## Task 1 - 191221

Determine all triples $(x, y, z)$ of real numbers that satisfy the following system of equations:

$$
\begin{aligned}
& x-\frac{1}{y}=1 \\
& y-\frac{1}{z}=1 \\
& z-\frac{1}{x}=1
\end{aligned}
$$
By pairwise subtraction of any two of the three equations, one immediately obtains the new system of equations

$$
x-y=\frac{z-y}{y z}, \quad x-z=\frac{x-y}{x y}, \quad y-z=\frac{x-z}{x z}
$$

and from this, by multiplying, the new equation

$$
(x-y)(x-z)(y-z)=\frac{(x-y)(x-z)(z-y)}{(x y z)^{2}}
$$

If here $(x-y)(x-z)(y-z) \neq 0$, then one could cancel by this expression, which would immediately lead to the contradiction $(x y z)^{2}=-1$. Therefore, it must be

$$
(x-y)(x-z)(y-z)=0
$$

i.e., at least one of the differences $x-y, x-z, y-z$ must be 0. By substituting into (*), one sees immediately that the other two differences must also vanish, i.e., $x=y=z$ must hold.

The rest is very simple: Using $x=y$, the first equation of the original system of equations becomes

$$
x-\frac{1}{x}=1 \quad \text{or} \quad x^{2}-x-1=0
$$

from which its two solutions are immediately obtained as

$$
x=y=z=\frac{1 \pm \sqrt{5}}{2}
$$