$1 \cdot 42$ Find all quadruples of real numbers $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ such that the sum of any one of the numbers and the product of the other three is equal to 2.
[Solution] Let $x_{1} x_{2} x_{3} x_{4}=d$,
so $\quad x_{i}+\frac{d}{x_{i}}=2,(i=1,2,3,4)$,
hence $x_{i}=1 \pm \sqrt{1-d},(i=1,2,3,4)$.
Clearly, $d \leqslant 1$. There are five cases:
(1) $d=(1+\sqrt{1-d})^{4} \geqslant 1 \geqslant d$,

then $d=1, x_{i}=1,(i=1,2,3,4)$.
$$
\text { (2) } \begin{aligned}
d & =(1-\sqrt{1-d})(1+\sqrt{1-d})^{3} \\
& =d(1+\sqrt{1-d})^{2},
\end{aligned}
$$

then $d=0$ (which is not possible to verify) or $d=1$ (same as (1)).
(3) $d=(1-\sqrt{1-d})^{2}(1+\sqrt{1-d})^{2}=d^{2}$. Then $d=0$ or 1 (same as (2)).
(4) $d=(1-\sqrt{1-d})^{3}(1+\sqrt{1-d})$
$$
=d(1-\sqrt{1-d})^{2} \text {, }
$$

then $d=0$ or 1 (same as (2)) or $d=-3$, in which case three of the $x_{i}$ are -1 and one is +3.
(5) $d=(1-\sqrt{1-d})^{4}$, then
$0 \leqslant d=\left(\frac{d}{1+\sqrt{1-d}}\right)^{4} \leqslant d^{4} \leqslant d$,
so $d=1$, same as $(1)$.
In summary, either $x_{1}=x_{2}=x_{3}=x_{4}=1$, or three of them are -1 and one is +3.