## Task $6 / 73$

Give all real solutions of the system of equations:

$$
\begin{array}{r}
x^{2}+y^{2}=8 \\
x^{2} y^{2}-z^{2}=16 \\
\sin \left(x^{2}-2 y\right)=z
\end{array}
$$
Assume $(x ; y ; z)$ is a real solution. Then it must be true that $(x-y)^{2} \geq 0$, so $8=x^{2}+y^{2} \geq 2 x y$ and $(x+y)^{2} \geq 0$, so $8=x^{2}+y^{2} \geq-2 x y$; consequently, $|x y| \leq 4$ and $x^{2} y^{2} \leq 16$.

However, due to (2), $x^{2} y^{2}=16+z^{2}$, which implies $z=0$, leading to $x^{2}=y^{2}=4$ from (1) and (2). Therefore, the only possible solutions are $(2 ; 2 ; 0),(2 ;-2 ; 0),(-2 ; 2 ; 0),(-2 ;-2 ; 0)$.

For $(2 ;-2 ; 0)$ and $(-2 ;-2 ; 0)$, (3) is not satisfied; indeed, $2 \pi<8=x^{2}-2 y<3 \pi$, so $\sin \left(x^{2}-2 y\right) \neq 0=z$.

However, the other two triples satisfy all the equations. Thus, $(2 ; 2 ; 0)$ and $(-2 ; 2 ; 0)$ are the only solutions to the system of equations.