prove that 15 divides $n^7+2n^5+4n^3+8n$ for any integer n 
Prove that for all $n\in\mathbb{Z}$, $15$ divides $n^7+2n^5+4n^3+8n$.

I know that I have to show that 5 divides $n^7+2n^5+4n^3+8n$ and that
3 divides $n^7+2n^5+4n^3+8n$, I also know that by FLT $n^5 \equiv n \pmod{5}$ and that $n^3 \equiv n \pmod{3}$, but I'm not sure on how to further proceed with this. Any help would be appreciated. 
Thanks For the Help guys, I figured it out on my own though!

$$n^3-n=n(n^2-1), n^5-n=n(n^2-1)(n^2+1)\implies n^5-n$$ is divisible by $(3,5)=15$
Now $$n^7+2n^5+4n^3+8n=n^5(n^2+2)+4n^3+8n\equiv 5n^3+10n\pmod{15}$$
We need  $15|(5n^3+10n)\iff3|(n^3+2n)$
But $n^3+2n=n^3-n+3n$
