Show $\sqrt 3$ is irrational using $3p^2=q^2$ implies $3|p$ and $3|q$ This is a problem from "Introduction to Mathematics - Algebra and Number Systems" (specifically, exercise set 2 #9), which is one of my math texts. Please note that this isn't homework, but I would still appreciate hints rather than a complete answer.
The problem reads as follows:

If 3p2 = q2, where $p,q \in \mathbb{Z}$, show that 3 is a common divisor of p and q.

I am able to show that 3 divides q, simply by rearranging for p2 and showing that
$$p^2 \in \mathbb{Z} \Rightarrow q^2/3 \in \mathbb{Z} \Rightarrow 3|q$$
However, I'm not sure how to show that 3 divides p.

Edit:
Moron left a comment below in which I was prompted to apply the solution to this question as a proof of $\sqrt{3}$'s irrationality. Here's what I came up with...
[incorrect solution...]
...is this correct?
Edit:
The correct solution is provided in the comments below by Bill Dubuque.

Does not exist $p,q$ integers such that  $3q^2=p^2$ because
$v_3{3q^2}=odd$ but $v_3{p^2}=even$
