Prove that $n^{2}-n$ is divisible by $2$ for every positive integer $n$.

We know that any positive integer $n$ is of the form $2m$ or $2m+1$,where $m$ is a non-negative integer.
Case $1$: If $n = 2m$,then
$n^{2} - n = (2m)^{2} - (2m) = 4m^{2} - 2m = 2(2m^{2} - m)$.
Since $2(2m^{2} - m)$ is a multiple of $2$,$n^{2} - n$ is divisible by $2$.
Case $2$: If $n = 2m + 1$,then
$n^{2} - n = (2m + 1)^{2} - (2m + 1) = (4m^{2} + 4m + 1) - 2m - 1 = 4m^{2} + 2m = 2(2m^{2} + m)$.
Since $2(2m^{2} + m)$ is a multiple of $2$,$n^{2} - n$ is divisible by $2$.
Thus,in both cases,$n^{2} - n$ is divisible by $2$ for every positive integer $n$.