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.
That’s not quite right. I’ll start you off.
$$m^2 = (3k+1)^2 = (3k+1)(3k+1) = 9k^2+6k+1.$$
What do you notice about the first two terms?