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:

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

$\blacksquare$

Also see

• Power of Sum Modulo Prime