Proving that $\sqrt{2}+\sqrt{3}$ is irrational This is from Spivak.

Prove that $\sqrt{2}+\sqrt{3}$ is irrational.

So far, I have this:

If $\sqrt{2}+\sqrt{3}$ is rational, then it can be written as $\frac{p}{q}$ with integral $p, q$ and in lowest terms. 

$$\sqrt{2}+\sqrt{3}=\frac{p}{q}$$
$$2\sqrt{6}+5 =\frac{p^2}{q^2}$$
$$(2\sqrt{6}+5)q^2=p^2$$

And that's about where I get stuck. In a similar question (prove that $\sqrt{2}+\sqrt{6}$ is irrational,) I was able to show that both $p,q$ had to be even which is impossible. I obviously can't apply this trick here. Any hints?
$(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=3-2=1$ is rational, so if the first factor were rational, so would be the second, and their difference $2\sqrt2$. But surely you know that the latter isn't rational.
From your very last step:
 $$\sqrt{6} = \frac{p^2 - 5q^2}{2q^2}$$ This a rational number- a contradiction, hence the answer.