Does the equation $x^m-1=y^n+y^{n-1}+...+1$ have only finitely many solutions $(x,y,m,n)$ where $x,y$ are prime powers with $y>2$ and $m,n$ are integers with $m,n>1$?


This question arose in the study a group theory problem, where I'm examining a "natural" subgroup $G$ of $S\_d$ where $d:=x^m-1$. I can show that $G$ is primitive and contains a $d$-cycle, so if $d$ is composite then known results imply that $G$ must be either $S\_d$ or $A\_d$ or a group between $\text{PGL}\_{n+1}(y)$ and its automorphism group, for some $(n,y)$ satisfying the above equation. I'm hoping to show that the last case almost never happens. The best I can do so far is the simple observation that if $m=2$ then the set of prime $x$'s for which the equation has a solution is a density-zero subset of the primes. But I'm hoping that a much stronger result is possible.



The equation $x^m-1=y^n+y^{n-1}+...+1$ in prime powers $x,y$
Mike Bennett's comments gives good reason to suspect that my original question is beyond what can be proved with existing techniques. Here I record one thing that can be proved (again thanks to Mike's comment), namely that there are no solutions with $x=4$. That is:
the equation $4^m-1=y^n+y^{n-1}+...+1$ has no solutions $(y,m,n)$ in which $y$ is a prime power with $y>2$ and $m,n$ are integers with $m,n>1$.


Suppose that $(y,m,n)$ is a solution. First note that $y$ is odd and $n$ is even.
For, if $y$ were even then we would contradict uniqueness of base-$2$ expansion by writing the number $4^m-1$ in two different ways as the sum of distinct powers of $2$, namely as both $2^{2m-1}+2^{2m-2}+...+1$ and $y^n+y^{n-1}+...+1$. Hence $y$ is odd, so the mod $2$ reduction of the equation $4^m-1=y^n+y^{n-1}+...+1$ implies that $n$ is even.


Next show that the polynomial $X^2-1-(Y^n+Y^{n-1}+...+1)$ is irreducible in $\mathbf{Q}[X,Y]$. If it were reducible then it must factor as $(X-f(Y))\cdot(X+f(Y))$ for some polynomial $f(Y)$, so that $f(Y)^2=1+(Y^n+Y^{n-1}+...+1)$ is a square in $\mathbf{Q}[Y]$. Rewrite this as $1+(Y^{n+1}-1)/(Y-1)$, or equivalently $(Y^{n+1}+Y-2)/(Y-1)$. Then
$f(Y)$ divides both $Y^{n+1}+Y-2$ and its derivative $(n+1)Y^n+1$, so $f(Y)$ also divides
$(n+1)\cdot(Y^{n+1}+Y-2) - Y\cdot((n+1)Y^n+1)=nY-2(n+1)$, whence $\deg(f)=1$ so $n=2\deg(f)=2$. But $Y^2+Y+2$ is not a square in $\mathbf{Q}[Y]$, contradiction.


Now we can apply Theorem 4.1(ii) of the paper "Estimates for the solutions of certain Diophantine equations by Runge's method" by A.Sankaranarayan and N.Saradha (Int.J.Number Theory 4 (2008), 475-493). According to this result (which uses the condition that $n$ is even), if $u,v$ are integers such that $u^2-1=v^n+v^{n-1}+...+1$, then
$|u|\le 2^{3n+4}$.  But we know that $(u,v)=(2^m,y)$ is a solution, so $2^m\le 2^{3n+4}$.  On the other hand we have $y^n<4^m$, so that $y^n<4^{3n+4}$ and thus $y<4^{3+(4/n)}$.  I checked via computer that there are no solutions with $n<44$. So assume $n\ge 44$, which implies that $y<73$.


Since $4^m=(y^n+y^{n-1}+...+y)+2$, we have $4^m\equiv 2\pmod{y}$, so the order of $2$ in $(\mathbf{Z}/y\mathbf{Z})^\*$ is odd. Since $y<73$, it follows that $y\in\{7,23,31,47,49,71\}$. Each of these values $y$ satisfies either $y\equiv -1\pmod{4}$ or $y\equiv -1\pmod{5}$. If $y\equiv -1\pmod{4}$ then, since $n$ is even, we obtain the contradiction
$4^m=(y^n+y^{n-1}+...+y)+2\equiv 2\pmod{4}$. Likewise, if $y\equiv -1\pmod{5}$ then
$(y^n+y^{n-1}+...+y)+2\equiv 2\pmod{5}$, so that $4^m\equiv 2\pmod{5}$, which is impossible.


(Note: the Sankaranarayan-Saradha result is a refinement of an earlier result by P.G.Walsh; if one uses Walsh's result, then the same proof works except that one must rule out $y=73$ by a different method. One way to do this is to note that the equation
$4^m-1\equiv 73^n+73^{n-1}+...+1\pmod{91}$ has no solutions in which $n$ is even.) 


