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?

Yes, we can give a simpler and more conceptual inductive proof. Notice that 
$\qquad\qquad  a(a^2+2)\, =\, a(a^2\!-\!1 + 3)\, =\, \color{#0a0}{(a-1)a(a+1)} + 3a$
so it suffices to show that one of any $\rm\color{#0a0}{3\ consecutive\ integers}$ is divisible by $3$
This has a simple inductive proof. Note that shifting such a sequence  by one simply replaces the old least element $\,\:\color{#C00}n\,$ by the new greatest element $\,\color{#C00}{n+3}$
$$ \begin{array}{}  \:\color{#C00}n &  n+1 &  n+2   \\
                    \to  &  n+1 &  n+2 &  \color{#C00}{n+3} \end{array}$$
Since $ \: \color{#C00}n\equiv \color{#C00}{n+3} \pmod{\! 3},\,$ the shift does not change the set of remainders $\bmod 3$  of the elements. Thus the remainders remain the same as in the base case $ \ 0,1,2\: =\: $ all possible remainders mod $ \,3.\,$ Therefore the sequence has an element with remainder $\,0,\,$ i.e. an element divisible by $ \,3.$ 
Remark $ $ The same method works to show that a sequence of $d$ consecutive integers contains a multiple of $d.\,$ Alternatively this can be proved by using division with remainder (which has a natural proof by induction), which is closely connected to the above method.
