If $x,y\in\mathbb{N},\varepsilon>0$ then are there infinitely many positive integer pairs $(n,m)$ s.t. $\vert\frac{x^n}{y^m}- 1\vert < \varepsilon?$

Question

Proposition: If $x,y\in\mathbb{N}_{\geq2}$ then for any $\varepsilon>0,$ there are infinitely many pairs of positive integers $(n,m)$ such that $$\frac{\left\lvert y^m-x^n \right\rvert}{y^m} < \varepsilon,$$

i.e. $\displaystyle\large{\frac{x^n}{y^m}} \to 1\ $ as these pairs $(m,n) \to (\infty,\infty).$

I think this is true, and I want to prove it. For all integers $n,$ we have

$$\frac{x^n}{y^{ {n\log_y x}}} = 1.$$

Therefore, we want to find integers $n$ such that $n\log_y x$ is, in some sense, extremely close to an integer.

This above question can also be stated as follows. If $x,y\in\mathbb{N}_{\geq2}$ and $x>y,$ then either $\ \displaystyle\limsup_{n\to\infty} \frac{x^n}{y^{\lceil n(\log_y x)\rceil}} = 1 $ or $\ \displaystyle\liminf_{n\to\infty} \frac{x^n}{y^{\lfloor n(\log_y x)\rfloor}} = 1. $

Can we use Dirichlet's approximation theorem to prove this, or the fact that $\{ n\alpha: n\in\mathbb{N} \} $ is dense in $[0,1]$ for irrational $\ \alpha\ ?$ Or do we have to use other tools?

Answer 1

Taking Alex K's suggestion in the comments, suppose we are given integers $x,y\in\mathbb{N}_{\geq2}.$

There are two cases: a trivial case, and a non-trivial case.

If $\exists $ integers $m,n$ such that $y^m = x^n.$ In this case $y^{km} = x^{kn}\ \forall\ k\in\mathbb{N},$ then we see that the result is trivially true.

Non-trivial case: suppose $\not\exists\ m,n\in\mathbb{N}$ such that $y^m = x^n.$ Since $y^m = x^n\implies \ln\frac{y^m}{x^n} = 0 \implies \ln y^m = \ln x^n \implies m\ln y = n\ln x \implies \frac{\ln y}{\ln x} = \frac{n}{m},$ it follows that $\not\exists\ m,n\in\mathbb{N}$ such that $\frac{\ln y}{\ln x} = \frac{n}{m},$ in other words, $\frac{\ln y}{\ln x}$ is irrational. Thus we may use the following consequence of Dirichlet's approximation theorem, which applies to irrational numbers: $\ \exists\ $ arbitrarily many pairs of positive integers $(m,n)$ such that $\frac{\ln y}{\ln x} = \frac{n}{m} + \frac{\alpha}{m^2}\ $ for some $\vert \alpha \vert < 1,\implies m\ln y - n\ln x = \frac{\alpha\ln x}{m},\implies e^{m\ln y - n\ln x} = e^{\frac{\alpha\ln x}{m}} \implies$

$$ (1)\qquad \frac{y^m}{x^n} = e^{\frac{\alpha\ln x}{m}}\ \text{ for some }\ \vert \alpha \vert < 1.$$

Since $(1)$ is true for arbitrarily many pairs of positive integers $(m,n),$ with both $m$ and $n\ $ $\to\infty$ in these pairs, we get the result:

$${\frac{y^m}{x^n}} \to 1\ \text{ as these pairs } (m,n) \to (\infty,\infty).$$

Answer 2

COMMENT (It is not an answer!).-You need the following two restrictions so that your proposition has any chance of being true:

$$(x,y)\ne(1,y)\text { with }y\ne1\\(x,y)\ne(x,1)\text { with }x\ne1$$ Clearly because if not you have respectively $$\frac{x^n}{y^m}=\frac{1}{y^m}\\\frac{x^n}{y^m}=x^n$$ and in both cases it is impossible your condition.

Note that for $(x,y)=(1,1)$ your condition is trivially verified.

Answer 3

Since $$ \left|\,e^x-1\,\right|\le\frac{|x|}{1-|x|}\tag1 $$ if $|n\log(x)-m\log(y)|\le\frac\epsilon{1+\epsilon}$, then $$ \begin{align} \left|\,\frac{x^n}{y^m}-1\,\right| &=\left|\,e^{n\log(x)-m\log(y)}-1\,\right|\tag{2a}\\ &\le\frac{|n\log(x)-m\log(y)|}{1-|n\log(x)-m\log(y)|}\tag{2b}\\[3pt] &\le\epsilon\tag{2c} \end{align} $$ Now, using Dirichlet's Approximation Theorem, we can find $n,m\in\mathbb{Z}$, arbitrarily large, so that $$ |n\log(x)-m\log(y)|\le\min\left(\frac{\log(x)}m,\frac{\log(y)}n\right)\tag3 $$ which allows us to make $\epsilon$ as small as we want.