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It occurred to me that this might be straight forward to prove. I wanted to check if my reasoning was correct.

I will be glad if anyone points out a mistake or a simpler, more general way to make the same argument.

Let:

  • $m > 1, n> 1$ be integers
  • $2^m > 3^n$

Claim:

$(2^m - 3^n) \nmid (3^n - 2^n)$

Argument:

(1) $2^m - 3^n \ne 3^n - 2^n$ since:

  • Assume $2^m - 3^n = 3^n - 2^n$
  • $2^n(2^{m-n} + 1) = 2\cdot3^n$
  • which is impossible for $n > 1$

(2) Assume that there exists an integer $a>1$ such that:

$$a(2^m - 3^n) = 3^n - 2^n$$

(3) It follows that:

$$a(2^n(2^{m-n} - 1)) + a2^n - a3^n = 3^n - 2^n$$

$$a(2^n(2^{m-n} - 1)) = (a+1)(3^n - 2^n)$$

(4) But then:

$$(2^n(2^{m-n} - 1)) = \frac{a+1}{a}(3^n - 2^n)$$

which is impossible

Edit:

I missed the case where $2^n | (a+1)$. If I can find an argument against this case, I will post it. Otherwise, my argument is incomplete.

Larry Freeman
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    Your example supports the claim that $2^m - 3^n$ does not divide $3^n - 2^n$. $55$ does not divide $5$. – Larry Freeman Oct 13 '20 at 01:26
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    Take the sum of the quotient and show it diverges? $r=1$ if it's geometric. – floor cat Oct 13 '20 at 02:03
  • @floorcat could you provide additional details. It sounds interesting. I am not clear how it would work. – Larry Freeman Oct 13 '20 at 02:38
  • Is there a typo? Did you mean $2^m-3^n|3^m-2^n$? – fleablood Oct 13 '20 at 06:33
  • @fleablood I meant $2^m - 3^n$ divides $3^n - 2^n$. It is one of those points that seems like it is easy to prove but then it's not. $2^5 - 3^3 = 5$ is less than $3^3 - 2^3 = 19$ – Larry Freeman Oct 13 '20 at 06:46
  • @LarryFreeman FYI, your question is somewhat related to the Generalization section of Wikipedia's "Catalan's conjecture" article, i.e., the "... conjecture that for every natural number $n$, there are only finitely many pairs of perfect powers with difference $n$". If true, this means there's only a relatively limited # of small values which $2^m - 3^n$ can take, indicating any counter-example may be difficult to find and, if it any exist, will quite likely involve very large $2^m$ and $3^n$ values. – John Omielan Oct 13 '20 at 13:05
  • @JohnOmielan, So, an argument, if it exists, could well be significantly involved. Thanks. I always find it interesting when what appears to be a simple problem is actually related to a deeper mathematical problem – Larry Freeman Oct 13 '20 at 15:11
  • @LarryFreeman If you can represent the quotient as a geometric series, the value of r which stands for the distance between terms is equal to 1, and thus the series diverges. I'm just spit balling ideas here, this may not work. – floor cat Oct 13 '20 at 20:21

1 Answers1

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If you narrow your question to Collatz cycles (which I suspect to be the case), that is a question I asked myself (see EDIT point here: Showing $3^iq-2^{i-p}\neq2^pq-1$, with $p:=\lceil i\,\log_23\rceil$, $q:=\left\{\frac{2^{i}\,3^{2^{p-i-2}}}{2^p3^i}\right\}$, $i>6$) in the case of a non-trivial "1-cycle" of the Collatz conjecture. Note that this adds an additional constraint on $m$.

To have a one cycle you start at odd $a = b2^n - 1$, you climb up to $b3^n - 1$ and than divide by $2^{m-n}$ to reach $a$ again:

$$b2^n - 1=\frac{b3^n - 1}{2^{m-n}}$$

In other words $$a=\frac{(a+1)\frac{3^n}{2^n}-1}{2^{m-n}}$$ or $$a(2^m-3^n)=3^n-2^n$$

In a cycle, $\frac{3^n}{2^m}<1$ and with the above formula we can write

$$0 < \frac{2^m}{3^n} - 1 = \frac{2^{m-n} -1}{b3^n} = \frac{2^{m-n} - 1}{2^{m-n}(b2^n - 1) + 1} < \frac {1}{2^n}$$

For an $x>1$ you have $0<\log x<x-1$ so you can write the above formula as

$$0 < m \log 2 - n \log 3 < \frac{2^m}{3^n} - 1 < \frac {1}{2^n}$$

But with the Rhin bound: $|m\log 2 - n\log 3| > 0.00218n^{-13.3}$ and some manual check (for $n<96$ as shown by Steiner in 1977), we can see this is not possible ($2^m-3^n\nmid3^n-2^n$ at least with the additional constraint on $m$)

Collag3n
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