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\begin{center} \vskip 1cm{\LARGE\bf 
On the Ternary Purely Exponential \\
\vskip .01in
Diophantine  Equation $(ak)^x+(bk)^y=((a+b)k)^z$ \\
\vskip .11in
for Prime Powers $a$ and $b$}
			
		\vskip 1cm 
		\large 
		{Maohua Le}\\
		
		Institute of Mathematics\\ 
		Lingnan Normal College\\
		Zhanjiang, Guangdong 524048\\ 
		China\\
		\href{mailto: lemaohua2008@163.com}{\tt lemaohua2008@163.com}\\
		\ \\
		{G\"{o}khan Soydan}\\
		Department of Mathematics \\
		Bursa Uluda\u{g} University\\ 
		16059 Bursa\\T\"urkiye\\
		\href{mailto: gsoydan@uludag.edu.tr}{\tt gsoydan@uludag.edu.tr}\\ 
	\end{center}
	
	\vskip .2 in
	
\begin{abstract} Let $k$ be a positive integer, and let $a,b$ be coprime
positive integers with $a, b>1$. In this paper, using a combination
of some elementary number theory techniques with classical results on the
Nagell-Ljunggren equation, the Catalan equation, and some new properties
of the Lucas sequence, we prove that if $k>1$ and $a,b >2$ are both prime
powers, then the equation $(ak)^x+(bk)^y=((a+b)k)^z$
has only one positive integer solution:  namely, $(x,y,z)=(1,1,1)$.
This proves some cases of a conjecture of Yuan and Han.
\end{abstract}

	\section{Introduction}
	\label{sec1} 
	Let
	$\mathbb{Z}$, $\mathbb{N}$, $\mathbb{P}$ be the sets of all
	integers, positive integers, and odd primes, respectively. Let
	$A,B,C$ be fixed positive integers with $A,B,C>1$. In
	recent decades, the ternary purely exponential Diophantine
	equation \begin{equation} \label{eq.1.1}
		A^x+B^y=C^z, \quad x,y,z\in\mathbb{N}
	\end{equation} has yielded very rich results, but some important
	problems about it are far from being solved (see\ \cite{LSS}). In
	recent 20 years, many authors have considered equation
	\eqref{eq.1.1} when $A$, $B$, and $C$ are Fibonacci \seqnum{A00045}
	or Lucas \seqnum{A000204} or Pell \seqnum{A000129} numbers in
	OEIS\ \cite{OEIS} or when $A$ and $B$ are Fibonacci or Lucas or
	Pell numbers (see\ \cite{AHRT,BrLu,DdLu,KoLu,RFLT,Si,ZaTo}).

	Let $k$ be a positive integer, and let $a,b$ be coprime
	positive integers with $a,b>1$. In this paper, we discuss
	\eqref{eq.1.1} for $(A,B,C)=(ak,bk,(a+b)k)$. Then \eqref{eq.1.1}
	can be written as \begin{equation}\label{eq.1.2}
		(ak)^x+(bk)^y=((a+b)k)^z,\quad x,y,z\in\mathbb{N}.
	\end{equation} Obviously, for any $a,b$ and $k$, \eqref{eq.1.2}
	has the solution $(x,y,z)=(1,1,1)$. A solution $(x,y,z)$
	of \eqref{eq.1.2} with $(x,y,z)\neq (1,1,1)$ is called
	exceptional. In 2018, Sun and Tang\ \cite{SuTa} proved that if
	$k>1$, $a,b>2$, and $(x,y,z)$ is an exceptional solution
	of \eqref{eq.1.2}, then either $x>z>y$ or $y>z>x$. On this
	basis, they further proved that if $k>1$ and $(a,b)\in\{(3,5),
	(5,8),(8,13),(13,21)\}$, then \eqref{eq.1.2} has no exceptional
	solutions. In the same year, Yuan and Han\ \cite{YuHa} proposed
	the following conjecture.
\begin{conjecture}\label{con} For any $k$, if $a,b>3$, then
\eqref{eq.1.2} has no exceptional solutions.  \end{conjecture}
This
conjecture is formally similar to Je\'smanowicz' conjecture
concerning Pythagorean triples (see\ \cite{Jes,SCDT}). So far, it has
only been solved for some very special cases. For example, the authors
of\ \cite{YuHa} proved that if $a$ and $b$ are squares with $b\equiv 4
\pmod{8}$, then \eqref{eq.1.2} has no solutions $(x,y,z)$ with $y>z>x$,
in particular, if $a$ is a square and $b=4$, then Conjecture \ref{con}
is true. Afterwards, using Baker's method, Le and Soydan\ \cite{LS}
proved that if $a$ and $b$ are squares with $a>64b^3$, then \eqref{eq.1.2}
has no solutions $(x,y,z)$ with $x>z>y$. It implies that if $a$ and $b$
are squares with $a>64b^3$ and $b\equiv 4 \pmod{8}$, then Conjecture
\ref{con} is true.

In this paper, using a combination of some elementary number theory
techniques with classical results on the Nagell-Ljunggren equation,
the Catalan equation and some new properties of the Lucas sequence
\seqnum{A000204}, we prove the following result.
	\begin{theorem}\label{theo.1.2}
		Let $r,s$ be positive integers, and let $p,q$ be
		distinct odd primes. If $k>1$ and $a,b$ satisfy one of
		the following conditions: \begin{enumerate} \item[\rm
		(i)] $(a,b)=(2^r,p^s)$ with $r>1$; or \item[\rm (ii)]
		$(a,b)=(p^r,2^s)$ with $s>1$; or \item[\rm (iii)]
		$(a,b)=(p^r,q^s)$, then \eqref{eq.1.2} has no solutions
		$(x,y,z)$ with $x>z>y$.
     \end{enumerate}
	\end{theorem} Since $a$ and $b$ are symmetric in \eqref{eq.1.2},
	by the first result mentioned in\ \cite{SuTa}, we can obtain
	the following corollary from Theorem \ref{theo.1.2} immediately.
	\begin{corollary}\label{cor.1.3}
		If $k>1$ and $a,b$ are prime powers, then Conjecture
		\ref{con} is true.
	\end{corollary} Finally, we briefly analyze the effect of
	the above mentioned results. For any enough large positive
	integer $N$, let $F(N)$ denote the number of pairs $(a,b)$
	such that $a,b\le N$ and they have been proved
	to hold for Conjecture \ref{con} with $k>1$. Clearly,
	by the mentioned results in\ \cite{LS,SuTa,YuHa} and
	Corollary \ref{cor.1.3}, we have $F(N)=4$, $F(N)=\sqrt{N}$,
	$N>F(N)>\sqrt{N}$, and $F(N)>N^2/(\log N)^2$, respectively.

\section{Preliminaries}
	\label{sec2} 
	Let us now
	recall that if $\alpha$ and $\beta$ are roots of a quadratic
	equation of the form $x^2-rx-s=0$ for nonzero coprime integers
	$r$ and $s$ and such that $\alpha/\beta$ is not a root of
	unity, then the sequence $(u_{\ell})_{\ell\ge 0}$ with general
	term \begin{equation*}
		u_{\ell}=\dfrac{\alpha^{\ell}-{\beta}^{\ell}}{\alpha-{\beta}}%
		\qquad {\text{for~all}}~\ell\ge 0
	\end{equation*} is called a \textit{Lucas sequence}
	\seqnum{A000204}. It can also be defined inductively as $
	u_0=0$, $u_1=1$, and $u_{\ell+2}=r\cdot u_{\ell+1}+s\cdot
	u_{\ell}$. When $\beta=1$, Lemmas
	\ref{lem.2.6}, \ref{lem.2.7}, and \ref{lem.2.8} below are three new
	properties we proved about the Lucas sequence.
	\begin{lemma}[\cite{Lj}]\label{lem.2.1}
		The equation \begin{equation*}
			\dfrac{X^m-1}{X-1}=Y^n,\,\,\,
			X,Y,m,n\in\mathbb{N},\,\,\,
			X>1,\,\,Y>1,\,\,m>2,\,\,n>1
		\end{equation*} has only the solution
		$(X,Y,m,n)=(3,11,5,2)$ with $2\mid n$.
	\end{lemma}

	\begin{lemma}[\cite{Mih}]\label{lem.2.2}
		The equation \begin{equation*}
			X^m-Y^n=1,\,\,\,
			X,Y,m,n\in\mathbb{N},\,\,\,X,Y,m,n>1
		\end{equation*} has only the solution
		$(X,Y,m,n)=(3,2,2,3)$.
	\end{lemma}

	Let $X,\ell,m,n$ be positive integers with $X>1$ and $\ell>1$,
	and let $p,q$ be odd primes. Three lemmas for divisibility are
	given directly below.

	\begin{lemma}[\cite{Nag}]\label{lem.2.3}
		If $X^n+1\equiv 0 \pmod{X^m+1}$, then $m\mid n$ and $n/m$
		is odd.
	\end{lemma}

	\begin{lemma}[\cite{BV}]\label{lem.2.4}
		Let $\ell$ be an odd prime. If $X\not \equiv 1
		\pmod{\ell}$, then every prime divisor $p$ of
		$(X^{\ell}-1)/(X-1)$ satisfies \begin{equation}
		\label{eq.2.1}
			p\equiv \modd{1} {2\ell}.
		\end{equation} If $X \equiv \modd{1} {\ell}$, then $\ell
		\mid \mid (X^{\ell}-1)/(X-1)$ and every prime divisor $p$
		of $(X^{\ell}-1)/\ell (X-1)$ satisfies \eqref{eq.2.1}.
	\end{lemma}

	\begin{lemma}[\cite{BV}]\label{lem.2.5}
		When $X \equiv 1 \pmod{p}$, $p^m \mid \mid
		(X^{\ell}-1)/(X-1)$ if and only if $p^m \mid \mid  \ell$.
	\end{lemma}

	\begin{lemma}\label{lem.2.6}
		If \begin{equation}\label{eq.2.2}
			\dfrac{X^{\ell}-1}{X-1}=p^n,\,\,2 \nmid \ell,
		\end{equation} then $\ell$ is an odd prime with
		\eqref{eq.2.1}.
	\end{lemma} \begin{proof}
		We now assume that $\ell$ is not an odd prime. Since
		$\ell>1$ and $2\nmid \ell$, $\ell$ has an odd prime
		divisor $q$ with $q<\ell$. Then we have $q\mid
		\ell$ and $\ell/q>1$. By \eqref{eq.2.2}, we get
		\begin{equation}\label{eq.2.3}
			p^n=\dfrac{X^{\ell}-1}{X-1}=\left(\dfrac{X^{\ell/q}-1}{X-1}\right)\left(\dfrac{(X^{\ell/q})^q-1}{X^{\ell/q}-1}\right),
		\end{equation} where $(X^{\ell/q}-1)/(X-1)$
		and $((X^{\ell/q})^q-1)/(X^{\ell/q}-1)$ are
		positive integers greater than 1. Since $X>1$,
		by \eqref{eq.2.3}, we have \begin{equation}\label{eq.2.4}
		\dfrac{X^{\ell/q}-1}{X-1}=p^f,\,\,\dfrac{(X^{\ell/q})^q-1}{X^{\ell/q}-1}=p^{n-f},\,\,f\in\mathbb{N},\,\,f<n.
		\end{equation} Further, by the first and the second
		equalities of \eqref{eq.2.4}, we get $X^{\ell/q}\equiv
		1 \pmod{p}$ and \begin{equation}\label{eq.2.5}
			0\equiv p^{n-f}\equiv
			\dfrac{(X^{\ell/q})^q-1}{X^{\ell/q}-1}\equiv
			(X^{\ell/q})^{q-1}+\cdots +X^{\ell/q}+1\equiv
			q \pmod{p}.
		\end{equation} Since $p$ and $q$ are odd primes,
		by \eqref{eq.2.5}, we obtain $q=p$. Hence by Lemma
		\ref{lem.2.4}, we see from \eqref{eq.2.4} that $p\mid \mid
		((X^{\ell/p})^p-1)/(X^{\ell/p}-1)=((X^{\ell/q})^q-1)/(X^{\ell/q}-1)$
		and \begin{equation}\label{eq.2.6}
			p= \dfrac{(X^{\ell/p})^p-1}{X^{\ell/p}-1}=
			(X^{\ell/p})^{p-1}+\cdots+ X^{\ell/p}+1>p,
		\end{equation} a contradiction. It implies that
		$\ell$ must be an odd prime. Moreover, using Lemma
		\ref{lem.2.4} again, if $X\equiv 1 \pmod{\ell}$, then
		from \eqref{eq.2.2} we can get $\ell=p$ and $n=1$, which
		is the same contradiction as \eqref{eq.2.6}. Therefore,
		we have $x\not \equiv 1 \pmod{\ell}$ and $p$ satisfies
		\eqref{eq.2.1}. The lemma is proved.
	\end{proof} \begin{lemma}\label{lem.2.7}
		If \begin{equation}\label{eq.2.7}
			X^{\ell}-1=2^mp^n,
		\end{equation} then one of the following three
		conclusions holds.  \begin{enumerate} \item[\rm (i)]
		$(p,X,\ell,m,n)=(3,5,2,3,1),(3,7,2,4,1),(5,9,2,4,1),(5,3,4,4,1),(3,17,2,5,2)$
		or\\ $(7,15,2,5,1)$.  \item[\rm
		(ii)]\begin{equation}\label{eq.2.8}
			\ell=2,\,\, m\ge 6,\,\, n=1,\,\,
			X=2^{m-1}+\zeta,\,\, p=2^{m-2}-\zeta,\,\,
			\zeta\in\{1,-1\}.
		\end{equation} \item[\rm (iii)]
		\begin{equation}\label{eq.2.9}
			\ell\,\, \text{is an odd prime},\,\,\,
			X-1=2^m,\,\, \dfrac{X^{\ell}-1}{X-1}=p^n,\,\,
			p\equiv \modd{1} {2\ell}.
		\end{equation}
	\end{enumerate} \end{lemma} \begin{proof}
		Obviously, by \eqref{eq.2.7}, we have $X>1$, $2\nmid X$
		and $p\nmid X$. When $2\mid \ell$, since $X^{\ell}\equiv
		1 \pmod{8}$, $\gcd(X^{\ell/2}+1,X^{\ell/2}-1)=2$
		and $p$ is an odd prime, by \eqref{eq.2.7}, we have
		\begin{equation}\label{eq.2.10}
			X^{\ell/2}+\zeta=2p^n,\,\,X^{\ell/2}-\zeta=2^{m-1},\,\,m\ge
			3,\,\,\zeta=((-1)^{(X-1)/2})^{\ell/2},
		\end{equation} where \begin{equation}\label{eq.2.11}
			\zeta\in\{1,-1\}.
		\end{equation} If $3\le m\le 5$, then from \eqref{eq.2.10}
		and \eqref{eq.2.11} we can easily obtain the conclusion
		\rm (i). If $m\ge 6$, by Lemma \ref{lem.2.2}, then from
		the second equality of \eqref{eq.2.10} we get $\ell=2$. In
		addition, eliminating $X^{\ell/2}$ from \eqref{eq.2.10},
		we have \begin{equation}\label{eq.2.12}
			p^n-2^{m-2}=\zeta.
		\end{equation} Since $m-2\ge 4$, using Lemma \ref{lem.2.2}
		again, we see from \eqref{eq.2.12} that $n=1$. Therefore,
		by \eqref{eq.2.10}, \eqref{eq.2.11}, and \eqref{eq.2.12},
		we obtain \eqref{eq.2.8} and conclusion \rm (ii)
		is proved.

		When $2\nmid \ell$, since $\ell>1$ and
		$\dfrac{X^{\ell}-1}{X-1}$ is an odd positive
		integer greater than 1, by \eqref{eq.2.7}, we have
		\begin{equation}\label{eq.2.13}
			X-1=2^mp^f,\,\,\dfrac{X^{\ell}-1}{X-1}=p^{n-f},\,\,f\in\mathbb{Z},\,\,0\le
			f<n.
		\end{equation} If $f>0$, then from the first equality
		of \eqref{eq.2.13} we get $X\equiv 1 \pmod{p}$. Hence,
		applying Lemma \ref{lem.2.5} to the second equality
		of \eqref{eq.2.13}, we obtain $p^{n-f}\mid \mid
		\ell$. However, since $\ell\ge p^{n-f},$ we get
		$p^{n-f}=(X^{\ell}-1)/(X-1)>\ell \ge p^{n-f},$
		a contradiction. So  $f=0$. Therefore, by
		\eqref{eq.2.13}, we get \begin{equation}\label{eq.2.14}
			X-1=2^m,\,\,\dfrac{X^{\ell}-1}{X-1}=p^n.
		\end{equation} Further, by Lemma \ref{lem.2.6},
		we see from the second equality of \eqref{eq.2.14}
		that $\ell$ is an odd prime with \eqref{eq.2.1}. Thus,
		we obtain \eqref{eq.2.9} and the conclusion \rm (iii)
		is proved. The proof is completed.
	\end{proof} Using the same method as in the proof of Lemma
	\ref{lem.2.7}, we can obtain the following lemma without
	difficulty \begin{lemma}\label{lem.2.8}
		If \begin{equation}\label{eq.2.15}
			X^{\ell}-1=p^mq^n,
		\end{equation} the one of the following six
		conclusions holds.  \begin{enumerate} \item[\rm (i)]
		\begin{equation}\label{eq.2.16}
			2\mid
			\ell,\,\,X^{\ell/2}+\zeta=p^m,\,\,X^{\ell/2}-\zeta=q^n,\,\,\zeta\in\{1,-1\}.
		\end{equation} \item[\rm (ii)]
		\begin{equation}\label{eq.2.17}
			2\nmid \ell,\,\,X=2,\,\,2^{\ell}-1=p^mq^n.
		\end{equation} \item[\rm (iii)]
		\begin{equation}\label{eq.2.18}
			\ell=p,\,\,m>1,\,\,X-1=p^{m-1},\,\,\dfrac{X^p-1}{X-1}=
			pq^n,\,\,q\equiv\modd{1} {2p}.
		\end{equation} \item[\rm (iv)]
		\begin{equation}\label{eq.2.19}
			\ell=q,\,\,n>1,\,\,X-1=q^{n-1},\,\,\dfrac{X^q-1}{X-1}=p^mq,\,\,p\equiv
			\modd{1} {2q}.
		\end{equation}
			\item[\rm (v)]
		\begin{equation}\label{eq.2.20}
			\ell \,\,\text{is an odd
			prime},\,\,X-1=p^m,\,\,\dfrac{X^{\ell}-1}{X-1}=q^n,\,\,q\equiv
			\modd{1} {2\ell}.
		\end{equation} \item[\rm (vi)]
		\begin{equation}\label{eq.2.21}
			\ell \,\,\text{is an odd
			prime},\,\,X-1=q^n,\,\,\dfrac{X^{\ell}-1}{X-1}=p^m,\,\,p\equiv
			\modd{1} {2\ell}.
		\end{equation}
	\end{enumerate} \end{lemma}

	\begin{proof}
		Since $p$ and $q$ are distinct odd primes, by
		\eqref{eq.2.15}, we have $2\mid X$. When $2\mid
		\ell$, since $\gcd(X^{\ell/2}+1,X^{\ell/2}-1)=1$, by
		\eqref{eq.2.15}, we can directly obtain \eqref{eq.2.16}
		and the conclusion \rm (i) is proved. When $2\nmid
		\ell$ and $X=2$, we see from \eqref{eq.2.15} that
		\eqref{eq.2.17} is clearly true and the conclusion
		\rm (ii) is proved. When $2\nmid \ell$ and $X>2$, by
		\eqref{eq.2.15}, we have \begin{equation}\label{eq.2.22}
			\begin{aligned}
				&X-1=p^fq^g,\,\,\dfrac{X^{\ell}-1}{X-1}=p^{m-f}q^{n-g},\,\,f,g\in\mathbb{Z},\,\,0\le
				f\le m,\\ &0\le g\le n,\,\, (f,g)\neq
				(0,0)\,\,\text{or}\,\,(m,n).
			\end{aligned}
		\end{equation}

		If $0<f<m$ and $0<g<n$, then from \eqref{eq.2.22} we get
		$X\equiv 1 \pmod{pq}$ and $0\equiv p^{m-f}q^{n-g}\equiv
		(X^{\ell}-1)(X-1)\equiv X^{\ell-1}+\cdots +X+1\equiv
		\ell \pmod{pq}$. It follows that $pq\mid \ell$ and
		$(X^{pq}-1)/(X-1)\mid (X^{\ell}-1)(X-1)$. Hence,
		by the second equality of \eqref{eq.2.22}, we have
		\begin{equation}\label{eq.2.23}
			\begin{aligned}
				&\dfrac{X^{pq}-1}{X-1}=\left(\dfrac{X^{p}-1}{X-1}\right)\left(\dfrac{(X^p)^q-1}{X^p-1}\right)=p^{f'}q^{g'},\,\,f',g'\in\mathbb{Z},\\
				&f'\ge 0,\,\,g'\ge 0,\,\,(f',g')\neq
				(0,0),
			\end{aligned}
		\end{equation} where $(X^p-1)/(X-1)$ and
		$((X^p)^q-1)/(X^p-1)$ are positive integers greater
		than 1. Further, since $X\equiv 1 \pmod{p}$,
		by Lemma \ref{lem.2.5}, we get $p\mid \mid
		(X^p-1)/(X-1)$. Furthermore, since $(X^p-1)/(X-1)>p,$ by
		\eqref{eq.2.23}, we have \begin{equation}\label{eq.2.24}
			\dfrac{X^p-1}{X-1}=pq^{g''},\,\,g''\in \mathbb{N}.
		\end{equation} Recall that $X\equiv 1 \pmod{q}$. By
		\eqref{eq.2.24}, we get $0\equiv pq^{g''}\equiv
		(X^p-1)/(X-1)\equiv X^{p-1}+\cdots +X+1\equiv p \pmod{q}$
		and $p=q$, a contradiction. Therefore, the case $0<f<m$
		and $0<g<n$ is impossible.

		If $0<f<m$ and $g=n$, then from \eqref{eq.2.22} we get
		\begin{equation}\label{eq.2.25}
			X-1=p^fq^n,\,\,\dfrac{X^{\ell}-1}{X-1}=p^{m-f}.
		\end{equation} By the first equality of \eqref{eq.2.25},
		we have $X\equiv 1 \pmod{p}$. In addition, by Lemma
		\ref{eq.2.6}, we see from the second equality of
		\eqref{eq.2.25} that $\ell$ is an odd prime with $p\equiv
		1 \pmod{2 \ell}$. But, since $X\equiv 1 \pmod{p}$,
		we get $0\equiv p^{m-f}\equiv (X^{\ell}-1)/(X-1)\equiv
		X^{\ell-1}+\cdots +X+1\equiv \ell \pmod{p}$ and $p=\ell$,
		a contradiction. Therefore, the case $0<f<m$ and $g=n$
		is impossible. Using the same method, we can eliminate
		the case $f=m$ and $0<g<n$.

		Recall that $(f,g)\neq (0,0)$ or $(m,n)$. According
		to the above analysis, we are left with only the cases
		\begin{equation}\label{eq.2.26}
			f=0,\,\,0<g\le n
		\end{equation} and \begin{equation}\label{eq.2.27}
			0<f\le m,\,\,g=0
		\end{equation} that have not yet been discussed.

		If \eqref{eq.2.26} holds, by \eqref{eq.2.22}, then we
		have \begin{equation}\label{eq.2.28}
			X-1=q^g,\,\,\dfrac{X^{\ell}-1}{X-1}=p^mq^{n-g},\,\,g\in\mathbb{N},\,\,g\le
			n.
		\end{equation} When $g=n$, by \eqref{eq.2.28}, we get
		\begin{equation}\label{eq.2.29}
			X-1=q^n,\,\,\dfrac{X^{\ell}-1}{X-1}=p^m.
		\end{equation} Applying Lemma \ref{lem.2.6} to the second
		equality of \eqref{eq.2.29}, $\ell$ is an odd prime with
		$p\equiv \modd{1} {2\ell}$. Hence, by \eqref{eq.2.29}, we
		get \eqref{eq.2.21} and the conclusion \rm(vi) is proved.

		When $g<n$, by the first equality of \eqref{eq.2.28}, we
		have $X\equiv 1\pmod{q}$. Hence, by Lemma \ref{lem.2.5},
		we see from the second equality of \eqref{eq.2.28}
		that $q^{n-g}\mid \mid \ell$. It follows that
		\begin{equation}\label{eq.2.30}
			\ell=q^{n-g}\ell_1,\,\,\ell_1\in\mathbb{N},\,\,q\nmid
			\ell_1.
		\end{equation} Then we have
		\begin{equation}\label{eq.2.31}
			\dfrac{X^{\ell}-1}{X-1}=\left(\dfrac{X^{\ell_1}-1}{X-1}\right)\prod_{j=1}^{n-g}\left(\dfrac{X^{q^j\ell_1}-1}{X^{q^{j-1}\ell_1}-1}\right),
		\end{equation} where $(X^{\ell_1}-1)/(X-1)$
		and $(X^{q^j\ell_1}-1)/(X^{q^{j-1}\ell_1}-1)$
		$(j=1,\cdots ,n-g)$ are positive integers with
		\begin{equation}\label{eq.2.32}
			q\nmid \dfrac{X^{\ell_1}-1}{X-1},\,\,q \mid \mid
			\dfrac{X^{q^j\ell_1}-1}{X^{q^{j-1}\ell_1}-1},\,\,j=1,\cdots
			, n-g.
		\end{equation} By \eqref{eq.2.31} and \eqref{eq.2.32},
		we get from the second equality of \eqref{eq.2.28}
		that \begin{equation}\label{eq.2.33}
			\dfrac{X^{\ell_1}-1}{X-1}=p^{m_0},\,\,m_0\in\mathbb{Z},\,\,m_0\ge
			0
		\end{equation} and \begin{equation}\label{eq.2.34}
			\dfrac{X^{q^j\ell_1}-1}{X^{q^{j-1}\ell_1}-1}=p^{m_j}q,\,\,m_j\in\mathbb{N},\,\,j=1,\cdots,n-g,
		\end{equation} where \begin{equation}\label{eq.2.35}
			m_0+m_1+\cdots +m_{n-g}=m.
		\end{equation} If $n-g>1$, then from \eqref{eq.2.34}
		we have $X^{q\ell_1}\equiv 1 \pmod{p}$ and $0\equiv
		p^{m_2}q\equiv (X^{q^2\ell_1}-1)/(X^{q\ell_1}-1)\equiv
		X^{q\ell_1(q-1)}+\cdots +X^{q\ell_1}+1\equiv q
		\pmod{p}$, whence we get $p=q$, a contradiction. So
		we have $n-g=1$. Further, if $n-g=1$ and $\ell_1>1$,
		by \eqref{eq.2.33} and \eqref{eq.2.34}, then $m_0$ is a
		positive integer, $X^{\ell_1}\equiv 1 \pmod{p}$, $0\equiv
		p^{m_1}q\equiv (X^{q\ell_1}-1)/(X^{\ell_1}-1)\equiv
		X^{\ell_1(q-1)}+\cdots +X^{\ell_1}+1\equiv q \pmod{p}$ and
		$p=q$, a contradiction. Therefore, if \eqref{eq.2.26}
		holds, then we have $n-g=1$ and $\ell_1=1$. By
		\eqref{eq.2.28},\eqref{eq.2.30},\eqref{eq.2.31},
		\eqref{eq.2.34} and \eqref{eq.2.35}, we get
		\eqref{eq.2.19} and the conclusion \rm (iv) is proved.

		Using the same method, as in the proof about the case
		\eqref{eq.2.26}, we can deduce that if \eqref{eq.2.27}
		holds, then there can only obtain \eqref{eq.2.18} or
		\eqref{eq.2.20}, and the conclusions \rm (iii) and \rm
		(v) are proved. To sum up, the proof is complete.
	\end{proof} For any positive integer $m$, let $\rad(m)$
	denote the product of all distinct prime divisors
	of $m$, and let $\rad(1)=1$. Obviously, $\rad(m)$
	is equal to the largest squarefree divisor of $m$.
	\begin{lemma}[\cite{SuTa}]\label{lem.2.9}
		If $k>1$ and $(x,y,z)$ is a solution of
		\eqref{eq.1.2} with $x>z>y$, then we have $$\rad(k)\mid
		b,\,\,b=b_1b_2,\,\,b_1,b_2\in\mathbb{N},\,\,b_1>1,\,\,\gcd(b_1,b_2)=1,$$
		$$b_1^y=k^{z-y}$$ and $$a^xk^{x-z}+b_2^y=(a+b)^z.$$
	\end{lemma} By Lemma \ref{lem.2.9}, we can obtain the following
	lemma immediately.  \begin{lemma}\label{lem.2.10}
		If $k>1$, $b$ is a prime power and $(x,y,z)$ is a
		solution of \eqref{eq.1.2} with $x>z>y$, then we have
		$$b^y=k^{z-y}$$ and $$a^xk^{x-z}+1=(a+b)^z.$$
	\end{lemma} 

	\section{Proof of Theorem \ref{theo.1.2}}
	We now assume
	that $k>1$ and $(x,y,z)$ is a solution of \eqref{eq.1.2}
	with $x>z>y$. We will prove that this solution does not
	exist in the following three cases.  

	\medskip\noindent
	\subsection{Case (i): $(a,b)=(2^r,p^s)$ with $r>1$.}
	Since
	$k>1$ and $p$ is an odd prime, Lemma \ref{lem.2.10}, we have
	\begin{equation}\label{eq.3.1}
		p^{sy}=k^{z-y}
	\end{equation} and \begin{equation}\label{eq.3.2}
		2^{rx}k^{x-z}+1=(2^r+p^s)^z.
	\end{equation} We see from \eqref{eq.3.1} that $k$ is a power
	of $p$. So we have \begin{equation}\label{eq.3.3}
		k^{x-z}=p^t,\,\,t\in\mathbb{N}.
	\end{equation} Substituting \eqref{eq.3.3} into \eqref{eq.3.2},
	we get \begin{equation}\label{eq.3.4}
		(2^r+p^s)^z-1=2^{rx}p^t.
	\end{equation} We find from \eqref{eq.3.4} that equation
	\begin{equation}\label{eq.3.5}
		X^{\ell}-1=2^mp^n,\,\,X,\ell,m,n\in\mathbb{N},\,\,X>1,\ell>1
	\end{equation} has a solution \begin{equation}\label{eq.3.6}
		(X,\ell,m,n)=(2^r+p^s,z,rx,t).
	\end{equation} Since $r\ge 2$ and $x>z>y\ge 1$, we have $x\ge
	3$ and $rx\ge 6$. Hence, by Lemma \ref{lem.2.7}, the solution
	\eqref{eq.3.6} must satisfy the conclusion (ii) or (iii)
	in this lemma.

	When \eqref{eq.3.6} satisfies the conclusion (ii) in Lemma
	\ref{lem.2.7}, by \eqref{eq.2.8} and \eqref{eq.3.6}, we have
	\begin{equation}\label{eq.3.7}
		z=2,\,\,t=1,\,\,2^r+p^s=2^{rx-1}+\zeta,\,\,p=2^{rx-2}+\zeta,\,\,\zeta\in\{1,-1\}.
	\end{equation} Since $z=2$ and $z>y\ge 1$, we get $y=1$. Hence,
	by \eqref{eq.3.1}, we have \begin{equation}\label{eq.3.8}
		k=p^s.
	\end{equation} Further, by \eqref{eq.3.3}, \eqref{eq.3.6} and
	\eqref{eq.3.8}, we get $p=p^t=k^{x-z}=p^{s(x-z)}=p^{s(x-2)}$,
	whence we obtain $s=1$ and $x=3$. Therefore, by
	the third and fourth equalities of \eqref{eq.3.7}, we have
	$$2^{3r-1}+\zeta=2^{rx-1}+\zeta=2^r+p^s=2^r+p=2^r+(2^{rx-2}+\zeta)=2^{3r-2}+2^r+\zeta,$$
	whence we get $2^r=2^{3r-1}-2^{3r-2}=2^{3r-2}$ and $r=1$,
	a contradiction.

	When \eqref{eq.3.6} satisfies the conclusion (iii) in Lemma
	\ref{lem.2.7}, by \eqref{eq.2.9} and \eqref{eq.3.6}, we have
	\begin{equation}\label{eq.3.9}
		z\,\, \text{is an odd prime},\,\,2^r+p^s-1=2^{rx},\,\,
		\dfrac{(2^r+p^s)^z-1}{2^r+p^s-1}=p^t,\,\, p\equiv
		1 \pmod{2z}.
	\end{equation} By the second equality of \eqref{eq.3.9}, we get
	\begin{equation}\label{eq.3.10}
		p^s=2^{rx}-2^r+1.
	\end{equation} If $t\ge s$, then from the third equality of
	\eqref{eq.3.9} we have \begin{equation*}
		0\equiv
		p^t\equiv\dfrac{(2^r+p^s)^z-1}{2^r+p^s-1}=\dfrac{2^{rz}-1}{2^r-1}\,\,\pmod{p^s}.
	\end{equation*} It implies that $(2^{rz}-1)/(2^r-1)$ is a positive
	integer satisfying \begin{equation}\label{eq.3.11}
		\dfrac{2^{rz}-1}{2^r-1}\ge p^s.
	\end{equation} Since $x>z$, by \eqref{eq.3.10} and
	\eqref{eq.3.11}, we have \begin{equation*}
		2^{r(z-1)+1}>\dfrac{2^{rz}-1}{2^r-1}\ge
		p^s=2^{rx}-2^r+1\ge 2^{r(z+1)}-2^r+1>2^{r(z+1)}-2^r
	\end{equation*} and $r\ge 2r-1$, a
	contradiction. So we have $t<s$. Then, since
	$z>2$, by the third equality of \eqref{eq.3.9}, we get
	$$p^s>p^t=\dfrac{(2^r+p^s)^z-1}{2^r+p^s-1}>\dfrac{(2^r+p^s)^2-1}{2^r+p^s-1}=2^r+p^s+1>p^s,$$
	a contradiction. Thus, the theorem holds for this case.

	\medskip\noindent
	\subsection{Case (ii): $(a,b)=(p^r,2^s)$ with $s>1$.}
	By Lemma
	\ref{lem.2.10}, we have \begin{equation}\label{eq.3.12}
		2^{sy}=k^{z-y}
	\end{equation} and \begin{equation}\label{eq.3.13}
		p^{rx}k^{x-z}+1=(p^r+2^s)^z.
	\end{equation} We see from \eqref{eq.3.12} that $k$ is a power
	of 2. So we have \begin{equation}\label{eq.3.14}
		k^{x-z}=2^t,\,\,t\in\mathbb{N}.
	\end{equation} Substituting \eqref{eq.3.14} into \eqref{eq.3.13},
	we get \begin{equation}\label{eq.3.15}
		(p^r+2^s)^z-1=2^tp^{rx}.
	\end{equation} We find from \eqref{eq.3.15} that \eqref{eq.3.5}
	has a solution \begin{equation}\label{eq.3.16}
		(X,\ell,m,n)=(p^r+2^s,z,t,rx).
	\end{equation} Since $rx\ge x\ge 3,$ by Lemma \ref{lem.2.7}, the
	solution \eqref{eq.3.16} only satisfies the conclusion (iii)
	in this lemma. Then, by \eqref{eq.2.9} and \eqref{eq.3.16},
	we have \begin{equation}\label{eq.3.17}
		z\,\, \text{is an odd prime},\,\,p^r+2^s-1=2^t,\,\,
		\dfrac{(p^r+2^s)^z-1}{p^r+2^s-1}=p^{rx},\,\, p\equiv
		1 \pmod{2z}.
	\end{equation} Further, since $rx\ge x>2$, applying Lemma
	\ref{lem.2.1} to the third equality of \eqref{eq.3.17}, we get
	\begin{equation}\label{eq.3.18}
		2\nmid x.
	\end{equation}

	By the first equality of \eqref{eq.3.17}, $z$ is an
	odd prime. Since $z>y\ge 1$, we have $\gcd(y,z)=1$ and
	\begin{equation}\label{eq.3.19}
		\gcd(y,z-y)=1.
	\end{equation} We see from \eqref{eq.3.12} and \eqref{eq.3.19}
	that $s\equiv 0 \pmod{z-y}$, \begin{equation}\label{eq.3.20}
		s=(z-y)s_1,\,\,s_1\in\mathbb{N}
	\end{equation} and \begin{equation}\label{eq.3.21}
		k=2^{s_1y}.
	\end{equation} Further, by \eqref{eq.3.14} and \eqref{eq.3.21},
	we have \begin{equation}\label{eq.3.22}
		t=s_1y(x-z),
	\end{equation} whence we get \begin{equation}\label{eq.3.23}
		y\mid t.
	\end{equation}

	By the second equality of \eqref{eq.3.17}, we have
	\begin{equation}\label{eq.3.24}
		p^r+2^s=2^t+1
	\end{equation} and \begin{equation}\label{eq.3.25}
		p^r\equiv -2^s \pmod{2^t+1}.
	\end{equation} Substituting \eqref{eq.3.24} into
	the third equality of \eqref{eq.3.17}, we get
	\begin{equation}\label{eq.3.26}
		p^{rx}=
		\dfrac{(2^t+1)^z-1}{(2^t+1)-1}=(2^t+1)^{z-1}+\cdots+(2^t+1)+1
	\end{equation} and \begin{equation}\label{eq.3.27}
		p^{rx}\equiv 1 \pmod{2^t+1}.
	\end{equation} Further, by \eqref{eq.3.18}, \eqref{eq.3.25}
	and \eqref{eq.3.27}, we have \begin{equation}\label{eq.3.28}
		2^{sx}+1\equiv 0 \pmod{2^t+1}.
	\end{equation} Applying Lemma \ref{lem.2.3} to \eqref{eq.3.28},
	we get $sx\equiv 0 \pmod{t}$ and \begin{equation}\label{eq.3.29}
		sx=tt_1,\,\,t_1\in\mathbb{N},\,\,2\nmid t_1.
	\end{equation} Hence, by \eqref{eq.3.20}, \eqref{eq.3.22} and
	\eqref{eq.3.29}, we have \begin{equation}\label{eq.3.30}
		(z-y)x=y(x-z)t_1.
	\end{equation} Furthermore, by \eqref{eq.3.19} and
	\eqref{eq.3.30}, we get \begin{equation}\label{eq.3.31}
		y\mid x.
	\end{equation} Therefore, by \eqref{eq.3.23}, \eqref{eq.3.26}
	and \eqref{eq.3.31}, we have \begin{equation}\label{eq.3.32}
		1=(2^t+1)^z-2^tp^{rx}=(2^t+1)^z-(2^{t/y}p^{rx/y})^y,
	\end{equation} where $2^{t/y}p^{rx/y}$ is a positive integer
	greater than 6. Since $z\ge 3,$ by Lemma \ref{lem.2.2}, we see
	from \eqref{eq.3.32} that \begin{equation}\label{eq.3.33}
		y=1.
	\end{equation} Thus, by \eqref{eq.3.20} and \eqref{eq.3.33},
	we get \begin{equation}\label{eq.3.34}
		s\ge z-1.
	\end{equation}

	On the other hand, we see from \eqref{eq.3.24} that
	\begin{equation}\label{eq.3.35}
		t>s
	\end{equation} and \begin{equation}\label{eq.3.36}
		p^r\equiv 1 \pmod{2^s}.
	\end{equation} Further, by \eqref{eq.3.26}, \eqref{eq.3.35}
	and \eqref{eq.3.36}, we have \begin{equation}\label{eq.3.37}
		1\equiv p^{rx}\equiv
		\dfrac{(2^t+1)^z-1}{(2^t+1)-1}=(2^t+1)^{z-1}+\cdots+(2^t+1)+1\equiv
		z \pmod{2^s}.
	\end{equation} Since $z>1$, by \eqref{eq.3.37}, we get
	\begin{equation}\label{eq.3.38}
		z\ge 2^s+1.
	\end{equation} Therefore, the combination of \eqref{eq.3.34} and
	\eqref{eq.3.38} yields $s\ge z-1\ge 2^s$, a contradiction. Thus,
	the theorem holds in this case.

	\medskip\noindent
	\subsection{Case (iii): $(a,b)=(p^r,q^s)$}
	By Lemma
	\ref{lem.2.10}, we have \begin{equation}\label{eq.3.39}
		q^{sy}=k^{z-y}
	\end{equation} and \begin{equation}\label{eq.3.40}
		p^{rx}k^{x-z}+1=(p^r+q^s)^z.
	\end{equation} We see from \eqref{eq.3.39} that $k$ is a power
	of $q$. So we have \begin{equation}\label{eq.3.41}
		k^{x-z}=q^t,\,\,t\in\mathbb{N}.
	\end{equation} Substituting \eqref{eq.3.41} into \eqref{eq.3.40},
	we get \begin{equation}\label{eq.3.42}
		(p^r+q^s)^z-1=p^{rx}q^t.
	\end{equation} We find from \eqref{eq.3.42} that the equation
	\begin{equation}\label{eq.3.43}
		X^{\ell}-1=p^mq^n,\,\,X,\ell,m,n\in\mathbb{N},\,\,X>1,\ell>1
	\end{equation} has a solution \begin{equation}\label{eq.3.44}
		(X,\ell,m,n)=(p^r+q^s,z,rx,t).
	\end{equation} Since $p^r+q^s\ge 8$, applying Lemma \ref{lem.2.8}
	to \eqref{eq.3.43} and \eqref{eq.3.44}, we only need to consider
	the following five subcases.

	\bigskip 
	\noindent{\it Subcase (iii)-1}:
	
	\begin{equation}\label{eq.3.45}
		2\mid
		z,\,\,(p^r+q^s)^{z/2}+\zeta=p^{rx},\,\,(p^r+q^s)^{z/2}-\zeta=q^t,\,\,\zeta\in\{1,-1\}.
	\end{equation}

	Since $rx\ge 3$, by Lemma \ref{lem.2.2}, we see from the
	second equality of \eqref{eq.3.45} that $z=2$. So we have
	\begin{equation}\label{eq.3.46}
		p^r+q^s+\zeta=p^{rx},\,\,p^r+q^s-\zeta=q^t.
	\end{equation} By the first equality of \eqref{eq.3.46}, we get
	\begin{equation}\label{eq.3.47}
		q^s=p^{rx}-p^r-\zeta=p^r(p^{r(x-1)}-1)-\zeta\ge
		p^r(p^{2r}-1)-1>p^r.
	\end{equation} However, by the second equality of \eqref{eq.3.46},
	we have $t>s$ and \begin{equation*}
		p^r=q^t-q^s+\zeta=q^s(q^{t-s}-1)-1\ge q^s(q-1)-1>q^s,
	\end{equation*} which contradicts \eqref{eq.3.47}. Therefore,
	this subcase can be eliminated.

	\bigskip 
	\noindent{\it Subcase (iii)-2}:
	\begin{equation}\label{eq.3.48}
		z=p,\,\,\,\,p^r+q^s-1=p^{rx-1},\,\,
		\dfrac{(p^r+q^s)^p-1}{p^r+q^s-1}=pq^t,\,\, q\equiv
		\modd{1} {2p}.
	\end{equation} If $t\ge s$, then from the third equality of
	\eqref{eq.3.48} we get \begin{equation}\label{eq.3.49}
		0\equiv pq^t\equiv \dfrac{(p^r+q^s)^p-1}{p^r+q^s-1}\equiv
		\dfrac{p^{rp}-1}{p^r-1} \pmod{q^s},
	\end{equation} where $(p^{rp}-1)/(p^r-1)$ is a positive
	integer. However, by \eqref{eq.3.48} and \eqref{eq.3.49}, we have
		\begin{align*}
			2p^{r(z-1)}=2p^{r(p-1)}&=p^{r(p-1)}\sum_{i=0}^{\infty}\dfrac{1}{2^i}>p^{r(p-1)}\sum_{j=0}^{p-1}\dfrac{1}{p^{r_j}}=\dfrac{p^{rp}-1}{p^r-1}
			\\ &\ge q^s=p^{rx-1}-p^r+1\ge p^{r(z+1)-1}-p^r+1,
		\end{align*}
	whence we get
		\begin{align*}
			p^r>p^r-1&\ge
			p^{r(z+1)-1}-2p^{r(z-1)}=p^{r(z-1)}(p^{2r-1}-2)\\
			&\ge p^{r(z-1)}(p-2)\ge p^{r(z-1)}\ge p^r,
		\end{align*}
	a contradiction. So we have $t<s$. Then, since $p$ is an
	odd prime, by the third equality of \eqref{eq.3.48}, we get
	\begin{equation*}
		pq^s>pq^t=\dfrac{(p^r+q^s)^p-1}{p^r+q^s-1}\ge
		\dfrac{(p^r+q^s)^3-1}{p^r+q^s-1}>(p^r+q^s)^2>pq^s,
	\end{equation*} a contradiction. Thus, this subcase can be
	eliminated.

	\bigskip
	\noindent{\it Subcase (iii)-3}:

	\begin{equation}\label{eq.3.50}
		z=q,\,\,p^r+q^s-1=q^{t-1},\,\,
		\dfrac{(p^r+q^s)^q-1}{p^r+q^s-1}=p^{rx}q,\,\, p\equiv
		\modd{1} {2q}.
	\end{equation}

	By \eqref{eq.3.50}, we have $p^r+q^s=q^{t-1}+1$ and
	\begin{equation*}
		p^{rx}q=\dfrac{(p^r+q^s)^q-1}{p^r+q^s-1}=\dfrac{(q^{t-1}+1)^q-1}{q^{t-1}}=\sum_{i=1}^{q-1}\binom{q}{i}q^{(t-1)(i-1)},
	\end{equation*} whence we get \begin{equation} \label{eq.3.51}
		p^{rx}=1+\sum_{i=2}^{q-1}\binom{q}{i}q^{(t-1)(i-1)-1}.
	\end{equation} On the other hand, by the second equality of
	\eqref{eq.3.50}, we have \begin{equation}\label{eq.3.52}
		s<t-1
	\end{equation} and \begin{equation}\label{eq.3.53}
		p^r=q^s(q^{t-s-1}-1)+1.
	\end{equation} Hence, by \eqref{eq.3.53}, we get \begin{equation}
	\label{eq.3.54}
		p^{rx}=1+\sum_{j=1}^{x}\binom{x}{j}(q^s(q^{t-s-1}-1))^j.
	\end{equation} The combination of \eqref{eq.3.51} and
	\eqref{eq.3.54} yields \begin{equation} \label{eq.3.55}
		\sum_{i=2}^{q-1}\binom{q}{i}q^{(t-1)(i-1)-1}=\sum_{j=1}^{x}\binom{x}{j}(q^s(q^{t-s-1}-1))^j.
	\end{equation} Let $q^\alpha\mid \mid x$, where $\alpha$ is a
	nonnegative integer. Notice that \begin{equation} \label{eq.3.56}
		q^{t-1}\mid \mid
		\sum_{i=2}^{q-1}\binom{q}{i}q^{(t-1)(i-1)-1},\,\,q^{\alpha+s}\mid
		\mid \sum_{j=1}^{x}\binom{x}{j}(q^s(q^{t-s-1}-1))^j.
	\end{equation} By \eqref{eq.3.55} and \eqref{eq.3.56},
	we have $t-1=\alpha+s$. Further, by \eqref{eq.3.52},
	we get $\alpha=(t-1)-s>0$. It implies that
	\begin{equation}\label{eq.3.57}
		q\mid x.
	\end{equation}

	Recall that $z>y$ and $z=q$ is an odd prime. Hence, $y$ and $z$
	satisfy \eqref{eq.3.19}. By \eqref{eq.3.19} and \eqref{eq.3.39},
	we have $s\equiv 0 \pmod{z-y}$ and \eqref{eq.3.20}. Further,
	by \eqref{eq.3.39} and \eqref{eq.3.41} we get $k=q^{s_1y}$
	and \begin{equation}\label{eq.3.58}
		t=s_1y(x-z)=s_1y(x-q).
	\end{equation} Furthermore, by \eqref{eq.3.57} and
	\eqref{eq.3.58}, we obtain \begin{equation}\label{eq.3.59}
		q\mid t.
	\end{equation} Therefore, by \eqref{eq.3.50}, \eqref{eq.3.57}
	and \eqref{eq.3.59}, we have 
		\begin{align*}
			p^{rx/q}q^{t/q}\in\mathbb{N},\,\,1&=(p^r+q^s)^q-p^{rx}q^t=(p^r+q^s)^q-(p^{rx/q}q^{t/q})^q\\
			&=(p^r+q^s-p^{rx/q}q^{t/q})\sum_{i=0}^{q-1}(p^r+q^s)^{q-i-1}(p^{rx/q}q^{t/q})^i>1,
		\end{align*}
	a contradiction. Thus, this subcase can be eliminated.

	\bigskip 
    \noindent{\it Subcase (iii)-4}:
	\begin{equation}\label{eq.3.60} z\,\,\text{is
	an odd prime},\,\,p^r+q^s-1=p^{rx},\,\,
	\dfrac{(p^r+q^s)^z-1}{p^r+q^s-1}=q^t,\,\, q\equiv 1 \pmod{2z}.
	\end{equation}

	We can eliminate this subcase by using the same method as in the
	proof of Subcase (iii)-2. An outline of this process is given
	below. If $t\ge s,$ then from the second and third equalities
	of \eqref{eq.3.60} we have \begin{equation*}
		0\equiv q^t\equiv \dfrac{(p^r+q^s)^z-1}{p^r+q^s-1}\equiv
		\dfrac{p^{rz}-1}{p^r-1} \pmod{q^s},
	\end{equation*} whence we get $(p^{rz}-1)/(p^r-1)\ge q^s$.
	So we have \begin{equation*}
		p^{r(z-1)+1}>\dfrac{p^{rz}-1}{p^r-1}\ge
		q^s=p^{rx}-p^r+1\ge p^{r(z+1)}-p^r+1>p^{r(z+1)-1},
	\end{equation*} a contradiction. It implies that $t<s$. Then,
	by \eqref{eq.3.60}, we get \begin{equation*}
		q^s>q^t=\dfrac{(p^r+q^s)^z-1}{p^r+q^s-1}>p^r+q^s+1>q^s,
	\end{equation*} a contradiction. Thus, this subcase can be
	eliminated.

	\bigskip 
    \noindent{\it Subcase (iii)-5}:
	\begin{equation}\label{eq.3.61}
	z\,\,\text{is an odd prime},\,\,p^r+q^s-1=q^t,\,\,
	\dfrac{(p^r+q^s)^z-1}{p^r+q^s-1}=p^{rx},\,\, p\equiv 1 \pmod{2z}.
	\end{equation}

	We see from the second equality of \eqref{eq.3.61} that $t>s$. So
	we have $p^r=q^s(q^{t-s}-1)+1$ and \begin{equation}\label{eq.3.62}
		p^r\equiv 1 \pmod{q^s}.
	\end{equation} By the second and third equalities of
	\eqref{eq.3.61}, we get \begin{equation}\label{eq.3.63}
		p^r+q^s=q^t+1
	\end{equation} and \begin{equation}\label{eq.3.64}
		p^{rx}=\dfrac{(p^r+q^s)^z-1}{p^r+q^s-1}=
		\dfrac{(q^t+1)^z-1}{(q^t+1)-1}.
	\end{equation} Further, by \eqref{eq.3.64}, we have
	\begin{equation}\label{eq.3.65}
		p^{rx}\equiv (q^t+1)^{z-1}+\cdots +(q^t+1)+1\equiv
		z\pmod{q^t}.
	\end{equation} Since $t>s$, by \eqref{eq.3.62} and
	\eqref{eq.3.65}, we get $z\equiv 1 \pmod{q^s}$. Furthermore,
	since $z>1$, we have \begin{equation}\label{eq.3.66}
		z\ge q^s+1.
	\end{equation}

	On the other hand, since $z$ is an odd prime with $z>y\ge 1$,
	$y$ and $z$ satisfy \eqref{eq.3.19}. Hence, by \eqref{eq.3.19}
	and \eqref{eq.3.39}, $s$ satisfies \eqref{eq.3.20}, whence we
	can obtain \eqref{eq.3.23}.

	Since $p^r+q^s>3,$ applying Lemma \ref{lem.2.1}
	to the third equality of \eqref{eq.3.61}, we have
	\begin{equation}\label{eq.3.67}
		2\nmid x.
	\end{equation} By \eqref{eq.3.64}, we have
	\begin{equation}\label{eq.3.68}
		p^{rx}\equiv
		\dfrac{(q^t+1)^z-1}{q^t+1}\equiv(q^t+1)^{z-1}+\cdots+(q^t+1)+1\equiv
		1 \pmod{q^t+1}.
	\end{equation} Since $p^r\equiv -q^s \pmod{q^t+1}$ by
	\eqref{eq.3.63}, we get from \eqref{eq.3.67} and \eqref{eq.3.68}
	that \begin{equation}\label{eq.3.69}
		0\equiv p^{rx}-1\equiv (-q^s)^x-1\equiv -(q^{sx}+1)
		\pmod{q^t+1}.
	\end{equation} Hence, by Lemma \ref{lem.2.3}, we see
	from \eqref{eq.3.69} that $x$ satisfies \eqref{eq.3.29} and
	\eqref{eq.3.30}. Further, by \eqref{eq.3.19} and \eqref{eq.3.30},
	$x$ satisfies \eqref{eq.3.31}. Therefore, by \eqref{eq.3.23} and
	\eqref{eq.3.31}, we get from the third equality of \eqref{eq.3.61}
	that \begin{equation}\label{eq.3.70}
		p^{rx/y}q^{t/y}\in\mathbb{N},\,\,(p^r+q^s)^z-(p^{rx/y}q^{t/y})^y=1.
	\end{equation} Since $z\ge 3$ and $p^r+q^s>3$, by Lemma
	\ref{lem.2.2}, we find from \eqref{eq.3.70} that $y$
	satisfies \eqref{eq.3.33}. Furthermore, by \eqref{eq.3.20}
	and \eqref{eq.3.33}, $s$ satisfies \eqref{eq.3.34}. However,
	the combination of \eqref{eq.3.34} and \eqref{eq.3.66} yields
	$s\ge z-1\ge q^s$, a contradiction. Thus, this subcase can
	be eliminated.

	To sum up, the theorem is proved.


	\section{\bf Acknowledgments} We thank an anonymous referee for
	carefully reading our paper and for giving such constructive
	comments which substantially helped improving the quality of
	the paper.
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	\end{thebibliography}

	\bigskip \hrule \bigskip \noindent {\it 2020 Mathematics Subject
	Classification:} Primary 11D61; Secondary 11B39.

	\noindent \emph{Keywords:} Lucas sequence, ternary purely
	exponential Diophantine equation, Nagell-Ljunggren equation,
	Catalan equation, prime power base, elementary number theory
	method.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A000204}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  April 25 2023;
revised versions received  April 28 2023; August 15 2023; August 18 2023.
Published in {\it Journal of Integer Sequences}, August 22 2023.

\bigskip
\hrule
\bigskip

\noindent
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