Find Integer Values for 1999 Distinct Real Solutions

[anemone]

Find all the integer values of $m$ for which the equation $\left\lfloor \dfrac{m^2x-13}{1999}\right\rfloor=\dfrac{x-12}{2000}$ has 1999 distinct real solutions.

 

[jvicens]

After analyzing the given equation, it can be rewritten as:
$$\left\lfloor \dfrac{m^2x-13}{1999}\right\rfloor = \dfrac{x-12}{2000}$$
$$\dfrac{m^2x-13}{1999} - 1 < \dfrac{x-12}{2000} \leq \dfrac{m^2x-13}{1999}$$
$$\dfrac{m^2x-12}{1999} < \dfrac{x-12}{2000} \leq \dfrac{m^2x-11}{1999}$$
$$\dfrac{2000m^2x-24000}{1999} < x \leq \dfrac{2000m^2x-22000}{1999}$$
$$\dfrac{2000m^2x-22000}{1999} - \dfrac{2000m^2x-24000}{1999} = \dfrac{2000m^2}{1999} > 0$$

Since $\dfrac{2000m^2}{1999}$ is always positive, the given equation will have 1999 distinct real solutions if and only if the range of $x$ is greater than 0. This means that the lower bound of $x$ must be greater than 0, which can be represented as:
$$\dfrac{2000m^2x-24000}{1999} > 0$$
$$2000m^2x - 24000 > 0$$
$$m^2x > 12$$

Since $m$ is an integer, we can rewrite the inequality as:
$$m^2x \geq 13$$

This means that for the given equation to have 1999 distinct real solutions, $m$ must satisfy the following conditions:
$$m^2 \geq 13$$
$$m \geq \sqrt{13}$$

Therefore, all the integer values of $m$ for which the equation has 1999 distinct real solutions are:
$$m = \sqrt{13} + 1, \sqrt{13} + 2, \sqrt{13} + 3, ...$$

In other words, any integer greater than or equal to $\