Saudi Arabia Mathematical Competitions 2012

Consider $x=y=1$. Then we get that $\frac{1+f^2(1)}{1+f(1)}$ is an integer, so $1+f(1)$ divides $2f(1)$. It follows $1+f(1)=2$, that is $f(1)=1$.

For $y=1$, we obtain $x^2+1$ divides $f^2(x)+1$. Therefore $x^2\le f^2(x)$, which means that $x\le f(x)$ for every positive integer $x$.

For $x=1$, we get $1+f(y)$ divides $1+y$, and hence $f(y)\le y$, for every positive integer $y$.

From the above inequalities it follows that the unique function is $f(x)=x$, $x\in S$.