Show that for any $a\in\mathbb{Z}$, $3 \mid(a^3 - a)$

Question

Show that for any $a\in\mathbb{Z}$, $3 \mid (a^3 - a)$

my solution is : if $a$ is a multiple of $3$ then $a^3-a$ is a multiple of $3$; if $\gcd(a,3)=1$ then by FLT $a^2\equiv 1 \pmod3$, hence $a^3-a = a(a^2-1)$ is a multiple of $3$.

Am I right?

Answer 1

Hint: you can use $a^3-a=a(a+1)(a-1)$ one between 3 consecutive numbers is divisible by 3

Answer 2

Since $3$ is prime, by Fermat's little theorem we have $a^3 \equiv a \pmod 3$. Thus $a^3 - a \equiv 0 \pmod 3$. We conclude that $3 \mid (a^3 - a)$.

Answer 3

Verify that $0^3 \equiv 0 \pmod{3}$, $1^3 \equiv 1 \pmod{3}$, and $2^3 \equiv 2 \pmod{3}$, and that's all there is to it. Fermat's little theorem is overkill for this.