Proof

We present two proofs of the first statement in Fermat's little theorem. The second statement follows easily, as each nonzero element in Z/pZ has an inverse.

First proof.

Second proof.

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First proof

Let x Z. Then (x + 1)^p equals

p k=0

p k

xk.

Since p is a prime, we find that p divides

p k

for all 0 < k < p. Hence we have

(x + 1)^p = x^p + 1 mod p.

Now with induction on x we easily obtain the theorem.

Second proof

For a = 0 mod p or a = 1 mod p , the statement is clear.

Let S = {1, 2, ..., p - 1} Z/pZ. By a previous theorem, for 1 < a < p the map

h: S -> S,    h(x) = a · x

is bijective (because a is invertible, the map x -> a^-1x is the inverse map of h).

This implies

_{x S} x = _{x S} h(x) = a^p-1_{x S} x .

It follows that p | a^p-1 - 1.