Proving a solution to a double recurrence is exhaustive

Here we take $W=2A+1.$

The structure of the $\mathbb Z$ automorphism group of an indefinite binary quadratic form is discussed in many places in many ways. In brief, the proper automorphism group is infinite cyclic along with $\pm 1,$ so one commonly switches attention to $PSL_2 \mathbb Z.$ Then, as always applies for Pell forms $x^2 - D y^2,$ we throw in a single improper automorph, $(x,y)\mapsto (x,-y).$ My favorite book that discusses the group is Binary Quadratic Forms by Duncan A. Buell. Another one, some differences in notation, also called Binary Quadratic Forms, by Buchmann and Vollmer.

J. H. Conway introduced the topograph method for indefinite binary quadratic forms in The Sensual Quadratic Form. There is a brief presentation in Elements of Number Theory by John Stillwell. I have taken a sort of overlap of the two presentations, in which I play down the middle coefficients of the forms but emphasize the specific solutions to $x^2 - D y^2 = n,$ along with incorporating the automorphism group of the form in the diagram. So, below are all solutions to what I called $W^2 - 8 B^2 = -7.$

Meanwhile, acting on column vectors, the generator of the automorphism group of the form is $$ M = \left( \begin{array}{rr} 3 & 8 \\ 1 & 3 \end{array} \right). $$

We call $M$ an automorphism (proper) of the quadratic form because $\det M = 1$ and $$ \color{red}{ M^T G M = G}, $$ where $$ G = \left( \begin{array}{rr} 1 & 0 \\ 0 & -8 \end{array} \right). $$

Note that $M$ is a 2 by 2 matrix, determinant $1$ and trace $6.$ By Cayley-Hamilton, we have $$ M^2 - 6 M + I = 0, $$ or $$ M^2 = 6M - I. $$ As a result, put all solutions $(W_n, B_n)$ in sequence, and we get $$ W_{n+2} = 6 W_{n+1} - W_n, $$ $$ B_{n+2} = 6 B_{n+1} - B_n. $$ As you can see in the diagram, one may see how the values for $-7$ actually split into two interleaved sequences according to these recurrences: $$ W_n = 5,31,181,... $$ and $$ W_n = 1,11,65,... $$ We also get two threads for the $B's,$ $$ B_n = 2,11,64,... $$ and $$ B_n = 1,4,23,... $$ Sometimes the two threads can be combined, usually not. As you can see in the diagram below, the two threads are caused by an improper automorph, this time let us call it $(x,y)\mapsto (-x,y).$

Alrighty, I thought it might help to fill in as much of the complete diagram as would fit on my original. Properly speaking, every edge of the infinite graph, actually a tree, should have either a $0$ or a positive integer and a little arrow. I put many of those in, using orange, which is a little too close to the red I used for each occurrence of $(-7).$ Can't win them all. Also, as we move away from the "river," that being the central line, we simply add coordinates. For example $(1,0)+(3,1)= (4,1).$ On the negative side,$(-2,1)+(-1,1)= (-3,2).$ Worth emphasizing that there are two types of phenomena that give finiteness for the number of "orbits" that give $-7,$ the second of which is the "climbing lemma," that the absolute values of the numbers only increase when we move away from the river. So, in the first layer of negative values, we get $-4,-7,-8,-7,-4,$ and so on. In the second layer, possibly hard to read, we get only $-23,-31,-31,-23,$ over and over. So that is it, any further layers on the negative side have absolute values strictly larger than $7.$