For all integers n and m, if n − m is even then n^3 − m^3 is even.

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Let $n$ and $m$ be integers such that $n−m$ is even. Therefore, $n−m = 2k$ for some $k ∈ \Z.$ Note that

$n^3-m^3=(n-m)(n^2+nm+m^2)$

$=2k(n^2+nm+m^2)=2(k(n^2+nm+m^2))$

Let $k(n^2+nm+m^2).$ Since $k(n^2+nm+m^2)$ is an integer, $n ^3 − m^3 = 2r,$ with $r ∈ \Z.$ Therefore $n ^3 − m^3$ is even.