Originally Posted by **ThePerfectHacker**

Just like CaptainBlack said I do not know what high school math is but try it like this:

Show that if it works for $\displaystyle n$ then it must work for $\displaystyle n+1$. Meaning if it works for 1 it must work for 2 but if it works for 2 then it works for 3 but if it works for 3 it must work for 4.... (This is called induction).

Thus, let $\displaystyle n^5-n$ be divisible by 30.

Then show that $\displaystyle (n+1)^5-(n+1)$ is divisible by 30. Use the binomial theorem

$\displaystyle (n+1)^5-(n+1)=n^5+5n^4+10n^3+10n^2+5n+1-n-1$

Thus,

$\displaystyle (n^5-n)+5n^4+10n^3+10n^2+5n$

But, $\displaystyle n^5-n$ is divisible by 30,

thus,

$\displaystyle 5n^4+10n^3+10n^2+5n$ if divisible by 30 then the whole expression is divisible by 30.

But

$\displaystyle 5n^4+10n^3+10n^2+5n$ is divisible by 5

$\displaystyle n^4+2n^3+2n^3+n$. Factor,

$\displaystyle n(n^3+2n^2+2n+1)$.