TITLE: Is my proof correct? Proof: $n^{2}$ is odd then $n$ is odd.
QUESTION [9 upvotes]: I don't think it is but I don't understand why.
Conjecture: 
$n^2 \text{ is odd} \Rightarrow n \text{ is odd}$ (*)
Proof:
In general,
$$ (A \Rightarrow B) \Leftrightarrow (\neg(B) \Rightarrow \neg(A)) $$
Therefore $(n^2 \text{ is  odd } \Rightarrow n \text{ is odd } \Leftrightarrow (\neg(n \text{ is odd}) \Rightarrow \neg(n^2 \text{ is  odd})) $
This is the same as $n \text{ is  even} \Rightarrow n^2 \text{ is even}$  (✝)
Isn't it?
If it is, then
$$\begin{align}
 n = 2m &\\
 &\Leftrightarrow n^2 = 4m^2 \\
 &\Leftrightarrow n^2 = 2(2m^2)\\
\end{align}$$
Therefore $n^2$ is even, by definition of even.
And, since we'e proved (✝), surely we have then proved (*). 
I'm a novice to proofs and this is one of my first ones, so I can't notice any obvious problems with it, yet it doesn't seem to me to prove anything - it seems to me to be proving something else entirely. Is it correct?

REPLY [2 votes]: Alt. hint: $\,n^2+n=n(n+1)$ is always even, since one of two consecutive integers must be even. Therefore $n^2$ and $n$ always have the same parity and, in particular, $n^2$ is odd iff $n$ is odd.