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: you have it, but all you need is your observation that:
$$
a^3-a=(a-1)a(a+1)
$$
since exactly one of any three consecutive integers is divisible by 3
addendum
PROOF (for Git Gud)
let $a$ be the smallest number for which $3$ does not divide $a(a+1)(a+2)$
clearly $a \gt 0$ since $3 | 0$
write $k=(a-1)a(a+1)$, so $k$ is divisible by $3$
since $3$ doesn't divide $a$ or $a+1$ we have $3|(a-1)$
but now $3$ also divides $(a-1)+3=a+2$, a contradiction.
Of course there are various properties of the integers which need to be demonstrated to fill out the details here, but i will leave these as an exercise.
A: 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 $p$. 
A: 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 $3$)
A: Actually it is a case of Fermat's Little Theorem: If $p$ is a prime number, for every $a \not\equiv 0 \mod p$, we have $a^{p-1}\equiv 1 \mod p$. As a consequence, multiplying both sides by $a$, we have  $a^p\equiv a\mod p$ if $a \not\equiv 0 \mod p$. It is also true if $a \equiv 0 \mod p$ for trivial reasons.
A general proof of Little Fermat:
Consider the ring of integers modulo $p$, $\mathbf Z/p\mathbf Z$. Since $p$ is prime, it is actually a field, and the non-zero elements of this field are the group of units $\bigl(\mathbf Z/p\mathbf Z\bigr)^\times$ of the field, of order $p-1$. The subgroup $\langle \bar a\rangle$ generated by one of its elements $\bar a$ has order a divisor of the order $p-1$ of  $\bigl(\mathbf Z/p\mathbf Z\bigr)^\times$ (this is Lagrange's theorem). Hence $\bar a^{p-1}=\bar 1$ in  $\mathbf Z/p\mathbf Z$,which means $a^{p-1}\equiv 1\mod p$ in $\mathbf Z$.
A: $a^3-a=(a-1)a(a+1)$ 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 $0$.
