TITLE: Proving $\sum_{k=1}^m{k^n}$ is divisible by $\sum_{k=1}^m{k}$ for $ n=2013$
QUESTION [5 upvotes]: I got an interesting new question, it's about number theory and algebra precalculus. Here is the question:

a positive integer $n$ is called valid if $1^n+2^n+3^n+\dots+m^n$ is divisible by $1+2+3+\dots+m$ for every positive integer $m$.

Prove that 2013 is valid
Prove that there are infinite positive integers which are not valid


Every little hint, contribution and recommendation would be very helpful. Sorry for my bad english. Thanks before.

REPLY [5 votes]: If $n$ is odd, then modulo $m+1$ we have $2(1^n + 2^n + \ldots + m^n) = (1^n+m^n) + (2^n+(m-1)^n) + \ldots + (m^n+1^n) \\ \equiv (1^n-1^n) + (2^n-2^n) + \ldots + (m^n-m^n) = 0 \pmod {m+1}$.
Also, since $m^n \equiv 0 \pmod m$, $2(1^n + 2^n + \ldots + m^n) \equiv 2(1^n + 2^n + \ldots + (m-1)^n) \equiv 0 \pmod m$
Since $m$ and $m+1$ are coprime, this shows that $2(1^n + \ldots + m^n)$ is a multiple of $m(m+1)$, and since $m(m+1)$ is even, $1^n + \ldots + m^n$ is a multiple of $m(m+1)/2 = 1+2+\ldots+m$.
If $n$ is even then $1+2^n \equiv 2 \pmod 3$, which is not a multiple of $1+2 = 3$