# Parity of Integer equals Parity of Positive Power

## Theorem

Let $p \in \Z$ be an integer.

Let $n \in \Z_{>0}$ be a strictly positive integer.

Then $p$ is even if and only if $p^n$ is even.

That is, the parity of an integer equals the parity of all its (strictly) positive powers.

## Proof

Proof by induction:

For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition:

- For all $p \in \Z$, $p$ is even if and only if $p^n$ is even.

First it is worth confirming that $P \left({0}\right)$ does not hold:

- $\forall p \in \Z: p^0 = 1$

which is not even whatever the parity of $p$.

$P \left({1}\right)$ is true, as this just says:

- $p$ is even if and only if $p$ is even

which is a tautology.

### Basis for the Induction

$P \left({2}\right)$ is the case:

- $p$ is even if and only if $p^2$ is even

which is demonstrated in Parity of Integer equals Parity of its Square.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

- $p$ is even if and only if $p^k$ is even

Then we need to show:

- $p$ is even if and only if $p^{k+1}$ is even

### Induction Step

This is our induction step:

Let $p$ be even.

By the induction hypothesis, $p^k$ is also even.

Then:

\(\displaystyle p\) | \(=\) | \(\displaystyle 2 r\) | for some $r \in \Z$ | ||||||||||

\(\displaystyle p^k\) | \(=\) | \(\displaystyle 2 s\) | for some $s \in \Z$ |

and so:

\(\displaystyle p^{k+1}\) | \(=\) | \(\displaystyle p \cdot p^k\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({2 r}\right) \left({2 s}\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2 \left({2 r s}\right)\) |

So $p^{k+1}$ is even.

Now suppose $p$ is not even (that is, odd).

By the induction hypothesis, $p^k$ is also not even (that is, odd).

Then:

\(\displaystyle p\) | \(=\) | \(\displaystyle 2 r + 1\) | for some $r \in \Z$ | ||||||||||

\(\displaystyle p^k\) | \(=\) | \(\displaystyle 2 s + 1\) | for some $s \in \Z$ |

and so:

\(\displaystyle p^{k+1}\) | \(=\) | \(\displaystyle p \cdot p^k\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({2 r + 1}\right) \left({2 s + 1}\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 4 r s + 2 \left({r + s}\right) + 1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2 \left({2 r s + r + s}\right) + 1\) |

So $p^{k+1}$ is not even (that is, odd).

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

Therefore:

- For all $n \in \Z_{>0}$, for all $p \in \Z$, $p$ is even if and only if $p^n$ is even.

$\blacksquare$