4. (USS) Find four real numbers $x_{1}, x_{2}, x_{3}, x_{4}$ such that the sum of any of the numbers and the product of the other three is equal to 2.
4. It is easy to see that all $x_{i}$ are nonzero. Let $x_{1} x_{2} x_{3} x_{4}=p$. The given system of equations can be rewritten as $x_{i}+p / x_{i}=2, i=1,2,3,4$. The equation $x+p / x=2$ has at most two real solutions, say $y$ and $z$. Then each $x_{i}$ is equal either to $y$ or to $z$. There are three cases: (i) $x_{1}=x_{2}=x_{3}=x_{4}=y$. Then $y+y^{3}=2$ and hence $y=1$. (ii) $x_{1}=x_{2}=x_{3}=y, x_{4}=z$. Then $z+y^{3}=y+y^{2} z=2$. It is easy to obtain that the only possibilities for $(y, z)$ are $(-1,3)$ and $(1,1)$. (iii) $x_{1}=x_{2}=y, x_{3}=x_{4}$. In this case the only possibility is $y=z=1$.  Hence the solutions for $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ are $(1,1,1,1),(-1,-1,-1,3)$, and the cyclic permutations.