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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