(a + b)p ≡ ap + bp (mod p)

Lemma 1:

if p is prime, then:

(a + b)^p ≡ a^p + b^p (mod p)

Proof:

(1) Using the Binomial Theorem (see Theorem here), we know that:



(2) Now, since p is a prime, it is clear that for each term p!/(m!)(p-m)!, p is not divisible by any term ≤ m or by any term ≤ p-m so that we have:

p!/(m!)(p-m)! = p*([(p-1)*...*p-m+1]/[m!])

(3) This shows that each of these terms is divisible by p and therefore:

[p!/(m!)(p-m)!]a^p-mb^m ≡ 0 (mod p)

(4) So that there exists an integer n such that:

(a + b)^p = a^p + np + b^p

(5) Since a^p + nb + b^p ≡ a^p + b^p (mod p), it follows that:

(a + b)^p ≡ a^p + b^p (mod p)

QED