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Author |
hbghlyj
Posted 2022-8-24 05:31
Let $r= \frac{a}{b} \; a \geqslant b$, and notice that if $r$ is a solution so is $\frac{1}{r}$.
Of course $(a,b)=1$, so we have the following:
$$(\frac{a}{b})^{\frac{1}{\frac{a-b}{b}}} = \frac{p}{q} \iff (\frac{a}{b})^{\frac{b}{a-b}} = \frac{p}{q} \iff a^bq^{a-b}=b^bp^{a-b}$$So let $r^{\alpha}$ (where $r$ is a prime number) fully divide $a$ ( we write this as $r^{\alpha} \parallel a$).Then what that means is that $r^{b\alpha} \parallel b^bp^{a-b}$ (we don't need to look at the case for when $r$ divides $q$ because we will get a contradiction), since $(a,b) = 1$ we now know that $ r^{b\alpha} \parallel p^{a-b} $, now that implies that if we say that $r^{\beta} \parallel p$ we have that $r^{(a-b)\beta} \parallel p^{a-b}$, now that implies $b\alpha = (a-b)\beta \iff a= \frac{\alpha + \beta}{\beta} b$, so that means that $ r= \frac{\alpha + \beta}{\beta}$.So now we have that $(\frac{\alpha + \beta}{\beta})^{\frac{\beta}{\alpha}} \in \mathbb{Q}$.
Now we have 2 cases:
$1$ case is when $\alpha \mid \beta$, this is fairly straightforward, set $\beta = \alpha t$, then we get that $r= \frac{t+1}{t}$, thus giving us also $r = \frac{t}{t+1}$
$2$ case if when $\frac{\alpha + \beta}{\beta} = t^{\frac{\alpha}{\beta}}$, so now we set $t=\frac{c}{g}$, and also we set $\alpha = d\alpha_1$ and $\beta = d\beta_1$, where $(\alpha_1,\beta_1)=1$.
So now we get the equation
$$ (\alpha_1+\beta_1)^{d\beta_1}g^{d\alpha_1} = \beta_1^{d\beta_1}c^{d\alpha_1}$$.
Using a similar procedure, we get that this doesn't have a solution
So the only solutions are:
$$\boxed{r \in \{ \frac{k+1}{k}, \frac{k}{k+1} \mid \forall k \in \mathbb{N}\}}$$ |
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