# Prime Power of Sum Modulo Prime

## Theorem

Let $p$ be a prime number.

Then:

$\forall n \in \N_{> 0}: \paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$

### Corollary

Let $p$ be a prime number.

Then:

$\forall n \in \N_{> 0}: \paren {1 + b}^{p^n} \equiv 1 + b^{p^n} \pmod p$

## Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

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

### Basis for the Induction

First from Power of Sum Modulo Prime we have that $\map P 1$ is true:

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

This is our basis for the induction.

### Induction Hypothesis

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

So this is our induction hypothesis:

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

Then we need to show:

$\paren {a + b}^{p^{k + 1} } \equiv a^{p^{k + 1} } + b^{p^{k + 1} } \pmod p$

### Induction Step

This is our induction step:

 $\ds \paren {a + b}^{p^k}$ $\equiv$ $\ds a^{p^k} + b^{p^k}$ $\ds \pmod p$ Induction Hypothesis $\ds \leadsto \ \$ $\ds \paren {\paren {a + b}^{p^k} }^p$ $\equiv$ $\ds \paren {a^{p^k} + b^{p^k} }^p$ $\ds \pmod p$ Congruence of Powers $\ds \leadsto \ \$ $\ds \paren {\paren {a + b}^{p^k} }^p$ $\equiv$ $\ds \paren {a^{p^k} }^p + \paren {b^{p^k} }^p$ $\ds \pmod p$ Basis for the Induction $\ds \leadsto \ \$ $\ds \paren {a + b}^{p^{k + 1} }$ $\equiv$ $\ds a^{p^{k + 1} } + b^{p^{k + 1} }$ $\ds \pmod p$

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

Therefore:

$\forall n \in \N_{> 0}: \paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$

$\blacksquare$