Question

Suppose $m,n$ are integers and $m=n^2-n$. Then show that $m^2-2m$ is divisible by $24$.

Answer 1

Given$:m=n^2-n$

$\Rightarrow m^2-2m={(n^2-n)}^2-2(n^2-n)$

$=n^4+n^2-2n^3-2n^2+2n$

$=n^4-2n^3-n^2+2n$

$=n^3(n-2)-n(n-2)$

$=(n^3-n)(n-2)$

$=n(n^2-1)(n-2)$

$=n(n-1)(n+1)(n-2)$

$=(n-2)(n-1)n(n+1)$

The above expression represents the product of $4$ consecutive numbers.

$\therefore$ it should be divisible by $24$ provided $n>2$.