 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2009/c/1/iii/solution

Combine the preceding congruences:

a^{p} \equiv a

and

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

. Therefore

a^{p} + (p - 1)! a \equiv a - a = 0

, and

(p - 1)! a^{p} + a \equiv - a + a = 0

. This proves both stated divisibilities for every integer $a$ , with no assumption that

a

is coprime to

p

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