24 divides a(a^2 - 1) when a is Odd

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

Then:
 * $24 \divides a \paren {a^2 - 1}$

where $\divides$ denotes divisibility.

Proof
First suppose that $a$ is not divisible by $3$.

Then from Square Modulo 24 of Odd Integer Not Divisible by 3:
 * $24 \divides \paren {a^2 - 1}$

from which the result follows immediately.

Now suppose that $3 \divides a$.

Then immediately:
 * $3 \divides a \paren {a^2 - 1}$

From Odd Square Modulo 8:
 * $8 \divides \paren {a^2 - 1}$

and so:
 * $8 \divides a \paren {a^2 - 1}$

We have from Coprime Integers: $3$ and $8$ that:
 * $3 \perp 8$

where $\perp$ denotes coprimality.

As we have that:
 * $8 \divides a \paren {a^2 - 1}$

and:
 * $3 \divides a \paren {a^2 - 1}$

it follows from Product of Coprime Factors that:
 * $24 \divides a \paren {a^2 - 1}$