Answer to Question #272351 in Discrete Mathematics for sheriii malik

Question

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

is even.

Answer 1

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.