<s>[INST] Show that if $a$ is an integer, then 3 divides $a^3 - a $ Show that if $a$ is an integer, then 3 divides $a^3 - a $
we can write, where $k$ is an integer;
$a^3 - a = 3k $
$a(a^2 - 1) = 3k $
Now if $a = k$ then
$a^2 -1 = 3$ and $a= \pm2 $ so $ a^3 - a = 24 = 3 \times 8$
If $ a $ is not equal to $k$;
then
$a(a^2 - 1) = a(a+1)(a-1) = 3k$
since  $a(a+1)(a-1)$ is the product of 3 consecutive integers, the expression is divisible by 3.
Is this ok?, just a check. I'm not really all that good at number theory.
 [/INST] Dividing $a$ by $3$ allows to write $a=3q+r$. Now
$$
(3q+r)^3=27q^3+9q^2r+3qr^3+r^3=3s+r^3
$$
collecting together all multiples of $3$ in the expansion. Dividing again the result by $3$ we have
$$
(3q+r)^3=3Q+R
$$
where $R$ is the remainder of the division by 3 of $r^3$ (because $3s$ is a multiple of $3$).
But there are only 3 possible remainders $r$, namely:


*

*$r=0$, thus $r^3=0$ and $R=0$,

*$r=1$, thus $r^3=1$ and $R=1$,

*$r=2$, thus $r^3=8$ and $R=2$.
This shows that $r=R$ always, so that $a^3-a$ divided by 3 has remainder $0$ (i.e. is divisible by $3$)
</s>