Brazilian Mathematical Olympiad

If $A = 0$, then we have $B \lfloor x \rfloor = B \lfloor y \rfloor$ which obviously has infinitely many solutions with $x \neq y$. So assume $A \neq 0$. Then we can write the equation as $\frac{B}{A}(\lfloor x \rfloor - \lfloor y \rfloor) = y - x$. If $x \neq y$, we can assume $x < y$. We cannot have $\lfloor x \rfloor = \lfloor y \rfloor$, since $y - x > 0$, so $\lfloor x \rfloor < \lfloor y \rfloor$. Write $x = n - \epsilon$, $y = n + d + \delta$, where $n$ is any integer, $d$ is a non-negative integer, $\delta \in [0, 1)$ and $\epsilon \in (0, 1]$. So $\frac{B}{A} = -\frac{d+\nu}{d+1}$, where $d$ is a non-negative integer and $\nu = \delta + \epsilon \in (0, 2)$.

We note first that $-2 < -\frac{d+\nu}{d+1} < 0$. So if there are solutions with $x \neq y$ and $A \neq 0$, then $\frac{B}{A}$ must belong to the open interval $(-2, 0)$.

Conversely, by taking $d = 0$ and suitable $\nu$ it is clear that $\frac{B}{A}$ can take any value in $(-2, 0)$. Thus a necessary and sufficient condition for solutions with $x \neq y$ is $A = 0$ or $\frac{B}{A} \in (-2, 0)$.