SILK ROAD MATHEMATICAL COMPETITION

Answer. The answer is no.

Let $f(x)=1+\frac{1}{x}$ and $f^{(n)}(x)=f(f(\dots f(x)))$ (where $f$ is applied $n$ times). For each $x\in S$ the set $S$ contains all numbers of the form $f^{(n)}(x)$. If $S$ is finite, then $f^{(k)}(x)=f^{(\ell )}(x)$ for some $k>\ell$. Since $f(a)=f(b)$ implies $a=b$, $f^{(k-\ell )}(x)=x$. Thus each $x\in S$ is a root of equation $f^{(n)}(x)=x$ for some positive integer $n$.

The functions $f^{(n)}(x)$ obviously are of the form $\frac{px+q}{rx+s}$ with some $p,q,r,s$ (it is immediate for $n=1$, and $f$ transforms every such function into a function of this form). Therefore, the equation $f^{(n)}(x)=\frac{px+q}{rx+s}=x$ is reduced to an equation of the form $r{x}^2+(s-p)x-q=0$ which is an equation of degree at most 2. This equation is not identical (since, for example, $f(-2)=\frac{1}{2}$ and therefore $f^{(n)}(-2)$ is positive for all positive integers $n$), hence it cannot have more than two real roots. However, the roots $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$ of $f(x)=x$ are also roots of $f^{(n)}(x)=x$ and therefore its only roots.

We see now that if $S$ is finite, it can contain at most two elements.

Second solution. There is an alternative way to prove that for every $x$ except $\phi =\frac{1+\sqrt{5}}{2}$ and $\bar{\phi }=\frac{1-\sqrt{5}}{2}$, $f^{(n)}(x)$ and $x$ would not coincide.

If $x<-1$, then $f(x)>0$ and therefore all $f^{(n)}(x)$ are positive.

If $0<x\le 1$, then $f(x)>1$, and all of $f^{(n)}(x)$ are greater than 1.

If $x>1$, then $|f(x)-\phi |=|1+\frac{1}{x}-1-\frac{1}{\phi }|=\frac{|x-\phi |}{x\phi }<|x-\phi |$, that is, the distance from $f^{(n)}(x)$ to $\phi$ decreases as $n$ increases.

Finally, if $-1<x<0$, while the terms of the sequence $x,f(x),f(f(x)),\dots$ remain in the interval $(-1,0)$, their distances between each term of the sequence and $\bar{\phi }$ would increase (indeed, $|f(t)-\bar{\phi }|=|1+\frac{1}{t}-1-\frac{1}{\bar{\phi }}|=\frac{|t-\bar{\phi }|}{t\bar{\phi }}>|t-\bar{\phi }|$), and when a term appears outside this interval, all the proceeding terms would also be outside of this interval.