Prime Power of Sum Modulo Prime

Theorem

Let $p$ be a prime number.

Then:

$\forall n \in \N^*: \left({a + b}\right)^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$

Corollary

Let $p$ be a prime number.

Then:

$\forall n \in \N^*: \left({1 + b}\right)^{p^n} \equiv 1 + b^{p^n} \pmod p$

Proof

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

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

Basis for the Induction

First from Power of Sum Mod Prime we have that $P(1)$ is true:

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

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

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

Then we need to show:

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

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({a + b}\right)^{p^k}\)	\(\equiv\)	\(\displaystyle a^{p^k} + b^{p^k}\)	\(\displaystyle \)	\(\displaystyle \pmod p\)	\(\displaystyle \)	Induction Hypothesis
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({\left({a + b}\right)^{p^k} }\right)^p\)	\(\equiv\)	\(\displaystyle \left({a^{p^k} + b^{p^k} }\right)^p\)	\(\displaystyle \)	\(\displaystyle \pmod p\)	\(\displaystyle \)	Congruence of Powers
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({\left({a + b}\right)^{p^k} }\right)^p\)	\(\equiv\)	\(\displaystyle \left({a^{p^k} }\right)^p + \left({b^{p^k} }\right)^p\)	\(\displaystyle \)	\(\displaystyle \pmod p\)	\(\displaystyle \)	Basis for the Induction
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({a + b}\right)^{p^{k+1} }\)	\(\equiv\)	\(\displaystyle a^{p^{k+1} } + b^{p^{k+1} }\)	\(\displaystyle \)	\(\displaystyle \pmod p\)	\(\displaystyle \)

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall n \in \N^*: \left({a + b}\right)^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$

$\blacksquare$

Proof of Corollary

Follows immediately by putting $a=1$.

$\blacksquare$

Also see

Compare with the Freshman's Dream.

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 2.6$: Example $42 \ (4)$