${2}^{p}+{3}^{p}={a}^{n}$ , then n=1 for any p ? Is there an integer $m\geq 1$ such that $2^m+3^m$ is a perfect power?
 A: If you really wanted to prove this (and I'm afraid that I'm not sure why you would), you could invoke a Theorem of Darmon and Merel for $n=2$ and $3$, check that there are no solutions with $p \leq 5$, say, and then write down the usual $(n,n,n)$ Frey curve, assuming $n \geq 5$ is prime (which leads to a weight $2$, level $6$ cuspidal newform and hence the desired contradiction).
Of course, this is an absurdly big hammer for such a problem and likely something {\it much} simpler works.
A: By the Fermat theorem: n is not divisible by p.
A: I was hoping to see a post by Gjergji Zaimi on this question.  My guess is he deleted whatever he might have had, from which I had hoped to learn something.  So I will post a start of an elementary approach in the hopes that he or someone else will finish it.
Consider the case that p is an odd positive integer.  Then a must be a multiple of 5, and either n is one, or else 2^p + 3^p is a multiple of 25, in which case p must be an odd multiple of 5 by considering the sum mod 25.  So in this case a^n is a multiple of 3025 if n is not 1.  This can probably be refined by looking at 2^p + 3^p mod 125, and hopefully considerations mod 11 may finish it off.
Now assume p=2q for some positive integer q.  If n were even we could represent 3^p by
(a  - 2^q)(a+2^q), which would give 3^p = 2^(q+1) + 1, which is not solvable in integers p and q.  So n must be odd, a^n must be 1 mod 8 for p sufficiently large, and so must a if n is not 1.  Again, more work needs to be done here.  It looks plausible to me that n must be 1 if a is an integer.
Gerhard "Email Me About System Design" Paseman, 2011.06.27
