Prove $m=3k+1 \quad m,k \in \mathbb Z \implies m^2=3l+1 \quad m,l \in \mathbb Z$

Question

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 $3m+1$?

You should not have $3k+1$ on the right side. Maybe you'd want $3m+1$ there, but I'd just write, as you have, $$(3k+1)^2 = \cdots = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$$ which is "throdd" — The Chaz 2.0, Oct 27 '13 at 21:31

score 3 · Answer 1 · answered Oct 27 '13 at 20:40

3

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

answered Oct 27 '13 at 20:40

Sasha

score 2 · Answer 2 · answered Oct 27 '13 at 20:40

2

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?

answered Oct 27 '13 at 20:40

Cameron Williams

2 Answers2