\[\boxed{\text{1039\ (1039).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[x^{2} - (a - 2)x - a - 1 =\]
\[= x^{2} - (a - 2) \cdot x - (a + 1),\]
\[по\ теореме\ Виета:\ \]
\[x_{1} + x_{2} = a - 2,\]
\[x_{1} \cdot x_{2} = - a - 1,\]
\[\left( x_{1} \right)^{2} + \left( x_{2} \right)^{2} =\]
\[= \left( x_{1} + x_{2} \right)^{2} - 2x_{1}x_{2} =\]
\[= a^{2} - 4a + 4 + 2a + 2,\]
\[a^{2} - 2a + 6 = 0,\]
\[a_{0} = - \frac{b}{2a} = 1.\]
\[Ответ:при\ a = 1.\]