## Task 23/83

Solve the system of equations for any real numbers $x ; y ; z$ :

$$
x+y+\sin ^{2} z=12 \quad(1) \quad ; \quad x y=36
$$
From (1) it follows that due to $\sin ^{2} z \geq 0$ the inequality $x+y \leq 12$ holds. (1a) Furthermore, it holds that

$$
(\sqrt{x}-\sqrt{y})^{2}=x-2 \sqrt{x y}+y \geq 0
$$

and with $x y=36$

$$
x-2 \sqrt{36}+y \geq 0 \rightarrow x+y \geq 12
$$

From (1a) and (2a) it follows that $x+y=12, y=12-x$. Substituting into (2), the quadratic equation $x^{2}-12 x-36=0$ is obtained, which has the double solution $x=6$. It follows further that $y=6, \sin ^{2} z=0, z=k \pi$ with $k \in \mathbb{Z}$. Therefore, there exists exactly one solution set $x=6, y=6, z=k \pi$ with $k \in \mathbb{Z}$.