2
$\begingroup$

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?

1 Answers 1

2

OK, I eventually solved it myself. The manipulation is the following:

$ \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).