If n is any even integer and m is any odd integer, then (n + 2)2 - (m - 1)2 is even.

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Since $n$ is even integer and $m$ is odd integer, then $n=2k$ and $m=2s+1$ for some $k,s\in \mathbb Z$ .

Then $(n+2)^2-(m-1)^2= (2k+2)^2-(2s+1-1)^2=2^2(k+1)^2-(2s)^2=4(k+1)^2-4s^2=4((k+1)^2-s^2)$ where $(k+1)^2-s^2$ is integer. Therefore, $(n+2)^2-(m-1)^2$ is even.

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