# Criterion for Wolstenholme Primes

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?

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$$\sum_{i=1}^{p-1} \frac{1}{i} = \frac{1}{2} (\sum_{i=1}^{p-1} \frac{1}{i} + \frac{1}{p-i}) = \frac{-p}{2} (\sum_{i=1}^{p-1} \frac{1}{i(i-p)})$$ $$= \frac{-p}{2} (\sum_{i=1}^{p-1} \frac{1}{i^2} +\frac{1}{i(i-p)} - \frac{1}{i^2})$$ $$= \frac{-p}{2} (\sum_{i=1}^{p-1} \frac{1}{i^2} +\frac{1}{i}(\frac{1}{i-p}-\frac{1}{i}))$$ $$= \frac{-p}{2} (\sum_{i=1}^{p-1} \frac{1}{i^2} +\frac{p}{i}\frac{1}{i(i-p)})$$ $$= \frac{-p}{2} (\sum_{i=1}^{p-1} \frac{1}{i^2} + p\frac{1}{i^2(i-p)})$$
Since $p | \sum_{i=1}^{p-1} \frac{1}{i^3}$ for $p>3$, we have $p | \sum_{i=1}^{p-1} \frac{1}{i^2(i-p)}$ for $p>3$ which implies: $p^3 | \sum_{i=1}^{p-1} \frac{1}{i}$ iff $p^2 | \sum_{i=1}^{p-1} \frac{1}{i^2}$.
This shows that the very first criterion, $p^2 | \sum_{i=1}^{p-1} \frac{1}{i^2}$ (which follows from the proof to Theorem 1 in this paper, see page 3) is equivalent to the nice criterion $H_{p-1}$ being divisible by $p^3$ (at least for $p>3$. $p=2,3$ can be verified seperately).