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**VonNemo19** Show that if $\displaystyle m^2=m\cdot{m}$ is even, then $\displaystyle m$ is even.

I think I've got it:

By definition, we can write $\displaystyle m^2=2n$ for some integer $\displaystyle n$. This implies that $\displaystyle m\cdot{m}=2n$. But, this means that $\displaystyle m\cdot{m}$ has a factor of 2. Therefore, $\displaystyle m$ has a factor of 2. So, by definition, $\displaystyle m$ is even.

Now, my gut instinct is that I butchered this. Can someone clean it up?