$3 \cdot 9$ The common solutions of the equations $y=(x+1)^{2}$ and $x y+y=1$ are
(A) 3 pairs of real numbers.
(B) 4 pairs of real numbers.
(C) 4 pairs of imaginary numbers.
(D) 2 pairs of real numbers, 2 pairs of imaginary numbers.
(E) 1 pair of real numbers, 2 pairs of imaginary numbers.
(6th American High School Mathematics Examination, 1955)
[Solution] Given $x y+y=1$, we have $y=\frac{1}{x+1}$,
thus
$$
\begin{array}{c}
\frac{1}{x+1}=(x+1)^{2} . \\
(x+1)^{3}-1=0, \\
x\left(x^{2}+3 x+3\right)=0 .
\end{array}
$$

or
Notice that $x^{2}+3 x+3=0$ has no real roots.
Therefore, the answer is $(E)$.