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Fermat's little theorem states that

if $n$ is a prime number then $a^n \equiv a \pmod n$.

Euler's theorem and Carmichael's theorem relax it from "if $n$ is a prime number" to

if $a$ and $n$ are coprime integers then $a^m \equiv a \pmod n$.

, where $m=\phi(n)+1$ for Euler's theorem and $m=\lambda(n)+1$ for Carmichael's theorem.

That theorems are pretty amazing and simplify modular calculations a lot.

But what if $a$ and $n$ aren't coprimes? Are there some generalisations to previous theorems for arbitrary $a$ and $n$, or for $a$ and $n$ with even more relaxed requirements?

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    $\begingroup$ That should be $m = \phi(n)+1$ and $\lambda(n)+1$. $\endgroup$ – Robert Israel Oct 13 '16 at 6:50
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If, for example, prime $p$ divides both $a$ and $n$ but $p^2$ divides $n$ and does not divide $a$, then $a^m \not\equiv a\; (\text{mod}\; n)$ for any $m > 1$.

On the other hand, if $n$ is squarefree then for any $a$, $a^{\lambda(n)+1} \equiv a \; (\text{mod}\; n)$.

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