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?
 A: 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.
A: 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)
A: I'd suggest writing the proof in a more linear fashion which is easier to read. Below is how I would approach the induction step.

Let $P(n)$ be the statement $3\mid n(n^2+2)$.
Proving $P(k)\implies P(k+1)$:
$$P(k)\implies 3\mid k(k^2+2)$$
$\implies$
$$3\mid k(k^2+2)+3(k^2+k+1)$$
$\implies$
$$3\mid k^3+3k^2+5k+3$$
$\implies$
$$3\mid(k+1)(k^2+2k+3)$$
$\implies$
$$3\mid(k+1)((k+1)^2+2)$$
$\implies$
$$P(k+1)$$
A: If a is divisible by 3, then 3 divides a, and therefore a(a^2 + 2). 
Otherwise, $a = 3k ± 1$, and 3 divides $(a^2 + 2)$ = $((3k ± 1)^2 + 2)$ = $((9k^2 ± 6k + 1) + 2)$ = $(9k^2 ± 6k + 3)$. 
A: Any number is congruent to $0$, $1$ or $2$ modulo $3$, and upon squaring you get any square is congruent to $0$, $1$ or $1$ modulo $3$. Now take an arbitrary number $a$. If $3$ divides $a$ then you're done, if not $a^2+2$ is, by the above, congruent to $0$ modulo $3$ since $1+2=3$, and so you're good, too. 
A: Work backwards.  It's far more natural.
$(k+1)((k+1)^2+2)=$
$(k+1)(k^2+2k+1+2)=$
$(k+1)(k^2+2)+(k+1)(2k+1)=$
$k (k^2+2)+(k^2+2)+(k+1)(2k+1)=$
$3m+(k^2+2)+(2k^2+3k+1)=$
$3m+3k^2+3k+3$.
It is sometimes easier to subtract results.
$(k+1)((k+1)^2+2)-k (k^2+2)=$
$k (k+1)^2+(k+1)^2+2k+2-k^3-2k=$
$k^3+2k^2+k +k^2+2k+1+2-k^3=$
$3k^2+3k+3$
is a multiple of 3.
But it's probably easiest not to use induction at all.
Let $a \equiv i \mod 3;i=0,\pm 1$
If $i=0$ then $3|a $ so $3|a (a^2+2) $
Otherwise  $i=\pm 1$
So $a^2+2\equiv i^2+2\equiv 1+2\equiv 0\mod 3$
So $3|a^2+2$ so $3|a (a^2+2) $.
