[Example 4.3.1] Find all quadruples $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ such that the sum of any one of the numbers and the product of the other three numbers is equal to 2.
The conventional approach to solving this problem is to solve the following system of equations:
$$
\left\{\begin{array}{l}
x_{1}+x_{2} x_{3} x_{4}=2, \\
x_{2}+x_{3} x_{4} x_{1}=2, \\
x_{3}+x_{4} x_{1} x_{2}=2, \\
x_{4}+x_{1} x_{2} x_{3}=2,
\end{array}\right.
$$

Although the system of equations is symmetric with respect to \(x_{1}, x_{2}, x_{3}, x_{4}\), it is still quite complicated to discuss.
We can start from the structural characteristics of the above system of equations, not considering the product of three numbers, but rather the product of four numbers as a whole, i.e., consider
$$
p=x_{1} x_{2} x_{3} x_{4} .
$$

Clearly, \(x_{1}, x_{2}, x_{3}, x_{4}\) are all non-zero; otherwise, if one of them is 0, for example, \(x_{1}=0\), then \(x_{2}, x_{3}, x_{4}\) would all be 2, which would result in the solution \((x_{1}, x_{2}, x_{3}, x_{4})=(0,2,2,2)\). Substituting this into any equation in the system does not fit, so \(x_{i} (i=1,2,3,4)\) are all non-zero. Using \(p\), the system of equations can be transformed into
$$
x_{i}+\frac{p}{x_{i}}=2, \quad i=1,2,3,4 .
$$

This can be transformed into a quadratic equation in \(x_{i} (i=1,2,3,4)\):
$$
x_{i}^{2}-2 x_{i}+p=0 \text {. }
$$

Equation (1) has at most two distinct real solutions. Considering the discriminant
$$
\Delta=4-4 p=4(1-p) \geqslant 0, \quad p \leqslant 1 .
$$

When \(p \leqslant 1\), equation (1) has real solutions
$$
x_{i}=1 \pm \sqrt{1-p} \text {. }
$$

We now classify \(p\):
(1) When \(p=1\), equation (1) has equal roots
$$
x_{1}=x_{2}=x_{3}=x_{4}=1 \text {. }
$$
(2) When \(p<1\), the roots of equation (1) are distinct.
If two of \(x_{1}, x_{2}, x_{3}, x_{4}\) are equal to \(1+\sqrt{1-p}\) and the other two are equal to \(1-\sqrt{1-p}\), then we observe the product of these four roots as a whole.
$$
p=x_{1} x_{2} x_{3} x_{4}=(1+\sqrt{1-p})^{2}(1-\sqrt{1-p})^{2}=p^{2} \text {, }
$$

At this point, \(p=1\), which contradicts \(p<1\).
If three of \(x_{1}, x_{2}, x_{3}, x_{4}\) are equal to \(1+\sqrt{1-p}\) and one is equal to \(1-\sqrt{1-p}\), or if three are equal to \(1-\sqrt{1-p}\) and one is equal to \(1+\sqrt{1-p}\), then
$$
\begin{aligned}
p=x_{1} x_{2} x_{3} x_{4} & =(1+\sqrt{1-p})^{3}(1-\sqrt{1-p}) \\
& =p(1+\sqrt{1-p})^{2} .
\end{aligned}
$$

Or
$$
\begin{aligned}
p=x_{1} x_{2} x_{3} x_{4} & =(1-\sqrt{1-p})^{3}(1+\sqrt{1-p}) \\
& =p(1-\sqrt{1-p})^{2} .
\end{aligned}
$$

Thus,
$$
(1+\sqrt{1-p})^{2}=1 \text {, }
$$

Or
$$
(1-\sqrt{1-p})^{2}=1 \text {. }
$$

Solving the above equations gives \(p=1\) or \(p=-3\). Given \(p<1\), we have \(p=-3\).
When \(p=-3\), solve the equation
$$
\begin{array}{c}
x_{i}^{2}-2 x_{i}-3=0, \\
x_{i}=3 \text { or } x_{i}=-1 .
\end{array}
$$

Therefore, this problem has five solutions:
$$
\begin{array}{c}
(1,1,1,1),(3,-1,-1,-1),(-1,3,-1,-1), \\
(-1,-1,3,-1),(-1,-1,-1,3) .
\end{array}
$$

In solving this problem, we did not solve for \(x_{i}\) locally, but rather solved for the product \(p\) of \(x_{1}, x_{2}, x_{3}, x_{4}\). After finding \(x_{1} x_{2} x_{3} x_{4}\) as a whole, it became easier to find \(x_{i} (i=1,2,3,4)\).