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 $x$.
 A: Hint: We have $\{x\} + \{1/x\} = 1$ if and only if $x$ and $1/x$ are not integers and, for some $n \in \Bbb Z$, we have
$x + 1/x = n$, which is to say $x \notin \Bbb Z$, $1/x \notin \Bbb Z$, and there is an $n \in \Bbb Z$ such that
$$
x^2 - nx + 1 = 0
$$
A: To expand on Omnom's answer, we can complete the square (or use the quadratic formula) and arrive at a general solution:
\begin{align}
\left(x- \frac{n}{2}\right)^2 &= \frac{n^2}{4} - 1\\
x - \frac{n}{2} &= \pm \sqrt{\frac{n^2}{4} - 1}\\
x &= \frac{n \pm \sqrt{n^2 - 4}}{2}
\end{align}
$x$ is non-integer and real when $|n| > 2$. Indeed, when $|n| > 2$, the difference between two square roots is at least $5$, so $n^2 - 4$ may not be a perfect square, thus $x$ is irrational and non-integral.
Finally, I remark about the $\pm$ sign. You will find that taking the reciprocal of the plus answer will give you the minus answer and vice versa, so that the sum $x + \frac{1}{x}$ is the same.
A: $\{x\}+\{\frac{1}{x}\}=1$
$x+\frac{1}{x}=1+[x]+[\frac{1}{x}]$
$x+\frac{1}{x}$ 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 $n\in\mathbb Z\setminus\{-2,-1,0,1,2\}$
