Prove that $f(n) = 3n^5 + 5n^3 + 7n$ is divisible by 15 for every integer $n$ 
*

*So far I have only been able to complete the base case for which I got the following:


$$f(n) = 3n^5 + 5n^3 + 7n$$
$$f(n) = 3(1)^5 = 5(1)^3 + 7(1)$$
$$f(n) = 3 + 5 + 7$$
$$15/15 = 1$$
From here I got a bit confused with the inductive step in terms of number manipulation, a hint that my professor gave me was $f(-n) = -f(n)$. Any further help is appreciated.

$$\bmod3\\
n^1:0,1,2\\
n^2:0,1,1\\
n^3:0,1,2\\
5n^3+7n:0,0,0$$
This show that $3|(3n^5+5n^3+7)$.
$$\bmod5\\
n^1:0,1,2,3,4\\
n^2:0,1,4,4,1\\
n^3:0,1,3,2,4\\
n^5:0,1,2,3,4\\3n^5+7n:0,0,0,0,0
$$
This show that $5|(3n^5+5n^3+7)$.
