Prove that for prime $p \gt 2$, $(p - 1)^{p - 1} \equiv p+1 \pmod{p^2}$.

Question

Prove that for prime $p \gt 2$, $(p - 1)^{p - 1} \equiv p+1 \pmod{p^2}$

My work

In

At least one well-defined cyclic subgroup of $(\mathbb{Z}/{p^2}\mathbb{Z})^\times$, for prime $p$

I examined the generators $[pq + 1], \text{where 0 } \lt q \lt p$. This cyclic group appears to be subgroup of a larger cyclic group generated by $[p-1]$.

I convinced myself that the title question is true using wolfram, but a proof along with some elaboration of the group theory underpinnings would be of interest.

I would have attempted to develop this further but feel I am missing something since these investigations are not generating any interest - any comments to set me on the right path will be appreciated.

and one can easily prove that for prime $p \gt 2$, the order of $[p - 1]$ in $(\mathbb{Z}/{p^2}\mathbb{Z})^\times$ is equal to $2p$. — CopyPasteIt, Aug 17 '20 at 18:58

score 4 · Accepted Answer · answered Aug 17 '20 at 15:38

4

Hint:

As $p-1$ is even, $$(p-1)^{p-1}=(1-p)^{p-1}$$

Use Binomial Expansion

$$(1-p)^{p-1}\equiv1-\binom{p-1}1p\equiv1-(p-1)p\equiv?\pmod{p^2}$$

answered Aug 17 '20 at 15:38

lab bhattacharjee

274,582

Prove that for prime $p \gt 2$, $(p - 1)^{p - 1} \equiv p+1 \pmod{p^2}$.

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