Prove $3|n(n+1)(n+2)$ by induction I tried proving inductively but I didn't really go anywhere. So I tried:

Let $3|n(n+1)(n+2)$.
Then $3|n^3 + 3n^2 + 2n \Longrightarrow  3|(n(n(n+3)) + 2)$

But then?

Assume $n(n+1)(n+2) = 3p$.
$$(n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)}{n} =3p\cdot\frac{n+3}{n}=3p+3\cdot\frac{3p}{n}$$
Notice $\frac{3p}{n}$ is also an integer by assumption.
With this induction step, the base case has to be $n=1$. Prove the case for $n=0$ separately.
