TITLE: Nonexistence of a strongly multiplicative increasing function with $f(2)=3$
QUESTION [6 upvotes]: Show that there does not exist a strictly increasing function
  $f:\mathbb{N}\rightarrow\mathbb{N}$ satisfying 
$$f(2)=3$$ $$f(mn)=f(m)f(n)\forall m,n\in\mathbb{N}$$

Progress:
Assume the function exists.  Let $f(3)=k$
Since $2^3 < 3^2$,
$$3^2=f(2)^3=f(2^3)<f(3^2)=f(3)^2=k^2$$
so $k>5$ and since $3^3 < 5^2$, then
$$k^3=f(3)^3=f(3^3)<f(2^5)=f(2)^5=3^5=243<343=7^3$$
so $k<7$ therefore $k=6$.
I've messed around with knowing $f(3)=6$ and $f(2)=3$ but I am stuck.

REPLY [8 votes]: Hint: suppose not.  You know what $f(2^k)$ must be.  You'll show that $f(3)$ can't be any natural number.  You have bounds since $f(2) < f(3) < f(4)$.  Start considering $f(3^j) = f(3)^j$ for some small values of $j$ and compare to $f(2^k)$ for some small values of $k$ to eliminate all possibilities for $f(3)$.