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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?

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    $\begingroup$ Clearly the value of $f(0)$ determines inductively all the values of $f(x)$ over the integers. Now, try to assign some values to $f(0)$ and see what happens. $\endgroup$ – Crostul Jan 2 at 9:35
  • $\begingroup$ And observe that you can select the value of $f(t)$, $t\in[0,1)$ any which way you want. Only if you also require continuity is there something to worry. $\endgroup$ – Jyrki Lahtonen Jan 2 at 9:58
  • $\begingroup$ Why did you tag polynomials ? $\endgroup$ – Claude Leibovici Jan 2 at 10:10
  • $\begingroup$ Yes according to the latest information, non constant polynomial cannot be periodic. I will edit it thanks $\endgroup$ – Ekaveera Kumar Sharma Jan 2 at 10:17
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The hint.

Prove that: $$f(x+4)=f(x).$$

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  • $\begingroup$ But not all functions of period $4$ are solution of the given equation, are they? If not, then you have not answered the question . $\endgroup$ – Kavi Rama Murthy Jan 2 at 12:20
  • $\begingroup$ @Kavi Rama Murthy It was the hint. $f(x)\notin\{0,1,2,3\}$ of course. If you want, you can write a full solution. $\endgroup$ – Michael Rozenberg Jan 2 at 12:22
  • $\begingroup$ Yes one of the functions that satisfy is $f(x)=\tan\left(\frac{\pi x}{4}\right)+2$ where $x \ne 4p-2$ ,$\forall $, $ p \in \mathbb{Z}$. Obviously Cot also satisfies. I will be happy if there are any other functions $\endgroup$ – Umesh shankar Mar 4 at 16:49
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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

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