A10 (7-4, Soviet Union) Find 4 real numbers $x_{1}, x_{2}, x_{3}, x_{4}$, such that the sum of any one of them and the product of the other 3 equals 2.
Solution 1 First, for the system of equations
$$
\left\{\begin{array}{l}
x_{1}+x_{2} x_{3} x_{4}=2, \\
x_{2}+x_{1} x_{3} x_{4}=2, \\
x_{3}+x_{1} x_{2} x_{4}=2, \\
x_{4}+x_{1} x_{2} x_{3}=2,
\end{array}\right.
$$
(2)
we have $x_{i} \neq 0(i=1,2,3,4)$. Otherwise, for example, if $x_{1}=0$, then from (1), $x_{2} x_{3} x_{4}=2$, and from (2), (3), (4), $x_{2}=x_{3}=x_{4}=2$, which is a contradiction. Similarly, $x_{2}, x_{3}, x_{4}$ are all non-zero.
Now from (1), (2), we get
$$
x_{3} x_{4}=\frac{2-x_{1}}{x_{2}}=\frac{2-x_{2}}{x_{1}},
$$

thus we have $2 x_{1}-x_{1}^{2}=2 x_{2}-x_{2}^{2}$, or
$$
\left(x_{1}-1\right)^{2}=\left(x_{2}-1\right)^{2},
$$

or
$$
\left|x_{1}-1\right|=\left|x_{2}-1\right| \text {. }
$$

By cyclic permutation, we get
$$
\left|x_{1}-1\right|=\left|x_{2}-1\right|=\left|x_{3}-1\right|=\left|x_{4}-1\right| .
$$

We discuss the following five cases:
(1) All $x_{i} \geqslant 1$, then from (5) we get

From (1) we derive
$$
\begin{array}{c}
x_{1}=x_{2}=x_{3}=x_{1} . \\
x_{1}+x_{1}^{3}=2 .
\end{array}
$$

Factoring gives
$$
\left(x_{1}-1\right)\left(x_{1}^{2}+x_{1}+2\right)=0 .
$$

Since the latter is always greater than zero, we have $x_{1}=x_{2}=x_{3}=x_{1}=1$, which indeed satisfies equations (1), (2), (3), (4).
(2) Exactly three $x_{i} \geqslant 1$, without loss of generality, assume $x_{1}<1, x_{2}, x_{3}, x_{4} \geqslant 1$. Then from (5) we get
$$
1-x_{1}=x_{2}-1=x_{3}-1=x_{4}=1 .
$$

Thus $x_{2}=x_{3}=x_{1}=2-x_{1}$. From (1) we get
$$
2-x_{2}+x_{2}^{3}=2 \text { or } x_{2}\left(x_{2}^{2}-1\right)=0 \text {. }
$$

This gives $x_{2}=1$ and $x_{1}=1$, a contradiction.
(3) Exactly two $x_{i} \geqslant 1$, without loss of generality, assume $x_{1}, x_{2}<1, x_{3}, x_{4} \geqslant 1$. From (5) it is easy to see $x_{1}=x_{2}, x_{3}=x_{4}=2-x_{1}$. From (3), we get
$$
2-x_{1}+x_{1}^{2}\left(2-x_{1}\right)=2 \text { or } x_{1}\left(x_{1}-1\right)^{2}=0 \text {. }
$$

This gives $x_{1}=1$, a contradiction.
(4) Exactly one $x_{i} \geqslant 1$. Without loss of generality, assume $x_{1}, x_{2}, x_{3}<1, x_{1} \geqslant 1$. Then from (5) we get
$$
x_{1}=x_{2}=x_{3}=2-x_{4} \text {. }
$$

From (4), we get
$$
2-x_{\mathrm{i}}+x_{1}^{3}=2 \text {, or } x_{1}\left(x_{1}^{2}-1\right)=0 \text {, }
$$

This gives $x_{1}=-1$, thus $x_{1}=x_{2}=x_{3}=-1, x_{4}=3$. Substituting back, we find that it indeed satisfies (1) (4). By cyclic permutation, we get four solutions, namely $(-1,-1,-1,3),(-1,-1,3,-1),(-1,3,-1,-1),(3,-1,-1,-1)$.
(5) All $x_{i}<1$, then from (5) we get
$$
x_{1}=x_{2}=x_{3}=x_{1},
$$

It is easy to see that $x_{1}=1$, which contradicts $x_{1}<1$.
Therefore, there are five solutions in total, which are $(1,1,1,1),(-1,-1,-1,3)$, $(-1,-1,3,-1),(-1,3,-1,-1),(3,-1,-1,-1)$.

Solution 2 Let $P$ represent $x_{1} x_{2} x_{3} x_{4}$, then each equation has the form (from Solution 1, we know $P \neq 0$ )
$$
x_{i}+\frac{P}{x_{i}}=2, \quad(i=1,2,3,4)
$$

and
$$
x_{i}=1 \pm \sqrt{1-P} . \quad(i=1,2,3,4)
$$

It is easy to see that $P \leqslant 1$. Now we discuss as follows:
(1) $P=1$, then $x_{1}=x_{2}=x_{3}=x_{4}=1$.
(2) $P<1$, and the four roots are not all the same, we discuss the following three cases.
(i) Two of $x_{i}(i=1,2,3,4)$ are $1+\sqrt{1-P}$, and the other two are $1-\sqrt{1-P}$, then
$$
\begin{aligned}
P & =x_{1} x_{2} x_{3} x_{1} \\
& =(1+\sqrt{1-P})^{2}(1-\sqrt{1-P})^{2} \\
& =P^{2},
\end{aligned}
$$

Thus $P=1$, a contradiction.
(ii) Three of the four numbers are $1+\sqrt{1-P}$, and the remaining one is $1-\sqrt{1-P}$, then
$$
\begin{aligned}
P & =x_{1} x_{2} x_{3} x_{4} \\
& =(1+\sqrt{1-\bar{P}})^{3}(1-\sqrt{1-P}) \\
& =P(1+\sqrt{1-\bar{P}})^{2},
\end{aligned}
$$

In this case, we still have $P=1$, a contradiction.
(iii) One of the four numbers is $1+\sqrt{1-P}$, and the other three are $1-\sqrt{1-\bar{P}}$, then
$$
\begin{aligned}
P & =x_{1} x_{2} x_{3} x_{1} \\
& =(1+\sqrt{1-P})(1-\sqrt{1-\bar{P}})^{3} \\
& =P(1-\sqrt{1-\bar{P}})^{2},
\end{aligned}
$$

Solving gives $P=-3$, thus one of $x_{i}$ is 3, and the other three are -1.