0
$\begingroup$

This question already has an answer here:

This question arose as an attempt to answer the following question Relaxed Collatz 3x+1 conjecture. I wanted to show that there is a solution of the equation $2^{k}=3^{z}(2n+1)-1$ for each $n\geq 2$, where $k,z,n\in\mathbb{N}$. But even a special case has put me in a dead end.

Are there infinitely many solutions of the equation $2^k=3^z-1$, when $z\rightarrow \infty$? First solution: $2^3 =3^2 -1$.

$\endgroup$

marked as duplicate by Benoît Kloeckner, j.c., Community Sep 16 '17 at 16:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ math.stackexchange.com/questions/958304/… $\endgroup$ – j.c. Sep 16 '17 at 15:48
  • 2
    $\begingroup$ This question is answered by the answers to Are there any solutions to $2^n-3^m=1$ as well as mathoverflow.net/a/29928 $\endgroup$ – j.c. Sep 16 '17 at 15:54
  • $\begingroup$ More generally, for fixed $a,b\in\mathbb Z$ with $|a|\ge2$ and $|b|\ge2$, there are only finitely many $k,z\in\mathbb N$ such that $$ a^k = b^z-1.$$ I understand why people closed this question, since the result is well known. On the other hand, the solution requires non-trivial results from Diophantine approximation (at least Thue 1909 result), and effective estimates for the problem may need linear forms in logs, so one could make the case that this question is appropriate for MO. $\endgroup$ – Joe Silverman Sep 16 '17 at 16:09
  • $\begingroup$ The question has two parts; only the second one (the special case) is a duplicate. The first part can be rewritten as $$ { 2^k+1\over w } = 3^z $$ with $w \ge 5 $ is odd. Of course, for every $k$ we find some $w$ giving a solution for $z$ (following from "Little Fermat" from where $k$ is a relatively simple function of $z$). However, whether every w can occur I don't know at the top of my head, but I think that actually many $w$ don't occur. $\endgroup$ – Gottfried Helms Nov 25 '17 at 11:49
4
$\begingroup$

The answer is no. Actually Catalan's conjecture, or Mihăilescu's theorem, suggests that the only solution of your equation $2^k = 3^z - 1$ is $k=3,z=2$.


As the comment below has mentioned, I missed another solution $k=z=1$.

$\endgroup$
  • 3
    $\begingroup$ Not "suggests" but "shows". $\endgroup$ – GH from MO Sep 16 '17 at 18:28
  • 3
    $\begingroup$ Don't forget about $k=z=1$ (as an answer to the original question). $\endgroup$ – Pace Nielsen Sep 16 '17 at 20:59