Sum of Residues Modulo $p^2$.

Question

Let $p$ be an odd prime. Prove that

$$ \sum_{k = 1}^{p-1} k^{2p-1} \equiv \frac{p(p+1)}2 \pmod{p^2}$$

Answer 1

@user8268 (currently deleted) answer has the right idea, though there were algebraic errors. The idea is to group for $1 \leq k \leq \frac{p-1}{2}$

\begin{align} k^{2p-1}+(p-k)^{2p-1} \equiv k^{2p-1}+[(2p-1)pk^{2p-2}-k^{2p-1}] &\equiv p(2p-1)k^{2p-2} \pmod{p^2} \\ &\equiv p(2p-1) \pmod{p^2} \end{align}

where in the last step we have used $p \mid k^{2p-2}-1$ by Fermat's little theorem.

Thus \begin{align} \sum_{k=1}^{p-1}{k^{2p-1}}=\sum_{k=1}^{\frac{p-1}{2}}{\left(k^{2p-1}+(p-k)^{2p-1}\right)} & \equiv \frac{p-1}{2}p(2p-1) \pmod{p^2} \\ & \equiv -\frac{p(p-1)}{2} \pmod{p^2}\\ &\equiv \frac{p(p+1)}{2} \pmod{p^2} \end{align}