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Fermat's little thoerem states, that $a^{p-1} \equiv 1 \mod p$, which results in the obvious truth of $a^p \equiv a \mod p$. My question is whether an analogous version of this statement exists for Euler's theorem, without the assumption of coprimality.

So when does $$a^{\varphi(m) + 1} \equiv a \mod m$$ hold true?

Obviously, this is true when $a \perp m$, which follows directly from Euler's theorem. It is also false if we take $a = 2, m = 4$. But it seems to be true for $m = 10$ and any $a$. Could this be true for any pair $(a, m)$ where $m$ is not a power of $a$, perhaps?

Maurycyt
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  • Welcome to Math SE. Note your congruence equation also doesn't work for $a = 2$, $m = 12$, so it's not always true just where $m$ is not a power of $a$. – John Omielan Nov 21 '22 at 20:10
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    Thanks for the welcome! Also, that sucks. But for both $a=2, m=4$ and $a=2, m=12$ it's true that $m$ is divisible by a power of $a$ greater than $1$ (and this does not hold for $a \in \mathbb{N}, m = 10$. But as fun as trying to salvage this by adding whatever rules I'd like is, I better start thinking about something less arbitrary. – Maurycyt Nov 21 '22 at 20:16

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