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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:

$$(3k+1)^2 = 3k+1$$

$$(3k+1)(3k+1) = 3k+1$$

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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 $3m+1$?

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**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?

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