21. How many integers $x$ satisfy the equation
$$
\begin{array}{l}
\left(x^{2}-x-1\right)^{x+2} \\
=1 ?
\end{array}
$$
(A) 2; (B) 3,
(C) 4; (D) 5;
(E) None of the above conclusions is correct.
21. When $a$ and $b$ are integers, there are three cases that can make $a^{b}=1$: I) $a=1$; II) $a=-1, b$ is even; III) $b=0, a \neq 0$. In this problem, $a=x^{2}-x-1$, $b=x+2$. We will examine these three cases in sequence:

I) $x^{2}-x-1=1, (x-2)(x+1)=0$, $x=2$ or $x=-1$.
II) $x^{2}-x-1=-1$, and $x+2$ is even, $2^{2}-x=0, x=0$ or $x=1$.
When $x=0$, $x+2$ is even, so $x=0$ is a solution; because when $x=1$, $x+2=3$ is odd, and $x=1$ is not a solution.
III) $x+2=0$ and $x^{2}-x-1 \neq 0, x=-2$. This is a solution, because $(-2)^{2}-(-2)-1=5 \neq 0$.
Thus, there are a total of 4 solutions.
(C) True.