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An implication of Fermat's Little Theorem is the following:

If $p$ is prime, and $a$ is not a multiple of $p$, then $$r \equiv s \pmod {p-1} \implies a^r \equiv a^s \pmod p.$$

I need this implication to prove the verification of the Elgamal signature, but I honestly do not see how to derive from Fermat's Little Theorem to this implication and I could not find any proof of this.

Any help would be much appreciated!

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Assume without loss of generality that $r\geq s$, so $r=s+k(p-1)$ for some $k\geq 0$. Then $a^r=a^sa^{k(p-1)}$. Can you see how to get from here to what you want?

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  • $\begingroup$ Yes, because $a^{k(p-1)}= a^{(p-1)^k} = 1^k = 1$. Thank you! $\endgroup$ – Joey Jun 10 '18 at 10:22

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