Divisibility of $n$ and $n^2$? We're trying to prove that if $n$ is a positive odd integer then $n^2$ must also be an odd integer using an indirect evidence.
My textbook states that:
If we imagine that $n^2$ is divisible by two then $n$ is divisible by two, thus we can't arrive at this conclusion with an odd number as a starting point.
I don't get this.
If $n^2$ is an even number it means it's divisible by two. By dividing $n^2$ by two we get the positive integer $x$.
$$\frac{n^2}{2}=x$$
$$n^2=2x$$
$$n=\sqrt{2x}$$
$$n=\sqrt{2}\sqrt{x}$$
$$\frac{n}{2}=\frac{\sqrt{2}\sqrt{x}}{\sqrt{2}\sqrt{2}}$$
$$\frac{n}{2}=\frac{\sqrt{x}}{\sqrt{2}}=\sqrt{\frac{x}{2}}$$
Exactly how does it follow that $n$ is divisible by two?
If I am missing something elementary then please point me to a tutorial, thanks.  
 A: If a prime number divides a product, then it divides one of the factors. It follows that, if $2$ divides $n^2=n \cdot n$, then $2$ divides $n$.
A: An arbitrary odd integer is given by $2k + 1, k \in \mathbb{Z}$. Therefore an odd integer squared is $(2k + 1)^2 = 4k^2 + 4k + 1$. For all $k, 2*k$ is an even integer so therefore is follows that so is $4*k$. Adding one to an even integer makes it odd by definition. Therefore $4k^2 + 4k + 1$ must give an odd integer
A: $2$ is a prime number and when a prime number $p|ab$ then $p|a$ or $p|b$ 
From this we have that $2|n^2=nn \Rightarrow 2|n$

You want to prove that if $a^2$ is even the $a$ is even.

So assume that $a$ is not even thus $a$ must be odd then $a=2k+1$ for some integet $k$ thus $$(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1=2s+1$$ for $s=2k^2+2k \in Z$ proving that $a^2$ is odd thus $a^2$ is not even which is a contradiction.
A: Facts:  
$$\color{Green}{n^2=n+n(n-1)}$$


*

*product of any two consecutive integers is always even. 

*sum of two even integers is always even.

*sum of odd integer and even integer is odd.
