jasperE3 wrote: Solve the equation $\sqrt{1+\{2x\}}=\lfloor x^2\rfloor+2\lfloor x\rfloor+3$ (where $\{a\}=a-\lfloor a\rfloor$). This implies $\{2x\}$ is an integer, and so is $0$ So $x=\frac n2$ for some integer $n$ and $\left\lfloor\frac{n^2}4\right\rfloor+2\left\lfloor\frac n2\right\rfloor+2=0$ If $n=2m$, this is $m^2+2m+2=0$, without any solution If $n=2m+1$, this is $m^2+3m+2=0$ and so $m\in\{-2,-1\}$ And solutions $\boxed{x\in\left\{-\frac 32,-\frac 12\right\}}$