33. Solve the system in real numbers:

$$
\left\{\begin{array}{l}
x_{1}+x_{2}=x_{3}^{2} \\
x_{2}+x_{3}=x_{4}^{2} \\
x_{3}+x_{1}=x_{5}^{2} \\
x_{4}+x_{5}=x_{1}^{2} \\
x_{5}+x_{1}=x_{2}^{2}
\end{array}\right.
$$
76.33. Let's first consider the case when all unknowns are non-negative. If we denote by $X$ and $Y$ the largest and smallest numbers in the set of unknowns, then it is obvious that $X^{2} \leqslant 2 X, Y^{2} \geqslant 2 Y$, i.e., for non-zero $X$ and $Y$ we get $2 \leqslant Y \leqslant X \leqslant 2$, from which we see that the system has two solutions: $x_{1}=x_{2}=\ldots=x_{5}=2$ and $x_{1}=x_{2}=\ldots=x_{5}=0$.

It remains to prove that there are no negative numbers among the $x_{i}$. Suppose, for example, $x_{3}$ is the smallest number in the set and $x_{3}<0$. Then, subtracting the second equation from the first, we get $x_{1}-x_{3}=x_{3}^{2}-x_{4}^{2}$. Since $x_{1}-x_{3} \geqslant 0$, then $\left|x_{3}\right| \geqslant\left|x_{4}\right|$ and, consequently, $x_{3}+x_{4}=x_{5}^{2} \leqslant 0$, i.e., $x_{5}=0$ (and only in the case if $x_{1}=x_{3}$). Then the fifth equation leads to an obvious contradiction.