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.

$a^3-a=(a-1)a(a+1)$ is always true and therefore is always the product of three consecutive integers.
Another way is to see that $a$ can be congruent to $0$, $1$ or $2$ $mod 3$. In each case we get that $a^3-a$ is congruent to $0$.
