Answer
See direct proof below.
Work Step by Step
Let $m, n\in \mathbb{Z}$ and $mn$ be even. Then $mn$ has a factor of 2. $mn=2\frac{mn}{2}$. By factoring, either $\frac{m}{2}$ or $\frac{n}{2}$ is an integer. Without loss of generality, let this be $\frac{m}{2}$. Because $m=2*\frac{m}{2}$, $\frac{m}{2}\in \mathbb{Z}$, $m$ has a factor of 2 and thus is even.
