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I was motivated by my previous question here to ask a more general question:

Suppose $\ x,y\in\mathbb{R}_{>1}\ $ and $\ \not\exists\ p,q\in\mathbb{N}\ $ such that $\ x^p = y^ q.\ $ Then is the sequence $\ (x^p)_{p\in\mathbb{N}}\ $ eventually bound away from $\ (y^q)_{q\in\mathbb{N}}\ $ by a constant? That is, does there exist $\ N,M \in\mathbb{N},\ \varepsilon>0,\ $ such that $\ \left\lvert x^n - y^m \right\rvert > \varepsilon\quad \forall\ n\geq N, m\geq M\ ?$

I guess this has something to do with transcendental number theory, Liouville numbers and irrationality measure, which I've touched on but am not well-versed in. But maybe there is a simple counter-example I'm overlooking. I don't see an easy counter-example with a Pisot number for $x$ and and integer for $y$, which was my first thought, although that might be because I don't know enough about Pisot numbers...

Adam Rubinson
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    If I’m not mistaken, if $z=\log_y{x}>1$ ($z$ is irrational), it’s formal that the statement can only hold if $z$ has “exponentially good rational approximations”, that is, for all $\eta>0$, there are infinitely many pairs of rationals $(p,q)$ $|z-p/q|\leq \eta x^{-q}/q$. In particular, the irrationality measure of $z$ is infinite, but the condition is much stronger: if $a_n$ is an increasing sequence of integers with $1=o(a_{n+1}-a_n)$, the usual rational approximations of $\sum_n{10^{-a_n}}$ work only if $10^{a_n-a_{n+1}}=O(x^{-10^{a_n}})$ ie we need $a_{n+1}-a_n \geq \Omega( 10^{a_n})$. – Aphelli Nov 16 '22 at 23:20
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    So define $a_1=2$, $a_n=n^{a_{n-1}}$, take $z=1+\sum_n{10^{-a_n}}$, let $x=1.000001$, $y$ be such that $\log_y(x)=z$, then you should have your counterexample. – Aphelli Nov 16 '22 at 23:24

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I’m expanding my comment into an answer. Assume $y < x$ and let $z=\ln_y{x}$: then $z>1$ and $z$ is irrational.

The condition is false iff there exist increasing sequences $u_n , v_n$ of integers such that $x^{u_n}=y^{v_n}+o(1)$ (*).

Now, if $u_n,v_n$ are increasing sequences of integers, $x^{u_n}+o(1)=y^{v_n}$ iff $v_n=zu_n+\ln_y\left(1+o(x^{-u_n})\right)$. But $\ln_y(1+u) \sim \ln{y} u$ as $u$ is small, so that $x^{u_n}=y^{v_n}+o(1)$ iff $zu_n-v_n=o(x^{-u_n})$, ie iff $|z-\frac{v_n}{u_n}| = o(x^{-u_n}/u_n)$.

Thus, your property is false for the pair $(x,y)$ iff for every $\eta>0$, there are infinitely many pairs of positive integers $(p,q)$ such that $|z-p/q|< \eta x^{-q}/q$.

This is an extremely close rational approximation: it implies of course that the irrationality measure of $z$ is infinite (thus $z$ is a transcendent number), but it is, in fact, much stronger.

Assume for instance $z=\sum_{n \geq 0}{10^{-a_n}}$, where $a_n$ is an increasing sequence of integers, $a_0=0$, and $a_{n+1}-a_n \rightarrow \infty$ (eg $a_n=n!$ for $n\geq 1$ and $a_0=0$, like Liouville’s number).

Take $v_n=\sum_{k \geq n}{10^{a_n-a_k}}\in 1+10\mathbb{Z}$ and $u_n=10^{a_n}$.

Then $(z-\frac{v_n}{u_n})\sim 10^{-a_{n+1}}$. To have the approximation property for these two sequences and some $x >1$, we would thus need $10^{-a_{n+1}} = o(x^{-10^{a_n}})/10^{a_n}$, that is, after taking logarithms, $a_n-a_{n+1}+(\log_{10}{x})10^{a_n} \rightarrow -\infty$, that is, $a_{n+1}-C10^{a_n} \rightarrow \infty$ for some $C>0$, or in other words $10^{a_n}=O(a_{n+1})$.

This is completely possible, (choose $a_0=0$ and $a_{n+1}=10^{a_n}$, then any $x<10$ will work – that is, be a counter-example to your question with $y=x^{1/z}$); but you can see that even Liouville’s number does not come close to satisfying this property.

Aphelli
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