Is there a solution for the equation x^m-y^n=k in which k > 1?

The Catalan conjecture state that $x^m-y^n=1$ has only the solution $x=3, m=2, y=2, n=3$. This conjecture was proved by Preda Mihailescu in 2004, but I want to know about the equation mentioned above. Is there a asolution of this?

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 As Gerry points out, you only want $m\geq 2,n\geq2$. Also, if you take any two perfect powers, they have a difference $k\geq 1$, so that there are many solutions $(x,m,y,n,k)$. One way to state the question you are asking is this: let $a_i$ be the $i$-th smallest perfect power, i.e., $(a)=(4,8,9,16,25,27,32,36,49,64,\dots$. Does $a_{i+1}-a_i$ tend to infinity? – Kevin O'Bryant Jun 27 2010 at 16:28

Pillai's conjecture is that for each $k$, there are only finitely many solutions.

The ABC conjecture implies Pillai's conjecture as follows. First, I state a form of the ABC conjecture. Given three relatively prime positive integers $A+B=C$, the quality of the triple $(A,B,C)$ is $\log(C)/\log(R)$, where $R$ is the product of the primes that divide $ABC$. For example, the quality of $(5,27,32)$ is $\log(32)/\log(30)$. One strong form of the ABC conjecture is that there are only finitely many triples (of relatively prime positive integers) with quality greater than $1.001$.

Now a solution to $x^m-y^n=k$ has $(A,B,C)=(k,y^n,x^m)$, and so $R\leq k y x$. Thus, the quality of the triple is at least $$\frac{m\log(x)}{\log(k)+\log(x)+\log(y)}\approx \frac{m\log(x)}{\log(k)+(\frac mn +1)\log(x)}.$$ As $x\to\infty$, this gives a quality approaching $\frac{mn}{m+n}$, and for $m n>4$ this is $\geq 1.2$, which can only happen finitely many times by ABC. Also, $\frac{mn}{m+n}=1$ if $m=n=2$, so that case has to be handled separately.

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As a compliment, I would give a reference [M. A. Bennett, Canad. J. Math. 53 (2001) 897-922] (math.ubc.ca/~bennett/B-CJM-Pillai.pdf) where one can find a history of the problem as well as the latest news on approaches for it. – Wadim Zudilin Jun 27 2010 at 11:51
I appreciate the flattery, and your comment serves as a complement, too! :) – Kevin O'Bryant Jun 27 2010 at 16:22
Kevin, I just realised the artistic typo in the "compliment". Anyway, I meant both meanings! :-) – Wadim Zudilin Jun 28 2010 at 10:11
wadim zudilin:thank you for your reference. – Hashem sazegar Jun 30 2010 at 11:26

There is a conjecture that for every positive natural number k there are just finitely many solutions of the above equation. As far as I know, this is open for k>1. In fact, Erdős conjectured that the difference between a full power x and the next full power is at least xc for some positive constant c.

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That would depend on $k$. I'm assuming you want all variables positive integers with $m>1$ and $n>1$. It may have been proved that for any given $k$ there are only finitely many solutions.

Edit: in view of the other answers, it appears my "may have been proved" was overly optimistic.

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