The functional equation $ f ( m n ) = f ( m ) f ( n ) $ does not have a strictly increasing solution with $ f ( 2 ) = 3 $. Show that there does not exist a function $ f : \mathbb N \to \mathbb N $ which satisfy

*

*$ f ( 2 ) = 3 $;

*$ f ( m n ) = f ( m ) f ( n ) $ for all $ m , n \in \mathbb N $;

*$ f ( m ) < f ( n ) $ whenever $ m < n $.

 A: From $f(9)>f(8)$ we have $f(3)^2>27$ so $f(3)\geq6$.
But now $f(2^8)=3^8<6^5\leq f(3^5)$, while $3^5=243<256=2^8$. This contradicts condition c.
Maybe I should explain a little bit how I thought this up: I was hoping for a contradiction like this, and the next thing I did was looking for powers of $2$ and $3$ that were quite close. First I tried with small things like $27$ and $32$ but that didn't work. So I looked for some bigger ones, and I was lucky to find some I know by heart, $256$ and $243$: Indeed these are quite close, and it worked!
A: (what follows is more-or-less the usual proof of Ostrowski theorem in the Archimedean case)
If $f$ satisfies (2) and (3): for any $a,b\in\mathbb N$, and any $n\in\mathbb N$,
$$b^n\leq a^m$$
whenever $$m\geq n\log b/\log a;$$
in particular, there is such an $m$ satisfying
$$m\leq n\log b/\log a +1.$$
From (2) and (3) we get $$f(b)^n\leq f(a)^m$$
i.e.
$$f(b)\leq f(a)^{\log b/\log a+1/n}.$$
Since it's true for every $n$, we have
$$f(b)\leq f(a)^{\log b/\log a}.$$
When we exchange $a$ and $b$, we get that the opposite inequality, hence we have equality
$$f(b)= f(a)^{\log b/\log a}=b^{\log f(a)/\log a}.$$
From (1) we thus have
$$f(b)=b^{\log3/\log2}.$$
But that's not an integer (e.g. for $b=3$).
edit: what this proof really shows is that if $f$ satisfies (2) and (3) then there is a $k\in\mathbb N$ such that $f(n)=n^k$. This is since (say set $a=2$, $b=n$) 
$$f(n)=n^{\log f(2)/\log2}$$
and that's an integer for all $n$'s only if the exponent is a positive integer.
