Functiοnal Εquation in Real Line

Question

Find all functions $f:\mathbb N\rightarrow\mathbb R $ such that $f(n)f(n+2)+f(n+1)=1$ for all positive integers $n\geq 1$.

My Work

I put $n=1,2$: $~f(1)f(3)+f(2)=1=f(2)f(4)+f(3)~$ so $~f(2)(f(4)-1)-f(3)(f(1)-1)=0$, but I can find anything.

Answer 1

Hint:

Starting with $n=0$, $f_0=a$, $f_1=b$, we can observe that the sequence $f_n$, \begin{align} f_{n+2}&=\frac{1-f_{n+1}}{f_n} , \end{align}

is periodic with period 5, $f(n+5)=f(n)$:

\begin{align} f_2&=\frac{1-b}a ,\\ f_3&=\frac{a+b-1}{ab} ,\\ f_4&=\frac{1-a}{b} ,\\ f_5&=a ,\\ f_6&=b ,\\ \dots \end{align}

The simplest solution would be a constant function \begin{align} f(x)&=a ,\\ a^2+a-1&=0 ,\\ a&=-\tfrac12\pm \tfrac{\sqrt{5}}2 . \end{align}