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With the following equation-
$\frac{(3^x-2^x)}{(2^{(x+y)}-3^x)} = z $ What $x,y,z$ (all integers) satisfy this?
The trivial solutions are- $X=1,y=1,z=1 ,X=2,y=1,z=-5 ,X=1,y=0,z=-1 $

This is my own “restatement” of the impossibility of the $k=1$ cycle for collatz sequences and it would be incredibly helpful if someone could prove it in this form (it should be possible to prove that no integer solutions for $x,y,z$ exists other than the trivials)

Italian
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Charlie D.
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  • Perhaps this discussion of mine suffices: https://go.helms-net.de/math/collatz/Collatz_1cycledisproof.pdf . I'm applying the approximation-bound given by G. Rhin (see also the answer of @Collag3n) – Gottfried Helms Dec 16 '21 at 22:22

1 Answers1

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A "1-cycle" in Collaz implies

$$\frac{2^m}{3^n} - 1 < \frac {1}{2^n}$$ $$2^m-3^n < (\frac {3}{2})^n$$ $$2^m-3^n < 1.5^n$$

but from this we know that ($n>5$)

$$2^m-3^n \ge 2^{\lceil n \cdot log_2(3)\rceil}-3^n> 2^n$$ $$2^m-3^n > 2^n>1.5^n \implies \text{contradiction}$$

So for the value of $m=x+y$ and $n=x$ that you consider (Collatz cycle where $m$ is believed to be equal to $\lceil n\cdot log_2(3)\rceil$), and outside the exception for $n=1$, we have

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

EDIT

Another approach for the first part in the context of a 1-cycle ($a,b$ from the link): $$\frac{3^n-2^n}{2^m-3^n}=a=b\cdot2^n-1\ge2^n-1$$ $$\implies 2^m-3^n \le \frac{3^n-2^n}{2^n-1}<\frac{3^n}{2^n}$$ $$\implies 2^m-3^n < 1.5^n$$

For negative values, use the same reasoning (but here $2^m<3^n$):

$$\frac{3^n-2^n}{2^m-3^n}=a=-b\cdot2^n-1\le -2^n-1$$ $$\implies 2^m-3^n \ge \frac{3^n-2^n}{-2^n-1}$$ $$\implies 3^n-2^m \le \frac{3^n-2^n}{2^n+1}<\frac{3^n}{2^n}$$ $$\implies 3^n-2^m < 1.5^n$$

but from the same link above ($n>2$ and $m$ is smaller than $\lceil n\cdot log_2(3)\rceil$)

$$2^{m+1}-3^n\le2^{\lceil n\cdot log_2(3)\rceil}-3^n<3^n-2^n$$ $$2\cdot 2^m-2\cdot 3^n<-2^n$$ $$3^n-2^m>2^{n-1}>1.5^n \implies \text{contradiction}$$

with the exception you found for $n\le2$

Collag3n
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  • I’m sorry what are m and n in this? – Charlie D. Dec 16 '21 at 11:55
  • @Charlie D. ,I used the notations from the links, but m=x+y and n=x – Collag3n Dec 16 '21 at 11:59
  • @Charlie D. , I hope the edit clarify what is in the link. In short, in the context of a 1-cycle, there is no value of $m\ne2$ allowing an integer solution to your fraction. Note that it does not mean there is no possible integer solution outside this context (when there is no constraint on $m$), see https://en.wikipedia.org/wiki/Catalan%27s_conjecture#Generalization – Collag3n Dec 16 '21 at 14:41
  • Sorry but this form has y as the number of divisions of two in the cycle, so if there is integers solutions outside that would be a spanner in the works. – Charlie D. Dec 16 '21 at 21:33
  • @Charlie D. in the context of a 1 cycle, $y$ is determined by $x$ ($y>=\lceil x\cdot log_2(\frac{3}{2})\rceil$), and with this constraint, there is no integer solution has shown. Outside the 1-cycle context (y can be anything), nobody knows if there is an integer solution or not, but this is not important if you are only interested in the 1-cycle context. 1 cycle forces $\frac{3^n-2^n}{2^m-3^n}\ge2^n-1$ which is needed to prove your point. Why not use it? – Collag3n Dec 16 '21 at 21:41
  • Oh wow I was confused before but this makes sense thanks! – Charlie D. Dec 16 '21 at 21:47