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 $6 | 32 + 10 = 42$.  
 A: 1) Show that $2|n^5+5n$.
2) Show that $3|n^5 + 5n$.
If we can show both, separately, it has to follow that $6=2*3|n^5 + 5n$.
1)$n^5 + 5n = n(n^4+5)$.  If $n$ is even so is $n(n^4+5)$.  If $n$ is odd then $n^5$ is odd and $n^5 + 5$ is even and $n(n^4+5)$ is even.
So $n^5 + 5n$ is even.
2) $n = 3m + i$ for some $m$ and $i = 1, 0, $ or $-1$.
a) $i = 0; n = 3m$ then $3|n$ and $3|n(n^4 + 5)$.
b) $i = \pm 1$ then 
$n^4 + 5$ = $(3m \pm 1)^4 + 5$
$ = (3^4m^4 \pm 4*3^3m^3 + 6*3^2m^2 \pm 4*3m + 1) + 5$
$= 3^4m^4 \pm 4*3^3m^3 + 6*3^2m^2 \pm 4*3m + 6$
$= 3(3^3m^4 \pm 4*3^3m^3 + 6*3m^2 \pm 4*m + 2)$
So $3|n^4 + 5$ and $3|n^5 + 5n$.
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So both $2|n^5+5n$ and $3|n^5 + 5n$ so $6|n^5+5n$.
A: 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.
A: 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.
A: $$n^5+5n=6n+(n-1)n(n+1)(n^2+1)$$
Now $6=3!$ divides the product of any three consecutive integers(why?)
