How to prove if $3\nmid n$, then $n^2$ has a remainder $1$ when divided by $3$? Lets say: 
$n^2=3k+1$ ; $n\neq3$.
I'm trying to prove this by induction, therefore:
$(n+1)^2=n^2+2n+1=3k+2n+2$ 
Any suggestion on how to move foward? 

If 3 does not divide $n$, that means that $n$ has remainder 1 or 2 by division with 3.
So either $n=3k+1$ or $n=3k+2$.
In both cases we have $n^2=(3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\mod 3$
or $n^2=(3k+2)^2=3(3k^2+4k)+4=3(3k^2+4k)+3\cdot 1+1\equiv 1\mod 3$
