XII OBM

Put $f^k(x)=\underset{k\text{ times}}{\underbrace{f(\dots (f(x))\dots )}}$. Then we have $f^k(x)=\frac{a_{k}x+b_k}{c_{k}x+d_k}$. From $f^{k+1}(x)=f(f^k(x))$ we get: $$\begin{array}{rlrlr}{a}_{k+1}& =a{a}_k+b{c}_k{b}_{k+1}& =a{b}_k+b{d}_k{c}_{k+1}& =c{a}_k+d{c}_k{d}_{k+1}& =c{b}_k+d{d}_k\end{array}$$ But we also have $f^{k+1}(x)=f^k(f(x))$, so $$\begin{array}{rlrlr}{a}_{k+1}& =a{a}_k+c{b}_k{b}_{k+1}& =b{a}_k+d{b}_k{c}_{k+1}& =a{c}_k+c{d}_k{d}_{k+1}& =b{c}_k+d{d}_k\end{array}$$ Comparing, we get $b{c}_k=c{b}_k$ and $(a-d)b_k=b(a_k-d_k)$. If $f^n(0)=0$, then $b_n=0$, so $b{c}_n=0$. We are given that $f(0)\ne 0$, so $b\ne 0$. Hence $c_n=0$. Also $b(a_n-d_n)=0$ and hence $a_n=d_n$. Hence $f^n(x)=x$.