# Divisibility of a certain power sum

Does there exist any even integers $m,n$ such that $m$ divides $1^n + 2^n + \cdots +(8m+2)^n$ ?

After my initial attempt below that i'm not so sure of, i feel that Bernoulli polynomials might be the best approach, but not sure how ?

My attempt: Suppose $m$ divides $1^n + 2^n + \cdots +(8m+2)^n$. Applying the factor theorem, we arrive at $1^n + 2^n = 0$, which is absurd ?

EDIT: For $m$ even, the question was settled by @Vrugthagel below, what about when $m$ is odd ?

Nope. Your sum is always odd for even $m$ and $n$. The number of terms in $1^n+2^n+\cdots (8m+2)^n$ is $8m+2$, so the number of odd terms is $4m+1$, thus, your sum is odd, and so it cannot be divisible by $m$ (an even number).
If $m$ is odd, then there are solutions: for example, $(m,n)=(3,2)$ or $(m,n)=(5,6)$. Also, $m=1$ would be a solution regardless of what $n$ is.
• Thank you very much Vrugtehagel, what about if $m$ is odd ? – User1 Jan 26 '16 at 11:17