4. First, let us realize that for any real solution $(x, y)$ of the given system of equations, the pairs $(x, -y), (-x, y)$, and $(-x, -y)$ are also its solutions. Therefore, it suffices to limit ourselves to solutions in the domain of non-negative real numbers. Furthermore, for any solution $(x, y)$, the pair $(y, x)$ is also a solution of the given system. We can therefore further assume that $0 \leqq x \leqq y$. $-\sin ^{2} \alpha$:

Let us first rewrite both equations of the system using the known formula $\cos ^{2} \alpha=1-$

$$
\begin{aligned}
& \sin ^{2} x+1-\sin ^{2} y=y^{2} \\
& \sin ^{2} y+1-\sin ^{2} x=x^{2}
\end{aligned}
$$

By adding both equations, we then get

$$
x^{2}+y^{2}=2 \text {. }
$$

By subtracting the second equation from the first, we get

$$
2 \sin ^{2} x-2 \sin ^{2} y=y^{2}-x^{2},
$$

or

$$
2(\sin x+\sin y)(\sin x-\sin y)=y^{2}-x^{2}
$$

Given the assumption $0 \leqq x \leqq y$, it follows from (1) that $0 \leqq x \leqq y \leqq \leqq \sqrt{2}<\frac{1}{2} \pi$, and since the sine function is non-negative and increasing in the interval $\left\langle 0, \frac{1}{2} \pi\right)$, we see that for such real numbers $x$ and $y$, the left side of equation (2) is non-positive, while the right side is non-negative. This means that $y^{2}-x^{2}=0$, which under the given assumptions gives $x=y$ and together with (1) we have $x=y=1$.

In the domain of non-negative real numbers, the given system of equations has a unique solution, which is $(x, y)=(1,1)$.

Conclusion. The given system of equations has exactly four solutions in the domain of real numbers. They are the following pairs: $(1,1),(1,-1),(-1,1)$, and $(-1,-1)$.

For a complete solution, award 6 points. For identifying the equality $x^{2}+y^{2}=2$ without further significant progress, award 1 point. For guessing all solutions (e.g., due to a flawed reasoning that the equality $x=y$ immediately follows from the symmetry of the system), award 1 point. Mentioning the symmetry of the variables $x$, $y$ or the possibility of changing their signs arbitrarily, award 1 point. These partial one-point awards cannot be cumulated by addition. If a student solves the problem in only one quadrant without finding solutions in the other quadrants, award 4 points.
4. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& \sin ^{2} x+\cos ^{2} y=y^{2}, \\
& \sin ^{2} y+\cos ^{2} x=x^{2} .
\end{aligned}
$$

The regional round of category A takes place

## on Tuesday, January 24, 2006

so that it starts in the morning and contestants have 4 hours of pure time to solve the problems. Contestants can earn 6 points for each problem, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing supplies, school MF tables, and calculators without a graphic display. These details will be communicated to the students before the competition begins.