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

If $f(1)=1$, then the number of integers $n$ (where $n \neq 1$) that satisfy $f(n)=n$ is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) infinitely many.
(30th American High School Mathematics Examination, 1979)
[Solution] Substituting $x=1$ into the functional equation, we get
$$
f(y+1)=f(y)+y+2 \text {. }
$$

Since $f(1)=1$, substituting $y=2,3,4, \cdots$ consecutively, we can see that for $y$ being a positive integer, $f(y)>0$. Therefore, for $y$ being a positive integer,
$$
f(y+1)>y+2>y+1 ;
$$

Thus, for integers $n>1$, $f(n)=n$ has no solution. Solving the equation for $f(y)$:
From $\circledast$ we have
$$
f(y)=f(y+1)-(y+2),
$$

Substituting $y=0,-1,-2, \cdots$ consecutively into this equation, we get
$$
\begin{array}{l}
f(0)=-1, \quad f(-1)=-2, \quad f(-2)=-2, \quad f(-3)=-1, \\
f(-4)=1 .
\end{array}
$$

Since $f(-4)>0$ and for $y0$.
Thus, for $y0$.
Therefore, for $n<-4, f(n) \neq n$.
Hence, $f(n)=n$ has only the solutions $n=1,-2$.
Therefore, the answer is $(B)$.