Define $g(m) = \frac{f(m)}{m}$ $g(m) = 1$ iff $m = 1$, (using iii) $p\mid f(p)$ for any prime $p$, (using i), and then $g(p) \in \mathbb{N}$ for prime $p$ From (ii) we have that $g(mn)g(d) = g(m)g(n) \text{ if } \gcd(m,n) = d$ Using this, we get that: (1) $g(mn) = g(m)g(n) \text{ if } \gcd(m,n) = 1$ (2) $g(md) = g(m) \text{ if } d\mid m$ $g(p^s) = g(p^{s-1}\cdot p) = g(p^{s-1})$ for any $s\ge 2 \text{ (using (2))}$, and from this follows that $g(p^s) = g(p)$ Then $g(\prod p_i^{a_i}) = \prod g(p_i) \text{ (using (1))}$ Now we need to find the value of $g(p)$ for $p$ prime. We will prove that it must be a power of $p$ $pf(f(p)) = f(p)^2 \Rightarrow g(p\cdot g(p)) = g(f(p)) = \frac{f(f(p))}{f(p)} = \frac{f(p)}{p} = g(p)$ If $q\mid g(p)$ for some prime $q\ne p$ then $g(p)g(q) \mid g(p\cdot g(p)) = g(p)$ since $p$ and $q$ are prime divisors of $p\cdot g(p)$. This implies $g(q) = 1$ which is a contradiction. Then $g(p) = p^{v_p}$ for some $v_p$, and since $g(p) \ne 1$ we have that $v_p\ge 1$. Write $m = \prod p_i^{a_i}$. Then $f(m) = m\cdot g(m) = m\prod g(p_i) = m\prod p_i^{v_{p_i}}$ And we are done, because this satisfies the conditions of the problem