Find all functions satisfying $f(x+1)=\frac{f(x)-5}{f(x)-3}$ Find all functions satisfying
 $$f(x+1)=\frac{f(x)-5}{f(x)-3}$$
My try:
We have $$f(x+1)=1-\frac{2}{f(x)-3}$$
Letting $g(x) =f(x+1)-3$
We get $$g(x+1)=-2-\frac{2}{g(x)}$$
Any clue here? 
 A: The hint.
Prove that:
$$f(x+4)=f(x).$$
A: Let $f(x)=g(x)+3$ ,
Then $g(x+1)+3=\dfrac{g(x)+3-5}{g(x)+3-3}$
$g(x+1)+3=\dfrac{g(x)-2}{g(x)}$
$g(x+1)+3=1-\dfrac{2}{g(x)}$
$g(x+1)=-2-\dfrac{2}{g(x)}$
Let $g(x)=\dfrac{h(x+1)}{h(x)}$ ,
Then $\dfrac{h(x+2)}{h(x+1)}=-2-\dfrac{2h(x)}{h(x+1)}$
$\dfrac{h(x+2)}{h(x+1)}=-\dfrac{2h(x+1)+2h(x)}{h(x+1)}$
$h(x+2)+2h(x+1)+2h(x)=0$
$h(x)=\theta_1(x)(-1+i)^x+\theta_2(x)(-1-i)^x$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period
$h(x)=\theta_1(x)e^{x\ln(-1+i)}+\theta_2(x)e^{x\ln(-1-i)}$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period
$h(x)=\theta_1(x)e^{\frac{x\ln2}{2}+\frac{3i\pi x}{4}}+\theta_2(x)e^{\frac{x\ln2}{2}-\frac{3i\pi x}{4}}$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period
$h(x)=\Theta_1(x)2^\frac{x}{2}\sin\dfrac{3\pi x}{4}+\Theta_2(x)2^\frac{x}{2}\cos\dfrac{3\pi x}{4}$ , where $\Theta_1(x)$ and $\Theta_2(x)$ are arbitrary periodic functions with unit period
$\therefore f(x)=\dfrac{\Theta_1(x+1)2^\frac{x+1}{2}\sin\dfrac{3\pi(x+1)}{4}+\Theta_2(x+1)2^\frac{x+1}{2}\cos\dfrac{3\pi(x+1)}{4}}{\Theta_1(x)2^\frac{x}{2}\sin\dfrac{3\pi x}{4}+\Theta_2(x)2^\frac{x}{2}\cos\dfrac{3\pi x}{4}}+3$ , where $\Theta_1(x)$ and $\Theta_2(x)$ are arbitrary periodic functions with unit period
$f(x)=\dfrac{\sqrt2\sin\dfrac{3\pi(x+1)}{4}+\Theta(x)\sqrt2\cos\dfrac{3\pi(x+1)}{4}}{\sin\dfrac{3\pi x}{4}+\Theta(x)\cos\dfrac{3\pi x}{4}}+3$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
