The classical Wolstenholme’s theorem states that for any prime $p \geq 5$,

$1+\frac{1}{2}+\frac{1}{3}+\dots + \frac{1}{p-1}\equiv 0 \pmod {p^2}.$

For a prime $p>5$ and $k\geq 1$, define: \begin{equation} S_p^{(k)} = \sum \frac{1}{(mi+n)^k}, 1\leq n,m\leq p-1 \end{equation} Then the following congruences hold:

$S_p^{(1)} \equiv 0 \pmod{p^4}$

$S_p^{(2)} \equiv 0 \pmod{p^3}$

$S_p^{(3)} \equiv 0 \pmod{p^2}$

$S_p^{(4)} \equiv 0 \pmod{p}$

I conjectured that on 21 March 2025 and proved on 22 March 2025. Here is this article on arxiv.

Immediately, Bo Jiang and Zhi-Wei Sun proved a conjecture from my article.