一、Multiple Choice Questions (7 points each, total 42 points)
1. The number of integer solutions to the equation $\left(x^{2}-x-1\right)^{x+10}=1$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
From $x+10=0$, we get $x=-10$. When $x+10$ is odd, from $x^{2}-x-1=1$, we get $x=-1$; when $x+10$ is even, from $x^{2}-x-1= \pm 1$, we get $x=0$ or $x=2$.

In summary, the integer solutions of the original equation are $x_{1}=-10, x_{2}=-1$, $x_{3}=0, x_{4}=2$