Prove that if $a$ is an integer, then $a(a^2+2)$ is divisible by $3$

Question

I've been working on this problem for a while.

I've been trying a contrapositive contradiction proof, as the other examples of direct proofs I've seen aren't very clear.

For contrapositive, I assume $3$ does not divide $a(a^2 + 2)$, and $a$ is an integer. Then $a(a^2 + 2) = 3k + 1$ or $a(a^2 + 2) = 3k + 2.$

I'm thinking through simplifying/rearranging $a(a^2 + 2) = 3k + 1$ into $(a^2 + 2a - 2) / 3 = k$ might be sufficient as a contradiction, because since $a$ is an integer, the top portion is also an integer (since integers are closed under addition/subtraction. So an integer divided by $3$ is sometimes not an integer. But I'm not sure if I'm on the right track or not.

Thank you in advance for any help!

Answer 1

If $3\mid a$, then $3\mid a(a^2+2)$. (Why?)

If $3\nmid a$, then $a\equiv \pm 1\pmod{3}$, so that $a^2\equiv 1\pmod{3}$. Now $$\begin{align} a^2+2&\equiv 1+2\pmod{3}\\ &\equiv 3\pmod{3}\\ &\equiv 0\pmod{3},\end{align}$$ which happens iff $3\mid (a^2+2)$. Hence $3\mid a(a^3+2)$.