elementary number theory - Prove that $sqrt 5$ is irrational

I have to prove that $\sqrt 5$ is irrational.

Proceeding as in the proof of $\sqrt 2$, let us assume that $\sqrt 5$ is rational. This means for some distinct integers $p$ and $q$ having no common factor other than 1,

$$\frac{p}{q} = \sqrt5$$

$$\Rightarrow \frac{p^2}{q^2} = 5$$

$$\Rightarrow p^2 = 5 q^2$$

This means that 5 divides $p^2$. This means that 5 divides $p$ (because every factor must appear twice for the square to exist). So we have, $p = 5 r$ for some integer $r$. Extending the argument to $q$, we discover that they have a common factor of 5, which is a contradiction.

Is this proof correct?

Answer

It is, but I think you need to be a little bit more careful when explaining why $5$ divides $p^2$ implies $5$ divides $p$. If $4$ divides $p^2$ does $4$ necessarily divide $p$?