Since nnn is even integer and mmm is odd integer, then n=2kn=2kn=2k and m=2s+1m=2s+1m=2s+1 for some k,s∈Zk,s\in \mathbb Zk,s∈Z.
Then (n+2)2−(m−1)2=(2k+2)2−(2s+1−1)2=22(k+1)2−(2s)2=4(k+1)2−4s2=4((k+1)2−s2)(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)(n+2)2−(m−1)2=(2k+2)2−(2s+1−1)2=22(k+1)2−(2s)2=4(k+1)2−4s2=4((k+1)2−s2) where (k+1)2−s2(k+1)^2-s^2(k+1)2−s2 is integer. Therefore, (n+2)2−(m−1)2(n+2)^2-(m-1)^2(n+2)2−(m−1)2 is even.
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