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 $(n,m) \to (\infty,\infty).$

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?

If you write it as $n\ln x-m\ln y \rightarrow 0$, and consider arbitrary $x,y \in \mathbb R$ doesn't this become just showing that you can approximate $\ln y / \ln x$ as a rational with arbitrary accuracy? — Alex K, Mar 31 '24 at 19:21
It isn't true that $x^n/y^m\to1$ as $(n,m)\to(\infty,\infty).$ — Thomas Andrews, Mar 31 '24 at 19:32
@ThomasAndrews How do you know that? Please provide a counter-example (and prove that it is indeed a counter-example) — Adam Rubinson, Mar 31 '24 at 21:30
Hint: If $x^n/y^m< 2,$ then $x^n/y^{2m}<2/y^m.$ So it can't converge to $1.$ — Thomas Andrews, Mar 31 '24 at 23:07
However, there might be sequence of numbers $(m_i){i=1}^\infty$ and $(n_i){i=1}^\infty$ such that $$x^{n_i}/y^{m_i}\to 1$$ as $i\to\infty.$ — Thomas Andrews, Mar 31 '24 at 23:14

score 0 · Answer 1 · answered Mar 31 '24 at 22:52

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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.

answered Mar 31 '24 at 22:52

Piquito

29,594

OK, thanks for the suggested ammendment. I updated the question accordingly. – Adam Rubinson Mar 31 '24 at 22:58

score 0 · Answer 2 · answered Mar 31 '24 at 23:29

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.

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

2 Answers2