📐 Fermat

If $p$ is prime and $p \\nmid a$, then $a^{p-1} \\equiv 1 \\pmod{p}$.

Proof: Let $G_p = \\{[1], \\ldots, [p-1]\\}$. Multiplying all elements by $[a]$ permutes $G_p$. Product of $aG_p$ equals $[a]^{p-1}$ times product of $G_p$. Cancel to get $[a]^{p-1} = [1]$.

From: intro-discrete

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