When is $\{b^2 - \{b-1\}_2\}_2=1$ with odd $b$? (The bracket-notation explained below) For the complete extraction of the factor $p$ and its powers from a natural number $n$
let's define the notation $$ \{n\}_p := { n \over p^{\nu_p(n)}} \tag 1$$
$ \qquad \qquad $ Here $\nu_p(n)$ means the p-adic valuation of $n$ with respect to prime $p$.
While looking at the question of existences of 2-step-cycles in the generalized Collatz-problem for $mx+1$ I arrived, considering general divisibility conditions, at the question for which odd $b$ the evaluation of $f(b)$
$$ f(b) = \{b^2 - \{b-1\}_2 \}_2  \overset{?}= 1 \tag 2
$$
fully reduces to $1$.
I know, that $b=3$ there is a solution (the famous $(3^2-1)/2^3=1$). Generating a list for sequential $b$ didn't suggest any useful pattern so far, so this didn't help me with a clue.
For me it looks suspiciously like the problem might be of wider interest instead only for the Collatz-cycle sub-problem, where it occured.

Q1: Can we find a modular/diophantine argument for the impossibility of more solutions?


Q2: Additionally, can we say something more, if we use some other prime factor
$p \ne 2$ for extraction $$ \small f_p(b) =  \{b^2 - \{b-1\}_p \}_p  \overset{?}= 1 \tag 3$$?

(Remark for Q2: One solution I found was $f_7(19)=1$, another one, but with even $b$, I've got $f_5(56)=1$  )

remark: I've asked this in MSE a couple of days before but except a short observation didn't get an answer
 A: For Q1:
Let $b=2^kt+1$ with odd $t$. Then equality $f(b)=1$ is equivalent to
$$(2^kt+1)^2-t=2^m$$
for some $m$ (notice that $m>2k$).
This is a quadratic equation in $t$. To have integer solutions, its discriminant must be a square, that is
$$2^{2k+2+m}-2^{k+2}+ 1=z^2$$
for some $z>0$.
There are no solutions with even $m$ since
$$(2^{k+1+m/2})^2 > 2^{2k+2+m}-2^{k+2}+ 1 > (2^{k+1+m/2}-1)^2.$$
For odd $m$, we have
$$2^{k+2} - 1 = 2^{2k+2+m}-z^2 = (2^{k+1+(m-1)/2}\sqrt{2}-z)(2^{k+1+(m-1)/2}\sqrt{2}+z) > (2^{k+1+(m-1)/2}\sqrt{2}-z)2^{k+2+(m-1)/2}(\sqrt{2}+1),$$
implying that
$$0 < \sqrt{2}-\frac{z}{2^{k+1+(m-1)/2}} < \frac{1}{2^{k+m-1}(\sqrt{2}+1)} < \frac{1}{(2^{k+1+(m-1)/2})^{3/2}}.$$
That is, $\frac{z}{2^{k+1+(m-1)/2}}$ represents an order $3/2$ rational approximation to $\sqrt{2}$, which suggests (although this is just a heuristics) that we should look for such fractions among convergents and semiconvergents to $\sqrt{2}$. Their denominators are given by Pell numbers (A000129) and sums of consecutive Pell numbers (A001333). The latter are odd, while it is easy to establish that the former contain just two powers of $2$ -- namely, $1$ and $2$.
PS. The known solution gives $z=11$, $k=1$, $m=3$ with the rational approximation $\frac{11}{8}$ to $\sqrt{2}$ not being a convergent or a semiconvergent. So, above is not a proof but a heuristic argument that such approximations are extremely rare, and cannot be obtained from the known sources (convergents and semiconvergents) for good rational approximations.
A: The question is equivalent to finding all squares which are of the form $2^a - 2^b + 1$. This MO answer discusses the related question which squares are of the form $2^a + 2^b + 1$, and presumably the techniques in the paper (which admittedly I haven't read) should be applicable in your case as well.
EDIT: The paper referenced in the MO answer is "The equations 2N ± 2M ± 2L = z2" by László Szalay, which should answer your question as well. Unfortunately, the link there doesn't work for me, but I hope this helps.
A: The article of L. Szalay as mentioned by user Random, gives the solution.
We can reformulate
$$ \begin{array} {} 
 \{b^2 - \{b-1\}_2 \}_2 & \overset{?}= 1 &&& (1.1)\\
  b^2 - \{b-1\}_2  & \overset{?}= 2^A & &&(1.2)\\ 
  2^B b^2 - (b-1)  & \overset{?}= 2^B 2^A & &&(1.3)\\ 
  4 \cdot 2^{2B} b^2 - 4\cdot 2^Bb + 1  & \overset{?}= 2^{2B+2} 2^A &- 2^{B+2}&+1& (1.4)\\ 
  (2\cdot 2^Bb - 1)^2  & \overset{?}= 2^{A+2B+2} &- 2^{B+2}&+1& (1.5)\\ \hline
   x^2 & = 2^n &- 2^m &+1& (2)\\
 \end{array}$$
Szalay gives the following solutions in theorem 2:

*

*$ \displaystyle (n,m, x)\in \{ (2t,t+1,2^t-1) \mid t \in \mathbb N, t \ge 2 \}$.
Inserting this gives by $x=2^t-1$ that $b=1$, which is of course a trivial solution for $f(b)=1$.


*$ \displaystyle  (n,m, x) \in  \{(t,t,1) \mid t \in \mathbb N, t \ge 1 \}$
Here, to have $x=1$ we must have $2\cdot 2^Bb - 1=1$ and $ 2^Bb =1 \implies b=1, B=0$
which is again the trivial solution, and


*$ \displaystyle  (n,m, x) \in \{(5,3,5),(7,3,11),(15,3,181) \}$
For $x=5$ we get $b=3$ which is the "known" solution, for $x=11$ we get again $b=3$ and finally for $x=181$ we get $b=91$ . By the latter we have in the lhs that $B=0$ but in the rhs we have with $m=3$ that $B=1$, which is in contradiction to the specific form in which my equation realizes Szalay's (with additional restrictions on $x$), so this is not an allowed case.
Conclusion: the two cases $b=1$ (trivial) and $b=3$ (known) are the only solutions for the equation in question.
Underlying problem solved: For the generalized Collatz-problem $ T(x): x \to \{mx+1 \}_2 $ this means:
there are exactly the following $2$-step-cycles with $b=T(a)$ and $a=T(b)$ when  $a=1$ is fixed:

*

*for $m$ of the form $m=2^k-1$  the trivial 2-step-cycle $a=b=1$  and

*for $m=5$  and $(a,b,m)=(1,3,5)$
(The known 2-step-cycles for $m=181$ are not counted because their smallest element $a$ is  $a \ne 1$)

Thanks to user "Random" for the reference to Szalay's article.
