Belarusian Mathematical Olympiad

The distance between the parabola $y = ax^2 + bx + c$ and the axis of abscissae is equal to
$$d = \frac{4ac - b^2}{4a}$$
Therefore, the problem is equivalent to the following problem: find the greatest and the smallest value of the expression above if positive integers $a$, $b$, $c$ satisfy the inequalities $1 \le a \le 50$, $1 \le b \le 50$, $1 \le c \le 50$ and $b^2 - 4ac < 0$.

We find the greatest value. We have
$$d = \frac{4ac - b^2}{4a} = c - \frac{b^2}{4a} \le 50 - \frac{b^2}{4a} \le 50 - \frac{1}{4a} \le 50 - \frac{1}{200}.$$
The first inequality follows from the inequality $c \le 50$, the second one follows from the inequalities $a \le 50$ and $b^2 \ge 1$. We see that all inequalities become equalities for $c = 50$, $b = 1$ and $a = 50$. Moreover, for these $a$, $b$, $c$ the inequality $b^2 - 4ac = 1^2 - 4 \cdot 50 \cdot 50 < 0$ is valid, too. Therefore the required greatest value is equal to $50 - 1/200$.

Now we find the smallest value. In the fraction above the numerator and the denominator are positive integers. We find the smallest possible value of the numerator. The numerator can be neither $1$ nor $2$. Indeed, in these cases we have either $b^2 = 4ac - 1$ or $b^2 = 4ac - 2$, i.e. $b^2 = 4(ac - 1) + 3$ or $b^2 = 4(ac - 1) + 2$. It follows that $b^2$ is congruent to either $2$ or $3$ modulo $4$, which is impossible.

So the smallest value of the numerator is $3$, i.e. $b^2 = 4ac - 3$. Now our next goal is to determine the greatest possible value of the denominator. From the equality $b^2 = 4ac - 3$ it follows that $b$ is odd, i.e. $b = 2k + 1$ for some nonnegative integer $k$. So $(2k+1)^2 = 4ac - 3$, i.e. $k^2 + k + 1 = ac$. It follows that $a$ and $c$ are odd. Thus the possible values of $a$ are $a = 49, 47, 45, \dots$. We verify these values (starting with the greatest one).

Set $a = 49$. We have $c = \frac{b^2+3}{4a} \le \frac{2503}{4 \cdot 49} < 13$. Now we must verify whether $4 \cdot 49 \cdot c - 3 = 196c - 3$ is a perfect square for some of $c = 1, 3, 5, 7, 9, 11$. First, for $c = 1, 5, 11$ the last digit of the number $196c - 3$ is either $3$ or $7$, which is impossible for a perfect square. Further, for $c = 3, 7$ and $9$ we respectively have $196c - 3 = 585, 1369$ and $1671$. We see that $1369 = 37^2$, therefore for $c = 7$, $b = 37$ and $a = 49$ the value of $d$ is equal to $\frac{3}{4 \cdot 49} = \frac{3}{196}$.

Hence, if the denominator is equal to $3$, then the smallest value of $d$ is $3/196$. If the denominator is no smaller than $4$, then we have $d = \frac{4ac - b^2}{4a} \ge \frac{4}{4 \cdot 50} = \frac{1}{50} > \frac{3}{196}$. Therefore, the required smallest value is $\frac{3}{196}$.