Finding all primes $p,$ such that for all integers $a$, $a^{25} \equiv a\pmod p$.

Question

Find all primes $p$ such that the following congruence holds for all integers $a$: $\quad a^{25}\equiv a\pmod{p}$.

I suspect there is a very simple solution, but I can't find it.

Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. — amWhy, Jan 04 '19 at 20:55

score 2 · Accepted Answer · answered Jan 04 '19 at 20:57

2

Hint:

If $a$ is not divisible by $p$, it is equivalent to $a^{24}\equiv 1\pmod p$.

By lil' Fermat, this implies that $p-1$ divides $24$.

answered Jan 04 '19 at 20:57

Bernard

1

LOOL Lil' Fermat xD – Cristian Baeza Jan 04 '19 at 21:01
Thanks for the reply. Yes, I now see that it holds for every $p$ where $p-1$ divides $24$. But I don't see why it can't hold for some other $p$ as well. – John Jan 04 '19 at 21:19
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Because it has to hold for every $a$, and in particular for the generator of the cyclic group $(\mathbf Z/p\mathbf Z)^\times$. This generator has order $p-1$. – Bernard Jan 04 '19 at 21:33

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