Austria 2014

Let $z>2$. We consider the function $g:S\to \mathbb{R}$ with $g(x)=x^2-z^2/x^2$. As $x\mapsto x^2$ and $x\mapsto -z^2/x^2$ are both increasing functions for $x>0$, $g$ is also increasing and obviously continuous. As $g(1)=1-z^2<0$ and $g(x)=z^2-1>1$, there is an $x_0\ge 1$ such that $g$ is a bijection from the interval $I=[x_0,z]$ to the interval $[1,z^2-1]$. For $x\in I$ and $y=z/x$, we have $$x\ge 1,y\ge 1,1\le x^2-y^2=g(x)\le z^2-1,$$ so that the functional equation yields $$f(z)=f(xy)=f(x^2-y^2)=f(g(x))$$ for all $x\in I$. We conclude that $f$ is constant on the interval $[1,z^2-1]$. As $z\ge 2$ was arbitrary, we conclude that $f$ is constant on all these intervals and therefore on $S$. On the other hand, every constant function $f:S\to S$ is a solution to the functional equation.