Power of Sum Mod Prime

Theorem

Let $p$ be a prime number.

Then:

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

Corollary

Let $p$ be a prime number.

Then:

$\left({1 + b}\right)^p \equiv 1 + b^p \pmod p$

Proof

From the Binomial Theorem:

$\displaystyle \left({a + b}\right)^p = \sum_{k=0}^p \binom p k a^k b^{p-k}$

Also note that:

$\displaystyle \sum_{k=0}^p \binom p k a^k b^{p-k} = a^p + \sum_{k=1}^{p-1} \binom p k a^k b^{p-k} + b^p$

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \forall k: 0 < k < p: \binom p k\)	\(\equiv\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \pmod p\)	\(\displaystyle \)	Binomial Coefficient of Prime
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \binom p k a^k b^{p-k}\)	\(\equiv\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \pmod p\)	\(\displaystyle \)	Definition of modulo multiplication
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \sum_{k=1}^{p-1} \binom p k a^k b^{p-k}\)	\(\equiv\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \pmod p\)	\(\displaystyle \)	Definition of modulo addition
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \sum_{k=0}^p \binom p k a^k b^{p-k}\)	\(\equiv\)	\(\displaystyle {a^p + b^p}\)	\(\displaystyle \)	\(\displaystyle \pmod p\)	\(\displaystyle \)	from above

$\blacksquare$

Proof of Corollary

Follows immediately by putting $a=1$.

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 2.6$: Example $42 \ (3)$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 23 \zeta$