10. Given the system of inequalities $\left\{\begin{array}{l}x^{2}-x+a-a^{2}<1\end{array}\right.$ has exactly two integer solutions, find the range of values for $a$.
10. The roots of the equation $x^{2}-x+a-a^{2}=0$ are $x=a$ or $1-a$. (1) When $a=\frac{1}{2}$, the original inequality system has no solution; (2) When $a<\frac{1}{2}$, the original inequality system becomes $\left\{\begin{array}{l}a<1-2 a\end{array}\right.$ Since the original inequality system has a solution, $1-2 a>0$, at this time $x$ is not an integer; (3) When $a>\frac{1}{2}$, the original inequality system becomes $\left\{\begin{array}{l}1-a<1-2 a\end{array}\right.$ Since $a>\frac{1}{2}$, so $1-a>1-2 a$, the solution to the original inequality system is $1-a<x<a$. Also, the original inequality system has only two integer solutions, so $1<a-(1-a) \leqslant 3$, which means $1<a \leqslant 2$.