75th Romanian Mathematical Olympiad

First, we have that $x\in \mathbb{R}\setminus (\mathbb{Z}\cup [0,1))$. The given equation is equivalent to: $$\frac{[x]+\{x\}}{[x]\{x\}}=-\frac{1}{x}\iff -x^2=[x]\{x\}.(\ast )$$ Since $-x^2\le 0$ and $x\ne 0$, we have $[x]\{x\}<0$, hence $[x]\le -1$. Because $[x]=-k$, with $k\in {\mathbb{N}}^{\ast }$, equation $(\ast )$ becomes $-x^2=-k(k+x)$, which is equivalent to $x^2-kx-k^2=0$, having solutions $x_{1,2}=\frac{k\pm k\sqrt{5}}{2}$. From $[x]=-k$, we obtain that the only possible solution is $x_1=\frac{k-k\sqrt{5}}{2}$. Finally, we must find the values of $k\in {\mathbb{N}}^{\ast }$ for which $\left[\frac{k-k\sqrt{5}}{2}\right]=-k$, which is equivalent to: $-k\le k\frac{1-\sqrt{5}}{2}<-k+1\iff 3k\ge k\sqrt{5}>3k-2$, from which we obtain $k<\frac{3+\sqrt{5}}{2}$, i.e., $k\in \{1,2\}$. From here we obtain the solutions $x_1^{\ast }=\frac{1-\sqrt{5}}{2}$ and $x_2^{\ast }=1-\sqrt{5}$, which satisfy the given equation.