Prove that the proof for $\sqrt{2}$ being irrational doesn't work for $\sqrt{4}$

Question

So I am supposed to show that the standard way of proving that $\sqrt{2}$ is irrational doesn't hold for $\sqrt{4}$. So making the assumption that $p$ and $q$ are natural numbers and any common factors have been cancelled we have: $\frac{p}{q}=\sqrt{4}\Rightarrow p^2=4q^2$, now this supposedly breaks down because if $p^2$ is divisible by 4 then it is divisible by an integer of the form $(4k+2)$ or $4k$. Doesn't this just leave us with two cases though and both of them lead to $p$ and $q$ having common factors?

Answer 1

We get, as in the proof for $\sqrt{2}$, that $p$ is even, say $p=2p_1$. Substitute and cancel a $4$. We get $p_1^2=q^2$, and there is nowhere to go except to conclude (if $q\gt 0$) that $q=1$.