Proof that 6 divides $a \in \mathbb{Z}, a(a^2 - 7)$ I am trying to prove a question from my tutorial sheet, is this an acceptable proof?
Six cases exist:
$$a,k \in \mathbb{Z}, a(a^2 - 7) = 6k
\\\text{Proof:}\\
a = 0 \mod 6 \longrightarrow a^2 = 0 \mod 6 \longrightarrow a^2 - 1\mod 6 = 5 \longrightarrow a(a^2 - 7) = 0\mod 6\\
a = 1 \mod 6 \longrightarrow a^2 = 1 \mod 6 \longrightarrow a^2 - 1\mod 6 = 0 \longrightarrow a(a^2 - 7) = 0\mod 6\\
a = 2 \mod 6 \longrightarrow a^2 = 4 \mod 6 \longrightarrow a^2 - 1\mod 6 = 3 \longrightarrow a(a^2 - 7) = 0\mod 6\\
a = 3 \mod 6 \longrightarrow a^2 = 3 \mod 6 \longrightarrow a^2 - 1\mod 6 = 2 \longrightarrow a(a^2 - 7) = 0\mod 6\\
a = 4 \mod 6 \longrightarrow a^2 = 4 \mod 6 \longrightarrow a^2 - 1\mod 6 = 3 \longrightarrow a(a^2 - 7) = 0\mod 6\\
a = 5 \mod 6 \longrightarrow a^2 = 1 \mod 6 \longrightarrow a^2 - 1\mod 6 = 0 \longrightarrow a(a^2 - 7) = 0\mod 6\\
$$
Therefore, since for all $a$, $a(a^2 - 7) = 0 \mod 6$, $6$ divides $a(a^2 - 7)$.
Is there an easier proof, or a better way to lay this out via $LaTeX$?

HINT:
$$a(a^2-7)= a(a^2-1)-6a=(a-1)a(a+1)-6a$$
Now the first term being the product of $3$ consecutive integers is divisible by $3!$
