Prove $m=3k+1 \quad m,k \in \mathbb Z \implies m^2=3l+1 \quad m,l \in \mathbb Z$ 
Suppose we call an integer "throdd" $\iff$ $m=3k+1$ for some integer
  $k$. Prove that the square of any throdd integer is throdd.

So here is what I have so far:
$$(3k+1)^2 = 3k+1$$
$$(3k+1)(3k+1) = 3k+1$$
Am I going in the right direction?
2nd part: $(3k+1)(3k+1)= 9k^2 + 6k + 1$
$3k(3k+ 2) +1$
So im confused because $(3k +2)$ is not m and $3k(3k+2) +1$ isnt in the form of $3m+1$?

HINT. You need to show that $(3k+1)^2$ can be written as $3 m +1$ for some $m$. You are real close.
