Iranian Mathematical Olympiad

Let $b=1$ we have $f^a(1)+f(a)$ divides $2f(a)$. Hence, $2f(a)=C(f^a(1)+f(a))$, for some positive integer $C$. Thus, $C=1$ and $f^a(1)=f(a)$. Whence, $f^a(b)=f^{a-1}(f(b))=f^{a-1}(f^b(1))=f^{a+b-1}(1)=f(a+b-1)$. By the same argument $f^b(a)=f(a+b-1)$. It then follows that $f(a+b-1)$ divides $f(ab)+b^2-1$. Interchanging $a,b$ to obtain $f(a+b-1)\mid b^2-a^2$. Hence, $f(2a)$ divides $2a+1$. If the function is injective, it follows that $$f^2(n)=f(f(n))=f(n+1).$$ Hence $f(n)=n+1$. This is indeed a solution. If the function is not injective then $f(r)=f(s)$ for some $r>s$. Thus, $f^r(1)=f^s(1)$. Hence, for each positive integer $m$; $f(m+r)=f^m(f^r(1))=f^m(f^s(1))=f(m+s)$. Hence, the function is periodic with a period of $r-s$. This implies that the function is indeed bounded. That is, for all $n$ we have $f(n)<M$. Choose a prime $p>M$ it follows that $f(p-1)$ divides $p$ and $f(p-1)<M<p$. Yielding $f(p-1)=1$. Whence, for all large enough $p$; $f(p-1)=f^{p-1}(1)=1$. Let $d$ be the smallest positive integer such that $f^d(1)=1$ by using the division algorithm, we can easily prove that $d$ divides $p-1$. This means that for all large enough $p$ we have $p\equiv 1(\mathrm{mod}d)$. This implies that $d\in \{1,2\}$. Thus $f(f(1))=1$. Hence, $f(2k)=f^{2k}(1)=1$ for all positive integers $k$. On the other hand, $f(2k+1)=f^{2k+1}(1)=f(1)$ for all non-negative integers $k$. Finally putting $a=b=2$ to obtain $f(1)$ divides $4$. Hence, $f(2k+1)\in \{1,2,4\}$. Whence, we have four functions satisfying the statement of the problem. $■$