This problem is a modified version of a problem from Australian mathematics competition 1984:
The problem is let $f:\mathbb Z^+ \to \mathbb Z^+$ be a function from positive integers to positive integers which satisfy the following three conditions.
- $f(2)=2$
- $f(mn)=f(m) f(n)$ for $m,n \in \mathbb Z^+$
- If $m>n$ then $f(m)>f(n)$ for $m,n \in \mathbb Z^+$
Find such an $f$ and prove it is the only function satisfying the above 3 conditions.
It can be proved $f(n)=n$ by induction very easily.
An alternative attempt would be:
Suppose $f(n)=A_1+A_2n+A_3n^2+A_4n^3+ \cdots$. Then, $$f(n \cdot n)=A_1+A_2n^2+A_3n^4+A_4n^6+...=(A_1+A_2n+A_3n^2+A_4n^3+...)^2 ,$$ and by compairing coefficents of similer powers of $n$s it can be showen that $A_i=0$ for $i\in {\mathbb Z^+-\{ 2 \} }$ and $A_2=1$ which leads to $f(n)=n$.
My problem is:
Is this alternative method valid, since it supposes the function to be polynomial but the expected function can be any function (which can't be expressed as a polynomial)? I have a doubt that this is not a valid method to prove this is the only function.
If at all, the second approach uses only the fact that $$f(n^2)=f(n)^2$$ There are many functions with this much weaker property, hence any proof that shows $f(n)=n$ for all $n$ on these grounds is faulty.
E.g., if we define $f(n)=nu^{17}$ where $u$ is the largest odd divisor of $n$, we have $f(n^2)=f(n)^2$ (and even $f(nm)=f(n)f(m)$) as well as $f(2)=2$, but the third condition of the problem statement (monotonicity) is broken.