4. Let three real numbers $x_{1}, x_{2}, x_{3}$ be such that the sum of any one of them and the product of the other two is 1. Then the number of ordered triples $\left(x_{1}, x_{2}, x_{3}\right)$ that satisfy this condition is ( ).
(A)1
(B) 3
(C) 5
(D) more than 5

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4.C.

From the problem, we know
$$
\left\{\begin{array}{l}
x_{1}+x_{2} x_{3}=1, \\
x_{2}+x_{1} x_{3}=1, \\
x_{3}+x_{2} x_{1}=1 .
\end{array}\right.
$$
I. If one of $x_{1}, x_{2}, x_{3}$ is 0, then the other two are 1, i.e., $(0,1,1), (1,0,1), (1,1,0)$ all satisfy the problem.
II. If $x_{1} x_{2} x_{3} \neq 0$, from (1) and (2) we get
$$
\begin{array}{l}
x_{3}=\frac{1-x_{1}}{x_{2}}, x_{3}=\frac{1-x_{2}}{x_{1}} . \\
\therefore \frac{1-x_{1}}{x_{2}}=\frac{1-x_{2}}{x_{1}}, \\
x_{1}-x_{1}^{2}=x_{2}-x_{2}^{2}, \\
\left(x_{1}-\frac{1}{2}\right)^{2}=\left(x_{2}-\frac{1}{2}\right)^{2} . \\
\therefore\left|x_{1}-\frac{1}{2}\right|=\left|x_{2}-\frac{1}{2}\right| .
\end{array}
$$

By symmetry, we get
$$
\left|x_{1}-\frac{1}{2}\right|=\left|x_{2}-\frac{1}{2}\right|=\left|x_{3}-\frac{1}{2}\right| \text {. }
$$
(1) When $x_{1} \leqslant \frac{1}{2}, x_{2} \leqslant \frac{1}{2}, x_{3} \leqslant \frac{1}{2}$,
$$
\begin{array}{l}
x_{1}=x_{2}=x_{3} . \\
\therefore x_{1}+x_{2}^{2}=1 .
\end{array}
$$

At this time, $x_{1}=\frac{-1 \pm \sqrt{5}}{2}$.
$\because \frac{-1+\sqrt{5}}{2}>\frac{-1+2}{2}=\frac{1}{2}$,
$\therefore x_{1}=x_{2}=x_{3}=-\frac{1}{2}-\frac{\sqrt{5}}{2}$.
$\therefore\left(-\frac{1+\sqrt{5}}{2},-\frac{1+\sqrt{5}}{2}, \cdots-\frac{1+\sqrt{5}}{2}\right)$ satisfies the problem.
(2) When two $x_{i} \leqslant-\frac{1}{2}, \cdots$ and one $x_{i}>\frac{1}{2}$, without loss of generality, let $x_{1} \leqslant \frac{1}{2}, x_{2} \leqslant \frac{i}{2}, x_{3}>\frac{1}{2}$.
At this time, $\frac{1}{2}-x_{1}=\frac{1}{2}-x_{2}=x_{3}-\frac{1}{2}$.
$$
\begin{array}{l}
\therefore x_{1}=x_{2}, x_{3}=1-x_{1} . \\
\therefore x_{1}+x_{1}\left(1-x_{1}\right)=1 .
\end{array}
$$

Solving, we get $x_{1}=1$. This contradicts the assumption $x_{1} \leqslant \frac{1}{2}$.
(3) When two $x_{i}>\frac{1}{2}$ and one $x_{i} \leqslant \frac{1}{2}$, without loss of generality, let $x_{1}>\frac{1}{2}, x_{2}>\frac{1}{2}, x_{3} \leqslant \frac{1}{2}$.
At this time, $x_{1}-\frac{1}{2}=x_{2}-\frac{1}{2}=\frac{1}{2}-x_{3}$.
$$
\therefore x_{1}=x_{2}, x_{3}=1-x_{1} \text {, }
$$

Thus, $x_{1}+x_{1}\left(1-x_{1}\right)=1$.
Therefore, $x_{1}=1$.
Hence, $(1,1,0),(1,0,1),(1,1,0)$.
(4) When all three $x_{i}>\frac{1}{2}$, $x_{1}=x_{2}=x_{3}$.

Then $x_{1}=\frac{-1+\sqrt{5}}{2}$.
Thus, $\left(\frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right)$ satisfies the problem.