number theory - Show that $3p^2=q^2$ 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.

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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.

Answer

Write $q$ as $3r$ and see what happens.