If $n$ is odd then $n^2$ is odd

Question

Suppose $n \in\mathbb Z$. If $n$ is odd then $n^2$ is odd.

My proof by contradiction is as follows but I'm pretty sure its wrong any help would be welcome.

If $n$ is odd then $n^2$ is even. If $n$ is odd $n = 2k + 1$ for some integer $k$. $$n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 +2k) + 1.$$ Let $m = 2k^2 + 2k$. Then $n^2 = 2m$ so by definition $n^2$ is even but this is a contradiction because $n^2$ is odd.

Answer 1

You actually don't need a proof by contradiction, and you've almost proved the result in your original post. For $n$ odd, i.e. $n = 2k +1$, we have $n^2 = 2(2k^2 + 2k) + 1 = 2m + 1$ is odd, $m=2k^2 + 2k$.

Answer 2

if $n$ is odd then $n(n-1)$ is even and so $n(n-1)+n$ is odd.

Answer 3

There is no need for proof by contradiction here; but if you wish:

Since $n$ is odd, then $n = 2k + 1$ for some integer $k$. Let's assume for the sake of contradiction that $n^{2}$ is even. Then, $n^{2} = (2k + 1)^{2} = 4k^{2} + 4k + 1 = 4(k^{2} + 1) + 1,$ which is an odd number, contrary to our assumption that $n^{2}$ is even. Therefore, $n^{2}$ must be odd.

This is pretty much what you had up to the second to last line of your proof where you obtained your contradiction, and so, $n^{2}$ must be odd. (You don't want your last line.) Do you see why proving this directly would have been more expedient?