37th Hellenic Mathematical Olympiad 2020

We fix $\kappa >1$. The idea for the solution of the problem is to find $\alpha ,\beta$ with respect to $\kappa$ such that the denominator of the fraction is equal to $1$. In other words, we seek polynomials $\alpha =P(\kappa )$, $\beta =Q(\kappa )$ such that $$\begin{gathered} P(\kappa {)}^2+{\kappa }^{2}Q(\kappa {)}^2-{\kappa }^{2}P(\kappa )Q(\kappa )=1. \\ P(\kappa {)}^2={\kappa }^{2}Q(\kappa )(P(\kappa )-Q(\kappa ))+1(1) \end{gathered}$$ If $m=\mathrm{deg}P(\kappa ),n=\mathrm{deg}Q(\kappa )$ with $m>n\ge 0$, then from (1) it follows that $$2m=2+m+n\iff m=n+2.$$ We will deal with the easiest case, by seeking polynomials of degrees $2$ and $0$, respectively, which satisfy relation (1). Finally, we find that $P(\kappa )={\kappa }^2-1$, $Q(\kappa )=1$ satisfy relation (1). We choose $\alpha ={\kappa }^2-1$, $\beta =1$ and then the expression is equal to $\frac{{\kappa }^2-1+1}{1}={\kappa }^2$, which is composite for all $\kappa >1$. Therefore, all values of $\kappa >1$ are not solutions.

For $\kappa =1$, we will prove that: $0<A(\alpha ,\beta ,\kappa )=\frac{\alpha +\beta }{{\alpha }^2+{\beta }^2-\alpha \beta }\le 2$, which means that the rational number $A(\alpha ,\beta ,1)$ cannot be a composite positive integer for all $\alpha ,\beta$. Indeed, we have $$\alpha +\beta >0\text{and}{\alpha }^2-\alpha \beta +{\beta }^2={\left(\alpha -\frac{\beta }{2}\right)}^2+\frac{3{\beta }^2}{4}>0,\text{ and}$$ $$\begin{gathered} \frac{\alpha +\beta }{{\alpha }^2+{\beta }^2-\alpha \beta }\le 2\iff 2{\alpha }^2+2{\beta }^2-2\alpha \beta \ge \alpha +\beta \iff (\alpha -\beta {)}^2+{\alpha }^2-\alpha +{\beta }^2-\beta \ge 0 \\ \iff (\alpha -\beta {)}^2+\alpha (\alpha -1)+\beta (\beta -1)\ge 0,\text{ which holds.} \end{gathered}$$