Number of solutions of $\left\{x\right\}+\left\{\frac{1}{x}\right\}=1$ Find the number of solutions of $$\left\{x\right\}+\left\{\frac{1}{x}\right\}=1,$$ where $\left\{\cdot\right\}$ denotes Fractional part of real number $x$.
My try:
When $x \gt 1$ we get
$$\left\{x\right\}+\frac{1}{x}=1$$ $\implies$
$$\left\{x\right\}=1-\frac{1}{x}.$$
Letting $x=n+f$, where $n \in \mathbb{Z^+}$ and $ 0 \lt f \lt 1$, we get
$$f=1-\frac{1}{n+f}.$$
By Hint given by $J.G$, i am continuing the solution:
we have
$$f^2+(n-1)f+1-n=0$$ solving we get
$$f=\frac{-(n-1)+\sqrt{(n+3)(n-1)}}{2}$$ $\implies$
$$f=\frac{\left(\sqrt{n+3}-\sqrt{n-1}\right)\sqrt{n-1}}{2}$$
Now obviously $n \ne 1$ for if  we get $f=0$
So $n=2,3,4,5...$ gives values of $f$ as
$\frac{\sqrt{5}-1}{2}$, $\sqrt{3}-1$, so on which gives infinite solutions.

Let $x:=n+f$. The equation is
$$f+\frac1{n+f}=1,$$ 
giving the solutions in $f$
$$f=\frac{\pm\sqrt{(n+1)^2-4}-n+1}2.$$
The negative sign cannot work, nor the negative $n$. Then $n\ge1$ is required, but $n=1$ yields $x=1$, which is wrong. Finally,
$$f=\frac{\sqrt{(n+1)^2-4}-n+1}2, \forall n>1.$$
