Is my proof correct? Proof: $n^{2}$ is odd then $n$ is odd. 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?
 A: Very well done. Yes, indeed, proving $n\text{ is even}\implies n^2\text{ is even}$ is much easier than the original statement.
A: Alternatively, if $n^2$ is odd, then $n^2-1$ is even. It can be expressed as:
$$n^2-1=(n-1)(n+1).$$
If $n$ is even, $n-1$ and $n+1$ are both odd, whose product is also odd and it contradicts the given condition ($n^2-1$ is even). Hence, $n$ is odd.
A: Your idea is correct, but it might be better to write the proof this way:
We prove the contrapositive statement, that if $n$ is even then $n^2$ is even. Since $n$ is even, $n=2k$ for some integer $k$. Then $n^2=4k^2=2(2k^2)$ with $2k^2$ an integer, so $n^2$ is even.
Notice that in your proof, you did not mention that $m$ or $2m^2$ are integers, which are important details. The last two biconditional arrows also obscure the picture, especially when the nature of $m$ is not specified. (You only need one direction in the proof, and the other direction might not even hold in some cases.)
A: 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.
A: Your proof is correct as has been pointed by others and perhaps it is the most simple way to go about this, but just for fun, here is somewhat more direct proof (trying to avoid proof by contrapositive or contradiction): 
If $n^2$ is odd, then $n^2+2n+1$ is even (odd + odd is even). Notice that this is just $(n+1)^2$. This means that $2$ divides $(n+1)^2$ and since $2$ is a prime, it follows $2$ divides $(n+1)$ and so $n$ is odd.
Another way to go about this is noticing that if $n^2$ is odd, then it can be written as $n^2=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k}$ where $p_i$ are odd primes. Now $n$ is divisor of $n^2$ and so it must also consist only of odd primes, hence $n$ is odd.
