<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] In fact, divisible by $6$. The product of three consecutive integers is divisible by $3!$. Might as well take them naturals. Note that the quotient  is
$$\frac{(a+1)a(a-1)}{1\cdot 2\cdot 3} = \binom{a+1}{3}$$
Similarly you can show that the product of $n$ consecutive integers is divisible by $n!$. 
Another important generalization is the fact discovered by Fermat
$p$ divides $a^p-a$ for any prime $p$. 
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