Can the following proof concerning irrationality of square roots be improved? Is the following proof too lengthy? Can it be made shorter, yet still be elaborate?

Prove that there is no rational number whose square is $\sqrt{12}$.

Suppose $p, q \in \Bbb Q$. Of course, both elements cannot both be even if they are expressible in a rational field. Suppose there was a rational number such that $$({p \over q})^2 = 12$$
then $p^2 = 12q^2$. This implies $p^2$ (and therefore $p$) is even since an odd number multiplied by itself gives an odd number. Since any number multiplied by an even number is even, this implies the RHS is even.
Thus, the RHS must be divisible by 4, so the result is $p^2 = 3q^2$. This implies that there is no rational whose square is $\sqrt{3}$, as the RHS isn't divisible by 4 when q is odd. Since $4\sqrt{3}$ yields an irrational number, there is no rational whose square is 12. QED.
 A: It's not quite correct as stated, since you move from $p^2 = 12 q^2$ to $p^2 = 3 q^2$. What you mean is to say is that $p = 2k$ for some $k$, so
$$(2k)^2 = 12 q^2 \implies k^2 = 3 q^2$$
Now assuming that you know $\sqrt{3}$ is irrational, this is pretty much the quickest proof, except for the proof that notes $\sqrt{12} = 2 \sqrt{3}$ as MJD pointed out. 
If you can't assume this, then you can essentially just repeat the proof: since $3 | k^2$, $3$ is a divisor $k$, so $9 | k^2 = 3 q^2$. Thus, $3 | q^2$ and $3 | q$; but now $q$ and $k$ have a common factor of $3$. Looking back at $p$, we see that $3$ is a divisor of $p$, contradicting the fact that $p$ and $q$ can be selected to have no common factors.
It's also fairly circuitous how you get that $p^2$ is even by invoking that "an odd times an odd is odd." The far more direct way is to note that $12 | p^2$, and $12$ is even. It's also unnecessary to note that the right hand side is even (which you seem to state as an implication of the left side being even), since it's divisible by $12$. So that paragraph should be cleaned up a bit.
A: You are probably not supposed to assume that $\sqrt{3}$ is irrational.  But let us first suppose that we can assume it. 
Then you can proceed as follows. Suppose to the contrary that $\sqrt{12}=\frac{a}{b}$ where $a$ and $b$ are integers. 
Then since $\sqrt{12}=2\sqrt{3}$, we have $\sqrt{3}=\frac{a}{2b}$. This contradicts the fact that $\sqrt{3}$ is not rational.

Now we give a start to a proof that does not use the irrationality of $\sqrt{3}$. Suppose that $\sqrt{12}$ is rational. Then there exist integers $a$ and $b$ such that $b\ne 0$ and $\left(\frac{a}{b}\right)^2=12$. Without loss of generality we may assume that $\gcd(a,b)=1$.
We then obtain the equation $a^2=12b^2$. Now for your contradiction, do not work with $2$, work with $3$. Note that $3$ divides $12b^2$, so $3$ divides $a^2$, and therefore $3$ divides $a$. Continue. 
