Number of solutions of $x+\frac{87}x=\lfloor x\rfloor+\frac{87}{\lfloor x\rfloor} $ where $x\in\mathbb{Q} , x\notin\mathbb{Z}$

Question

How many solutions the equation $x+\frac{87}x=\lfloor x\rfloor+\frac{87}{\lfloor x\rfloor} $ has in rational non integer sets of numbers?

$1)\text{zero}\quad\quad\quad\quad\quad\quad\quad2)1\quad\quad\quad\quad\quad\quad\quad3 )2\quad\quad\quad\quad\quad\quad\quad4)87\quad\quad\quad\quad\quad\quad\quad5)\text{infinity}$

So the solution can't be an integer therefore $\lfloor x\rfloor\ne x$ and generally we can say $\lfloor x\rfloor< x$. also denominator of the fractions can't be zero so $\lfloor x\rfloor \neq0$ hence $x\notin[0,1)$

I wrote the equation as:

$$x-\lfloor x\rfloor=\frac{87}{\lfloor x \rfloor}-\frac{87}{x}=\frac{87(x-\lfloor x\rfloor)}{x\lfloor x\rfloor}$$

But can't see a way to continue.

Answer 1

You're on the right track.

Hint 1: Divide both sides by $(x - \lfloor x \rfloor)$

Hint 2: Note that for $\lfloor x\rfloor \geq 0$, $$ \lfloor x\rfloor^2 \leq \lfloor x\rfloor x \leq \lfloor x\rfloor(\lfloor x\rfloor + 1). $$

Answer 2

$$x+\frac{87}x=\lfloor x\rfloor+\frac{87}{\lfloor x\rfloor}$$

Let $x = n + \delta$ where $x \in \Bbb Z$ and $\delta \in (0,1)$

$$n+\frac{87}n=(n+\delta)+\frac{87}{n+\delta}$$

$$(n+\delta)(n^2 + 87) = n((n+\delta)^2 + 87) $$

$$n^3 + \delta n^2 + 87n + 87 \delta = n^3 + 2 \delta n^2 + \delta^2 n + 87n$$

$$\delta n^2 + \delta^2 n - 87\delta = 0$$

$$\delta n = 87-n^2$$

If $n > 0$

$$ 0 \lt 87 - n^2 < n$$

$$n \in \{1,2,\dots, 9\} \wedge n \in \{9, 10, \dots \}$$ $$ n = 9$$

If $n < 0$

$$ n < 87-n^2 < 0$$

$$n \in \{-9, -8, \dots, -1\} \wedge n \in \{\dots -12,-11, -10\}$$

$$n \in \emptyset$$

For $n = 9$ we get $\delta = \dfrac{87-n^2}{n} = \dfrac 23$.

So the only solution is $x = 9\frac 23 = \frac{29}{3}$.