Alternate way to Prove or disprove $6\mid n(n+1)(n+2)$ This is my proof, I'm wondering if I'm correct, and how to do without induction.
My Work
Basis Step
$$\frac{(1)(2)(3)}{6} = 1$$
Inductive Hypothesis
Assume that $\dfrac{k(k+1)(k+2)}{6} = d$ where $d \in \mathbb Z$
Induction Step
We must show $\frac{(k+1)(k+2)(k+3)}{6}$ is an integer
$$\frac{(k+1)(k+2)(k+3)}{6} = \frac{(k+1)(k+2)}{6} \cdot (k+3)$$
We know that:
$\dfrac{(k+1)(k+2)}{6} = \dfrac{1}{k}\cdot\dfrac{k(k+1)(k+2)}{6}$ by our inductive hypothesis we know this to be an integer
Therefore, by the closure property of integers under multiplication and addition, $\dfrac{(k+1)(k+2)}{6} \cdot (k+3)$ must be integer.

Without induction:
At least one of $n, n+1, n+2$ is even and exactly one is divisible by three. Hence $2 \ | \ n(n+1)(n+2)$ and $3 \ | \ n(n+1)(n+2)$. As $2$ and $3$ are co-prime, this means $6 \ | \ n(n+1)(n+2)$.
