Cesko-Slovacko-Poljsko 2013

a) Let $s=x^2-rx$ and $t=x^3-rx$ be rational. Then $$x^2=s+rx,$$ $$x^3=x^2\cdot x=(s+rx)x=sx+r{x}^2=sx+r(s+rx)=(r^2+s)x+rs,$$ consequently $$t=x^3-rx=((r^2+s)x+rs)-rx=(r^2-r+s)x+rs.$$ if $(r^2-r+s)\ne 0$ (which is $x^2-rx+r^2-r\ne 0$), then $$x=\frac{t-rs}{r^2-r+s}$$ is rational. We conclude that the given statement holds iff the equation $$x^2-rx+r^2-r=0(1)$$ has no irrational roots. For the sake of completeness, note that if (1) has a rational root then $s$ and $t$ are rational as well: $$s=x^2-rx=r-r^2\text{and}t=0\cdot x+rs=rs=r(r-r^2).$$ If the determinant $D=r(4-3r)$ of (1) is less or equal 0, then (1) has no real solutions or a solution $x=\frac{r}{2}$, which is rational. Since $D\le 0\iff (r\ge \frac{4}{3}$ or $r\le 0)$ the statement under consideration is true.

b) According to a) it suffices to show that $D>0$ and $\sqrt{D}$ is irrational. We have $$D=r(4-3r)=\frac{p}{q}\left(4-\frac{3p}{q}\right)=\frac{p(4q-3p)}{q^2}>0.$$ But $p\nmid 4q$ thus $p\mid p(4q-3p)$ and $p^2\nmid p(4q-3p)$, that is $p(4q-3p)$ is not a perfect square, that is $\sqrt{D}$ is irrational.