It's true that $(n) (n + 1) (2n + 1)$ is a multiple of 6?

Question

Show that for all n ∈ N, it is true that $(n) (n + 1) (2n + 1)$ is a multiple of 6, that is: $$\exists m\in \mathbb{N}: n (n + 1) ( 2n + 1) = 6m$$

I think I can prove it by induction, but I don't know how proceed.

Answer 1

One of $n$ and $n+1$ is a multiple of $2$.

Then, either:

$n$ is a multiple of $3$, or

$n$ is $1$ more than a multiple of $3$, in which case $2n+1$ is a multiple of $3$, or

$n$ is $2$ more than a multiple of $3$, in which case $n+1$ is a multiple of $3$.

In any case, you have a factor which is even, and a factor which is a multiple of $3$, thus the expression is a multiple of $6$.

Answer 2

Yes, do the base case, $0*1*1 = 0 = 6*0$.

Now for the inductive case, assuming it works for n:

\begin{equation*} \begin{split} (n+1)(n+2)(2(n+1)+1) & = 6 + 13 n + 9 n^2 + 2 n^3 \\ & = (n + 3 n^2 + 2 n^3) + (6+12n+6n^2) \\ & = n(n+1)(2n+1) + (6+12n+6n^2) \end{split} \end{equation*}

Because $n$, is an integer $(n+1)(n+2)(2(n+1)+1)$. This concludes the proof.

Answer 3

Try to prove by induction that $$1^2+2^2+\cdots+n^2=\frac16n(n+1)(2n+1)$$ then as both sides are integers we know that $6$ divides $n(n+1)(2n+1)$.

Answer 4

Just look at it mod 6 and plug in 0 through 5.

You don't have to be fancy.