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?

• Hints: $x^2-x-2=(x-1/2)^2-\dots$. Draw the graphs of these functions. Commented Jan 4, 2015 at 23:24
• I wanted to know the exact solution, not just a guess from the graph... Commented Jan 4, 2015 at 23:30
• It's math.SE policy to require that you give more context and/or show some of your work when asking questions. What have you tried? Did any of the help you got on your last question give you ideas for this one?
– aes
Commented Jan 4, 2015 at 23:32
• A "guess from the graph" will tell you what region the floors are in, permitting you to solve the equation algebraically and get the exact answer. Commented Jan 4, 2015 at 23:37
• See useful techniques. Commented Jan 5, 2015 at 0:15

Your system of equations isn't quite right (for example, $x$ may be the smaller of the two).

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 $\epsilon$).

$$\left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor \tag{1.}$$

$$\left \lfloor y \right \rfloor = \left \lfloor x \right \rfloor \iff 0 \le y-x < 1$$ 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.

1. If $$x \lt -1$$, then $$y-x \gt 1$$.
2. If $$0\le x \le 1$$, then $$-2 \ge y-x \ge -3$$.
3. If $$1\le x \le 2$$, then $$-3 \le y-x \le -2$$.
4. 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

$$x \in \left(-1, -\dfrac{\sqrt 5 - 1}{2} \right] \cup \left[\dfrac{1+\sqrt{17}}{2}, \dfrac{1+\sqrt{21}}{2} \right)$$