Solving the Floor Function Equation

[juantheron]

Solve the equation $\mid x \mid+\sqrt{\lfloor x \rfloor+\sqrt{1+\{x\}}} = 1$

where $\lfloor x \rfloor = $ floor function

 

[Evgeny.Makarov]

Obviously, |x| <= 1, and x = 0 is one solution. Further, {x} >= 0 for all x, so x cannot be positive. If -1 <= x < 0, then |x| = -x, floor(x) = -1 and {x} = x + 1. This gives an equation that has two solutions: one simple and one complicated.