Prove that if $n^2$ is odd then $n$ is odd? Here is my solution:
I assume $n^2$ is odd 
then I put $n^2$= $(2x-1)^2$, now I am taking root square for both sides:
$\sqrt{(n^2)}$ = $\sqrt{(2x-1)^2}$ $\Rightarrow$ $n = (2x-1)$ $\Rightarrow$ $n$ is odd since R.H.S has -1 of an even number.
Is my solution a good proof? 
 A: Not a correct proof because if $n^2$ is odd, then it doesn't necessarily take the form $(2k - 1)^2$. In fact, that's what you are required to prove. 
Your assumption should be $\exists$ $k \in \mathbb N$, such that $n^2 = 2k - 1$.
However, this isn't a very fruitful approach. The classical solution to this is to work by contraposition.
Suppose that $n$ is even, then we can write $n = 2k$. Then, $n^2 = 4k^2 = 2(2k^2)$, so it is even. This gives that if $n^2$ is odd, then $n$ is odd.
A: You only need to note if $n$ is even, $n^2$ is even !
A: You accidentally assume the conclusion... what you really know is that $n^2 = 2x-1$ for some integer $x$. Look at prime factorizations instead. How many factors of $2$ does $n^2$ have? Then how many could $n$ have?
A: To complete a proof differently from the one others have suggested.
Suppose $n^2$ is odd, then $n^2=2m-1$ and $(n+1)^2=2(m-n)$
Now $2$ is prime and $2\mid (n+1)^2$ so $2\mid n+1$ therefore $n+1=2r$ (for some integer $r$) whence $n=2r-1$ and $n$ is odd.
But the easier route is to show that the square of an even number is even, so an odd number can't be the square of an even number. So if an odd number is a square it must be the square of an odd number.
A: on contrary suppose that $n$ is even and $n=2k$
So $n^2=(2k)^2=4k^2$
$\implies 2|n^2$ 
which shows that $n^2$ is even, which contradicts the fact that $n^2$ is odd.
