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 $x$.

Solution 1:

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 $$ x^2 - nx + 1 = 0 $$

Solution 2:

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.