What a Proof Is · Medium problem

$n^2 + 5n = n(n + 5)$. If $n$ is even, $n$ is an even factor; if $n$ is odd, $n + 5$ is even. Either way the product is even.

For $n = 1, 2, 3, \ldots, 10$ the values $6, 14, 24, \ldots, 150$ are all even, so it holds for all $n$.

Let $n = 2k$. Then $n^2 + 5n = 4k^2 + 10k = 2(2k^2 + 5k)$, which is even.

$n^2 + 5n = n(n + 5)$, and since $n + 5 > n$ the product is even.