Can there be a function $f\colon\mathbb Q_{+}^{*}\to \mathbb Q_{+}^{*}$ such that $f(xf(y))=\frac{f(f(x))}{y}$?

Question

Problem: Can an $f$ function be created where:$$f\colon\mathbb Q_{+}^{*}\to \mathbb Q_{+}^{*}$$ The function is defined on the set of fully positive rational numbers and is achieved: $\forall(x,y)\in \mathbb Q_{+}^{*}\times\mathbb Q_{+}^{*},f(xf(y))=\frac{f(f(x))}{y}$

This question is similar to one of the Olympiad questions that I was very passionate about and used several ideas to solve this problem, but I did not arrive at any result from one of them by using the basic theorem in arithmetic that states that there is a corresponding application between $(\mathbb Q_{+}^{*})$and $(\mathbb Z^{\mathbb N})$ where: $$\left\{\mathbb Z^{\mathbb N} =\text{ A set of stable sequences whose values ​​are set in} \quad\mathbb Z\right\}$$ This app is defined like this $$\varphi\colon\mathbb Z^{\mathbb N}\to \mathbb Q_{+}^{*} ,(\alpha_n)_{n\in\mathbb N}\longmapsto \prod_{n\in\mathbb N} P_n^{\alpha_n}$$ Where:$$\mathbb P=\left\{P_k:k\in\mathbb N\right\}\text{ is the set of prime numbers} $$ And put $x=\prod_{n\in\mathbb N}P_n^{\alpha_n},\quad y=\prod_{n\in\mathbb N }P_n^{\beta_n},\text{and}\quad $ $$f(\prod_{n\in\mathbb N}P_n^{\alpha_n})=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)$$

Found in the latter$$f(xf(y))=\frac{\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)}{\left(\prod_{n\in\mathbb N}P_n^{\beta_{2n}}\right)}=\frac{f(x)}{y}$$ However, this did not help me create this method

I need an idea or suggestion to solve this problem if possible and thank you for your help

Note: $(\alpha_n)_{n\in\mathbb N}\quad \text{is a stable sequence}\leftrightarrow \forall n\in\mathbb N ,\exists n_0\in\mathbb N :\left( n\geq n_0 \quad \alpha_{n}=0\right) $

Answer 1

For $x=1$ we get $f(f(y))=f(f(1))/y$. For $y=1$ we get $f(xf(1))=f(f(x))$. And so if we replace $y$ with $x$ in the first equation and compare both we get

$$f(f(1))/x=f(xf(1))$$

We now replace $x$ with $x\cdot f(1)^{-1}$ and we get

$$f(x)=\frac{f(f(1))\cdot f(1)}{x}$$

We conclude that our function is of the form:

$$f(x)=\frac{c}{x}$$

for some constant $c\in\mathbb{Q}^*_+$. Note that $f(f(x))=x$. We now evaluate both sides of your original equation:

$$f(xf(y))=f(x\cdot\frac{c}{y})=\frac{c}{x\cdot\frac{c}{y}}=\frac{y}{x}$$ $$\frac{f(f(x))}{y}=\frac{x}{y}$$

These are clearly different, e.g. when $x=1,y=2$, regardless of $c$. Hence no such function exists.

Answer 2

There is no such function.

Proof: Specifically for $x = 1$, we obtain that $f(f(y)) = \frac{f(f(1))}{y}$ for all $y$.

Now, let $a \in \mathbb Q$, $a > 0$ and write $y = \frac{y}{f(a)} \cdot f(a)$. We then obtain (using the first equality)

$$f(y) = f\left(\frac{y}{f(a)} \cdot f(a)\right) = \frac{f(f(\frac{y}{f(a)}))}a = \frac{f(f(1)) \cdot f(a)}{a \cdot y}.$$

Thus we have a formula for $f(y)$, which is independent of $a$. Specifically for $a = y$, one obtains

$$f(y) = \frac{f(f(1)) \cdot f(y)}{y^2},$$

hence $y^2 = f(f(1))$ for all $y$, a contradiction.

Answer 3

we put $$x=\prod_{n\in\mathbb N}P_{n}^{\alpha_{n}}\quad ,y=\prod_{n\in\mathbb N}P_n^{\beta_n}$$ where $(x.y)\in\mathbb Q_{+}^{*}\times\mathbb Q_{+}^{*}$ and $$f(\prod_{n\in\mathbb N}P_n^{\alpha_n})=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)$$ We find : \begin{align*} xf(y)=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n}^{\beta_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\beta_{2n}}\right)\\ &=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n}+\beta_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{\alpha_{2n+1}-\beta_{2n}}\right)\\ \end{align*} $\implies$ \begin{align*} f(xf(y))&=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}-\beta_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}-\beta_{2n+1}}\right)\\ &=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n}^{-\beta_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\beta_{2n+1}}\right)\\ &=\frac{\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)}{\left(\prod_{n\in\mathbb N}P_{n}^{\beta_{n}}\right)}\\ &=\frac{f(x)}{y}\\ \end{align*} So $f(f(xf(y)))=f(\frac{f(x)}{y})\implies\left (f(f(xf(y)))=f(xf(y)) \wedge f(\frac{f(x)}{y})=\frac{f(f(x))}{f(y)} \wedge f(xf(y))=\frac{f(x)}{y}\right)\implies f(x)\text{is apl linear and } f=id_{\mathbb Q_{+}^{*}}\text{and}f(xf(y))=\frac{f(x)}{y}$ But $f$ cannot achieve. Any of these wicker for $(x,y)\in\mathbb Q_{+}^{*}$