# $a^m+k=b^n$ Finite or infinite solutions?

Given positive integers k,a,b, is there a finite or infinite number of solutions in positive integers $m,n>1$, to $a^m+k=b^n$?

Pillai's conjecture states that each positive integer occurs only finitely many times as a difference of perfect powers (Only k given, a,m,n,b are variables) . It is an open problem.

What are known lower bounds on f(d) defined as how many times d, for d=1,2,3... occurs as a difference of perfect powers?

Catalan's conjecture is the theorem that f(1)=1

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Do you mean $a^{m+1}$ or $a^{m}+1$? –  Thomas E. Feb 6 '12 at 10:27
In spite of the name, Catalan's conjecture is actually a theorem. Your second question differs from Pillai's conjecture by fixing $a$ and $b$, which is likely to make the answer always "finite". You should be clearer about which question exactly you are asking. –  Marc van Leeuwen Feb 6 '12 at 10:53
Please avoid double use of letters for different things. Why not use "d" for the difference instead of "a" (which already means one of the two bases in the problemdescription)? –  Gottfried Helms Feb 6 '12 at 11:04
Is it required that m and n are different? If it is permitted that m = n = 2, a non-zero lower bound can be found for most k as follows. Find the prime factors of k, then use these to find all pairs p, q such that pq = k and p > q. If k is even, consider only those where p, q are both even. Then use each such pair to find a series of q consecutive odd integers with mean p and therefore summing to k. The sum of any such series must be the difference between two squares, ie betwen ((p+q)/2)^2 and ((p-q)/2)^2. –  Adam Bailey Feb 10 '12 at 16:10

For your first question, if we suppose that $a, b \geq 2$ and $k$ are all fixed, then there are at most two solutions in positive exponents $m$ and $n$. This follows from lower bounds for linear forms in logarithms (and probably other approaches). As for Pillai's conjecture, it's wide open still (modulo developments on the ABC conjecture). It is still unknown, by way of example, whether there are only finitely many perfect powers differing by $2$.
Standard conjectures would imply that your function $f(d)$ is zero for "most" $d$ (if we agree to avoid $m=n=2$).