1. Given that for every pair of real numbers $x, y$, the function $f$ satisfies $f(x)+f(y)=f(x+y)-x y-1$. If $f(1)=1$, then the number of integers $n$ that satisfy $f(n)=n$ is $\qquad$ .
1. 2 Detailed: Let $y=1$ we get $f(x)+f(1)=f(x+1)-x-1$, i.e., $f(x+1)=f(x)+x+2$. Let $x=0$ we get $f(1)=f(0)+2$. Knowing that $f(1)=1$, we have $f(0)=-1$. When $n \in \mathbf{N}_{+}$, $f(n)=\sum_{k=1}^{n}[f(k)-f(k-1)]+f(0)=$ $\sum_{k=1}^{n}(k+1)+f(0)=\frac{n(n+3)}{2}-1$. Similarly, $f(-n)=-\frac{n(-n+3)}{2}-1$. Therefore, $f(n)=\frac{n(n+3)}{2}-1, n \in \mathbf{Z}$. Let $f(n)=n$, solving gives $n=-2$ or $n=1$.