Suppose that $\tan{\alpha} = p/q$, where $p$ and $q$ are integers and $q \neq 0$. Prove that the number $\tan{\beta}$ for which $\tan{2\beta} = \tan{3\alpha}$ is rational only when $p^2+q^2$ is the square of an integer.
This is a pretty easy question to do if you are willing to do a lot of computations. With $\tan{\alpha} = p/q$, just simplify $\tan{2\beta} = \tan{3\alpha}$ using the sum of angles tangent formula. You will get a quadratic in $\tan{\beta}$ and you will have to show that $\sqrt{(p^2+q^2)^3}$ is rational iff $p^2+q^2$ is a perfect square.
There is an easier way to solve this question, though, and is given by the solution below. What I don't get is why $2\gamma = \alpha$. How did they arrive at that and did they use the fact that $\tan{2\beta} = \tan{3\alpha}$?
Solution
Easy. $2\beta = 3\alpha \Rightarrow 2(\beta-\alpha) = \alpha$. So rename $\gamma = \beta - \alpha$.