Prove $a^2$ is even $\implies$ $a$ is even. It was proven via contradiction by my friend.
Here is the proof in question, which uses proof by contradiction
Assume towards a contradiction that $a$ is odd. Let $a = 2k+1$, then $a^2 = 4k^2 + 4k + 1 = 2(2k^2+2k)+1$ Therefore $a^2$ is odd. Contradiction.
To me, this proof in question seems incorrect because it shows that if $a$ is odd then $a^2$ is odd. We have shown that the converse of what we want to contradict is a contradiction. I feel like this proof by contradiction is not mathematically correct or vigorous enough, and for it to be correct we would have to start from $a^2$ being even, working our way down to $a$ being odd, and then finding a contradiction.
Am I correct? Or am I just confused about something?