Discrete mathematics, divisibility How can I prove that for all $n\in\mathbf{N}$ that $6 | n^5 + 5n$?
I tested for $n = 2$ and got $6 | 32 + 10 = 42$.  

Work modulo 6:
$$\begin{align*}
\text{if $n \equiv 0 \mod 6$,}&\qquad n^5+5n\equiv 0,\\
\text{if $n \equiv 1 \mod 6$,}&\qquad n^5+5n\equiv 1+5=6\equiv0,\\
\text{if $n \equiv 2 \mod 6$,}&\qquad n^5+5n\equiv 32+10\equiv2+4\equiv0,\\
\text{if $n \equiv 3 \mod 6$,}&\qquad n^5+5n\equiv 3+3\equiv0,\\
\text{if $n \equiv 4 \mod 6$,}&\qquad n^5+5n\equiv 4+2\equiv0,\\
\text{if $n \equiv 5 \mod 6$,}&\qquad n^5+5n\equiv 5+1\equiv0.\\
\end{align*}
$$
So $n^5+5n$ is always a multiple of 6.
