4. Find four real numbers $x_{1}, x_{2}, x_{3}, x_{4}$, such that each, when added to the product of the others, equals two.
Solution. Let $x_{1} x_{2} x_{3} x_{4}=p$. Then our system of equations will take the form:

$$
x_{i}+\frac{p}{x_{i}}=2, \quad i=1,2,3,4
$$

The case when one of the sought numbers is zero leads to a contradiction. Indeed, substituting zero for $x_{1}$, we get

$$
x_{2} x_{3} x_{4}=2, x+2=x+3=x+4=2 \text {. }
$$

Thus, all $x_{i}$ are roots of the same quadratic equation $x_{i}^{2}-2 x_{i}+p=0$. Therefore, among them, there can only be two different ones. Let's consider three cases.

(a) $x_{1}=x_{2}=x_{3}=x_{4}=m, m+m^{2}=2$. Due to the monotonicity of the function $m+m^{2}$, we have only one real solution $m=1$.

(b) Three of the sought numbers are equal, and the fourth is not equal to them. Let $x_{1}=x_{2}=x_{3}=m$, $x_{4}=n, m \neq n$

$$
\left\{\begin{array}{l}
m+m^{2} n=2 \\
n+m^{3}=2
\end{array}\right.
$$

From this, $(m-n)\left(1-m^{2}\right)=0$. Since $m \neq n$, we get that $1-m^{2}=0$. Note that $m \neq 1$, because otherwise, from the system of relations, it immediately follows that $n=1$, which contradicts $m \neq n$. Therefore, $m=-1$, from which it follows that $n=3$. This gives us four more solutions.

(c) $x_{1}=x_{2}=m, x_{3}=x_{4}=n, m \neq n$,

$$
\left\{\begin{array}{l}
m+m n^{2}=2 \\
n+n m^{2}=2
\end{array}\right.
$$

From this, $(m-n)(1-m n)=0 \Rightarrow m n=1$. Then $m+n=2, m=n=1$. Contradiction.

Thus, either $x_{1}=x_{2}=x_{3}=x_{4}=1$, or one of the $x_{i}=3$, and the rest are equal to $-1$.