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Given two positive integers $a,b$, and an odd prime $p$, I want to know whether the number of solutions to the following equation is finite:

$X^2=a+bp^{Y}$

where $X,Y$ are variables and are integers.

I checked with google, and in the case $b=1$ this seems to follow from a result of A Baker on logrithmetic forms (1966).

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  • $\begingroup$ I vaguely remember something like this was in the old book by Sierpinski "Number Theory". Nice book $\endgroup$
    – Victor
    Dec 30, 2013 at 14:30

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Yes, there are only finitely many solutions. Even more is true. For fixed nonzero $a$ and $b$, the equation $$ x^2 = a + bz^y $$ has only finitely many solutions in integers $(x,y,z)$ with $y\ge3$. See the article on the Ramanujan-Nagell equation (and its generalizations) https://en.wikipedia.org/wiki/Ramanujan-Nagell_equation.

It is conjectured that there are only finitely many solutions $(x,y,z,w)\in\mathbb{Z}^4$ to $$ x^w = a + bz^y $$ with $\max(w,y)\ge3$ and $\min(w,z)\ge2$, but this is only known for a few small values of $a$ and $b$, such as $a=1$ and $b=-1$, which is Catalan's equation.

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  • $\begingroup$ Just a remark that equations of the shape $x^n + c = yz^k$ where $x,y,z$ are integral variables, $c$ is a fixed integer, and $n > k$ are positive integers have been investigated by D.R. Heath-Brown. In particular, he showed that if $-X \leq x \leq X$ then there are at most $O(X^{1 - \delta})$ solutions provided $k \geq (5d+4)/9$. $\endgroup$
    – Stanley Yao Xiao
    Dec 30, 2013 at 15:49

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