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).

1. 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?

1. 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?

• try it for Pell type equation, such as $x^2 - 19 y^2 = 85.$ Under the (oriented) automorphism group there are four orbits of solutions. A representative for each orbit can be picked by inequalities; it ought to be more clear, with $a=1, b=0, c = -19$ just what inequalities are meant. – Will Jagy May 23 '17 at 17:49
• 170^2 - 19 39^2 = 1 u^2 - 19 v^2 = 85 Tue May 23 10:49:27 PDT 2017 u: 16 v: 3 ratio: 5.33333 SEED KEEP +- u: 79 v: 18 ratio: 4.38889 SEED KEEP +- u: 92 v: 21 ratio: 4.38095 SEED BACK ONE STEP 79 , -18 u: 497 v: 114 ratio: 4.35965 SEED BACK ONE STEP 16 , -3 – Will Jagy May 23 '17 at 17:50
• @WillJagy: Would you be so kind as to elaborate a little? – José Hdz. Stgo. May 23 '17 at 18:27
• I have some more errands to do, but I can get to this later. I do not know Landau's specific inequalities but I can likely work it out – Will Jagy May 23 '17 at 19:12
• requested Landau book from library, might get it this week. If not, Tuesday, as Monday is a holiday here. It is not difficult to set up inequalities such that: every representation $ax^2 + bxy+ cy^2 = k$ can be acted on by the automorphism group ( an "equivalent" solution, in the same "orbit") to produce a solution that satisfies the inequalities; next, no two solutions that satisfy the inequalities are "equivalent" under the group. After fiddling yesterday, I suspect Landau's way of doing this is different from mine. – Will Jagy May 24 '17 at 15:27

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.
• Thanks a lot for taking the time to leave a reply to this question of mine. In the case of negative discriminants, the $w$'s count the number of automorphs of any primitive form $ax^{2}+bxy+cy^{2}$ of disc. $d$; this number turns out to be exactly equal to the number of solutions of the equation $t^{2}−du^{2}=4$. Now, I have a question: in the beginning of your reply, did you mean to write "a more algebraic number theoretic"? – José Hdz. Stgo. May 28 '17 at 2:11
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 $y.$