How do I show that the terms of this sequence matches with the fixed points of this function?

Question

$a_n = \frac{(\sqrt{(n^{2} + 4)} - n)}{2}$ and I have function $f:(0,1) -> \mathbb{R}$ is defined by $f(x) = \frac{1}{x} - \lfloor\frac{1}{x}\rfloor$ how would I go about showing that the fixed points of $f$ are the terms of the sequence of $a_n$?

Answer 1

Let $x\in (0,1)$ be a fixed point of $f$. Since $1/x>1$, $n:=\lfloor\frac{1}{x}\rfloor\geq 1$. Therefore $x=f(x)=1/x-n$, that is $x^2+nx-1=0$ Solving this quadratic equation we get that $x=a_n$.