Mongolian Mathematical Olympiad

Answer: $f(x)=x+1$ and $g(x)=1/x$. The pair above is a solution. In order to prove that there is no other solution, fix $x\in {\mathbb{R}}_{>0}$ and denote $g^0=x$ and $g^n=g(g^{n-1})$ for $n\ge 1$. We have $g(x)(1+g(g(x)))=f(g(x))=f(x)g(x)=x(1+g(x))g(x)$. Since $g(x)\ne 0$, we have $1+g^2=x(1+g)$. This can be rewritten as $$g{g}^2-1=(xg-1)(1+g).$$ Then $g^2{g}^3-1=(g{g}^2-1)(1+g^2)=(xg-1)(1+g)(1+g^2)$ and more generally $$g^n{g}^{n+1}-1=(g^{n-1}{g}^n-1)(1+g^n)=(xg-1)(1+g)(1+g^2)\dots (1+g^n)$$ for any $n\ge 1$ by induction. There is $n>m$ with $g^n=g^m$ by assumption. Then we have $$(xg-1)(1+g)\dots (1+g^m)((1+g^{m+1})\dots (1+g^n)-1)=0.$$ Since $g^k>0$, we must have $xg-1=0$. It follows that $g=1/x$ and $f=x+1$.