22
votes

Accepted

### reference for: no finite set of positive (integer) binary quadratic forms represents all primes

This is indeed correct; I don't know a reference, but here's a proof. Let ${\mathcal D}$ be a finite set of $K$ negative fundamental discriminants. We want to show that the set of primes not ...

21
votes

Accepted

### Is the diophantine equation $3x^2+1=py^2$ always solvable for each prime $p\equiv 13\pmod{24}$?

Updated on 2019/08/21: I prove Conjectures 1-3 below. I will use the usual terminology and notations for binary quadratic forms. In particular, I will use the first two pages of Pall: Discriminantal ...

18
votes

Accepted

### Infinite cyclic subgroups of $\text{SL}_2(\mathbb{Z})$

The maximal infinite cyclic subgroups are of the form you mentioned (to get all infinite cyclic subgroups you also need to include their finite index subgroups). This follows from Theorem 1.4 in ...

16
votes

Accepted

### Primes of the form $x^2 + y^2 + 1$

Iwaniec (Acta Arith. 1972) showed that the number of such primes has the order $x/(\log x)^{3/2}$. His result applies more generally to translates of binary quadratic forms. One can also show that ...

15
votes

### $x^2+7y^2=2^n$ and sums of four squares

The Diophantine equation $x^2 + 7y^2 = 2^n$ is "trivial" in the sense that its solutions can be described in a very simple way (see boxed formula near the end for solutions in odd numbers ...

11
votes

Accepted

### Equidistribution of CM points in the principal genus

This is known, and follows from Theorem 2 in Harcos and Michel's paper The subconvexity problem for Rankin-Selberg $L$-functions and equidistribution of Heegner points. II (Invent. math., vol. 163, ...

10
votes

### Equidistribution of CM points in the principal genus

Vesselin Dimitrov gave a nice answer, but let me point out that for the OP's question one does not need the result of Harcos-Michel (2006). Instead, the original work of Duke (1988) or alternatively ...

10
votes

Accepted

### A conjecture for primes $p\equiv\pm1\pmod5$

Let us consider $\mathbb{Q}(\sqrt{5})$ as a subfield of $\mathbb{R}$. Let us also consider the positive fundamental unit $\epsilon:=(1+\sqrt{5})/2$, whose square generates the group of totally ...

10
votes

Accepted

### The mean value of $y \log{y}$ over the ordinates of the CM points

Let $g : \Gamma \backslash \mathbb{H} \to \mathbb{C}$ be any bounded continuous function. Duke's theorem states that
\[\frac{1}{h(D)} \sum_{A \in \mathrm{Cl}_K} g(z_A) = \frac{1}{\mathrm{vol}(\Gamma \...

9
votes

### Equivalence of binary quadratic forms over $\operatorname{GL}_2(\mathbb{Q}_p)$ or $\operatorname{GL}_2(\mathbb{Z}_p)$

The number of classes of (regular) binary quadratic forms over $\mathbb{Q}_p$ equals $7$ when $p\neq 2$, and it equals $15$ if $p=2$. See Section IV.2.3 of Serre: A course in arithmetic.
The number ...

8
votes

Accepted

### Representation of integers by principal binary quadratic forms

There are precisely two such proper divisors of $k$, and their product equals $-k$. This result is essentially due to Gauss, see Theorem 1 in Pall: Discriminantal divisors of binary quadratic forms, J....

8
votes

Accepted

### Representation of two related integers by the same binary quadratic form

This question can be answered by general theory, at least when $\Delta$ is a fundamental discriminant (so it's not a square times a smaller discriminant). Assuming this, the general procedure for ...

8
votes

Accepted

### On binary quadratic forms which are not proper subforms of another binary quadratic form

Yes. It is about equivalence over $SL_2 \mathbb Z.$ Your form $f$ (primitively) represents some value not divisible by the fixed prime $p.$ Indeed, from original coefficients $\langle a,b,c \rangle$ ...

8
votes

### asymptotics of numbers represented by certain indefinite binary quadratic forms

This follows from Bernays' generalization of Landau's theorem, the original reference is this:
P. Bernays, Uber die Darstellung von positiven, ganzen Zahlen durch die primitiven,
binaren ...

7
votes

### The mean value of $y \log{y}$ over the ordinates of the CM points

Can we do this by reducing it to a sum over ideals?
By Duke's theorem, we can easily deal with the low $y$ terms. Thus we can adjust the boundary condition to $-a < b \leq a < \sqrt{- D /3 }$.
...

5
votes

Accepted

### A class number estimate

This is from Buell, Binary Quadratic Forms, pages 109-119.
Note that $h(-16) = h(-4) = 1.$
When $p$ is an odd prime, we use the Legndre symbol in
$$ h(-4p^2) = \frac{p - (-1|p)}{2} $$
After that, ...

5
votes

Accepted

### heights of ideal classes and reduction theory for Bhargava cubes

No, it is not true that $M(D) = o(|D|^{3/2})$. For example,
if $D = -4abcd$ where $a,b,c,d$ are
odd, pairwise coprime and of roughly equal size, then
the forms $ab\,x^2+cd\,y^2, ac\,x^2+bd\,y^2, ad\,...

4
votes

### Is the diophantine equation $3x^2+1=py^2$ always solvable for each prime $p\equiv 13\pmod{24}$?

Here's a second proof of Claim 1 (I guess the others can be taken care of similarly) that perhaps explains a little bit better what is going on. Assume first that $3x^2 + 1 = py^2$ has an integral ...

4
votes

### On the equivalence of a pair of binary quadratic forms

The exact class number formula was given by Jorge Morales in the following paper:
Jorge Morales. The classification of pairs of binary quadratic forms. Acta Arith., 59(2):105–121, 1991.
In ...

4
votes

### On the equivalence of a pair of binary quadratic forms

We can show that $ h(1,1,n) $ increases like $ n $-th powers of 2 : indeed, in the half-plane presentation of the set of positive definite quadratic forms, the euclidean form is identified to $ ...

4
votes

Accepted

### Ray class groups through binary quadratic forms

I guess that the answer is no. The binary forms with discriminant
$\Delta \cdot f^2$ describe ring class fields modulo $f$ (see Cox's book), so by taking the limit as $f \to \infty$ something like ...

4
votes

### On the notion of primary representation of a natural number by a quadratic form

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 ...

4
votes

### Is the set of integers represented by a quadratic form of non-fundamental discriminant a subset of the rep. set of a form of fundamental discriminant?

This is true, of course. The book you want is Binary Quadratic Forms by D. A. Buell. He talks about how, given some discriminant $\Delta$ and class number(primitive forms) $h(\Delta),$ we can predict $...

4
votes

Accepted

### A cubic equation, and integers of the form $a^2+32b^2$

Update 2. Now the proof should be more readable. I deleted some content because it is replaced by more elegant version.
The equation is unsolvable. The proof requires two theorems
Theorem 1. Let $p = ...

3
votes

Accepted

### Distribution of 'square classes' of binary quadratic forms

The standard definition for a form to belong to the principal genus of forms with fundamental discriminant $d$ is that the primes $p$ coprime to $d$ that
the form $Q$ represents satisfy $(d_1/p) = \...

3
votes

### Classifying binary quadratic forms by the set of representable squares

No, the set $S_f$ is empty for lots of forms. For example, if $a$ is not a square modulo $b$, then $ax^2+by^2$ is never a square (consider the equation modulo $b$). Thus, $S_f$ is empty for $f(x,y)=6x^...

3
votes

Accepted

### Clarification regarding a claim in Heilbronn’s 1934 paper

One can see more directly that $a^H$ has a representation by the principal form with $y\neq 0$ (hence also with $y>0$). Indeed, let
$$\omega:=\begin{cases}\sqrt{d},&d\equiv 0\pmod{4}; \\
(1+\...

3
votes

### Representation of integers by principal binary quadratic forms

As noted in an earlier answer referring to Pall's 1969 paper in the Journal of Number Theory, it is unlikely that there is a simple way to describe all the divisors of $k$ that are represented by the ...

3
votes

### Integers representable as binary quadratic forms

you don't seem to be mentioning Gauss composition. You have a genus of forms, equivalent to $\langle 1,8,27 \rangle,$ then $\langle 3,8,9 \rangle,$ then $\langle 9,8,3 \rangle.$ These are convenient ...

3
votes

### A cubic equation, and integers of the form $a^2+32b^2$

details, details. From $x$ odd and
$$
x^4-32x-16 = (x^2 + 4)^2 - 2(2x+4)^2.
$$
we see that $x^4-32x-16$ is not divisible by any prime $q \equiv 3,5 \pmod 8.$ That is, $x^2 + 4$ is also odd. ...

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