Solving the equation $\{x\}+\{\frac{1}{x} \}=1$ Solve the equation $$\large\{x\}+\left\{\dfrac{1}{x}\right\}=1$$
given $x\in\mathbb R\setminus \{0\}$ and $\{x\}$ denotes the fractional part of $x$.

$\{x\}+\{\frac{1}{x}\}=1$
$x+\frac{1}{x}=1+[x]+[\frac{1}{x}]$
$x+\frac{1}{x}$ is an integer.
let $n=x+\frac{1}{x}$
$n=1+[x]+[n-x]=1+[x]+n+[-x]$
$[x]+[-x]=-1$
$x\not\in\mathbb Z$
$n=x+\frac{1}{x}$
$x^2-nx+1=0$
since x is real, discriminant $\geq 0$
$n^2\geq 4$
$|n|\geq 2$
$|n|=2 \iff |x|=1 \implies x\in\mathbb Z$ , contradiction
when $n>2$ ,
$0 < \frac{n-\sqrt{n^2 - 4}}{2} < 1$ or $n-1 < \frac{n+\sqrt{n^2 - 4}}{2} < n$ , so $x\not\in\mathbb Z$
when $n<-2$ ,
$n < \frac{n-\sqrt{n^2 - 4}}{2} < n+1$ or $-1 < \frac{n+\sqrt{n^2 - 4}}{2} < 0$ , so $x\not\in\mathbb Z$
the solns are $x=\frac{n\pm\sqrt{n^2 - 4}}{2}$ , where $n\in\mathbb Z\setminus\{-2,-1,0,1,2\}$
