Does there exist a function such that $f(a)f(b)=f(a^2b^2)?$

Question

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.

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My attempt:

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

Answer 1

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

Answer 2

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.

Answer 3

The existing answer by Michael is a nice first proof, but I feel it's quite a deus ex machina.

Therefore, let me sketch an approach to solve the general situation, providing more insight.

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So suppose that for all $a \ne b$, we have: $$f(a^nb^n) = f(a)f(b)$$

Then in particular, for all $a,b,c,d$, $a \ne b,c \ne d$:

$$a^nb^n = c^nd^n \implies f(a)f(b) = f(c)f(d)$$

Rewriting and using that $n$th powers are injective on the natural numbers:

$$\frac ac = \frac db \leadsto ab = cd \leadsto f(a)f(b) = f(c)f(d) \leadsto \frac{f(a)}{f(c)} = \frac{f(d)}{f(b)}$$

In particular, for any $x,y \ge 2$:

$$\frac{ax}x = \frac{ay}y \leadsto \frac{f(ax)}{f(x)} = \frac{f(ay)}{f(y)}$$

That is, there exists some $\lambda_a$ such that for all $x$, $f(ax) = \lambda_a f(x)$.

Because $a,x,y$ were completely arbitrary, we can derive inductively from $$f(aa^n) = \lambda_a f(a^n) = \lambda_{a^n}f(a)$$ that $\lambda_{a^n} = \lambda_a^n$. From here on, we can apply the following basic algebra:

\begin{align*} f(a^{3n}) &= \lambda_a^{3n-1} f(a) = f(a)f(a^2) = \lambda_a f(a)^2 \leadsto f(a) = \lambda_a^{3n-2}\\ f(a^{4n}) &= \lambda_a^{4n-1} f(a) = f(a)f(a^3) = \lambda_a^2 f(a)^2 \leadsto f(a) = \lambda_a^{4n-3} \end{align*} Hence, $\lambda_a^{n-1} = \frac{f(a)}{f(a)} = 1$, and we conclude that:

$$f(a) = \lambda_a^{3n-2} = \lambda_a (\lambda_a^{n-1})^3 = \lambda_a$$

which contradicts that $f(a) \in \Bbb N_{>1}$. Therefore, such an $f$ cannot exist.

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I came to this approach by writing down the conditions on values of the form $f(a^2b^2)$ and recognising that this was reminiscent of the construction of the rational numbers.