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.

Although you asked for induction, it's much easier to simply split this into cases. We'll first prove that it's divisible by $3$.\begin{align}
n\equiv 0\mod 3&\Rightarrow $f(n)\equiv 3n^5+5n^3+7n\equiv 0+0+0\equiv 0\mod 3\\
n\equiv 1\mod 3&\Rightarrow $f(n)\equiv 3n^5+5n^3+7n\equiv 3+5+7\equiv 0\mod 3\\
n\equiv 2\mod 3&\Rightarrow $f(n)\equiv 3n^5+5n^3+7n\equiv 3(-1)^5+5(-1)^3+7(-1)\equiv 0\mod 3
\end{align}
And so $f(n)\equiv 0\mod 3$ for every $n$, so $3|f(n)$. I suppose you try $5|f(n)$ as an excercise for yourself, as I'm sure you're able to. Let me know if you encounter any problems.
I should probably also note Fermat's Little Theorem, which states that for any prime $p$, the equivalence $a^p\equiv a\mod p$ holds for all integer $a$. This means you can reduce your polynomial to (let's take $\mod 3$ again) to \begin{align}
f(n)&=3n^5+5n^3+7n\\
&\equiv 3\cdot n^3\cdot n^2+5n^3\cdot n^3\cdot n+7n\\
&\equiv 3\cdot n\cdot n^2+5n\cdot n\cdot n+7n\\
&\equiv 3n^3+5n^3+7n\\
&\equiv 3n+5n+7n\\
&\equiv 15n \mod 3
\end{align}
and from $f(n)\equiv 15n\mod 3$ we can easily see that $$f(n)\equiv 3\cdot 5n\equiv 0\mod 3$$
Hope this helped!
