Question

If $n$ is greater than $2$, show that $n^5-5n^3+4n$ is divisible by $120$.

Answer 1

$n^5-5n^3+4n=n(n^4-5n^2+4)=n(n^4-4n^2-n^2+4)=n\left\{n^2\right(n^2-4)-1(n^2-4\left)\right\}=n(n^2-1)(n^2-4)=n(n-1)(n+1)(n-2)(n+2)=(n-2)(n-1)n(n+1)(n+2)$

Product of any $r$ consecutive integers is divisible by $r!$

We have five consecutive integers so it will be divisible by $5!=120$

Hence proved.