Konigsberg wrote: Find all real numbers $x$ such that $\lfloor x^2-2x \rfloor+2\lfloor x \rfloor=\lfloor x \rfloor^2$. (For a real number $x$, $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$.) Setting $x=n+y$ with $n\in\mathbb Z$ and $y\in[0,1)$, equation becomes $\lfloor y^2+2(n-1)y \rfloor=0$ $y=0$ is always a solution. If $y>0$, equation is $y\ge -2(n-1)$ and $y^2+2(n-1)y-1<0$ If $n<1$, no solution (first inequality is always false) If $n=1$, any $y\in(0,1)$ is solution If $n>1$, first inequality is always true and second becomes $y\in(0,\sqrt{n^2-2n+2}-n+1)$ Hence the result : $\boxed{x\in\mathbb Z_{\le 0}\cup\left(\bigcup_{n=1}^{+\infty}[n,1+\sqrt{n^2-2n+2})\right)}$