## Task A-4.4.

Determine all triples $(x, y, z)$ of real numbers for which

$$
\begin{aligned}
& \left(x^{2}+1\right) y=z^{2}+1 \\
& \left(y^{2}+1\right) z=x^{2}+1 \\
& \left(z^{2}+1\right) x=y^{2}+1
\end{aligned}
$$
## First solution.

If we multiply all three given equations, we see that

$$
\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right) y z x=\left(z^{2}+1\right)\left(x^{2}+1\right)\left(y^{2}+1\right)
$$

Notice that for all real numbers $t$, $t^{2}+1 \geqslant 1>0$, so from the above equation, we conclude that $x y z=1$.

Since $x^{2}+1>0$ and $z^{2}+1>0$, we conclude that $y>0$, and similarly $z>0$ and $x>0. \quad 1$ point

Without loss of generality, assume that $z=\min \{x, y, z\}$.

Assume that $z1>z^{2}$ (since $x y z=1$ ).

Consider the first equation

$$
z^{2}+1=\left(x^{2}+1\right) y \geqslant 2 x y=x y+x y>z^{2}+1
$$

where in the first inequality we use the fact that for any real number $x, x^{2}+1 \geqslant 2 x$, and in the second the fact that $x y>z^{2}$ and $x y>1$. Thus, the case where $z<1$ is impossible.

If $z \geqslant 1$, then the only possibility is $x=y=z=1$ (since $x y z=1$ ).

By substitution, we see that $x=y=z=1$ is indeed a solution, and the only solution to the given system.