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*



When is $\{b^2 - \{b-1\}_2\}_2=1$ with odd $b$? (The bracket-notation explained below)
The question is equivalent to finding all squares which are of the form $2^a - 2^b + 1$. [This MO answer](https://mathoverflow.net/questions/126284/binary-expansion-of-squares) 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.


