Why does this sum mod out to 0? In making up another problem today I came across something odd. I've been thinking it over and I can't exactly place why it's true, but after running a long Python script to check, I haven't yet found a counter example.
Why is $\sum_{n=1}^{m}{n^m}\equiv 0\mod m$ true for all odd $m \ge 3$? The script showed me that each term for odd $m$ is equivalent to $n$ when taken $\mod m$ (until term $m$), and so the sum would be $\frac{m(m-1)}{2}$ which is obviously $0 \mod m$. What I am unable to understand is why $n^m\equiv n \mod m$ only for odd $m$.

If it is odd, each n can be paired with -n. So we get $n^m+(-n)^m=n^m-n^m=0$
