1. For every pair of real numbers $x, y$, the function $f$ satisfies the equation
$$
f(x+y)-f(x)-f(y)=1+xy,
$$

and $f(1)=1$. Then, the number of integers $n$ for which $f(n)=n (n \neq 1)$ is ( ) .
(A) 0
(B) 1
(C) 2
(L) 3
-1 (B).
Let $x=1$, then
$$
\begin{array}{l}
f(1)+f(y)=f(y+1)-y-1, \\
f(y+1)-f(y)=y+2 .
\end{array}
$$

Let $y=1,2,3, \cdots, y-1$ and add all the equations to get
$$
f(y)=\frac{y^{2}+3 y-2}{2} \text {. }
$$

Thus, for all natural numbers $y$, equation (2) holds. If $f(y)=y$, we get $y^{2}+3 y-2=2 y, y=-2, y=1$. Therefore, for $n \in N$ and $n \neq 1$, $f(n)=n$ has no solution.

Taking $y=0$ in formula (1) gives $f(0)=-1$. Taking $y=-1$ gives $f(-1)=-2$. Further, we get $f(-2)=-2, f(-3)=-1, f(-4)=1$. From equation (1), for $y<0$. Thus, for $n<-4$, $f(n) \neq n$. Therefore, only when $n=-2$, $f(-2)=-2$.