Winter Mathematical Competition

Writing the equation in the form
$$(x-1)(x^2 + (1-a)x - a + a^2) = 0,$$
we obtain $x_1 = 1$. Let $x_2$ and $x_3$ be the roots of the quadratic equation. If $1$ is the second term of the progression then $x_2 + x_3 = 2$, giving $a - 1 = 2$, i.e. $a = 3$. When $a = 3$ the roots of the quadratic equation are not real.
If $x_1 = 1$ is not the second term we may assume that $1 + x_2 = 2x_3$, which together with $x_2 + x_3 = a - 1$ implies $3x_3 = a$. Therefore $\frac{a}{3}$ is a root of the quadratic equation, i.e.
$$\left(\frac{a}{3}\right)^2 + (1-a)\frac{a}{3} - a + a^2 = 0.$$
Thus, $a = 0$ or $a = \frac{6}{7}$. When $a = 0$ we obtain $x_2 = -1$, $x_3 = 0$ and when $a = \frac{6}{7}$ we find $x_2 = -\frac{3}{7}$ and $x_3 = \frac{2}{7}$. Hence the desired values are $a = 0$ and $a = \frac{6}{7}$.