Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$ If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.
By induction:
Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$
If $a=1$ then, $1\in S$
So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$
If $k(k^2+2)=3m$ then,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&3m+3(k^2+k+1)\\=&3(m+k^2+k+1)\end{align*}$
Also,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&k^3+2k+3k^2+3k+3\\=&k^3+2k^2+k^2+2k+3k+3\\=&k^2(k+2)+k(k+2)+3(k+1)\\=&(k+2)(k^2+k)+3(k+1)\\ =&k(k+2)(k+1)+3(k+1)\\=&(k+1)(k(k+2)+3)\\ =&(k+1)(k^2+2k+1+2)\\ =&(k+1)((k+1)^2+2)\\ =&3(m+k^2+k+1) \end{align*}$
where $n=m+k^2+k+1$
Therefore,
$3\mid(k+1)((k+1)^2+2)$
Can I do it simplier using induction?

Here is an alternative proof that doesn't use induction - just for fun! 
Let $a \in \mathbb{Z}$. Observe that $a^3 - a = a(a^2 - 1) = (a-1)a(a+1)$ is the product of three consecutive integers, hence divisible by three. Adding $3a$ to the initial expression does not alter divisibility by $3$.
In particular, $a^3 - a + 3a = a^3 + 2a = a(a^2 + 2)$ is divisible by $3$, which means that it is still an integer after dividing by $3$. QED
(A ridiculous version of the above: $a(a^2+2)$ is the sum of $a^3 - a$ and $3a$, where the latter addend is a multiple of three by observation, and the former is divisible by three by Fermat's Little Theorem; so, their sum is divisible by three, and its quotient by three is an integer. QED)
