Prove that $6|n(n+1)(2n+1)$, $n\in\mathbb{Z}$ I am trying to solve this question without modular arithmetic.

Let $n\in\mathbb{Z}$. Prove that $6$ divides $n(n+1)(2n+1)$.

Using induction:
$\text{For }n=1
\\P(1) = 1(1+1)(2(1)+1)=6
\\\text{For }n=k
\\P(k)=k(k+1)(2k+1)
$
$P(k)$ is divisible by $6$ so $k(k+1)(2k+1)=6h, h\in\mathbb{Z}$. Once by the induction hypothesis, $P(k)$ and $P(k+1)-P(k)$ being divisible by $6$ so $P(k+1)$ will be divisible by $6$. as well.
$P(k+1)= (k+1)(k+1+1)(2(k+1)+1) = (k+1)(k+2)(2k+3)$
So,
\begin{align}
P(k+1)-P(k) & =(k+1)(k+2)(2k+3)-k(k+1)(2k+1) \\
& = 2k^3+9k^2+13k+6 - (2k^3+3k^2+k) \\
& = 6k^2+12k+6 \\
& = 6(k^2+2k+1) \\
& = 6(k+1)^2
\end{align}
Therefore, $P(k+1)$ is true and $6$ divides $n(n+1)(2n+1)$.
But once $n\in\mathbb{Z}$ the proof is not complete. I was wondering about considering $6|(-n)(-n+1)(-2n+1)$, $n\in\mathbb{Z}$ but I am not really sure how to proceed. It doesn't seem to be the right approach. How can I keep doing that or what would be a good proof? Proofs without modular arithmetic are preferable, but if you got a nice proof it would be welcomed. I've seen similar questions, but they considered $n\in\mathbb{N}$.
 A: Hint $ $ You proved $\,6\mid P(k\!+\!1)-P(k),\,$ thus $\ 6\mid P(k\!+\!1)\color{#0a0}\Leftarrow\!\color{#c00}\Rightarrow 6\mid P(k).\,$ You used the $\,\color{#0a0}{{\rm direction}}\ \ 6\mid P(k\!+\!1)\color{#0a0}\Leftarrow 6\mid P(k)\,$ to inductively ascend its truth from $\,k\,$ to $\,k\!+\!1.\,$ Now use the reverse $\rm\color{#c00}{{\rm direction}\ (\Rightarrow)}$ to inductively descend its truth from $\,k\!+\!1\,$ to $\,k,\,$ so to all negatives too.
Remark $ $ You may find it helpful to first prove the following Lemma for any integer predicate $p$.
Lemma $ $  (Bidrectional Induction) $ $ Suppose that  $\ p(k\!+\!1)\!\iff\! p(k)\ $ is true for all $\,k\in \Bbb Z$.
Then $\,p(k)\,$ is true for all $\,k\in \Bbb Z\!\iff\! p(k)\,$ is true for some $\,k_0\in\Bbb Z$.
OP is the special case $\ p(k) := 6\mid P(k),\, $ and $\,k_0 = 0$.
A: For $n$ such that $(n,6)=1$, we could apply Euler's theorem, with $\varphi(6)=2$.  We get $n^2\equiv1\bmod6$.  Then look at $n(n+1)(2n+1)\equiv2n^3+3n^2+n\equiv3n+3\equiv3(n+1)\equiv0\bmod6$, since $n$ is odd.
When $n\equiv2;3;4;0\bmod6$, we get $n(n+1)(2n+1)\equiv0\bmod6$ by calculation.
A: We can use The Division Algorithm:
$n = 6m+r$ with $0\le r <m$, So $r \in \{0,1,2,3,4,5\}$
$$n = 6m; n=6m+1; n=6m+2; n=6m+3; n=6m+4; n=6m+5 $$
Now you have to just plug these values to:
$$\frac{n(n+1)(2n+1)}{6}$$
And you will get an integer every time, which means that $6 \mid n(n+1)(2n+1)$
