Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|improve this question
    
I vaguely remember something like this was in the old book by Sierpinski "Number Theory". Nice book –  Victor Dec 30 '13 at 14:30

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
Thanks very much! –  Tao Feng Dec 30 '13 at 14:21
    
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 $k \geq (5d+4)/9$. –  Stanley Yao Xiao Dec 30 '13 at 15:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.