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:
\(\ds p\) | \(=\) | \(\ds 2 r\) | for some $r \in \Z$ | |||||||||||
\(\ds p^k\) | \(=\) | \(\ds 2 s\) | for some $s \in \Z$ |
and so:
\(\ds p^{k+1}\) | \(=\) | \(\ds p \cdot p^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \left({2 r}\right) \left({2 s}\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \left({2 r s}\right)\) |
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:
\(\ds p\) | \(=\) | \(\ds 2 r + 1\) | for some $r \in \Z$ | |||||||||||
\(\ds p^k\) | \(=\) | \(\ds 2 s + 1\) | for some $s \in \Z$ |
and so:
\(\ds p^{k+1}\) | \(=\) | \(\ds p \cdot p^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \left({2 r + 1}\right) \left({2 s + 1}\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4 r s + 2 \left({r + s}\right) + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \left({2 r s + r + s}\right) + 1\) |
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.
$\blacksquare$