3. For every 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 $n$ that satisfy $f(n)=n$ (where $n \neq 1$) is
A. 0
B. 1
C. 2
D. 3
3. $\mathrm{B}$ Hint: Let $x=1, f(y+1)=f(y)+y+2(*)$. For $y$ being a positive integer, $f(y+1)>y+2>y+1$; also from * we have $f(y)=f(y+1)-(y+2)$, it is easy to see that for $y>0$. The original functional equation $f(n)=n$ has only the solutions $n=1,-2$.