Direct Proof of Parity of Roots So in my CS class regarding proofs, the professor wrote something interesting. His idea was:
If $n^2$ is odd then $n$ is odd.
However, his reason for this being true is that $\sqrt{2n+1}$ is also odd, but provided that as an axiom and provided no proof.
I understand how to proof this using the contrapositive, however is there a direct proof of (if $n^2$ is odd then $n$ is odd)?
 A: Since $n^2-n=n(n-1)$ is always even (one of $n$ or $n-1$ must be even) and $n^2$ is odd, $n$ is odd.
($n^2-(n^2-n)=n$, and odd-even=odd)
A: The reason that I give my students that the direct proof that if $n^2$ is odd is then $n$ is odd is much more challenging than the proof of the contrapositive is the following:
The square root function is hard to get a handle on.  The most direct proof would be to say that since $n^2$ is odd, $n^2=2k+1$ for some integer $k$ and then look at $\sqrt{2k+1}$.  The problem with this method is that it ends up trying to prove:
If $a$ is odd, then $\sqrt{a}$ is odd, which is not true (because $\sqrt{a}$ might be an integer).  So, the main challenge is that it's hard to work the assumption that $n^2$ is a square of an integer into the assumptions.
There are direct proofs, see @Akababa's answer +1, but they don't use the "expected" direct approach ("expected" in terms of basic proof courses).
A: If $n^2$  is odd then:
$n^2 - 1 = (n - 1)(n + 1)$ is even.
RHS : (n +1) = (n - 1) + 2
1) If (n - 1) is odd then (n + 1) is odd
2) If (n -1) is even then (n + 1) is even.
So in our equation 
$n^2 -  1 = (n - 1)(n + 1)$ both factors on the RHS are 
1) both odd, or 
2) both even.
Since the product $ n^2 - 1$ is even, only  option 2) remains.
If $n-1$, $n+1$ are even then $n$ is odd.
A: Write $n=2m+r$ with $r=0$ or $r=1$.
Then $n^2=4m^2+4m+r^2$ and $n^2$ is odd iff $r=1$.
But then $n$ is odd.
