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Theorem 5.25 in Apostol's Analytic Number Theory book (Wolstenholme's theorem) states that the following holds for any prime $p \ge 5$ : $$\sum_{k=1}^{p-1} \dfrac{(p-1)!}{k} = 0 \mod p^2.$$ I read and read the proof but I never found somewhere when $p=2, 3$ are invalid. That is, in the process of proof there is nowhere when the proof fails for $p=2, 3$. However, the statement does not hold for $p=2, 3$ and holds for $p=5$. Perhaps in order for $S_1$ to appear in $g(x)$, $p=2$ has been put aside, but $S_{p-2}=S_1$ for $p=3$ still is valid. So how is the proof valid for $p=2,3$ as well but the Wolstenholme's identity is not valid for $p=2,3$?

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  • $\begingroup$ Just take $p=2$ or $p=3$ and go through the proof line by line until it breaks down. In either case the sums (or binomial symbols, or factorials) will all be very simple. In the Wkipedia proof of an equivalent version, at some point there is an explicit division by $3$, for instance. Without seeing the proof you have in mind, there's not a lot more to say. $\endgroup$
    – lulu
    Oct 8, 2023 at 11:55
  • $\begingroup$ In this proof the problem comes from the sum of squares formula, which has a $6$ in the denominator. $\endgroup$
    – lulu
    Oct 8, 2023 at 12:02

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The issue is just before the end of the proof, where an equation is taken modulo $p^3$. If $p \ge 5$, then this will cancel all terms except $-S_{n-1}p$, but if $p=3$, then $p^{p-1} = 9$ is not divisible by $p^3 = 27$ and will not be cancelled either.

In that case, you get $3S_1 \equiv 3^2 = 9$ mod 27, so $S_1 \equiv 3$ mod 9, which is confirmed by the actual sum being $S_1 = 3$.

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