Does there exist a function such that $f(a)f(b)=f(a^2b^2)?$ Given $S=\{2,3,4,5,6,7,\cdots,n,\cdots,\} = \Bbb N_{&gt;1}$, 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 $f(2)=2^2,f(3)=5^2,f(5)=11^2,\cdots?$
$$f(a^6)f(a^5)f(a^3)f(a^2)f(a)=f(a^{22})f(a^3)f(a^2)f(a)\\
=f(a^{50})f(a^2)f(a)\\=f(a^{104})f(a)\\
=f(a^6)f(a^5)f(a^3)f(a^2)f(a)=f(a^6)f(a^5)f(a^3)f(a^6)\\
=f(a^6)f(a^5)f(a^{18})\\
=f(a^6)f(a^{46})=f(a^{104})$$
Let $f:S\mapsto S $ be $ f(x)=y$ where $y$ is the smallest number in $S$ so that $y^k=x$ with $k\in\{1,2,3...\}. $ This function "drops" exponents.
It would work like this $f(36)=f(216)=f(6)=6$

Suppose $a=c^d$ and $b=e^f$ where $c\neq h^i,  e\neq j^k $with $h,j \in S$ and $i,k \in \{1,2,3...\}$ $. Let $g=gcd(d,f)$

Then $f(a^2b^2)=f((ab)^2)=f(ab)=f(c^de^f)=f((c^\frac{d}{g}d^\frac{f}{g})^g)=f(c^\frac{d}{g}d^\frac{f}{g})=c^\frac{d}{g}d^\frac{f}{g}$ 

$=f(c^{\frac{d}{g} g})f(d^{\frac{f}{g}g})=f(c^d)f(d^f)=f(a)f(b)$ 

showing that the function has the desired property.