2. The integer solutions of the equation $\left(x^{2}+x-1\right)^{x+98}=1$ are ( ) in number.
(A) 2
(B) 3
(C) 4
(D) 5
2. (C).

From $\left\{\begin{array}{l}x+98=0, \\ x^{2}+x-1 \neq 0\end{array}\right.$ we get $x_{1}=-98$,
From $x^{2}+x-1=1$ we get $x_{2}=1, x_{3}=-2$,
From $\left\{\begin{array}{l}x^{2}+x-1=-1, \\ x+98 \text { is even }\end{array}\right.$ we get $x_{4}=0$.