Wolstenholme Theorem is a nice theorem that states that every prime $p >3$ satisfies:
$\binom{2p}{p} \equiv 2 \pmod {p^3}$
A Wolstenholme prime is a prime $p$ such that $\binom{2p}{p} \equiv 2 \pmod {p^4}$. There are several equivalent criteria for this. I was able to prove myself or find proofs on the internet that any of following congruences is equivalent to $p$ being a Wolstenholme prime: $\sum_{i=1}^{p-1} i^{-2} \equiv 0 \pmod {p^2}$ $\sum_{1 \le i < j \le p-1} \frac{1}{ij} \equiv 0 \pmod {p^2}$ $p\sum_{1 \le i < j \le p-1} \frac{1}{ij} + \sum_{i=1}^{p-1} \frac{1}{i} \equiv 0 \pmod {p^3}$
The proofs are expanding $\binom{2p}{p} = 2\frac{\prod_{i=1}^{p-1} (i+p)}{\prod_{i=1}^{p-1} i}$ in different ways. But in Wikipedia I found the nicest criterion:
$\sum_{i=1}^{p-1} \frac{1}{i} \equiv 0 \pmod {p^3}$
This congruence is implies by the last 2 criteria, but I am not sure how it implies one of the criteria. Is there an elementary proof of this?