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In the Wikipedia article about the ElGamal signature scheme it is written, that Fermat's little theorem is used in the following proof of correctness:

From the signature generation in ElGamal we can derive, that:

$$H(m) \equiv xr + sk \pmod {p-1}$$

Then - the article states - Fermat's little theorem implies the following:

$$\begin{align} g^{H(m)} &\equiv g^{xr}g^{ks} &\pmod p\\ &\equiv (g^x)^r(g^k)^s &\pmod p\\ &\equiv (y)^r(r)^s &\pmod p \end{align}$$

The proof makes sense to me, i am just wondering, in what way Fermat's little theorem is used here, instead of just regular properties of exponentiation.

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    $\begingroup$ Hint: apply the definition of $H(m)\equiv xr+sk\pmod{p-1}$ so that you remove the modulo. You should then see how establishing $g^{H(m)}\equiv g^{xr}g^{ks}\pmod p$ involves Fermat's little theorem. $\endgroup$ – fgrieu Jan 24 '16 at 20:24
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Thanks to fgrieu's comment above and the following quote from here (PDF):

Theorem:

Let $p$ be a prime and let $a$ be a number not divisible by $p$. Then if $$ r \equiv s \pmod {p − 1} $$ we have $$ a^r \equiv a^s \pmod p$$ In brief, when we work $\mod p$, exponents can be taken $\mod{p − 1}$.

I (think i) understood, how the first (implicit) step relies on Fermat's little theorem: $$\begin{align} g^{H(m) \pmod{p-1}} &= g^{xr+sk \pmod{p-1}} \\ g^{H(m)} &\equiv g^{xr+sk} \pmod p \end{align}$$

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For $c\ne0$, the definition of $a\equiv b\pmod c$ is: $\exists d\in\mathbb Z$ such that $a=b+cd$.

Applying that definition to $H(m)\equiv xr+sk\pmod{p-1}$, we have that $\exists d\in\mathbb Z$ such that $H(m)=xr+sk+(p-1)d\;\text{ (equ. 1)}$.

Fermat's little theorem is that for $p$ prime and $g\not\equiv0\pmod p$, it holds that $g^{p-1}\equiv1\pmod p\;\text{ (equ. 2)}$.

We can now compute $$\begin{align} g^{H(m)} &\equiv g^{xr+sk+(p-1)d}&\pmod p&&\text{ (by equ. 1)}\\ &\equiv {(g^x)}^r\;{(g^k)}^s\;(g^{p-1})^d&\pmod p&&\text{ (rearanging)}\\ &\equiv {(g^x)}^r\;{(g^k)}^s\;1^d&\pmod p&&\text{ (by equ. 2)}\\ &\equiv {(g^x)}^r\;{(g^k)}^s&\pmod p&&\text{ (simplifying)}\\ &\equiv y^r\;r^s&\pmod p&&\text{ (by definition of }y\text{ and }r\text{)} \end{align}$$

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