When proving that there is not rational number $m/n$ equal to $\sqrt{2}$, why does $m$ and $n$ must be not both even?

Question

I've read this on Rudin's Principles of Mathematical Analysis:

1.1 Example We now show that the equation

$$p^2=2$$

is not satisfied by any rational $p$. If there were such a $p$, we could write $p=m/n$ where $m$ and $n$ are integers that are not both even. Let us assume this is done, then it implies

$$m^2=2n^2$$

My doubt is: Why not both even? I was thinking that it has some relation with $m^2=2n^2$ but this premise is given before the enunciation of $m^2=2n^2$. I guess there is some property for rational numbers that has some relation with this, but I'm unaware of. The question may be trivial but I can't figure it out. I guess I understand the rest of the proof quite well, but I'm stuck at this statement.

Answer 1

If they are both even, we can divide the equation by $4$, getting $m'^2=2n'^2$ with $m=2m'$ and $n=2n'$. We can keep doing this until at least one is not even. As each has a finite number of factors of $2$, we can only do it a finite number of times.

Answer 2

Suppose that $m$ and $n$ are even, so that $m=2p$ and $n=2q$. Then $m^2/n^2=p^2/q^2=2$, so the proof can be done supposing $p,q$ are not both even.

Answer 3

All rational numbers have this property... The most simple form of $p=m/n$ where $m$ and $n$ are not both even, so the fraction can no more be simplified... Of course it can be simplified in some cases where $m$ and $n$ are both not even but that has nothing to do with the proof. This property was used smartly to prove that there is not rational number $p$ such as $p^2=2$.