4. Solve the system of equations

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
x^{3}+y^{5}=1
\end{array}\right.
$$
Answer: $(0 ; 1),(1 ; 0)$.

Solution. Subtract the first equation from the second:

$$
x^{2}(x-1)+y^{2}\left(y^{3}-1\right)=0 .
$$

From the first equation of the system, it follows that $x \leqslant 1$ and $y \leqslant 1$. Therefore, $x^{2}(x-1) \leqslant 0$ and $y^{2}\left(y^{3}-1\right) \leqslant 0$. The sum of two non-negative numbers is zero if and only if both are zero. If the first term in the left-hand side of equation (*) is zero, then $x=0$ or $x=1$. In the first case, from the original system, it follows that $y=1$. And in the second case, obviously, $y=0$.

Evaluation. 14 points for a correct solution. If the answer is guessed but not proven that there are no other solutions, 2 points.