When is $\{x\}=\frac{1}{x}$?

Question

Let $\{x\}$ denote the fractional part of $x$. For example, $\{1.23\}=0.23$. I’m looking for positive real solutions to $\{x\}=\frac{1}{x}$. Right off the bat, I recognised that $\phi=\frac{1+\sqrt5}{2}$ is a solution, as $\phi-1=\frac{1}{\phi}$. I tried Wolfram Alpha, and it displays a few solutions before the page fully loads in, but after the page loads fully, the only solution it displays is $x\approx-12.0827625302982$. $\phi$ is one of the solutions Wolfram Alpha displays before the page loads in. Does anyone know what the other solutions are, and how to find them?

Answer 1

Simply from the graph of $\frac1x$ and $\{x\}$, it should be clear that, on every interval $(n, n+1]$ for $n\in\mathbb N$, there is exactly one solution to the equation $\frac1x = \{x\}$. *

Now, let's focus on one of those intervals. On that interval, we can write $x$ as $x=n + \{x\}$ for some $n$, then you see that in order for $\{x\} = \frac1x$ to hold, the following must hold:

$$\{x\} = \frac1x\\ x-n = \frac 1x\\ x^2-nx - 1 = 0.$$

---

* in fact, this statement can quite easily be shown to be true. For $n=1$, equality is achieved at $x=\phi$, as you showed. For $n\geq 2$, the equality is achieved somewhere inside the interval $[n, n+\frac12]$:

Let $f(x)=\frac1x$ and $g(x)=\{x\}$. Note that $f$ is a decreasing function on $(n,n+1)$, with $\frac{1}{n}>f(x)>\frac{1}{n+1}$ for all $x\in (n,n+1)$. Also, note that $g(n) = 0 < f(n)$, and $g(n+\frac12) = \frac12 > f(x)$. Also note that on $[n,n+\frac12]$, the functions $f$ and $g$ are continuous, so by the bisection theorem, they must intersect.