Romanian Mathematical Olympiad - Final Round

a) The discriminant of the equation is
$$\Delta = (a+b-1)^2 - 4(ab-a-b) = (a-b)^2 + 2a + 2b + 1 > 0,$$

b) If $x_1 < x_2$ are the two solutions of the equation, from Vi\`ete's first relation, $x_1 + x_2 = 1 - a - b \in \mathbb{Z}$, so $x_1 \in \mathbb{Z} \iff x_2 \in \mathbb{Z}$.
Let $f(x) = x^2 + (a+b-1)x + ab - a - b$, $x \in \mathbb{R}$. Since $f(1) = ab > 0$, $f(-a) = -b < 0$, $f(-b) = -a < 0$, using the sign of the quadratic function we obtain $x_1 < -b \leq -a < x_2 < 1$, so both solutions are non positive.
Thus, $a \geq 1 - x_2$, $b \geq 1 - x_2$ and since $f(x_2) = 0$, we obtain $(a - 1 + x_2) \cdot (b - 1 + x_2) = 1 - x_2 \geq 1$. We get that $d_1 = a - 1 + x_2$ and $d_2 = b - 1 + x_2$ are natural divisors of $1 - x_2$ and $d_1 d_2 = 1 - x_2$, $a = d_1 + d_1 d_2$, $b = d_2 + d_1 d_2$. We obtain $2a - b = 2d_1 + d_2(d_1 - 1) \geq 2d_1 > 0$, thus $b < 2a$.