Functions!

Note that $F(x) = x - \lfloor x \rfloor = \{x\}$ , the fractional part of $x$ ; $\implies F(x) + \dfrac 1{F(x)} = \{x\} + \dfrac 1{\{x\}}$ . Since $\{x\}, \dfrac 1{\{x\}} >0$ , we can apply AM-GM inequality . And $\{x\} + \dfrac 1{\{x\}} \ge 2$ , that is $F(x) + \dfrac 1{F(x)}$ has a minimum value of 2 and never equal to 1, therefore the number of solution is $\boxed{0}$ .

Any equation of positive real variable x of the form x+1/x has a minimum value of 2. Thus there are no solutions of F(x)+1/F(x)=1.

Rearrange $F(x) + \frac{1}{F(x)} = 1$ to obtain $F(x)^2 - F(x) + 1 = 0$ which does not have integer solutions.