An odd number squared is odd.

If $n$ is odd, $n^{2}$ is odd.

Proof

Let $n$ be odd. Then $\exists k \in Z$ such that $n = 2 k - 1$ .
Then

\begin{aligned} n^{2} & = (2 k - 1)^{2} \\ = 4 k^{2} - 4 k + 1 \\ = 4 k^{2} - 4 k + 2 - 1 \\ = 2 (2 k^{2} - 2 k + 1) - 1 \end{aligned}

and $n^{2} = 2 m - 1$ for an integer $m$ , so $n^{2}$ is odd.

QED.