Can we always find arbitrarily close powers of $a$ and $b$?

Question

This post was motivated by this other post.

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I'm aware of Pillai's conjecture (yet to be proven) which states that the gaps in the sequence of perfect powers tend to infinity. However, what happens if we consider their ratios? Can we make them arbitrarily close to $1$?

I would actually like to know something even stronger:

Given $a, b$ positive integers, can we always find positive integers $m,n$ such that $\dfrac{a^m}{b^n}$ is as close to $1$ as we want?

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$$a^m\approx b^n\iff \dfrac{m}{n}\approx \log_a b$$ Of course we can find a fraction as close of $\log_a b$ as we want, i.e., given $\varepsilon>0$, we can find $m, n$ positive integers such that $$0<\dfrac{m}{n}-\log_a b<\varepsilon\implies 0<m-n\log_a b<n\varepsilon \implies 1<\dfrac{a^m}{b^n}<a^{n\varepsilon}$$ Can we get rid of the $n$ dependence on the RHS?

Answer 1

If $a,b$ are positive integers and $\log_ba$ is irrational the set $A=\{k+n\log_ba\colon n\in\mathbb{N},k\in\mathbb{Z}\}$ is dense in $\mathbb{R}$.

Given $\alpha\in\mathbb{R}^+$ and $\varepsilon>0$, since the function $t\mapsto b^t$ is continuous at $\log_b\alpha$ there's $\delta>0$ such that $|x-\log_b\alpha|<\delta\implies|b^x-\alpha|<\varepsilon$.

Now we choose $k\in\mathbb{Z}$ and $n\in\mathbb{N}$ such that $|k+n\log_ba-\log_b\alpha|<\delta$, and since there are infinitely many of them, we can choose them in such a way that $b^{|k|}>\alpha$, then we have that $|b^ka^n-\alpha|<\varepsilon$, this implies that $b^k<(\alpha+\varepsilon)/a^n\le\alpha$ and thus $k<0$ taking $m=-k$ we have $$ \left|\frac{a^n}{b^m}-\alpha\right|<\varepsilon. $$ We conclude that the set $\{\frac{a^n}{b^m}\colon n,m\in\mathbb{N}\}$ is dense in $\mathbb{R}$. (Please let me know if you find some mistake).

Answer 2

If we ever have an equality, $a^m = b^n$, then we are done. So assume there is no such solution. Then $\log_a b$ will be irrational: $\log_a b = m/n$ would give $a^m = b^n$.

Then you can use Dirichlet's approximation theorem, which says that for any irrational $\alpha$ there are infinitely many $m,n$ with $|\alpha - m/n| < 1/n^2$.

Letting $\epsilon = 1/n^2$ in your inequalities above, we have $$0<\dfrac{m}{n}-\log_a b< 1/n^2 \implies 1<\dfrac{a^m}{b^n}<a^{1/n}.$$

Since the inequality is satisfied for infinitely many $n$, we can take $n$ arbitrarily large, which gives the desired result.

edit: I see now that this is similar to what mjqxxxx has mentioned in the comments.