
> 
> **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 $
$$f(\prod\_{n\in\mathbb N}P\_n^{\alpha\_n})=\left(\prod\_{n\in\mathbb N}P\_{2n}^{\alpha\_{2n+1}}\right)\left(\prod\_{n\in\mathbb N}P\_{2n+1}^{-\alpha\_{2n}}\right)$$


.
\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\*}


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) $
> 
> 
> 



can there be a function $f:\mathbb Q_{+}^{*}\longmapsto\mathbb Q_{+}^{*}$ such that $f(xf(y))=\frac{f(f(x))}{y}$?
There is no function $f\colon Q\to Q$ such that
$$f(xf(y))=\frac{f(f(x))}y \tag{1}$$
for all $x$ and $y$ (in $Q$), where $Q:=\mathbb Q\_{+}^{\*}$.


Indeed, for $x=1$ equality (1) is
$$f(f(y))=\frac{f(b)}y,$$
where $b:=f(1)$. Replacing here $y$ by $x$, from (1) we get
$$f(xf(y))=\frac{f(b)}{xy}.$$
This with $y=1$ yields
$$f(xb)=\frac{f(b)}{x},$$
or
$$f(z)=\frac cz$$
for $c:=bf(b)$ and all $z\in Q$. Now (1) becomes $y/x=x/y$ for all $x,y$ in $Q$, which is clearly false.


