250. How many real solutions does the system have

$$
\left\{\begin{array}{c}
x+y=2 \\
x y-z^{2}=1 ?
\end{array}\right.
$$
250. From the first equation we get $x=2-y$; substituting into the second:

$$
2 y-y^{2}-z^{2}=1
$$

or

$$
z^{2}+y^{2}-2 y+1=0, \text { i.e. } z^{2}+(y-1)^{2}=0
$$

Each term of the last equality is non-negative and, consequently, equals zero. Hence,

$$
z=0, \quad y=1
$$

and, therefore,

$$
x=1 .
$$

Thus, the system has a unique real solution.