Prove for all M and N, if M and M-N are even, then N is even.
Assume, by contradiction, that $\displaystyle N$ is odd. Then, $\displaystyle N=2k+1$ for some $\displaystyle k \in \mathbb{N}$
We also know that $\displaystyle M = 2r$ for some $\displaystyle r \in \mathbb{N}$. Then, $\displaystyle M-N = 2r - (2k+1) = 2(r-k) + 1$ which is an odd number, thus $\displaystyle M-N$ is odd, in contradiction, and so N is even.
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Or:
$\displaystyle N = M + (-)(M - N)$
And the sum of two even integers is even.