$\liminf |2^m - 3^n|$

Question

In order to make it clear, I ask three questions:

1. Does $|2^m - 3^n|<10^6$ have any integers solution for $m>20$ ?
2. Is $ \liminf |2^m - 3^n|$ infinite ?
3. Is $ \liminf |2^m - 3^n|/m$ finite ?

Answer 1

The answer to question 1 is yes, since $2^{21}-3^{13}=502829$.

The answer to question 3 is no, and hence the answer to question 2 is yes. The following argument is adapted from Terry Tao's blog post.

By Baker's theorem applied to $n=1,\lambda_1=\log_2 3,\beta_0=\frac{m}{n},\beta=1$, we have for some constant $C$ $|\frac{m}{n}-\log_2 3|>H^{-C}$, where $H=\max\{m,n\}$.

We have $|2^m-3^n|=2^m|1-2^{n(\log_2 3-\frac{m}{n})}|$. Assume first $|\log_2 3-\frac{m}{n}|<1$. Since $|2^x-1|>c|x|$ for some constant $c>1$ assuming $|x|<1$, we get $|2^m-3^n|>2^m\cdot cn|\log_2 3-\frac{m}{n}|>2^m\cdot cn\cdot H^{-C}$. We also get $\frac{m}{n}<3,n>\frac{m}{3}$, so $H>\frac{m}{3}$. Applying this, $|2^m-3^n|>2^m\cdot cn\cdot H^{-C}>2^m\cdot c\cdot\frac{m}{3}\cdot(\frac{m}{3})^{-C}=2^m\cdot c'\cdot m^{-C+1}$, hence $|2^m-3^n|/m>2^m\cdot c'\cdot m^{-C}$, which can be arbitrarily large for large $m$.

For $|\log_2 3-\frac{m}{n}|\geq 1$ we have $|1-2^{n(\log_2 3-\frac{m}{n})}|$ greater than some fixed constant $d$, so $|2^m-3^n|>2^md$ which again can be arbitrarily large.

Answer 2

The problem can empirically examined with the use of the continued fraction of $ \beta = \log_2(3)$ For each $n$ we can find the optimal $m_n$ by $m_{n,\text{lo}}=\lfloor n \cdot \beta \rfloor$ resp. $m_{n,\text{hi}}=\lceil n \cdot \beta \rceil$ and the optimal $n$ can be found by the convergents of the continued fraction of $\beta$.
The best results -as I assume- are collected in the work of Wadim Zudilin, for instance you can find a preview of "A new lower bound for ||(3/2)^k||" at W. Zudilin's homepage.

I've a small discussion and a couple of - I think: instructive - pictures at my webspace which indicate a dependence of the smallest differences $|2^{m_n}-3^n|$ from the parameter $n$ (as also indicated by @Wojowu's answer), up to $n \gt 10^{1000}$ - maybe this is helpful or at least useful for visual intuition. Here is one picture

A short review of the formulae empirically up to $n \approx 3.78 \cdot 1e29$ (which is pretty near the right bottom edge in the above picture) suggests, that possibly $ \lim \sup g(n)=\Large{{|2^{m_n}-3^n|\over 3^n }} \small \cdot 3n $ might be finite, see the following table

     n                  g(n)  
     1.00000000000      1.00000000
     5.00000000000      0.80246914
     41.0000000000      1.41804877
     306.000000000      0.93889533
     15601.0000000      0.85156907
     79335.0000000      0.87222344
     190537.000000      0.03687320
     10781274.0000      0.39482048
     171928773.000      0.92285323
     397573379.000      0.12624166
     6586818670.00      0.20003757
     137528045312.      0.37077074
  5.40930392448E12      1.06988637
  1.15717186888E13      0.44771035
  4.31166034847E14      1.72524065
  5.75093460288E15      1.83493581
  1.30441933148E17      0.70175324
  3.97560349370E17      0.18091377
  4.64028225930E18      0.37542422
  2.74441332064E19      0.83355872
  7.76921173599E19      0.79349425
  2.05632218873E20      0.05490846
  3.11509610182E22      0.03729717
  1.75219014922E24      1.03849316
  1.71940427683E26      1.32651409
  3.47354084703E26      1.21289278
  2.43842505160E27      1.38110746
  1.13220034315E28      0.80968184
  8.81376024008E28      0.50404350
  3.78155609260E29      0.35681677

This is the link to the essay: (essay)

[update]: Here is a picture of the function $ g(n)=\Large {{2^{m_{n,hi}} - 3^n \over 3^n }} \small \cdot 3n $ which seems to indicate a bounded interval which does not vanish in the limit of $n,m_n \to \infty$ up to $n \approx 10^{525} $ (at the convergents at the continued fraction) :