Proving $3\mid p^3 \implies 3\mid p$ I want to prove $3\mid p^3 \implies 3\mid p$ (Does it?)
The contrapositive would be $3 \nmid p \implies 3 \nmid p^3$ I believe.
$3\nmid p \implies p = 3q + r$ ($0<r<3$), so $p^3 = 27q^3+27q^2r+9qr^2+r^3$
Dividing by $3$ we get $3(9q^3 + 9q^2r + 3qr^2) + r^3$
Is this correct so far? How do I finish the proof please? Do I need to show that $r^3$ can never be a multiple of $3$?

$$p^3\bmod3=(p\bmod3)^3\bmod 3,$$ so that 
$$p\bmod3=0\to p^3\bmod3=0$$
$$p\bmod3=1\to p^3\bmod3=1$$
$$p\bmod3=2\to p^3\bmod3=2$$
