A Problem for IMO, which does not deserve to be considered / Проблем за ИОМ, који не заслужује да буде разматран

                    In the CRUX mathematics magazine, [2000 : 24] the review of a mathematics book, related to the International Olympiads, appeared.

In addition to the given problems, there is a list of problems that are discussed in the competition jury, but which no longer see the light of day.

          Today we will deal with Problem #2 from the list in the previous image.

"Suppose $\tan x =\frac{p}{q}$, where $p$ and $q$ are positive integers and $q \neq 0.$

Prove that the number $\tan y$ for wich $\tan 2y=\tan 3x$ is rational only

      when $p^2+q^2$ is a perfect square."                                                                                                                        

          The solution is too simple for the problem to be at the level of the problems given at the IOM (even from the early 60's).

Solution CiP

                    Let $\tan x =\frac{p}{q}.$ With the triple angle formula $\tan 3x=\frac{3\tan x-\tan^3 x}{1-3\tan^2x}$ we obtain $\tan 3x=\frac{p(3q^2-p^2)}{q(q^2-3p^2)}.$

     We also apply the double angle formula $\tan 2y=\frac{2\tan y}{1-\tan^2y}.$

          The  equation $\tan 2y=\tan 3x$ is written

$$\frac{2\tan y}{1-\tan^2y}=\frac{p(3q^2-p^2)}{q(q^2-3p^2)}.$$

Or equivalent

$$p(3q^2-p^2)\cdot \tan^2y+2q(q^2-3p^2) \cdot \tan y-p(3q^2-p^2)=0.$$

In order for the previous equation, with the unknown $\tan y$, to admit rational number solutions, we must have $\Delta$ to be a perfect square. Here $\Delta=b^2-4ac$ is the discriminant of the quadratic equation $a\cdot t^2+b\cdot t+c=0.$ Since $b=2\cdot b'$ is an even number, it is enough to consider the discriminant "halved" $\Delta '=(b')^2-ac$. For us
$$\Delta ' =q^2(q^2-3p^2)^2+p^2(3q^2-p^2)^2=$$

$$=q^6-6p^2q^4+9p^4q^2+9p^2q^4-6p^4q^2+p^6=$$

$$=p^6+3p^4q^2+3p^2q^4+q^6=(p^2+q^2)^3=(p^2+q^2)^2 \cdot (p^2+q^2).$$

          Now it is obvious that $\Delta '$ is a perfect square if and only if $p^2+q^2$ is a perfect square.

$\blacksquare$