The functional equation $ f ( m n ) = f ( m ) f ( n ) $ does not have a strictly increasing solution with $ f ( 2 ) = 3 $.

Question

Show that there does not exist a function $ f : \mathbb N \to \mathbb N $ which satisfy

1. $ f ( 2 ) = 3 $;
2. $ f ( m n ) = f ( m ) f ( n ) $ for all $ m , n \in \mathbb N $;
3. $ f ( m ) < f ( n ) $ whenever $ m < n $.

Answer 1

(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.

Answer 2

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!