Find all pairs of positive rational numbers $x,y$ such that $x^y=(2y)^x$ https://ift.tt/eA8V8J

Find all pairs of positive rational numbers $x,y$ such that $x^y=(2y)^x$.

My work. We consider two possible cases.

I) Let $y=2^a$, where $a \in \mathbb{Z}$. Then $x^{2^a}=2^{(a+1)x}$. Then $x=2^b$, where $b \in \mathbb{Z}$. Then $2^{b2^a}=2^{(a+1)2^b}$. Then $b2^a=(a+1)2^b$. Then I found all the solutions $(a,b)=\left(-1, 0 \right); (3,2); (3,1)$. Therefore $(x,y)=\left(1, \frac{1}{2} \right); (4,8); (2,8)$.

II) Let $y \ne 2^a$, where $a \in \mathbb{Z}$.

Definition. Let $p$ be some prime number. We shall say that a rational number $A$ is $p$-free if $A=\frac{r}{q}$, where $r,q \in \mathbb{Z}$ and $r \not\equiv 0 (\mod p)$, $q \not\equiv 0 (\mod p)$.

Since $y \ne 2^a$ then there is a prime $p \ge 3$ such that $y$ is not $p$-free. Let $y=p^kn$, where $p$ is a prime number, $k \in \mathbb{Z}$, $k \ne 0$, $n$ is a $p$-free positive rational number. Let $x=p^tm$, where $t \in \mathbb{Z}$ ($t$ may be zero), $m$ is a $p$-free positive rational number. Then $p^{ty}m^y=p^{kx}(2n)^x$. Since $m$ and $2n$ are $p$-free then $ty=kx$. Then $m^y=(2n)^x$. Then $ty=kx \Rightarrow tp^kn=kp^tm \Rightarrow m=\frac{tnp^{k-t}}{k}$. Then $$m^y=(2n)^x \Rightarrow \left( \frac{tp^{k-t}}{k} \right)^y=2^xn^{x-y} \Rightarrow $$ $$\left( \frac{tp^{k-t}}{k} \right)^{p^kn}=2^{ p^kn \frac{t}{k}}n^{ p^kn \frac{t-k}{k}}$$ Easy to see that $k \ne t$. Then $$np^k=\left(\frac{1}{2} \right)^{\frac{t}{t-k}} \left(\frac{t}{k} \right)^{\frac{k}{t-k}}$$ Let $S=\frac{t}{2k}$. Since $ np^k=y$ then $$y=\frac{1}{2}S^{\frac{1}{2S-1}}$$. Easy to see that $S>0$. We need to find all positive rational numbers $S$ such that the number $S^{\frac{1}{2S-1}}$ is rational. I have a hypothesis.

Hypothesis: If $S$ is a positive rational number then a number $S^{\frac{1}{2S-1}}$ is rational if and only if $\frac{1}{2S-1}=c$, where $c \in \mathbb{Z}$, $c \ne 0$, $c \ne -1$. If this hypothesis is true, then $$y=\frac{1}{2^{c+1}} \left( \frac{c+1}{c} \right)^c$$ $$x=\frac{1}{2^{c+1}} \left( \frac{c+1}{c} \right)^{c+1}$$.

Edited. In the comments found a counterexample. Therefore, the hypothesis is not true.

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