Prove that $10|n+3n^3+7n^7+9n^9$ Prove that $10|n+3n^3+7n^7+9n^9$ for every $n\in \mathbb N$
Only what i see that 10=5*2 and both number is free numbers, and if I show that $5|n+3n^3+7n^7+9n^9$ and $2|n+3n^3+7n37+9n^9$ that I prove, since 5 and 2 is free numbers i can use Fermat's little theorem such that  $5|n^5-n$ and $2|n^2-n$ but i can not see that somehow help me, do you have some idea?

Hint:
$$n+3n^3+7n^7+9n^9\equiv n+3n^3-3n^7-n^9\equiv -(n^4-1)(n)(n^4+3n^2+1)$$
mod 10, so prove
$$2|n(n^4-1)$$
and 
$$5|n^4-1$$
when $n\neq5k$ and $5|n$ when $n=5k$.
