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 $6 | 32 + 10 = 42$.  

Use modular arithmetic and solve for the finite number of cases $n \equiv 0, 1, 2, 3, 4, 5 \pmod{6}$.
For example, $4 \cdot 4 = 16 \equiv 4 \pmod{6}$, so that $4^5 \equiv 4 \pmod{6}$, and $n^5 + 5n \equiv 4 + 20 \pmod{6} \equiv 24 \pmod{6} \equiv 0 \pmod{6}$.
The five other cases are similar.
