<s>[INST] 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?
 [/INST] If a is divisible by 3, then 3 divides a, and therefore a(a^2 + 2). 
Otherwise, $a = 3k ± 1$, and 3 divides $(a^2 + 2)$ = $((3k ± 1)^2 + 2)$ = $((9k^2 ± 6k + 1) + 2)$ = $(9k^2 ± 6k + 3)$. 
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