Solve for $x$ in the equation containing ${\lfloor{x}\rfloor}$ and $\{x\}$

Question

Calculate all possible values of $x$ satisfying,

$$\frac{\lfloor{x}\rfloor}{\lfloor{x-2}\rfloor}-\frac{\lfloor{x-2}\rfloor}{\lfloor{x}\rfloor}=\frac{8\{x\}+12}{\lfloor{x}\rfloor \lfloor{x-2}\rfloor}$$

where $\{x\}$ stands for fractional part of $x$.

My Attempt:

$${\lfloor{x}\rfloor}^2-{\lfloor{x-2}\rfloor}^2={8\{x\}+12}$$

${\lfloor{x}\rfloor}^2-{\lfloor{x-2}\rfloor}^2$ is an integer, so ${8\{x\}+12}$ must also be an integer, i.e. $8\{x\}$ must be an integer.

As, $0\leq{\{x\}}\lt{1}$, Therefore the only integer values of $8\{x\}$ will be,

$$\{x\}=0 \implies 8\{x\}=0 \implies 8\{x\}+12=12$$

$$\{x\}=\frac{1}{2} \implies 8\{x\}=4 \implies 8\{x\}+12=16$$

$$\{x\}=\frac{1}{4} \implies 8\{x\}=2 \implies 8\{x\}+12=14$$

$$\{x\}=\frac{1}{8} \implies 8\{x\}=1 \implies 8\{x\}+12=13$$

On calculating , we get that $12$,$16$ can be expressed as,

$$16=5^2-3^2 \implies 5\leq{x}\lt 6$$

but $\{x\}=\frac{1}{2}, \therefore x=5+\frac{1}{2}=\frac{11}{2}$

$$12=4^2-2^2 \implies 4\leq{x}\lt 5 $$

but $\{x\}=0, \therefore x=4+0=4$

Therefore the only solution set is,

$$\bigg\{4,\frac{11}{2}\bigg\}$$

Is this the correct approach? I am afraid that I am missing some values.

Answer 1

hint

Observe that

$$\lfloor x-2 \rfloor =\lfloor x\rfloor -2$$

the equation becomes

$$\lfloor x \rfloor=2\{x\}+4$$

Answer 2

Decompose $x=n+f$. Then

$$\frac n{n-2}-\frac{n-2}n=\frac{8f+12}{n(n-2)}$$ or

$$n^2-(n-2)^2=8f+12$$

or

$$f=\frac{n-4}2.$$

The possible values for $n$ are $4$ and $5$ (so that $0\le f<1$).

Answer 3

From Fatima's approach you should have something like $ (\lfloor x \rfloor-4)/2=\{x\}$. So the only solution to this $\lfloor x \rfloor=4$ and $\lfloor x \rfloor=5$, with $\{x\}=0$ and $\{x\}=1/2$, respectively.