How to solve $ \left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor $ I need some help to solve the next equation:
$$ \left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor $$
Where $ \left \lfloor \cdot \right \rfloor $ is the floor function.
What I've tried:
$$ x^2 - x - 2 - x < 1 $$
$$ x^2 - x - 2 \leq x < x^2 - x - 1 $$
When I try to solve this system, I don't get the right solution. Is it right what I'm doing? How can I do to solve this?
 A: Your system of equations isn't quite right (for example, $x$ may be the smaller of the two).
Here's one algebraic way to go about this:
To solve $\lfloor f(x) \rfloor = \lfloor x \rfloor$ write $x = n + \epsilon$ where $0\leq \epsilon <1$ and $n$ is an integer so that $\lfloor x\rfloor = n$.
Then the condition $\lfloor f(x)\rfloor = n$ is $n \leq f(n+\epsilon) < n+1$ (and the above conditions on $n$ and $\epsilon$).
A: $$\left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor \tag{1.}$$
Let's start with the observation that 
$\left \lfloor y \right \rfloor = \left \lfloor x \right \rfloor
  \iff 0 \le y-x < 1$ or $-1 \lt y-x \le 0$
The proof follows from the observation that, for any integer n,
$$ n \le x \le y < n+1 \iff n-x \le 0 \le y-x < 1  $$
Lets let $y = x^2-x-2$. One case where $(1.)$ must be true is when
\begin{align}
   y &= x \\
   x^2-x-2 &= x \\
   x^2-2x-2 &= 0 \\
   x &= 1 \pm \sqrt 3
\end{align}
In this case, $\left \lfloor y \right \rfloor=\left \lfloor x \right \rfloor = -1$
 or   $\left \lfloor y \right \rfloor =\left \lfloor x \right \rfloor = 2$.
Note that $y-x = x^2-2x-2$ is a decreasing function for $x \in (-\infty, -1]$ and an increasing function for $x \in [-1,-\infty)$. Note also that 
$$y-x = (x-1)^2 - 3$$ and $$x-y = 3-(x-1)^2$$
We consider four cases.


*

*If $x \lt -1$, then $y-x \gt 1$.

*If $0\le x \le 1$, then $-2 \ge y-x \ge -3$.

*If $1\le x \le 2$, then  $-3 \le y-x \le -2$.

*If $x \gt 3$, then $y-x \gt 1$.


In all four cases, $\left \lfloor x^2 - x - 2 \right \rfloor \ne \left \lfloor x \right \rfloor $
That leaves us with two possibilities.
If $\left \lfloor x \right \rfloor=\left \lfloor y \right \rfloor = -1$, then the possible values for $x$ are $x=-1 + h$ where $0 \le h < 1$.
Then we need to solve
\begin{array}{c}
   && \left \lfloor x^2 - x - 2 \right \rfloor &= &\left \lfloor x \right \rfloor \\
   && \left \lfloor h^2-3h \right \rfloor &=& -1 \\
   && \left \lfloor-1+(h^2-3h+1) \right \rfloor &=& -1 \\
   0  &\le  &h^2-3h+1 &<& 1 \\
   -1  &\le  &h^2-3h &<& 0 \\
   \dfrac 54  &\le &h^2-3h+\dfrac 94 &<& \dfrac{9}{4} \\
   \dfrac 54  &\le &\left(h-\dfrac 32\right)^2 &<& \dfrac{9}{4} \\
   \dfrac{\sqrt 5}{2}  &\le &\dfrac 32 - h &<& \dfrac 32 \\
   \dfrac{\sqrt 5-3}{2}  &\le& -h &<& 0 \\
   0 &<& h &\le& \dfrac{3 - \sqrt 5}{2} \\
   -1 &<& x &\le& -1 + \dfrac{3 - \sqrt 5}{2}\\
   -1 &<& x &\le& \dfrac{1 - \sqrt 5}{2}\\
\end{array}
If $\left \lfloor x \right \rfloor=\left \lfloor y \right \rfloor = 2$, then the possible values for $x$ are $x=2 + h$ where $0 \le h < 1$.
\begin{array}{c}
   && \left \lfloor x^2 - x - 2 \right \rfloor &= &\left \lfloor x \right \rfloor \\
   && \left \lfloor h^2+3h \right \rfloor &=& 2 \\
   && \left \lfloor 2+(h^2+3h-2) \right \rfloor &=& 2 \\
   0  &\le  &h^2+3h-2 &<& 1 \\
   2  &\le  &h^2+3h &<& 3 \\
   \dfrac{17}{4}  &\le &h^2+3h+\dfrac 94 &<& \dfrac{21}{4} \\
   \dfrac{17}{4}  &\le &\left(h+\dfrac 32\right)^2 &<& \dfrac{21}{4} \\
   \dfrac{\sqrt{17}}{2}  &\le& h+\dfrac 32 &<& \dfrac{\sqrt{21}}{2} \\
   \dfrac{\sqrt{17}-3}{2}  &\le& h &<& \dfrac{\sqrt{21}-3}{2}\\
   2 + \dfrac{\sqrt{17}-3}{2}  &\le& x &<& 2+\dfrac{\sqrt{21}-3}{2} \\
   \dfrac{1+\sqrt{17}}{2}  &\le& x &<& \dfrac{1+\sqrt{21}}{2}
\end{array}
So the solution set is
$$x \in \left(-1, -\dfrac{\sqrt 5 - 1}{2} \right] \cup 
  \left[\dfrac{1+\sqrt{17}}{2}, \dfrac{1+\sqrt{21}}{2} \right)$$
