Unable to prove that $n^5-n$ is a multiple of 30. It is a problem in sec. 2.2 of the book titled by Griffin Harriet, and although have no counter-example till now as below:
$n(n-1)(n+1)(n^2+1) \implies$ if $n=4$, then $n-1=3, n+1=5, n^2+1=17$, and $4*3*5*17=30*2*17$.
Basically, the issue is how to show that :
$(n-1)(n)(n+1)(n^2+1)$ for $n$ being odd or even, is a multiple of $30$.
Let us take two cases: (a)$n$ is odd, i.e. of the form $2k+1$, for a suitable integer $k$.
So, $(2k)(2k+1)(2k+2)(2k+1)$, as $n^2+1$ will also be an odd number.
This leads to a final expression of the form: $2k+2$.
But, not able to prove its being a multiple of $30$.
(b) Need not take even case, as the final even form is proved by one term being even; & still not able to prove it being a multiple of $30$.
So, try by smaller values, and that would not serve the cause until use induction (strong case might be needed, as need consider all lower value for the case of $k=1$).
Am I correct, & should proceed by strong induction?

Modulo $5$ we have $$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)\equiv n(n^2-1)(n^2-4) =n(n-1)(n+1)(n-2)(n+2)$$ and the product of five successive integers  is divisible by $5$, while $n(n^2-1)=n(n-1)(n+1)$ is divisible by $6$.
