n. \end{align} $$
Hence, since
$\gcd (j,j-1)=1$
and
$(p,s)\ne (2,1)$
, by (3.3), (3.4) and (3.6),
$$ \begin{align} \binom{n}{j}p^{s(j-1)}\equiv n(n-1)\binom{n-2}{j-2}\frac{p^{s(j-1)}}{(j-1)j}\equiv 0 \pmod{p^{t+1}} \quad \mbox{for all } j=2,\ldots,n. \end{align} $$
Therefore, by (3.3) and (3.7),
$$ \begin{align} p^t\,\|\,(n-1)+\sum_{j=2}^n\binom{n}{j}p^{s(j-1)}. \end{align} $$
$$ \begin{align} t=r. \end{align} $$
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 ([Reference Matveev24, Reference Mignotte and Voutier25])
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
$$ \begin{align} D&=[\,{\mathbb{Q}}(\alpha_1,\alpha_2,\alpha_3):{\mathbb{Q}}\,],\quad D'=[\,{\mathbb{R}}(\alpha_1,\alpha_2,\alpha_3):{\mathbb{R}}\,], \kern1pt\qquad\end{align} $$
$$ \begin{align} A_j&\ge \max\{D\mathop{\mathrm{h}}\nolimits(\alpha_j),|\log \alpha_j|\} \quad \mbox{for } j=1,2,3, \qquad\qquad\qquad\, \end{align} $$
$$ \begin{align} B&\ge \max\bigg\{b_j\frac{A_j}{A_1}\,\bigg|\,j=1,2,3\bigg\}, \qquad\qquad\qquad\qquad\qquad\qquad\kern1pt\end{align} $$
$$ \begin{align} C&=\frac{5\times 16^5}{6D'}e^3(7+2D')\bigg(\frac{3e}{2}\bigg)^{D'}(26.25+\log(D^2\log(e D))).\ \ \end{align} $$
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 [Reference Leung, Ma and Schmidt19], the proposition holds for
$(p,s)=(2,1)$
. We can therefore assume that
$(p,s)\ne (2,1)$
. By (2.3),
$$ \begin{align} (p^r-1)(p^s+1)^n=p^{rm+s}+(p^r-p^s-1). \end{align} $$
Since
$p^r-p^s-1>0$
, taking the logarithms of both sides of (3.14),
$$ \begin{align} \log(p^r-1)+n\log(p^s+1)=(rm+s)\log p+\log \bigg(1+\frac{p^r-p^s-1}{p^{rm+s}}\bigg). \end{align} $$
Since
$\log (1+\varepsilon )<\varepsilon $
for any
$\varepsilon>0$
, by (3.15),
$$ \begin{align} \begin{array}{ll} &0 <\log(p^r-1)+n\log(p^s+1)-(rm+s)\log p\\[6pt] &\quad =\log \bigg(1+\dfrac{p^r-p^s-1}{p^{rm+s}}\bigg)<\dfrac{p^r-p^s-1}{p^{rm+s}}. \end{array} \end{align} $$
Take
$$ \begin{align} \alpha_1=p^r-1,\quad \alpha_2=p^s+1,\quad \alpha_3=p,\quad b_1=1,\quad b_2=n,\quad b_3=rm+s \end{align} $$
and
$$ \begin{align} \varLambda=\log(p^r-1)+n\log(p^s+1)-(rm+s)\log p. \end{align} $$
By (3.16) and (3.18), we have
$\varLambda>0$
and
$$ \begin{align} (rm+s)\log p+\log \varLambda<\log(p^r-p^s-1)<\log(p^r-1). \end{align} $$
In order to apply Lemma 3.2, by (3.10), (3.11) and (3.17), we can choose the following parameters.
$$ \begin{align} D& =D'=1, \end{align} $$
$$ \begin{align} A_1=\log(p^r-1),\quad A_2& =\log(p^s+1),\quad A_3=\log p. \end{align} $$
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
$$ \begin{align} C<1.691\times 10^{10}. \end{align} $$
Applying Lemma 3.2 to (3.17) and (3.18), by (3.20)–(3.22),
$$ \begin{align} \log \varLambda&>-1.691\times 10^{10}(\log(p^r-1))(\log(p^s+1))(\log p)\nonumber\\ &\quad\times \bigg(1.406+\log \bigg(\dfrac{(rm+s)\log p}{\log(p^r-1)}\bigg)\bigg). \end{align} $$
Substituting (3.23) into (3.19), we get
$$ \begin{align} 1+1.691\times 10^{10}(\log(p^s+1))(\log p)\bigg(1.406+\log\bigg(\dfrac{(rm+s)\log p}{\log(p^r-1)}\bigg)\bigg)>\dfrac{(rm+s)\log p}{\log(p^r-1)}. \end{align} $$
Hence, since
$(p,s)\ne (2,1)$
and
$p^s+1 \ge 4$
, by (3.23), we can calculate that
$$ \begin{align} \frac{(rm+s)\log p}{\log(p^r-1)}<1.501\times 10^{12}(\log(p^s+1))(\log p)(\log \log(p^s+1)). \end{align} $$
On the other hand, by (3.16),
$$ \begin{align} \dfrac{(rm+s)\log p}{\log(p^r-1)}&>\bigg(1-\dfrac{p^r-p^s-1}{p^{rm+s}\log(p^r-1)}\bigg)+\dfrac{n\log(p^s+1)}{\log(p^r-1)}\nonumber\\ &>\dfrac{n\log(p^s+1)}{\log(p^r-1)}. \end{align} $$
Since
$\log p \le (\log p^r)/2$
for
$r \ge 2$
, the combination of (3.25) and (3.26) yields
$$ \begin{align} n&<1.501\times 10^{12}(\log p)(\log(p^r-1))(\log \log(p^s+1))\nonumber\\ &<7.505\times 10^{11}( \log p^r)^2(\log \log p^r). \end{align} $$
Further, since
$(p,s)\ne (2,1)$
, by Lemma 3.1, we have
$n>p^r$
. Hence, by (3.27),
$$ \begin{align} p^r<7.505\times 10^{11}(\log p^r)^2(\log \log p^r). \end{align} $$
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),
$$ \begin{align} m'<1.501\times 10^{12}(\log(p^r-1))(\log(p^s+1))(\log \log(p^s+1)). \end{align} $$
Since Proposition 3.3 implies that
$p^s \le p^{r-1}<1.718\times 10^{15}$
, we see from (4.1) that
$$ \begin{align} m'<6.702\times 10^{15}. \end{align} $$
On the other hand, we deduce from Lemma 3.1 and (3.26) that
$$ \begin{align} m'>\frac{n\log(p^s+1)}{\log p}>\frac{p^r \log(p^r+1)}{\log p}. \end{align} $$
Now, by (3.16),
$$ \begin{align} 0<n-m'\kappa+\mu<AB^{-m'}, \end{align} $$
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 [Reference Dujella and Pethő11, Lemma 5a] if the middle term of inequalities (4.4) is multiplied by
$-1$
, the lemma is proved in the same way as [Reference Dujella and Pethő11, 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
$$ \begin{align} (p,r,s)\in \{(2,3,2),(2,4,2),(2,4,3),(3,2,1)\}. \end{align} $$
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.
Acknowledgement
The authors thank the anonymous referee for suggesting the addition of a connection between Theorem 1.1 and the results in [Reference Leung, Ma and Schmidt19], which led to Corollary 1.2.
