
06.2. The real numbers $x, y$ and $z$ are not all equal and satisfy

$$
x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}=k
$$

Determine all possible values of $k$.


Solution. Let $(x, y, z)$ be a solution of the system of equations Since

$$
x=k-\frac{1}{y}=\frac{k y-1}{y} \quad \text { and } \quad z=\frac{1}{k-y}
$$

the equation

$$
\frac{1}{k-y}+\frac{y}{k y-1}=k
$$

to be simplified into

$$
\left(1-k^{2}\right)\left(y^{2}-k y+1\right)=0
$$

is true. So either $|k|=1$ or

$$
k=y+\frac{1}{y}
$$

The latter alternative, substituted to the original equations, yields immediately $x=y$ and $z=y$. So $k= \pm 1$ is the only possibility. If $k=1$, for instance $x=2, y=-1$ and $z=\frac{1}{2}$ is a solution; if $k=-1$, a solution is obtained by reversing the signs for a solution with $k=1$. So $k=1$ and $k=-1$ are the only possible values for $k$.
