<s>[INST] 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?
 [/INST] To do it inductively, we also need a base case: $0 \times 1 \times 2=0$ is divisible by $3$.  Then, if $3$ divides $k(k+1)(k+2)=k^3+3k^2+2k$ then
\begin{align*}
(k+1)(k+2)(k+3) &= k^3+6k^2+11k+6 \\
 &= (k^3+3k^2+2k)+3(k^2+3k+2).
\end{align*}
We know $k^3+3k^2+2k$ is divisible by $3$, by the inductive hypothesis, and $3(k^2+3k+2)$ is divisible by $3$ since $k^2+3k+2$ is an integer.  Therefore, by induction, $k(k+1)(k+2)$ is divisible by $3$ for all $k \geq 0$.
Note, if we want to prove it for negative $k$ too, we'd need to perform "induction in reverse".
</s>