Proof question: If $4 \not \mid a^2$, then $a$ is odd.

Question

Would somebody mind checking this proof? I'm not entirely confident in it, but I can't think of another way to approach it. Specifically, I'm not sure if the portion I bolded actually proves the statement true.

Proof:

Let $a$ be an integer. We will prove the contrapositive statement , "if a is even, then $4 \mid a^2$." By definition of even, $a=2k$. Substituting, we get $4 \mid (2k)^2$. We compute:

$4 \mid (2k)^2$

$4 \mid 4k^2$

$4 \mid 2(2k^2)$.

Now, let $j=2k^2$. Therefore, $4 \mid 2j$. Then, by definition of division, $2=4a$. Hence, $4 \mid a^2$, as desired. QED

Thanks.

Answer 1

Your proof is okay up until the last few lines.

So you have shown so far that $4|4k^2$ -- this is sufficient to prove your result, you don't have to take it any further!

How do we know this holds and gives the desired result? If $a|b$, then $\exists n \in \Bbb Z$ such that $b = an$. Here, $b = 4k^2, a = 4$. There thus exists an integer $n$ such that $4k^2 = 4n$. That very integer is $k^2$. Since $a^2=4k^2$ for some $k$, then $4|a^2$, concluding the proof.