
Problem 1. Find all values of the real parameters $p$ and $q$, for which the roots of the equations $x^{2}-p x-1=0$ and $x^{2}-q x-1=0$ form (in a suitable order) an arithmetic progression with four members.


Solution. Denote by $x_{1}, x_{2}$ the roots of the equation $x^{2}-p x-1=0$ and by $y_{1}, y_{2}$ the roots of $x^{2}-q x-1=0$. It is clear that for all $p$ and $q$ the numbers $x_{1}, x_{2}, y_{1}, y_{2}$ are real and $x_{1} x_{2}=y_{1} y_{2}=-1$. Assume that $x_{1}<0<x_{2}$ and $y_{1}<0<y_{2}$. If four numbers $a, b, c$ and $d$ form an arithmetic progression then the numbers $d, c, b, a$ form an arithmetic progression too. So, we can assume that $x_{1}, x_{2}, y_{1}, y_{2}$ in a suitable order form an increasing arithmetic progression. If $x_{1}<y_{1}$ (the case $y_{1}<x_{1}$ is analogous) then there are two possibilities:

I. The arithmetic progression is $x_{1}, y_{1}, y_{2}, x_{2}$. Then $x_{1}+x_{2}=y_{1}+y_{2}$, from where $p=q$, i.e. $x_{1}=y_{1}, x_{2}=y_{2}$ which is impossible.

II. The arithmetic progression is $x_{1}, y_{1}, x_{2}, y_{2}$. Then $x_{2}-x_{1}=y_{2}-y_{1}$, from where $\sqrt{p^{2}+4}=\sqrt{q^{2}+4}$, i.e. $p^{2}=q^{2}$ and since $p \neq q$, then $p=-q \neq 0$.

In the same way we have $y_{1}=\frac{x_{1}+x_{2}}{2}=\frac{p}{2}$ and therefore $\frac{p^{2}}{4}-\frac{p q}{2}-1=0$. Since $p=-q$, then $p^{2}=\frac{4}{3}$ and $p= \pm \frac{2}{\sqrt{3}}$. Hence $(p, q)=\left(\frac{2}{\sqrt{3}},-\frac{2}{\sqrt{3}}\right),\left(-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right)$.
