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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.2010-08-16

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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 $f(n)=n$.

You can read a proof by Pierre Pbornsztein here.

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    Also, thank you ShreevatsaR for the correction. I did intend to write that.2010-08-16