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.

Try to write out the factorization of the right and left handed sides.
Now compare the order of the 3 on the left and right side, one of them is equal, forcing the other side to become odd. Contradiction.
