http://www.mathlinks.ro/Forum/viewtopic.php?p=475349#p475349 They like taking problems from different competitions

Thanks Megus wrote: They like taking problems from different competitions That's easier

I think it's connected with Fermat's Last Theorem though it's a water problem and can be second-killed without it!

Say $3^k+5^k=n^m$. $n$ is clearly even and so $4 | 3^k+5^k$, and so $k$ is odd. Taking (mod $3$) we see that $n^m \equiv 5^k \equiv 2$ (mod $3$). So $m$ must be odd. Also, taking (mod $16$) we see that $3^k+5^k \equiv 8$ (mod $16$) when $k$ is odd. And so we have $m=3$. Suppose $k \neq 1$. Then $5^k \equiv n^3$ (mod $9$). From this, $3 | k$. But taking (mod $7$) we get that $3^k+5^k \equiv 0,1,-1$ (mod $7$). And so $k$ cannot be divided by $3$. Contradiction. Therefore $k=1$ is the only answer ($3+5=8$).