Square Modulo 24 of Odd Integer Not Divisible by 3/Proof 2

Proof
Let $a$ be as asserted.

We have that:
 * $3 \nmid a$

which means that either of the following hold:

Suppose $a = 3 k_1 + 1$.

As $a$ is odd, it must be the case that $k_1$ is even.

Hence:

If $k$ is even, then:
 * $\exists r \in \Z: k = 2 r$

Hence:
 * $a^2 = 24 r \paren {3 k + 1} + 1$

and so:
 * $24 \divides a^2 - 1$

If $k$ is odd, then $3 k + 1$ is even, and:
 * $\exists r \in \Z: 3 k + 1 = 2 r$

Hence:
 * $a^2 = 24 r k + 1$

and so:
 * $24 \divides a^2 - 1$

Suppose $a = 3 k_2 + 2$.

As $a$ is odd, it must be the case that $k_2$ is also odd.

Hence:

If $k$ is even, then:
 * $\exists r \in \Z: k = 2 r$

Hence:

and so:
 * $24 \divides a^2 - 1$

If $k$ is odd, then $\paren {3 k + 5}$ is even, and:
 * $\exists r \in \Z: 3 k + 5 = 2 r$

Hence:

and so again:
 * $24 \divides a^2 - 1$

All cases have been accounted for.

Hence the result.