360 divides a^2 (a^2 - 1) (a^2 - 4)

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

Then:
 * $360 \divides a^2 \paren {a^2 - 1} \paren {a^2 - 4}$

where $\divides$ denotes divisibility.

Proof
By Difference of Two Squares:


 * $a^2 \paren {a^2 - 1} \paren {a^2 - 4} = a \paren {a - 2} \paren {a - 1} a \paren {a + 1} \paren {a + 2}$

We have that $a - 2, a - 1, a, a + 1, a + 2$ are $5$ consecutive integers.

Hence from Product of 5 Consecutive Integers is Divisible by 120:
 * $120 \divides a \paren {a^2 - 1} \paren {a^2 - 4}$

and so:
 * $120 \divides a^2 \paren {a^2 - 1} \paren {a^2 - 4}$