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 $3$?

Since $0\lt r\lt3$ so either $r=1$ or $r=2$.
If $r=1$ then $r^3=1$ . So $3(9q^3+9q^2r+3qr^2)+r^3=3(9q^3+9q^2r+3qr^2)+1$ which leaves remainder $1$ when divided by $3$
And similarly, if $r=2$ then $r^3=8$ . So 
$3(9q^3+9q^2r+3qr^2)+r^3=3(9q^3+9q^2r+3qr^2)+8=3(9q^3+9q^2r+3qr^2)+3\cdot2+2=3(9q^3+9q^2r+3qr^2+2)+2$
which leaves remainder $2$ when divided by $3$.
It follows that both of the two expressions are not divisible by $3$ and you reach to a contradiction.
