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.
 A: Below is a conceptual proof of the irrationality of square-roots. It shows that this result follows immediately from unique fractionization -- the uniqueness of the denominator of any reduced fraction -- i.e. the least denominator divides every denominator. This in turn follows from the key fact that the set of all possible denominators of a fraction is closed under subtraction so comprises an ideal of $\,\mathbb Z,\,$ necessarily principal, since $\,\mathbb Z\,$ is a $\rm PID$. We can eliminate this highbrow language to obtain the following conceptual proof at high-school level.
Theorem $\ $ Let $\;\rm n\in\mathbb N.\;$ Then $\;\rm r = \sqrt{n}\;$ is $\rm\color{#90f}{integral}$ if rational.
Proof $\ $ Consider the set $\rm D$ of all  possible denominators $\,\rm d\,$ for $\,\rm r, \,$ i.e. $\,\rm D = \{ d\in\mathbb Z \,:\: dr \in \mathbb Z\}$. Note $\,\rm D\,$ is $\rm\color{#0a0}{closed\ under\ subtraction\!\!:}$ $\rm\, d,e \in D\:\! \Rightarrow\:\!dr,\,er\in\mathbb Z \:\!\Rightarrow\:\! (d-e)\,r = dr - er \in\mathbb Z.\,$
Further $\,\rm d\in D \,\Rightarrow\, \color{#c00}{dr\in D}\ $ by $\rm\ (dr)r = dn\in\mathbb Z, \,$ by $\,\rm r^2 = n\in\mathbb Z.\,$ Thus by the Lemma below,
with $\,\rm d =$  least positive natural in $\rm D,\,$ we infer that $\ \rm d\mid \color{#c00}{dr}, \ $  i.e. $\rm\ \color{#90f}{r = (dr)/d \in\Bbb Z}.\  $ QED
Lemma $\ $ Suppose $\,\rm D\subset\mathbb Z \,$ is $\rm\color{#0a0}{closed\  under\ subtraction}$ and  $\,\rm D\,$ contains a nonzero element.
Then $\rm D \:$ has a positive element and the least positive element of $\,\rm D\,$ divides every element of $\,\rm D$.
Proof $\rm\,\ \ 0 \ne d\in D \,\Rightarrow\, d-d = 0\in D\,\Rightarrow\, 0-d = -d\in D,\, $ so $\rm D$ contains a positive integer. Let $\,\rm d\,$ be the least positive integer in $\,\rm D.\,$ Since $\rm\: d\mid n \!\iff\! d\mid{-}n,\,$ if $\rm\ c\in D\,$ is $\rm\color{#90f}{not}$ divisible by $\,\rm d\,$ then we
may assume $\,\rm c\,$ is positive, and the least such. Then $\rm\, c-d > 0 \,$ is in $\rm D,\,$ $\rm\color{#90f}{not}$ divisible by $\,\rm d\,$
and smaller than $\,\rm c,\,$ contra leastness of $\,\rm c.\,$ So $\,\rm d\,$ divides every element of $\,\rm D.\ $ QED
The proof of the theorem exploits the fact that the denominator ideal $\,\rm D\,$ has the special property that it is $\rm\color{#c00}{closed}$ under multiplication by $\rm\,\color{#c00}r.\: $ The fundamental role that this property plays becomes clearer when one learns about Dedekind's notion of a conductor ideal. Employing such yields a trivial one-line proof of the generalization that a Dedekind domain is integrally closed since conductor ideals are invertible so cancellable. This viewpoint serves to generalize and unify all of the ad-hoc proofs of this class of results - esp. those proofs that proceed essentially by descent on denominators. This conductor-based structural viewpoint is not as well known as it should be - e.g. even some famous number theorists have overlooked this. See my post here for further details.
A: Write $q$ as $3r$ and see what happens.
A: Moron's answer certainly covers your question, but as someone who's not your instructor I'd like to see a few more details in your 'proof' of the first half - can you be more specific about how $q^2/3 \in \mathbb{Z} \Rightarrow 3|q$?  While that's easy, it's not necessarily trivial, and you've elided some details there...
A: Think about how many times each prime factor must appear on each side of the equation, if you were to break p and q into their prime factorizations. The left side has a 3 in it, how many must the right side have, at least?
A: Here we go. $3p^2=q^2$ implies that $3$ divides $q$, since $3$ is prime and if a prime divides a product, it divides one of the factors. But then, if $3$ divides $q$, then we also have that $3^2$ divides $q^2$. Hence, by factoring out the 9 on the rhs, we can cancle the 3 on the left hand side and still be left with a three. i.e $3\alpha=p^2$. But then, $3$ divides p, as required.
A: 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.
A: Does not exist $p,q$ integers such that  $3q^2=p^2$ because
$v_3{3q^2}=odd$ but $v_3{p^2}=even$
