How to solve $\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$? I am stuck with this equation. All I could do is this:
$\lfloor x \rfloor$ = $\lfloor n + m \rfloor$ such that $n \in N$ and $m<1$.
We get:
$\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$
$\lfloor n + m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + \lfloor m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + 0 + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + \lfloor \frac{1}{n+m} \rfloor = 1$
From here on I have no idea what to do!
Edit: It is easy to see that any value $1<x<2$ satisfies the equation, but can I find all the solutions?

First, we can notice that for $
x>1,$ $\lfloor \frac{1}{x} \rfloor=0$
Then x solution and $ x>1$  if and only if $ x$ strictly between 1 and 2
for $0<x<1 : \lfloor x \rfloor=0$ , then $x$ solution if and only if $\lfloor\frac{1}{x} \rfloor=1$ which can easily be solved using the definition of the floor function , then
$x$ solution if and only if  $x$ strictly between $0.5 $ and $1$
Clearly if $x$ negative x is not a solution .
