58th Ukrainian National Mathematical Olympiad

It follows from the given condition that $x\ne 0$, therefore, equation can be rewritten as $2+\frac{3y}{2x}=\left[1+\frac{y^2}{x^2}\right]$. Let $k=\frac{3y}{2x}$ is an integer. Since $[1+\frac{y^2}{x^2}]=1+\frac{y^2}{x^2}$, the following holds: $$1+k=\left[\frac{4k^2}{9}\right]\iff 1+k\le \frac{4k^2}{9}<2+k.$$ $$\{\begin{array}{l}4k^2-9k-9\ge 0,\\ 4k^2-9k-18<0,\end{array}$$ $$\left(\frac{9-3\sqrt{41}}{8};-\frac{3}{4}\right]\cup \left[3;\frac{9+3\sqrt{41}}{8}\right).$$ Therefore, $k=-1$ or $k=3$. So, $\frac{3y}{2x}=-1$ or $\frac{3y}{2x}=3$, and thus, all possible solutions are in the answer.