Problem 9.1. Consider the system $\left\lvert\, \begin{aligned} & x^{2}+y^{2}=a^{2}+2 \\ & \frac{1}{x}+\frac{1}{y}=a\end{aligned}\right.$, where $a$ is a real number.

a) Solve the system for $a=0$.

b) Find all $a$, for which the system has exactly two solutions.

Svetlozar Doychev, Sava Grozdev
9.1. a) If $a=0$, then $x=-y$ and hence $2 x^{2}=2$. It follows that $(x, y)=$ $(1,-1)$ or $(x, y)=(-1,1)$.

b) We know from a) that $a=0$ is one of the desired numbers. Let $a \neq 0$. Setting $x+y=p, x y=q$, we have $p=a q$ and $p^{2}-2 q=a^{2}+2$. Then $a^{2} q^{2}-2 q-a^{2}-2=0$ and hence $(p, q)=(-a,-1)$ or $(p, q)=\left(\frac{a^{2}+2}{a}, \frac{a^{2}+2}{a^{2}}\right)$. Note that these pairs of numbers are different. The first case $x+y=-a$, $x y=-1$ leads to the quadratic equation $z^{2}+a z-1=0$ which has two distinct real roots $z_{1}$ and $z_{2}$. It follows that $(x, y)=\left(z_{1}, z_{2}\right)$ and $(x, y)=\left(z_{2}, z_{1}\right)$ are
solution of the given system. Thus we have to find all $a$ for which the second case is impossible. This means that the discriminant of the quadratic equation $z^{2}-\frac{a^{2}+2}{a} z+\frac{a^{2}+2}{a^{2}}=0$ is negative, i.e., $a \in(-\sqrt{2}, \sqrt{2}) \backslash\{0\}$. So the answer to b) is $a \in(-\sqrt{2}, \sqrt{2})$.