Does there exist a function such that $f(a)f(b)=f(a^2b^2)?$ 
Given $S=\{2,3,4,5,6,7,\cdots,n,\cdots,\} = \Bbb N_{>1}$, prove whether there exists a function $f:S\to S$, such that for any positive $a,b$:
  $$f(a)f(b)=f(a^2b^2),a\neq b?$$

This is 2015 APMO problem 2 (this event ended yesterday: see APMO), maybe I think didn't exist such a function, but I can't prove it.

My attempt:
Consider $p_{i}$ is $i^{th}$ prime,such $f(2)=2^2,f(3)=5^2,f(5)=11^2,\cdots?$

$$f(a^6)f(a^5)f(a^3)f(a^2)f(a)=f(a^{22})f(a^3)f(a^2)f(a)\\
=f(a^{50})f(a^2)f(a)\\=f(a^{104})f(a)\\
=f(a^6)f(a^5)f(a^3)f(a^2)f(a)=f(a^6)f(a^5)f(a^3)f(a^6)\\
=f(a^6)f(a^5)f(a^{18})\\
=f(a^6)f(a^{46})=f(a^{104})$$
