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...

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 $$f(3)^5=f(3^5)=f(243) From $$(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) so we conclude that $$f(5)=25$$. Therefore, there is only one possible value for $$f(30)$$, namely $$f(2)f(3)f(5)=4\times 9\times 25 = 900$$.