Can there be integer solutions (please PROVE) [closed]

Solution 1:

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 $n\le2$