 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-4/6e/solution

p

a

a^{p} \equiv a (mod p); p ∤ a ⟹ a^{p - 1} \equiv 1 (mod p) .

p

(p - 1)! \equiv - 1 (mod p) .

All the following modular congruences are taken modulo the indicated prime.

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