## Task 1.

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

$$
\begin{aligned}
& x(x y-1)=2(y z-1) \\
& y(y z-1)=2(z x-1) \\
& z(z x-1)=2(x y-1)
\end{aligned}
$$
## Solution.

Multiply all three given equations and we see that

$$
x y z(x y-1)(z y-1)(z x-1)=8(x y-1)(y z-1)(z x-1)
$$

If $x y-1=0$, then from the first equation we conclude that $y z-1=0$, and then from the second that $z x-1=0$. Similarly, we conclude if $y z-1=0$ or if $z x-1=0$. Therefore, if one of the numbers $x y-1, y z-1, z x-1$ is equal to 0, then all three are.

If $x y=1, y z=1$ and $z x=1$, by multiplying these three equations we conclude that $x^{2} y^{2} z^{2}=1$, and since $x^{2} y^{2}=1$, we conclude that $z^{2}=1$ and similarly that $x^{2}=1$ and $y^{2}=1$.

If $z=1$, then from the last equation it follows that $x=1$, and from the second that $y=1$. Similarly, if $z=-1$, then $x=y=-1$.

Now assume that $(x y-1)(y z-1)(z x-1) \neq 0$. Then we see that $x y z=8$.

Multiply the first equation by $x$ and we see that $x^{3} y-x^{2}=2(x y z-x)$, or

$$
x^{3} y=15+(x-1)^{2} \geqslant 15
$$

We conclude that $x^{3} y \geqslant 15$, from which we conclude that $x y>0$, and since $x y z=8$, we conclude that $z>0$. Similarly, we conclude that $x>0$ and $y>0$.

If we assume that $x y>0, y z-1>0$ and $z x-1>0$.

Assume that $x>2$, then

$$
2 y z-2=2(y z-1)=x(x y-1)>2(2 y-1)=4 y-2,
$$

which means that $z>2$. Then from the second equation in the same way we conclude that $y>2$. This is impossible because $8=x y z>2 \cdot 2 \cdot 2=8$.

Therefore, it must be $x \leqslant 2, y \leqslant 2$ and $z \leqslant 2$, and since $x y z=8$, we conclude that the only possibility is $x=y=z=2$. Thus, we have shown that the system has exactly three solutions: $(x, y, z)=(1,1,1)$, $(x, y, z)=(-1,-1,-1)$ and $(x, y, z)=(2,2,2)$.