Advanced discrete functional equation

Question

The problem is:

Let $f$ be a function taking the positive integers to the positive integers, such that

(i) $f$ is increasing (i.e. $f(n + 1) > f(n)$ for all positive integers $n$)

(ii) $f(mn) = f(m) f(n)$ for all positive integers $m$ and $n,$ and

(iii) if $m \neq n$ and $m^n = n^m,$ then $f(m) = n$ or $f(n) = m.$ Find the sum of all possible values of $f(30).$

Using what the problem gives us, I was able to get that $f(1)=1, f(2)=4, f(4)=16,$ and $f(8)=64$ so I got some bounds for what $f(3)$ and $f(5)$ could be. Also, I found that $f(5) < 4f(3).$ I guess I could use casework to find all possible triples $(f(2), f(3), f(5))$ and then find all possible products of them and sum these up but I'm sure there's an easier way...

Thanks in advance.

Answer 1

Using (i), (ii), and the fact that $f(2)=4$, we may derive the following inequalities for $f(3)$: $$ f(3)^2=f(3^2)=f(9)>f(8)=f(2^3)=f(2)^3=64\Rightarrow f(3)>8 \tag{1} $$ and $$ f(3)^5=f(3^5)=f(243)<f(256)=f(2^8)=f(2)^8=65536 \Rightarrow f(3)\leq 9. \tag{2} $$ From $(1)$ and $(2)$ we conclude that $f(3)=9$. Similarly, $$ f(5)^2=f(5^2)=f(25)>f(24)=f(2^3\times 3)=f(2)^3f(3)=576 \Rightarrow f(5)>24 \tag{3} $$ and $$ f(5)^3=f(5^3)=f(125)<f(128)=f(2^7)=f(2)^7=16384 \Rightarrow f(5)\leq 25, \tag{4} $$ so we conclude that $f(5)=25$. Therefore, there is only one possible value for $f(30)$, namely $f(2)f(3)f(5)=4\times 9\times 25 = 900$.