3

Pillai's conjecture is a generalization of Catalan's conjecture. It's say that for fixed positive integers $A, B, C$ the equation $Ax^n - By^m = C$ has only finitely many solutions $(x,y,m,n)$ with $(m,n) ≠ (2,2)$.
But if we fix $A=B=1$ and $x=2,y=3$ then we have :
$$2^n-3^m=C$$

My question : Have this equation only finitely many solution for any fixed integer $C$?

For $C=1,-1$ you can see this answer.

Community
  • 1
  • This could be rephrased as "does there exist an infinite subset of $\mathbb N\times\mathbb N$ on which $(n,m)\mapsto 2^{n+2}-3^{m+2}$ is bounded?" – Thomas Andrews Feb 05 '16 at 17:49

3 Answers3

5

Yes, this follows from the following more general result by taking $P = \{2,3\}$:

Let $P$ be a finite set of primes, and let $S$ be the set of natural numbers whose prime factors are entirely contained in $P$. Then the gaps between elements of $S$ grows to infinity, that is for each $k$ there are only finitely many distinct pairs of $x,y \in S$ such that $|x-y| < k$.

In other words, the set of $r$-smooth numbers for any fixed $r$ is very sparse. This result can be deduced easily from standard theorems on finiteness for high-genus polynomial Diophantine equations. It was previously discussed on this site here:

Gap between smooth integers tends to infinity (Stoermer-type result)?

And some of the details were described in my answer to a related question:

https://math.stackexchange.com/a/725149/30402

Community
  • 1
Erick Wong
  • 26,108
3

Not an answer, but a start.

You can show that, if there is such a $C$ with infinitely many solutions, then there is a constant $K$ such that for all $m,n$ in your solution set:

$$\left|\frac m n-\log_2 3\right|<\frac{K}{n3^n}$$

This means that the continued fraction expansion for $\log_2 3$ must have lots of absolutely huge coefficients.

I think you might be able to prove the converse - if there is such a $K$ and infinitely many $m,n$ satisfying this inequality, then there is a $C$ with infinitely many solutions.

Thomas Andrews
  • 188,543
  • 1
    Baker’s Theorem shows that in fact $|m \log 2 - n \log 3| \gg n^{-C}$ for some constant $C$. – Erick Wong Jan 07 '19 at 19:52
  • @ErickWong : ... so that we could make the chain-inequality $n^{-C} \ll | m\log2-n \log 3 | < \log 2 \cdot K/3^n $ but which must be wrong from some lower bound of $n$ given some constants $C$ and $K$ ? – Gottfried Helms Oct 30 '20 at 03:28
  • @GottfriedHelms Yes, hence only finitely many solutions. But from my answer it can be deduced without the power of Baker’s theorem using older technology. – Erick Wong Nov 02 '20 at 18:30
0

The following paper, Pillai's conjecture revisited by Michael A. Bennet answers this question as yes, there is at most one solution for any c in your question apart from c=-1,5,13 which each have 2 solutions. https://reader.elsevier.com/reader/sd/pii/S0022314X02000495?token=5A3168779EC465B20F7BDA91EA5E568C90D1F648FCF17363B85AD8B8698813DC689A589AECE08EEE82C09486EAD4FAB6&originRegion=eu-west-1&originCreation=20220121224632

SFA
  • 395
  • 1
    While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review – Elliot Yu Jan 22 '22 at 01:32