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?
 A: In general, to solve $\lfloor a \rfloor + \lfloor b \rfloor =c$ you can consider the pairs of values that $a$ and $b$ can take.
In this case your possibilities are very small: $a=0, b=1$ or $a=1, b=0$.
Can you proceed from there?
A: Hints:

*

*For $x>0$ and $x\neq 1$, one of the two values, $x$ or $\frac1x$, is between $0$ and $1$, so the floor of that value will be $0$.

*If $x$ is a solution, then so is $\frac1x$. This means you can first only focus on $x>1$, to get half of the solutions, then know that this is just half of the solutions.

A: 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 .
A: We can rule out $x<0$. If $x\geqslant 2$, then $1-\lfloor x \rfloor \leqslant -1$, which is problematic. If $x\leqslant 1/2$, then $\lfloor \frac{1}{x} \rfloor = 1-\lfloor x \rfloor$ is again impossible. For any $x\in (1/2, 2)\setminus \{1\}$ we have
$$ \left\lfloor x \right\rfloor + \left\lfloor\frac{1}{x} \right\rfloor =1. $$
