Show $5n^3+7n^5 \equiv 0$ mod 12 for all integers $n$. I've been stuck on this question for quite a while now. I could obvious show this by setting $n$ to all numbers from 0-11, however this is not very efficient. 
Note $5n^3+7n^5 = n^3(5+7n^2)$. We could use the Chinese remainder theorem, we show $n^3(5+7n^2) \equiv 0 $ mod 3 and mod 4, but this doesn't seem to be any easier. What would be the best way to solve this problem?

The only quadratic residue mod $3$ is $1$.
The only quadratic residue mod $4$ is also $1$.
Therefore:
\begin{align*}
n^3 &\equiv n^5 \pmod{3} &\qquad n^3 &\equiv n^5 \pmod{4}
\end{align*}
and so
$$ n^3 \equiv n^5 \pmod{12} $$
Applying this to the original equation:
$$ 5n^3 + 7n^5 \equiv 12n^3 \equiv 0 \pmod{12} $$
