Prove that if $n^2$ is divided by 3, then also $n$ can also be divided by 3. $n\in \Bbb N$
Prove that if $n^2$ is divided by 3, then also n can also be divided by 3.

I started solving this by induction, but I'm not sure that I'm going in the right direction, any suggestions?
Hint: $p|ab \implies p|a$ or $p|b$ if $p$ is a prime
nnn + 2*n = n*(nn+2) = n{(n*n -1) + 3} = n*3 + {n*(n-1)*(n+1)} ==> P + Q

Here 
  P = n*3 ==> divisible by 3
  Q = (n-1)n(n+1) ==> this is multiplication of three consecutive number and hence it is divisible by 3
==> P + Q is divisible by 3