Can the following proof concerning irrationality of square roots be improved?

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.