N (n + 1) (2n + 1) over 6 is Integer

Theorem
Let $n \in \Z$ be an integer.

Then $\dfrac {n \paren {n + 1} \paren {2 n + 1} } 6$ is also an integer.

Proof
This is equivalent to proving that $n \paren {n + 1} \paren {2 n + 1}$ is a multiple of $6$.

There are $6$ cases to consider:


 * $(1): \quad n \equiv 0 \pmod 6$: we have $n = 6 k$


 * $(2): \quad n \equiv 1 \pmod 6$: we have $n = 6 k + 1$


 * $(3): \quad n \equiv 2 \pmod 6$: we have $n = 6 k + 2$


 * $(4): \quad n \equiv 3 \pmod 6$: we have $n = 6 k + 3$


 * $(5): \quad n \equiv 4 \pmod 6$: we have $n = 6 k + 4$


 * $(6): \quad n \equiv 5 \pmod 6$: we have $n = 6 k + 5$