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

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 $0\le f<1$).
