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:
$$\frac{n-2}{n+1}=\bigg|\frac{2-n}{3}\bigg|$$.
$$3\frac{n-2}{n+1}=\bigg|2-n\bigg|$$
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 $\delta\in\langle0,1\rangle$:
$$\left\lfloor\frac{(n+\delta)-2}{(n+\delta)+1}\right\rfloor=\bigg|\frac{2-(n+\delta)}{3}\bigg|$$
$$3\left\lfloor\frac{(n+\delta)-2}{(n+\delta)+1}\right\rfloor=\bigg|{2-(n+\delta)}\bigg|$$
$$3\bigg(1 – \left\lfloor\frac{3}{1+n+\delta}\right\rfloor\bigg)=\bigg|{2-n-\delta}\bigg|$$
Now put $\alpha=n+\delta$ and $\beta=-n-\delta$
$$3\bigg(1 – \left\lfloor\frac{3}{1+\alpha}\right\rfloor\bigg)=\bigg|{2+\beta}\bigg|$$
$$3=\bigg|2+\beta\bigg|+3\left\lfloor\frac{3}{1+\alpha}\right\rfloor$$
$$\beta=1-3\left\lfloor\frac{3}{1+\alpha}\right\rfloor$$
When we insert for $\alpha=n+\delta$, we obtain
$$\beta=1-3\left\lfloor\frac{3}{1+n+\delta}\right\rfloor.$$
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 $\alpha$?
Best Answer
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.