Where is the flaw in my logic in this collatz conjecture idea? Using this article from math stack exchange ( open problem - What does proving the Collatz Conjecture entail? - Mathematics Stack Exchange fourth answer down that starts "I think, for the question"), we can see that the problem to figure out if there are no non-trivial cycles in the Collatz conjecture is solving $\frac{3^N-2^N}{2^S-3^N} = a$ for $ N,S,a \in Z$.
Since a must be an integer, we can add one to our fraction, and still have integers. We add $\frac{2^S-3^N}{2^S-3^N}$ and get $\frac{3^N-2^N+2^S-3^N}{2^S-3^N}$ which reduces to $\frac{2^S-2^N}{2^S-3^N}$. Since the numerator has to be even no matter what and the denominator has to be odd no matter what, the denominator must be 1 to get an integer for a. The only possible solution (using Catalan's conjecture - Catalan's conjecture - Wikipedia) is $S=2,N=1,a=2$. Since we added 1 to a earlier, this shows that the only possible cycle ending for the Collatz conjecture is 1.
I'm guessing there is a flaw here somewhere, but I figured I'd put it out there to find out what it is.
 A: Well, as @MiloBrandt has already mentioned, that formula was about the so-called 1-cycles, a special simple form of general cycles.
But your operation gives a nice aspect, which helps to make the disproof of 1-cycles a bit simpler.
We start with a supposed solution for an 1-cycle of length $N$ odd steps and $S$ as sum of the involved exponents of $2$:
$$ a = { 3^N - 2^N \over 2^S - 3^N } \tag {1.1}
$$
Now you add $1$ to each side
$$ a + 1 = { 3^N - 2^N \over 2^S - 3^N } +1 ={ 3^N - 2^N + 2^S-3^N \over 2^S - 3^N } ={ 2^S - 2^N  \over 2^S - 3^N } \tag {1.2}
$$
Now we can extract the factor $2^N$ on the rhs and divide:
$$ {a + 1 \over 2^N} = { 2^{A-1} - 1 \over 2^S - 3^N } \qquad \Tiny \text{ where } A-1=S-N \tag {1.3}
$$
Because the denominators have no common factor ,$a+1$ must be a $k$'th multiple of $2^N$ , so the initial (and smallest by definition) element of a 1-cycle must have the form $a=2^N\cdot k -1$ (with $k \gt 0$). So, for a 1-cycle of length $N=10$ the smallest element must be equal or larger than $1023$ (actually of the form $ 1023 + k \cdot 2^{10}$).

From another formula, i.e. that for the general cycle on elements $a_1,a_2,...a_N$ we must have
$$ 2^S = (3+1/a_1)(3+1/a_2) \cdots (3+1/a_N) \tag {2.1}$$
we can for each given $N$ determine some (roughly) mean value $\alpha$ of all $a_k$ by
$$ 2^{S/N} = (3+ 1/\alpha) \\ 
   \alpha = {1 \over 2^{S/N}-3  } \tag {2.2}$$
Here (by (2.1)) $2^S$must always be larger than $3^N$, and we take the smallest solution for $S$:
$$ S= \lceil N \cdot \log_2(3)  \rceil = 16 \tag {2.3}$$
This gives for the mean-value $\alpha = 1/(2^{1.6}-3) \approx 31.81 $ (!).
But by the definition of the cycle (be it general or 1-cycle) $a_1$ is the smallest element and thus it must also be $a_1 \lt \alpha$.
Since for the 1-cycle of length $10$ the smallest element $a_1\ge 1023$ is required, we cannot have a 1-cycle of length $N=10$.

This style of consideration can be generalized, and it can be shown that the discrepancy from the requirements for $a_1$ in a 1-cycle of length $N$ ($a_1=2^N k -1$) and the upper bound $\alpha$ for $a_1$ (by 2.3) for a general cycle of length $N$ even increases (heuristically) and this can then lead to the proof that no (nontrivial) 1-cycle can exist at all in the positive numbers.
(A bit more explanation I have in a small treatize 1cycledisproof )
