On the notion of primary representation of a natural number by a quadratic form This "discussion" has to do with some of the material we can find in pages 183-186 of the translation into English of the first part of E. Landau's Vorlesungen über Zahlentheorie (published by the Chelsea Publishing Co. in 1966).


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*Let $k \in \mathbb{N}$ and $F=\{a,b,c\}$, where $a>0$, be an integral binary quadratic form of discriminant $d>0$ such that $F(x,y)=k$ for some $x, y \in \mathbb{Z}$. We say that the previous representation of $k$ by $F$ is primary if


$$2ax+(b-\sqrt{d})y>0, \quad 1 \leq \frac{2ax+(b+\sqrt{d})y}{2ax+(b-\sqrt{d})y} <\varepsilon^{2}$$
where $\varepsilon$ is the fundamental solution of the equation $t^{2}-du^{2}=4$.
I believe that Landau defines a primary representation of a natural number $k$ by a form $F$ in this somewhat peculiar way because it will allow him to single out, for every $\ell$ such that
$$\ell^{2} \equiv d \pmod{4k}, \quad 0\leq \ell < 2k,$$
exactly one (canonical?) primary representation of $k$.
What can you tell me about this appreciation of mine? Is it dead wrong? How do you usually motivate this definition when you teach out of this famous book?


*In pages 184-185, Landau proves the following:


Let us set 
$$w= \left\{ \begin{array}{cc} 
1 & \text{ if } d > \,\,\,\, 0 \\
2 & \text{ if } d < -4\\
4 & \text{ if } d=-4\\
6 & \text{ if } d=-3
\end{array}\right.$$
Then, for every $\ell$ such that
\begin{eqnarray}\ell^{2} \equiv d \pmod{4k},  \quad 0\leq \ell <2k,\end{eqnarray}
there are exactly $w$ proper primary representations of $k$ to which $\ell$ is associated (at this point of his exposition, Landau has already taught how to associate  to any proper representation of $k$ by a form a unique solution of $\ell^{2} \equiv d \pmod{4k}, 0\leq \ell <2k$, cf. Thm. 201 on page 180). 
I consider that, for negative $d$'s, this assertion can be proved more succinctly as follows.
Let us fix and $\ell$ such that $\ell^{2} \equiv d \pmod{4k}, 0 \leq \ell < 2k$. Then, if $m:= \frac{\ell^{2}-d}{4k}$ and $G :=\{k,l,m\}$, it follows that $G \sim F_{i}$ for exactly one form $F_{i}$ in a complete set of representatives of the equivalence classes of primitive and positive-definite forms of discriminant $d$. Then, it is not difficult to convince oneself that there is a bijective correspondence between the proper primary (proper primary = proper, in the case of negative discriminants) representations of $k$ to which $\ell$ is associated and the matrices $g\in \mathrm{SL}_{2}(\mathbb{Z})$ such that $gF_{i} = G$. Hence, since any such $g$ can be written in the form $g_{0}^{-1}A$ for a fixed $g_{0} \in \mathrm{SL}_{2}(\mathbb{Z})$ such that $g_{0}G = F_{i}$ and any automorph $A$ of $F_{i}$, it does follow that the number of proper representations of $k$ to which $\ell$ is associated is equal to  the number of different automorphs of $F_{i}$ (which is exactly $w$).
I suppose that Landau's proof turns out to be a wee bit long-winded because he is trying to handle simultaneously the cases $d<0$ and $d>0$, right? 
Please, let me thank you in advance for your attentive replies, comments, bibliographical suggestions, etc.
 A: I'm going to take a more algebraic number theoretic approach to this question. The solutions to $x^2-dy^2=k$ are the integers of norm $k$ in the number field $\mathbb{Z}[\sqrt(d)]$ if $d$ is a fundamental discriminant. Such integers are acted on by the unit group. To have a unique solution one must mod out by this action, but Landau singles out a unique representative instead. Those strange numbers $w$ are just the number of torsion units in various quadratic fields.
A: I still need to go to the post office. However, yesterday I drew pictures of $x^2 - 19 y^2 = 85$ in Conway's topograph method. The three tree diagrams below give, together, four solutions to $x^2 - 19 y^2 = 85$ that are distinct modulo the action by the "automorph" 
$$  (x,y) \mapsto (170 x + 741y, 39 x + 170 y) $$ or its inverse
$$  (x,y) \mapsto (170 x - 741y, -39 x + 170 y) $$
My guess is that Landau's pair of inequalities pick out either the four solutions below (85 in pink, $x/y$ in green) or the first two along with negating either $x$ or $y.$ 
Back in a bit.



