How is this set of solutions for exponential diophantine equations been found?(Generalized Collatz mx+1, 2-odd-step-cycle) I'm doctoring many weeks on the problem of the existence and set of solutions of 2-step-cycles in the generalized Collatz-problem, written in the Syracuse-form:
$$  Y_m(a): b =  {ma+1 \over 2^A}  \qquad\qquad  a,b,m \in Z \text{ odd }, A =\nu_2(ma+1)
$$
We discuss the existence of 2-step-cycles of the form
$$   b = {ma+1 \over 2^A } \qquad a = {mb+1 \over 2^B } \qquad \qquad  B =\nu_2(mb+1)
$$
To see, for which numbers $m,a,b$ solutions exist, I've constructed an equation
$$  a \cdot b = ({ma+1 \over 2^A })({mb+1 \over 2^B }) \\
    2^{A+B} = (m+{1 \over a })(m+{1 \over b }) 
$$
$$
    2^S = (m+{1 \over a })(m+{1 \over b })  \qquad \qquad \text{always } S = A+B \tag 1
$$
$$ \small \text{Note: this equality is necessary but -in the generalized version $mx+1$ } \\ \small \text{with $m \in \mathbb Z /2\mathbb Z$ - not sufficient to have a 2-step-cycle defined.}
$$
I was unable to solve this problem in general. To get a clue, I reduced complexity, assuming $a=1$ and $b \ne 1$. Still I could only solve the further reduced problem with $b=3$, finding that $m=5$ is required, (and $S=5, A=1, B=4$ follows) and that this is the only solution.
Reduced, but still unsolvable for me so far:
$$  2^S=2\cdot 4^T = (m+1)(m+{1 \over b })  \qquad \qquad \text{always } S = A+B =2T+1 \text{ odd} \tag 2
$$
What I can do so far, is, that $m$ is odd, so $m=2m'+1$ and $m'+1$ must be a multiple of $b$ such that for instance $m'+1=k \cdot b$ with some $k$ and we get
$$  2\cdot 4^T = (2m'+2)(2m'+1+{1 \over b })  \\
 2\cdot 4^T = 2(m'+1)(2m'+1+{1 \over b })  \\
 4^T = (m'+1)(2m'+1+{1 \over b })  \\
$$ getting
$$ 4^T = k\cdot(2b(m'+1)-b+1) = k\cdot(2kb^2-b+1)  \tag 3
$$
Here $k$ must be a perfect power of $2$.
But now, fiddling starts and I don't get progress...
Today I fed this problem, leaving $b$ indeterminate, to Wolfram Alpha and got the following set of solutions:
W/A:    m=11  S=7  b=-3  (m'=5  T=3)    check   128=2∙6∙(10+1+1/-3)=4∙(33-1)
W/A:    m=-3  S=3  b=-1  (m'=-2 T=1)    check   8=2∙-1∙(-3+1/-1)
W/A:    m=3   S=3  b=-1  (m'=1  T=1)    check   8=2∙-2∙(-3+1)
W/A:    m=5   S=5  b=3   (m'=2  T=2)    check   32=2∙3/3∙(15+1)

Hmm. Not only W/A found this solutions, it seems this are also all possible solutions. (Note, btw, that the solution with $m=11$ is a solution for (eq 2) but $b=-3$ is not actually an iteration of $a=1$, so in fact it is of course not a solution in this problem.)
My question: How can this result be derived in this generality? And how is it then determined that the set of solutions is finite?
 A: This is how far my own approach brought me. Don't know whether this road arrives at its goal, though...
First I show, that for each $S$ there is a finite (and small) number of solutions.
From (eq 1) in my question,
$$   2^S = (m+\frac1a)(m+\frac1b) \tag 1
$$
we can derive an upper bound for $a$. First we do not loose generality, if we fix $a<b$ so there is some mean value $a_m$ in between, which also satisfies
$$   2^S = \left(m+\frac1{a_m}\right)^2  \tag 2
$$
and
$$   a_m = {1\over 2^{S/2}-m} \tag 3
$$
Next we find that $m$ (being odd) is uniquely determined by a given $S$.

*

*($S$ even:) We see in (eq 2) , that if $S=2T$ is even, then $2^{T}$ is a perfect power of $2$ and $2^T - \frac1{a_m}=m \le 2^T-1$. $m$ must be smaller than $2^T$ and because it must be integral we have, that $m=2^T-1$ .
By this $m$ is automatically odd, and then $\frac1{a_m}=1=a_m$ , and it follows $a=b=1$. Thus we have the trivial cycle of one odd step, and no ("primitive") two-step-cycle exists.


*($S$ odd:) If $S=2t$ is odd, we have still from (eq 2) $2^{t}-\frac1{a_m} = m \gt 2^t-1$ but now we have a unique integer between $2^t$ and $2^t-1$ which is then $m$ by which we have: $m=\lfloor 2^t \rfloor $ . Here it can happen, that $m$ is even, and in this case that odd $S$ has no 2-step-cycle in $b>a>0$ integer.
So we know, we need only look at a subset of the odd $S$ whose elements allow $m$ being odd. On the other hand, there is no reason to assume that this subset is finite, so we must consider that we have to check an infinite number of $S$ - if we don't find one better argument against the existence of 2-step-cycles for $S \gt S_0$ with some $S_0$ a large value.
To actually find some 2-step-cycle with integer $0<a<b$ we have the crude upper bound $a \lt a_m$ in (eq 3), which heuristically is often small, even in most cases allow only $a=1$ as possible candidate. In other cases, if for some odd $S=2t$, the distance $2^t-m=\frac1{a_m}$ is small, we have a large $a_m$ and thus a large upper bound for $a$ and our consecutive checking for odd $a$ in the range $1..a_m$ needs still finite, but for (surely) infinitely many large $S$ also unmanageable long time for the checking. Because the runlength of consecutive zeros in the binary representation of $\sqrt 2$ determine the value of $a_m$ and that runlengthes are unbounded when we walk into that infinite bit-sequence we can expect arbitrary large $a_m$, at least bounded by some function on $S$. (For that latter functional bound I've not seen anything in literature and have only some numerical tests up to $S$ of some million, done on my own.)
Well, I've found some options to reduce the range $1..a_m$ as searchspace for $a$ : having some nontrivial lower and upper bounds in that range. But this alone does not allow to assume something like a lower bound for $S$ , above which no more 2-step-cycle can occur.
So surely one must go back to and deeper into the algebraical properties in (eq 1) to find a general statement about the existence of 2-step-cycles in the $mx+1$-problem. So, the "fiddling" as mentioned in my question-text, began/begins... (So that's my surprise that the W|A-result seems to suggest, that they have something like this)
