Fermat's little theorem

Given a prime number $𝑝$ and number $𝑎 \in ℕ$ , then $𝑎 𝑝 - 1 \equiv 𝑝 1$ and $𝑎 𝑝 \equiv 𝑝 𝑎$ . #m/thm/num

Proof

Let $𝑎 = 𝑚 𝑝 + 𝑟$ where $0 \leq 𝑟 < 𝑝$ , so $𝑎 \equiv 𝑝 𝑟$ , and it suffices to show $𝑟 𝑝 - 1 \equiv 𝑝 1$ . Then $𝑟 \in ℤ \times 𝑝$ , and since the order of an element divides the order of a group, $𝑎 𝑝 - 1 \equiv 1$ .

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