Solve the equation $$\left\lfloor\frac{x-2}{x+1}\right\rfloor=\bigg|\frac{2-x}{3}\bigg|.$$

My attempt

Let us consider the case where $x$ is an integer, such that $x=n$, then we have:



we get:

$$(n + 1) |n - 2| + 6 = 3 n \ \ \ \ ( n\ne-1)$$

This gives the solutions $n=2$ and $n=-4$

Now, assume that $x$ is not an integer, then we need to solve for $x=n+\delta$, where $\delta\in\langle0,1\rangle$:



$$3\bigg(1 - \left\lfloor\frac{3}{1+n+\delta}\right\rfloor\bigg)=\bigg|{2-n-\delta}\bigg|$$

Now put $\alpha=n+\delta$ and $\beta=-n-\delta$

$$3\bigg(1 - \left\lfloor\frac{3}{1+\alpha}\right\rfloor\bigg)=\bigg|{2+\beta}\bigg|$$



When we insert for $\alpha=n+\delta$, we obtain


With $0<\delta<1$ then we get $\beta= \langle -2, -5\rangle$. Since $\beta=-n-\delta$, then this gives that the solutions to the non-integer form of $x$ are all real numbers in the interval of $n=x=\langle 1,2\rangle$.

Since $\beta=-\alpha$ one could have tried to solve the problem with only the alpha variable. However, I wasn't able to extract $\alpha$ from the floor fraction. So calling them beta and alpha gave the opportunity to study each case separately. But here I wonder if it should have been solved also for the case of isolating $\alpha$?

  • 1
    $\begingroup$ Personally, my first line of attack on all such problems is to assume that $~x = P + r ~: ~P \in \Bbb{Z}, ~0 \leq r < 1.~$ Then, I simply solve for all satisfying ordered pairs $~(P,r).~$ $\endgroup$ Aug 7, 2023 at 12:21

2 Answers 2


Let us consider the case where x is an integer, such that x=n, then we have: $$ \frac{n-2}{n+1}=\bigg|\frac{2-n}{3}\bigg|$$

Here you actually assumed that $\frac{x-2}{x+1}$ is an integer as well, which simplifies situation. But with this assumption the rest seems OK, although I don't know why you write $n > 0$.

The part about $x$ not being an integer is unnecessasry, since $x$ must be an integer: \begin{align} \mathbb Z &\ni \left\lfloor\frac{x-2}{x+1}\right\rfloor=\bigg|\frac{2-x}{3}\bigg| \\ &\implies \qquad|2 - x| \in 3\mathbb Z \\&\implies \qquad x \in \mathbb Z \quad \land \quad x \operatorname{mod} 3 = 2. \end{align}

Here is another approach: We can substitute $x =3k + 2$ for unknown $k \in \mathbb Z$ to get $$ \left\lfloor\frac{3k}{3k+3}\right\rfloor=1+\left\lfloor-\frac{1}{k+1}\right\rfloor=|k| $$ On the left-hand side we have number which can only be $0$, $1$ or $2$, since $\left\lfloor-\frac{1}{k+1}\right\rfloor \in \{-1, 0, 1\}$ for any $k \in \ \mathbb Z\setminus \{-1\}$ (assumption $k \neq -1$ corresponds to $x \neq -1$). This gives finitely many possibilities $k \in \{-2, 0, 1, 2\}$ from which arise possible solutions $x \in \{-4, 2, 5, 8\}$. You can check one-by one which are actual solutions.

  • 1
    $\begingroup$ Esgeriath only asserts that the solutions of the equation must lie in $\{3k+2: k \in \{-2,0,1,2\}\}$ not that all four are solutions. In fact for $k\geq 0$ the equation $\lfloor 3k/(3k+3)\rfloor = |k|$ becomes $0=|k|$, hence $k=0$ or $x=2$ is the only solution with $k\geq 0$. Similarly if $k=-m<-1$ (as $k=-1$ if and only if $x=-1$) then the equation becomes $\lfloor m/(m-1)\rfloor=m$, and since $\lfloor m/(m-1)\rfloor \leq m/(m-1)\leq m$ with equality if and only if $m-1=1$, $k=-2$, or $x=-4$ is the only solution with $k<0$. $\endgroup$
    – krm2233
    Aug 7, 2023 at 12:13

$$\lfloor\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}\rfloor=\mid\frac{\mathrm{2}−{x}}{\mathrm{3}}\mid \\ $$ $$\mathrm{3}{n}={x}−\mathrm{2} \\ $$ $$\lfloor\frac{\mathrm{3}{n}}{\mathrm{3}{n}+\mathrm{3}}\rfloor=\pm{n} \\ $$ $$\lfloor\frac{-\mathrm{1}}{{n}+\mathrm{1}}\rfloor=\pm{n}−\mathrm{1}=\pm\mathrm{1} \\ $$ $${n}=\frac{\mathrm{1}\pm\mathrm{1}}{\pm\mathrm{1}}=\left\{-\mathrm{2},\mathrm{0},\mathrm{2}\right\} \\ $$ $${x}=\mathrm{2}+\mathrm{3}{n}=\left\{-\mathrm{4},\mathrm{2},\mathrm{8}\right\} \\ $$


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