4. The system of inequalities $\left\{\begin{array}{l}x^{2}-x+a-a^{2}<1\end{array}\right.$ has exactly two integer solutions, then the range of values for $a$ is $\qquad$
4. $11-2 a$. From the first inequality, we get $(x-a)(x-(1-a))\frac{1}{2}$, the solution is $1-a0$, at this time the original inequality has no integer solutions, which is a contradiction; when $a>\frac{1}{2}$, the solution set of the original inequality is $(1-a, a)$, where there are exactly two integers equivalent to $1<a-(1-a) \leqslant 3$, solving this yields $1<a \leqslant 2$.