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Number theory Difficulty 8.0 National Olympiad, round 2 Prove it Hong Kong

Find, with reasons, all functions f:NNf : \mathbb{N} \to \mathbb{N} such that
(i) f(n)=1f(n) = 1 if and only if n=1n = 1,
(ii) if d=gcd(n,m)d = \text{gcd}(n, m), then f(nm)=f(n)f(m)f(d)f(nm) = \frac{f(n)f(m)}{f(d)}, and
(iii) f2004(n)=nf^{2004}(n) = n for every nNn \in \mathbb{N}, where f2004(n)=f(f(f(n)))2004f^{2004}(n) = \underbrace{f(f(\cdots f(n)))}_{2004},
or show that they do not exist.

Solution

There is no such function.
Putting n=m=2n = m = 2 in (ii), we obtain
f(4)=f(2)f(2)f(2)=f(2). f(4) = \frac{f(2)f(2)}{f(2)} = f(2).
Then by condition (iii), we have
2=f2004(2)=f2003(f(2))=f2003(f(4))=f2004(4)=4. 2 = f^{2004}(2) = f^{2003}(f(2)) = f^{2003}(f(4)) = f^{2004}(4) = 4.
This is a contradiction. So there is no such function.

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