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  "body": "\n## Proof\n\nWe argue by cases on whether $p$ divides $a$.\n\n1. Let $p$ be a [[prime number|0SN0MAEKYX]] and let $a$ be an integer.\n2. Case $p \\mid a$. Write $a = p s$ with an integer $s$. Since $p \\ge 2$, we have $a^p - a = p s (a^{p-1} - 1)$.\n3. So $p$ [[divides|096PW7Q014]] $a^p - a$, and $a^p \\equiv a \\pmod p$ by the definition of [[congruence|0TPW25CBN6]].\n4. Case $p \\nmid a$. By [[Fermat's little theorem|0M25PRN3WQ]], $a^{p-1} \\equiv 1 \\pmod p$.\n5. Since $a \\equiv a \\pmod p$, [[compatibility with multiplication|072D5G4SK6]] gives $a^{p-1} \\cdot a \\equiv 1 \\cdot a \\pmod p$, that is, $a^p \\equiv a \\pmod p$.\n6. In both cases, $a^p \\equiv a \\pmod p$.\n\n## Idea\n\nWhen $p$ divides $a$, both sides are multiples of $p$. Otherwise, multiply Fermat's congruence $a^{p-1} \\equiv 1$ by $a$.\n",
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