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I’ve been working on the Collatz Conjecture, and I believe I’ve reduced it to a more tractable problem. Unless there are some errors I’ve overlooked, I have managed to reduce the Collatz Conjecture to an exponential Diophantine equation(EDE). In an attempt to solve it, I’ve looked at linear forms in logarithms( but have been unsuccessful) and the proof of Catalans conjecture ( which again I could glean no pathway to solving this EDE).

$\forall k \in \mathbb{N} \exists a, b, c, m, n \in \mathbb{N} \cup \{0\} :$ $$\frac{3^{a} - 2^{b}}{2^{c}} = 1 - \frac{3^{m}}{2^{n}}k$$

Any tips, pathways to solutions, or outright solutions would be greatly appreciated! If you want to see how I reduced it to this EDE, then I’ll be happy to show you.

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    $\begingroup$ I fixed one formatting issue here, but all the subscripts make this hard to read, just as with your previous question. $\endgroup$
    – user44143
    Commented Apr 10, 2023 at 5:49
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    $\begingroup$ I am interested how you reduced to this EDE. $\endgroup$
    – SARTHAK GUPTA
    Commented Apr 10, 2023 at 6:12
  • $\begingroup$ I’ll add the link to vixra and my Quora space. Vixra: vixra.org/abs/2304.0050. My Quora space : mathscape.quora.com/… $\endgroup$
    – John Eaton
    Commented Apr 10, 2023 at 11:36
  • $\begingroup$ I’ve also tried looking at S-unit equations. Still doing some research though. $\endgroup$
    – John Eaton
    Commented Apr 17, 2023 at 2:38
  • $\begingroup$ How do the values of $a,b,c,m,n,k$ relate to "solutions" to the Collatz conjecture? Are we to assume you are proposing every solution to your equation represents a loop containing $m+n$ steps, of which $m$ are odd and $n$ even? If so, the number of solutions to your equation for which $k\in\Bbb Z_2^\times$ (i.e. a 2-adic unit) are enumerated by the base $2$ Lyndon words of length $n+m$. This is because the functions orbits are topologically conjugate to the tent map. $\endgroup$
    – Robert Frost
    Commented Jun 5, 2023 at 12:05

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