# Finding Extra Condition for a function to satisfy $f(n)=n$

Given a function $f$ defined on the set of all natural numbers $\mathbb{N}$ with three conditions:

1. If $m,n$ relatively prime, then $f(mn) = f(m)f(n)$.

2. $f$ strictly increasing.

3. $f(2) = 2$.

Find a 4th condition such that the result will be that $f(n)$ must equal $n$ for any natural number $n$. (Of course all the conditions together are needed. Your 4th condition should not make any of these three conditions redundant.)

This is posted in my university website:

[1]: http:// mathstat.uohyd.ernet.in/noticeboard/generaldetails.php?id=12

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Maybe we shouldn't answer this here, if there's a prize waiting for students at University of Hyderabad for solving it. –  Rasmus Aug 16 '10 at 16:44

In fact, it's a theorem of Erdős that every non-decreasing multiplicative function $f:\mathbb{N}\to\mathbb{R}$ is $f(n)=n^c$ for some constant $c$. This implies that those three conditions are already enough to prove that $f(n)=n$.
Yes, actually "multiplicative" means that $f(mn)=f(m)f(n)$ is true for $m,n$ relatively prime. If $f(mn)=f(m)f(n)$ for all $m,n$ then $f$ is said to be "completely multiplicative". For more on this read en.wikipedia.org/wiki/Multiplicative_function –  Jorge Miranda Aug 16 '10 at 17:06