Divisibility of an expression Need some guidance
How to prove that $9\cdot n^9+7\cdot n^7+3\cdot n^3+n$ is divisible by $10$.
I've tried transforming the expression by adding $n^9$ and $-n^9$ in order to make a multiple of 10 but no use. I've even tried math. induction but got stuck, maybe this can not be proven. Are there any other methods?
Thanks in  advance
 A: Let $P$ be a polynomial: $P(n)=9n^9+7n^7+3n^3+n$.
$P(n)$ is even for any $n$, because if $n$ is even then you're looking at a sum of $4$ even numbers, if it's odd at a sum of $4$ odd numbers.
Then you only have to prove that $5|P(n)$.
Notice that $n^5\equiv_5 n$ for any $n$. (Proved at the end)
Then $P(n)\equiv_5 -n^9-3n^7+3n^3+n \equiv_5 -n^4n^5-3n^2n^5+3n^3+n \equiv_5 -n^5 - 3n^3 + 3n^3 + n = \ \ n - n^5 \equiv_5 0$
Here's a nice proof of $5|n^5-n$:
$$n^5-n=n(n^4-1)=n(n^2-1)(n^2-1)=n(n^2-1)(n^2+5-4)=n(n^2-1)(n^2-4)+5n(n^2-1) = (n-2)(n-1)n(n+1)(n+2)+5(n-1)n(n+1)$$
Which is a sum of a product of five consecutive integers and something divisible by $5$.
A: It is sufficient to check 2 and 5 divide this expression.
Since, you have an even number of terms with same parity, 2 divides it. 
$$9\cdot n^9+7\cdot n^7+3\cdot n^3+n \cong 4n^9 + 2n^7 + 3 n^3 +n\  mod\  5 $$ 
If $gcd(n, 5) =1 $, then by Fermat little theorem: 
$$n^ 4 \cong 1\  mod \ 5$$
Therefore, $4n^9 + 2n^7 + 3 n^3 +n \cong 4n + 2n^3 + 3 n^3 +n \cong 5n + 5n^ 2 \ mod \ 5$
Note: If $gcd(n, 5) \neq 1 \Rightarrow gcd(n, 5) = 5 $, which makes divisibility by 5 trivial.
