Inductive proof that $n(n-1)(n+1)$ is divisible by $6$ I am trying to prove that $n(n-1)(n+1)$ is divisible by $6$ for all $n$ in $\mathbb{N}$.
My attempt:
The result certainly holds for $n=0$. Suppose now that $n > 0$. Assume that $P(k)$ is true for all $k<n$. In particular $P(n-1)$ is true. Thus $(n-1)(n-2)n$ is divisible by 6. But
$n(n-1)(n+1) = (n-1)(n-2)n + 3n(n-1)$
Now I don't know how to proceed from here. It is not immediately apparent to me that the right hand side of the expression is divisible by 6.
 A: Does it have to be by induction?  The solutions suggested by the other answers are more straightforward.
Anyway...
Basis: $(1)(0)(2) = 0$ is divisible by $6$ (when $n = 1$).
Induction: Suppose $k = n(n-1)(n+1)$ is divisible by $6$.  Then
\begin{align}
(n+1)n(n+2) & = (n+1)n[(n-1)+3] \\
            & = k+3n(n+1)
\end{align}
and provided $n(n+1)$ is always even, this quantity is always also divisible by $6$.  We show this secondary claim, also by induction (!).
Basis: $(1)(2) = 2$ is even (when $n = 1$).
Induction: Suppose that $j = n(n+1)$ is even.  Then
$$
(n+1)(n+2) = j+2(n+1)
$$
is also even.
A: Hint : either $n$ or $n-1$ is even.
A: Hint: 


*

*if you multiply 2 consecutive integers together , one of them is going to be even (contains a factor of two)

*if you multiply 3 consecutive integers together, one of them is going to contain a factor of 2, and one of them is going to contain a factor of 3
A: $n(n+2)(n+2)=$ 6$n \choose 3$, $n \choose 3$ is always a natural number.
A: Since the OP already has an answer, here is one more approach for "fun" (in the spirit of killing a fly with a sledgehammer).
The product of the first two terms is $n^2 - n$ which is divisible by $2$, and the product of all three terms is $n^3 - n$ which is divisible by $3$, as follows in each case from Fermat's Little Theorem. 
Since $\gcd(2,3) = 1$, divisibility by each implies the expression is divisible by $2 \cdot 3 = 6$. QED.
