3.5 The number of real solutions to the system of equations $\left\{\begin{array}{l}x+y=2, \\ x y-z^{2}=1 .\end{array}\right.$ is
(A) 1 .
(B) 2 .
(C) 3 .
(D) infinitely many.
(Shanghai Junior High School Mathematics Competition, 1987)
[Solution] From the given, we have $x+y=2, xy=1+z^{2}$.
Thus, $x, y$ are the two roots of the equation $t^{2}-2t+(1+z^{2})=0$. The discriminant of this equation is
$$
\Delta=4-4(1+z^{2})=-4z^{2} \leqslant 0,
$$

which is equal to zero if and only if $z=0$. When $\Delta=0$, the equation has real roots, and only $x=y=1$ satisfies the system of equations, so the system has only one solution.
Therefore, the answer is $(A)$.