Collatz Conjecture: Does it follow that since there is no $1$-cycle, if $ab > 1$, then $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ cannot be an integer? For me, one of the most interesting results of the Collatz Conjecture is that if an $m$-cycle exists, then $m > 68$.  So, there are no non-trivial $1$-cycles, $2$-cycles, ..., up to $68$-cycles.  The details behind this argument can be found in this paper by Simons & De Weger (2003).
Let:

*

*$\nu_2(x)$ be the 2-adic valuation of $x$.

*$u = 3^kt - 1$

*$b = \nu_2(u)$

*$a = \nu_2(\frac{u}{2^b}+1)$
If a $1$-cycle exists, then the maximum value of the cycle is equal to $3^kt - 1$ where $k,t$ are positive integer and the lowest value of the cycle is equal to $2^kt - 1$ characterized by the following:
$$u = \left(\frac{3}{2}\right)^a\left(\frac{u}{2^b} + 1\right)-1$$
So that:
$$\left(\frac{2}{3}\right)^a(u+1) = \frac{u}{2^b} + 1$$
And:
$$\frac{u(2^{a+b} - 3^a)}{3^a2^b} = \frac{3^a - 2^a}{3^a}$$
Which reduces to:
$$\frac{u}{2^b} = \frac{3^a - 2^a}{2^{a+b} - 3^a}$$
It seems to me that since no non-trivial $1$-cycle exists, it follows that if $a,b$ are positive integers where $ab > 1$, $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is not a positive integer.
Here's the argument:
(1) Assume that there exists positive integers $a,b$ with $ab > 1$ such that $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is an integer.
(2)  Let $x = \frac{3^a - 2^a}{2^{a+b} - 3^a}$
(3)  Let $u = 2^bx$
(4)  Let $k = a$, $t = \left(\frac{1}{2^a}\right)\left(\frac{u}{2^b}+1\right)$ which are both integers.
(5)  So it follows that $u = 3^kt - 1$
(6)  But then we have a contradiction since no $1$-cycle exists.  Therefore, we can reject our assumption at step(1).
Is this argument valid?  Does the result from Simons & De Weger (2003) prove that with the given assumptions, $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is not an integer?

Edit:  I added the clarification that $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ cannot be a positive integer.
@rukhin makes a great point.  For positive integers $a=2$ and $b=1$, $\dfrac{3^a - 2^a}{2^{a+b} - 3^a} = -5$
This is correlated with the negative $1$-cycle with $k=1$ and $t=\left(\frac{1}{4}\right)\left(\frac{-10}{2} + 1\right) = -1$
The standard collatz sequence defined as:

*

*$T(x) = \begin{cases}
\frac{1}{2}(3x + 1), && \text{if }x\text{ is odd}\\
\frac{1}{2}x, && \text{if }x\text{ is even}\\
\end{cases}$

*$T_1(x) = T(x)$

*$T_{c+1}(x) = T(T_c(x))$
So that:
$$-1, -2, -1, \dots$$
 A: Say you start with $n_0=2^kt-1$, apply $a$ times the Collatz function $f(n)=\frac{3n+1}{2}$, than $b$ times the Collatz function $f(n)=\frac{n}{2}$, you can easily see that you end up with $$n=\frac{3^a}{2^{a+b}}n_0+\frac{3^a-2^a}{2^{a+b}}$$
or
$$n\cdot{2^{a+b}}-3^a\cdot n_0=3^a-2^a$$
and in the case of a cycle $n=n_0$
$$n=\frac{3^a-2^a}{2^{a+b}-3^a}$$
Note: $a=k$, the value $a = \nu_2(\frac{u}{2^b}+1)$ you gave is for the next iteration
Now in case of a cycle, it has been shown that (you can find it by noticing that $\frac{1}{2}<\frac{3^a}{2^{a+b}}<1$):
$$a+b=\lceil a\cdot \log_2(3)\rceil $$
So indeed, if there is such an integer $n=\frac{3^a-2^a}{2^{a+b}-3^a}$, it must be part of a 1-cycle, and you can conclude there is none (outside the trivial one), but it would only be valid within the above constraint on $a+b$.
Also notice that the question has already been raised.....by you
Is this argument valid for showing given integers $n>1$ and $2^m > 3^n$, $(2^m - 3^n) \nmid 3^n - 2^n$
Also note that a m-cycle does not mean m successive 1-cycles, so there is no such constraint on $a+b$ within the m-cycle (but in this case you don't have $n=n_0$ or the equation you mentioned)
