Problem - 4268
Let $m$ and $n$ be positive integers, $m$ be odd, and $(m, 2^{n} - 1)=1$. Show that $\displaystyle\sum_{k=1}^{m}k^n$ is a multiple of $m$.
Note that $\{1, 2, \cdots, m\}$ is a complete residue system modulo $m$. Meanwhile, because $m$ is odd, thus $(m, 2)=1$. This means that $\{2, 4,\cdots, 2m\}$ be a complete residue system modulo $m$ too. This means that $$\begin{array}{l} \sum_{k=1}^{m}k^n \equiv \sum_{k=1}^{m}(2k)^n\equiv 0\pmod{m}\\ \implies \sum_{k=1}^{m}(2k)^n - \sum_{k=1}^{m}(k)^n\equiv(2^n-1)\sum_{k=1}^{m}k^n\equiv 0\pmod{m}\end{array}$$
Because $(m, 2^n-1)=1$, therefore we must have $\displaystyle\sum_{k=1}^{m}k^n\equiv 0\pmod{m}$.