Prove that $(a+b)^p \equiv a^p+b^p \mod p$

Question

Prove that $$(a+b)^p \equiv a^p+b^p \mod p$$

My try: I tried expanding this and then applying modulo to this expression but it did not work.

Answer 1

Suppose $a,b$ are integers, and $p$ is prime.

By Fermat's little Theorem,

$$x^p \equiv x \pmod{p}$$

for all integers $x$, hence

$$(a+b)^p \equiv a+b \pmod{p}$$

and

$$a^p + b^p \equiv a+b \pmod{p}$$

Therefore

$$(a+b)^p \equiv a^p + b^p \pmod{p}$$

Answer 2

Hint:

Prove that if $p$ is prime, $p$ divides the binomial coefficients $\dbinom pk$ for all $1\le k\le p-1$, and apply the binomial theorem modulo $p$.