Parity of Integer equals Parity of Positive Power

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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.

$\blacksquare$