Regular ternary triangular forms

TL;DR

This paper classifies all primitive regular ternary triangular forms, proving that exactly 49 such forms exist, which represent all locally represented positive integers.

Contribution

It provides a complete classification of primitive regular ternary triangular forms, establishing the exact number as 49.

Findings

Exactly 49 primitive regular ternary triangular forms exist.

Each such form represents all positive integers locally represented.

The classification completes the understanding of regular ternary triangular forms.

Abstract

An integer of the form $T_x=\frac{x(x+1)}2$ for some positive integer $x$ is called a triangular number. A ternary triangular form $aT_{x}+bT_{y}+cT_{z}$ for positive integers $a,b$ and $c$ is called regular if it represents every positive integer that is locally represented. In this article, we prove that there are exactly 49 primitive regular ternary triangular forms.

Tables

Table 1. Table 1. Some values of a i ​ j subscript 𝑎 𝑖 𝑗 a_{ij}

	1	2	3	4	5	6	7	8	9	10	11
1	1	1	1	1	1	1	1	1	1	1	1
2	1	1	1	1	1	1	1	1	1	1	1
3	1	1	1	1	1	1	1	1	1	1	1
4	2	1	1	1	1	1	1	1	1	1	1
5	2	1	1	1	1	1	1	1	1	1	1
6	2	2	1	1	1	1	1	1	1	1	1
7	2	2	1	1	1	1	1	1	1	1	1
9	2	2	2	1	1	1	1	1	1	1	1
19	4	3	3	2	2	2	1	1	1	1	1
20	4	3	3	2	2	2	2	1	1	1	1
25	4	3	4	3	2	2	2	2	1	1	1
26	4	4	4	3	2	2	2	2	1	1	1
29	6	4	4	3	3	2	2	2	1	1	1
32	6	5	4	3	3	2	2	2	2	2	1
35	6	5	4	4	3	3	2	2	2	2	1
41	7	5	4	4	4	3	3	2	2	2	2
47	8	5	4	5	4	3	3	3	2	2	2
49	8	5	4	5	4	3	3	3	2	2	2
83	13	9	7	6	7	5	5	4	3	3	3
314	41	29	22	16	13	11	10	12	11	11	9

$(a,k)$	(10,5)	(9,5)	(8,7)	(7,7)	(6,7)	(5,9)	(4,13)	(3,29)
$u_k$	$\ge 5$	$\ge 5$	$\ge 15$	$\ge 11$	$\ge 8$	$\ge 17$	$\ge 31$	$\ge 164$

$a$	10	9	8	7	6	5	4	3
$c$	$\le 29$	$\le 26$	$\le 47$	$\le 41$	$\le 35$	$\le 49$	$\le 83$	$\le 314$

Table 4. Table 2. Inverse image of λ p subscript 𝜆 𝑝 \lambda_{p} -transformations

Cases	Triangular forms in ${\lambda }_p^{-1}\left(\Delta (a,p^{r}b,p^{s}c)\right)$
$r=s=0$	$\begin{array}{c}\Delta (p^{2}a,b,c),\Delta (a,p^{2}b,c),\Delta (a,b,p^{2}c),\\ \Delta (p^{2}a,p^{2}b,c),\Delta (p^{2}a,b,p^{2}c),\Delta (a,p^{2}b,p^{2}c)\end{array}$
$r=0,s=1$	$\Delta (pa,pb,c),\Delta (a,p^{2}b,p^{3}c),\Delta (p^{2}a,b,p^{3}c),\Delta (a,b,p^{3}c)$
$r=0,s\ge 2$	$\Delta (a,p^{2}b,p^{s+2}c),\Delta (p^{2}a,b,p^{s+2}c),\Delta (a,b,p^{s+2}c)$
$r=s=1$	$\Delta (pa,b,p^{2}c),\Delta (pa,p^{2}b,c),\Delta (pa,b,c),\Delta (a,p^{3}b,p^{3}c)$
$r=1,s\ge 2$	$\Delta (pa,b,p^{s+1}c),\Delta (a,p^{3}b,p^{s+2}c)$
$r\ge 2$	$\Delta (a,p^{r+2}b,p^{s+2})$

Table 5. Table 4. Regular ternary triangular forms

${\Delta }_1=\Delta (1,1,1)$,	${\Delta }_2=\Delta (1,1,2)$,	${\Delta }_3=\Delta (1,1,3)$,
${\Delta }_4=\Delta (1,1,4)$,	${\Delta }_5=\Delta (1,1,5)$,	${\Delta }_6=\Delta (1,1,6)$,
${\Delta }_7=\Delta (1,2,2)$,	${\Delta }_8=\Delta (1,2,3)$,	${\Delta }_9=\Delta (1,2,4)$,
${\Delta }_{10}=\Delta (1,1,9)$,	${\Delta }_{11}=\Delta (1,3,3)$,	${\Delta }_{12}=\Delta (1,2,5)$,
${\Delta }_{13}=\Delta (1,1,12)$,	${\Delta }_{14}=\Delta (1,3,4)$,	${\Delta }_{15}=\Delta (2,2,3)$,
${\Delta }_{16}=\Delta (1,1,18)$,	${\Delta }_{17}=\Delta (1,3,6)$,	${\Delta }_{18}=\Delta (2,3,3)$,
${\Delta }_{19}=\Delta (1,2,10)$,	${\Delta }_{20}=\Delta (1,1,21)$,	${\Delta }_{21}=\Delta (1,4,6)$,
${\Delta }_{22}=\Delta (1,5,5)$,	${\Delta }_{23}=\Delta (1,3,9)$,	${\Delta }_{24}=\Delta (1,3,10)$,
${\Delta }_{25}=\Delta (1,3,12)$,	${\Delta }_{26}=\Delta (1,4,9)$,	${\Delta }_{27}=\Delta (1,6,6)$,
${\Delta }_{28}=\Delta (3,3,4)$,	${\Delta }_{29}=\Delta (1,5,10)$,	${\Delta }_{30}=\Delta (1,3,18)$,
${\Delta }_{31}=\Delta (1,6,9)$,	${\Delta }_{32}=\Delta (2,3,9)$,	${\Delta }_{33}=\Delta (3,3,7)$,
${\Delta }_{34}=\Delta (2,3,12)$,	${\Delta }_{35}=\Delta (1,3,27)$,	${\Delta }_{36}=\Delta (1,9,9)$,
${\Delta }_{37}=\Delta (1,3,30)$,	${\Delta }_{38}=\Delta (2,5,10)$,	${\Delta }_{39}=\Delta (1,9,12)$,
${\Delta }_{40}=\Delta (2,3,18)$,	${\Delta }_{41}=\Delta (1,5,25)$,	${\Delta }_{42}=\Delta (3,7,7)$,
${\Delta }_{43}=\Delta (2,5,15)$,	${\Delta }_{44}=\Delta (1,6,27)$,	${\Delta }_{45}=\Delta (1,9,18)$,
${\Delta }_{46}=\Delta (1,9,21)$,	${\Delta }_{47}=\Delta (1,21,21)$,	${\Delta }_{48}=\Delta (5,6,15)$,
${\Delta }_{49}=\Delta (3,7,63)$.

Equations

$$f(x_1,x_2,\dots ,x_n)=\sum _{i,j=1}^n{a}_{ij}{x}_i{x}_j(a_{ij}=a_{ji}\in \mathbb{Z})$$

$$\Delta (a_1,a_2,a_3)=\Delta (a_1,a_2,a_3)(x_1,x_2,x_3):=a_1\frac{x_1(x_1+1)}{2}+a_2\frac{x_2(x_2+1)}{2}+a_3\frac{x_3(x_3+1)}{2}$$

$$\Delta (a_1,a_2,a_3)(x_1,x_2,x_3)=N$$

$$\Delta (1,1,1),\Delta (1,1,2),\Delta (1,1,4),\Delta (1,1,5),\Delta (1,2,2),\Delta (1,2,3),\Delta (1,2,4).$$

$$L=⟨Q({\mathbf{x}}_1),\dots ,Q({\mathbf{x}}_n)⟩.$$

$$\Delta (a_1,a_2,\dots ,a_k)=\Delta (a_1,a_2,\dots ,a_k)(x_1,x_2,\dots ,x_k)=a_1{T}_{x_1}+a_2{T}_{x_2}+\cdots +a_k{T}_{x_k}$$

$$a_1{T}_{x_1}+a_2{T}_{x_2}+\cdots +a_k{T}_{x_k}=n$$

$$T(n,⟨a_1,a_2,\dots ,a_k⟩)=\left\{(z_1,z_2,\dots ,z_k)\in {\mathbb{Z}}^k:a_1{T}_{z_1}+a_2{T}_{z_2}+\cdots +a_k{T}_{z_k}=n\right\}$$

$$a_1{T}_{x_1}+a_2{T}_{x_2}+\cdots +a_k{T}_{x_k}=n\text{ is soluble in }\mathbb{Z}$$

$$a_1(2x_1-1{)}^2+a_2(2x_2-1{)}^2+\cdots +a_k(2x_k-1{)}^2=8n+a_1+a_2+\cdots +a_k$$

$$R(n,f)=\left\{(x_1,x_2,\dots ,x_k)\in {\mathbb{Z}}^k:f(x_1,x_2,\dots ,x_k)=n\right\}\text{and}r(n,f)=|R(n,f)|.$$

$$R_{\mathbf{d}}(n,f)=\left\{(x_1,x_2,\dots ,x_k)\in R(n,f):x_i\equiv d_i(\mathrm{mod}2)\text{for any}1\le i\le k\right\}$$

$$n\stackrel{2}{⟶}⟨a_1,a_2,\dots ,a_k⟩$$

$$n\stackrel{2}{\nrightarrow }⟨a_1,a_2,\dots ,a_k⟩$$

$${\Lambda }_m(L)=\{x\in L:Q(x+z)\equiv Q(z)(\mathrm{mod}m)\text{for any}z\in L\}.$$

$${\lambda }_p(L)\simeq \{\begin{array}{ll}⟨a,b,c⟩& \text{if }m=n=0,\\ ⟨pa,b,p^{n-1}c⟩& \text{if }1=m\le n,\\ ⟨a,p^{m-2}b,p^{n-2}c⟩& \text{if }1<m\le n.\end{array}$$

$${\lambda }_p(\Delta (a,b,c))=\Delta (a^{\prime },b^{\prime },c^{\prime }),$$

$$p^{2}a{T}_x+p^{2}b{T}_y+p^{s}c{T}_z=n$$

$$8n+p^{2}a+p^{2}b+p^{s}c⟶\text{gen}(⟨p^{2}a,p^{2}b,p^{s}c⟩).$$

$$8\left(n+\frac{p^2-1}{8}a+\frac{p^2-1}{8}b\right)+a+b+p^{s}c⟶\text{gen}(⟨a,b,p^{s}c⟩).$$

$$p^{2}a{\left(\frac{x}{p}\right)}^2+p^{2}b{\left(\frac{y}{p}\right)}^2+p^{s}c{z}^2=8n+p^{2}a+p^{2}b+p^{s}c$$

$$⟨1,-1⟩⟶L_p\text{or}{L}_p\simeq ⟨1,-{\Delta }_p⟩\perp ⟨p{ϵ}_p⟩$$

$$⟨a,b,c⟩\simeq ⟨1,-{\Delta }_p⟩\perp ⟨p{ϵ}_p⟩\text{ over }{\mathbb{Z}}_p,$$

$$d{g}_1+a+b\equiv 3c(\mathrm{mod}{p}^2).$$

$$T$$

$$=\{p_1<p_2<\cdots <p_t\}.$$

$$T^{\prime }=T-\{3\}=\{q_1<q_2<\cdots <q_{t^{\prime }}\}.$$

$$24k+a+b\nrightarrow ⟨a,b⟩$$

$$24k+a+b+c⟶\text{gen}(⟨a,b,c⟩).$$

$$q_1{q}_2\cdots q_{t^{\prime }}\le abc\le c^3\le (3k{)}^3\le (3q_1^2(t^{\prime }+1)2^{t^{\prime }-2}{)}^3.$$

$$r_8{r}_9\cdots r_{t^{\prime }+1}>(3(t^{\prime }+1)2^{t^{\prime }-2}{)}^3.$$

Full text

This work of the second author was supported by the National Research Foundation of Korea (NRF-2017R1A2B4003758).

1. Introduction

A quadratic homogeneous polynomial

[TABLE]

is called an integral quadratic form. Throughout this article, we always assume that $f$ is positive definite, that is, $f(x_{1},x_{2},\dots,x_{n})>0$ for any non-zero real vector $(x_{1},x_{2},\dots,x_{n})\in\mathbb{R}^{n}$. Let $R$ be any ring containing ${\mathbb{Z}}$. For an integer $N$, if there is a solution $(x_{1},x_{2},\dots,x_{n})\in R^{n}$ of the equation $f(x_{1},x_{2},\dots,x_{n})=N$, then we say that $N$ is represented by $f$ over $R$. In particular, if $N$ is represented by $f$ over the $p$-adic integer ring ${\mathbb{Z}}_{p}$ for any prime $p$, then we say that $N$ is locally represented by $f$. From the definition, note that any integer that is represented by $f$ over ${\mathbb{Z}}$ is locally represented by $f$. However, it is known that the converse is not true, in general. A positive definite integral quadratic form is called regular if the converse is also true, that is, it represents every integer over ${\mathbb{Z}}$ that is locally represented.

Dickson [9] who initiated the study of regular quadratic forms first coined the term regular. Jones and Pall [16] gave the list of all 102 primitive diagonal regular ternary quadratic forms. Watson proved in his thesis [26] that there are only finitely many equivalence classes of primitive positive definite ternary regular forms. Jagy, Kaplansky and Schiemann [14] succeeded Watson’s study on regular quadratic forms and provide the list of 913 candidates of regular positive definite integral ternary quadratic forms. All but 22 of them are already proved to be regular at that time. Recently, the second author [21] proved the regularities of 8 ternary quadratic forms among remaining 22 candidates. A conditional proof for the remaining 14 candidates under the Generalized Riemann Hypothesis was given by Lemke Oliver [20]. Note that there are infinitely many regular positive definite integral quaternary quadratic forms (for this, see [12]).

Now we look into the representations of ternary triangular forms. An integer of the form $T_{x}=\frac{x(x+1)}{2}$ for some positive integer $x$ is called a triangular number. For positive integers $a_{1},a_{2},a_{3}$ with $a_{1}\leq a_{2}\leq a_{3}$, a polynomial of the form

[TABLE]

is called a ternary triangular form. We say an integer $N$ is represented by the triangular form ${\Delta}(a_{1},a_{2},a_{3})$ if

[TABLE]

has an integer solution $(x_{1},x_{2},x_{3})\in{\mathbb{Z}}^{3}$. Similarly to the quadratic form case, if Equation (1.1) has a solution in ${\mathbb{Z}}_{p}$ for any prime $p$, then we say that $N$ is locally represented by ${\Delta}(a_{1},a_{2},a_{3})$. We say ${\Delta}(a_{1},a_{2},a_{3})$ is regular if it represents every integer that is locally represented. Note that any ternary triangular form is regular if it represents all positive integers. Such a ternary triangular form is said to be universal.

Gauss’ Eureka Theorem says that every positive integer is a sum of at most three triangular numbers, which implies that the ternary triangular form ${\Delta}(1,1,1)$ is universal. In 1862, Liouville classified all universal ternary triangular forms, and they are, in fact, the following seven forms:

[TABLE]

As mentioned above, these universal triangular forms are regular. In 2013, Chan and Oh [7] proved that there are only finitely many regular ternary triangular forms. In 2015, Chan and Ricci [8] proved the finiteness of regular ternary triangular forms in a more general setting. They actually proved that for any given positive integer $c$, there are only finitely many inequivalent positive ternary regular primitive complete quadratic polynomials with conductor $c$. From this follows the finiteness of regular ternary $m$-gonal forms. Note that an integer of the form $\frac{(m-2)x^{2}-(m-4)x}{2}$ for some integer $x$ is called an $m$-gonal number, and a (regular) ternary $m$-gonal form is defined similarly.

In this article, we prove that there are exactly $49$ regular ternary triangular forms. In the previous papers [7] and [8], the authors use Burgess’ estimation on character sums (for this, see [4] and [11]) to prove the finiteness of regular ternary triangular forms. It seems to be quite difficult to find an explicit upper bound of the discriminant of regular ternary triangular forms by using Burgess’ estimation. In this article, we use a purely arithmetic method to find such an explicit and effective upper bound of the discriminant of regular ternary triangular forms, and finally, we classify all regular ternary triangular forms.

A ${\mathbb{Z}}$-lattice $L$ is a finitely generated free ${\mathbb{Z}}$-module equipped with a non-degenerate symmetric bilinear form $B$ such that $B(L,L)\subset{\mathbb{Z}}$. The corresponding quadratic map $Q$ is defined by $Q(\mathbf{v})=B(\mathbf{v},\mathbf{v})$ for any $\mathbf{v}\in L$.

Let $L={\mathbb{Z}}\mathbf{x}_{1}+{\mathbb{Z}}\mathbf{x}_{2}+\cdots+{\mathbb{Z}}\mathbf{x}_{n}$ be a ${\mathbb{Z}}$-lattice. The quadratic form $f_{L}$ corresponding to $L$ is defined by $f_{L}(x_{1},x_{2},\dots,x_{n})=\sum B(\mathbf{x}_{i},\mathbf{x}_{j})x_{i}x_{j}$. Furthermore, the corresponding symmetric matrix $M_{L}$ is defined by $M_{L}=(B(\mathbf{x}_{i},\mathbf{x}_{j}))$, which is called the matrix presentation of $L$. If $L$ admits an orthogonal basis $\{\mathbf{x}_{1},\dots,\mathbf{x}_{n}\}$, we call $L$ diagonal and simply write

[TABLE]

For any odd prime $p$, ${\Delta}_{p}$ denotes a non-square unit in ${\mathbb{Z}}_{p}$.

Any unexplained notations and terminologies can be found in [19] or [24].

2. Preliminaries

A nonnegative integer of the form $T_{x}=\frac{x(x+1)}{2}$ for some positive integer $x$ is called a triangular number. For example, $0,1,3,6,10,15,\cdots$ are triangular numbers. Since $T_{x}=T_{1-x}$, $T_{x}$ is a triangular number for any integer $x$. For positive integers $a_{1},a_{2},\dots,a_{k}$ with $a_{1}\leq a_{2}\leq\dots\leq a_{k}$, we call a polynomial of the form

[TABLE]

a $k$-ary triangular form. For a triangular form ${\Delta}={\Delta}(a_{1},a_{2},\dots,a_{k})$, we define $d({\Delta}(a_{1},a_{2},\dots,a_{k}))=a_{1}a_{2}\cdots a_{k}$, which is called the discriminant of the triangular form $\Delta(a_{1},a_{2},\dots,a_{k})$. A triangular form ${\Delta}(a_{1},a_{2},\dots,a_{k})$ is called primitive if $\gcd(a_{1},a_{2},\dots,a_{k})=1$. Unless stated otherwise, we always assume that

every triangular form is primitive.

For an integer $n$ and a $k$-ary triangular form ${\Delta}(a_{1},a_{2},\dots,a_{k})$, we say that $n$ is represented by ${\Delta}(a_{1},a_{2},\dots,a_{k})$ if the Diophantine equation

[TABLE]

has an integral solution. In this case, we write $n\longrightarrow{\Delta}(a_{1},a_{2},\dots,a_{k})$. We also define

[TABLE]

and $t(n,\langle a_{1},a_{2},\dots,a_{k}\rangle)$ to be the cardinality of the above set.

A triangular form ${\Delta}(a_{1},a_{2},\dots,a_{k})$ is called universal if it represents every positive integer, that is,

[TABLE]

for any positive integer $n$. A triangular form ${\Delta}(a_{1},a_{2},\dots,a_{k})$ is called regular if it globally represents every integer which is locally represented. In other words, ${\Delta}(a_{1},a_{2},\dots,a_{k})$ is regular if for any integer $n$ such that $a_{1}T_{x_{1}}+a_{2}T_{x_{2}}+\cdots+a_{k}T_{x_{k}}=n$ is soluble in ${\mathbb{Z}}_{p}$ for any prime $p$, the diophantine equation $a_{1}T_{x_{1}}+a_{2}T_{x_{2}}+\cdots+a_{k}T_{x_{k}}=n$ is soluble in ${\mathbb{Z}}$. As shown in [7], any primitive triangular form is universal over ${\mathbb{Z}}_{2}$.

Note that a triangular form ${\Delta}(a_{1},a_{2},\dots,a_{k})$ represents $n$ if and only if the Diophantine equation

[TABLE]

is soluble in ${\mathbb{Z}}$. This equivalence shows how the representation of a triangular form is transformed into the representation of a diagonal quadratic form with congruence conditions. Now, we can reformulate the regularity in a practical way. A ternary triangular form ${\Delta}(a,b,c)$ is regular if the following implication holds: for any positive integer $n$, if $ax^{2}+by^{2}+cz^{2}=8n+a+b+c$ is soluble in ${\mathbb{Z}}_{p}$ for any odd prime $p$, then there exist odd integers $x,y$ and $z$ such that $ax^{2}+by^{2}+cz^{2}=8n+a+b+c$.

Let $f(x_{1},x_{2},\dots,x_{k})$ be a positive definite integral quadratic form of rank $k$ and let $n$ be an integer. We define

[TABLE]

We say that $n$ is represented by $f$ if $r(n,f)>0$. For a vector $\mathbf{d}=(d_{1},d_{2},\dots,d_{k})\in({\mathbb{Z}}/2{\mathbb{Z}})^{k}$, we also define

[TABLE]

and $r_{\mathbf{d}}(n,f)=|R_{\mathbf{d}}(n,f)|$.

For an integer $n$ and a diagonal quadratic form $\langle a_{1},a_{2},\dots,a_{k}\rangle$, we write

[TABLE]

if there is a vector $(x_{1},x_{2},\dots,x_{k})\in{\mathbb{Z}}^{n}$ with $(x_{1}x_{2}\cdots x_{k},2)=1$ such that $a_{1}x_{1}^{2}+a_{2}x_{2}^{2}+\cdots+a_{k}x_{k}^{2}=n$. We also use the notation

[TABLE]

if there does not exist such a vector $(x_{1},x_{2},\dots,x_{k})\in{\mathbb{Z}}^{n}$. Under these notations, the followings are all equivalent:

1. (i)

$n\longrightarrow{\Delta}(a_{1},a_{2},\dots,a_{k})$; 2. (ii)

$t(n,\langle a_{1},a_{2},\dots,a_{k}\rangle)>0$; 3. (iii)

$r_{(1,1,\dots,1)}(8n+a_{1}+a_{2}+\cdots+a_{k},\langle a_{1},a_{2},\dots,a_{k}\rangle)>0$; 4. (iv)

$8n+a_{1}+a_{2}+\cdots+a_{k}\overset{2}{\longrightarrow}\langle a_{1},a_{2},\dots,a_{k}\rangle$.

Let $L$ be a ${\mathbb{Z}}$-lattice and let $m$ be a positive integer. Watson transformation of $L$ modulo $m$ is defined by

[TABLE]

We denote by $\lambda_{m}(L)$ the primitive ${\mathbb{Z}}$-lattice obtained from $\Lambda_{m}(L)$ by scaling $L\otimes{\mathbb{Q}}$ by a suitable rational number. Let $p$ be an odd prime. Let $L=\langle a,p^{m}b,p^{n}c\rangle$ be a ternary ${\mathbb{Z}}$-lattice, where $(abc,p)=1$ and $0\leq m\leq n$. Then one may easily check

[TABLE]

For a ternary triangular form ${\Delta}(a,b,c)$ and an odd prime $p$, we define

[TABLE]

where $\langle a^{\prime},b^{\prime},c^{\prime}\rangle\simeq\lambda_{p}(\langle a,b,c\rangle)$.

Lemma 2.1**.**

Let $p$ be an odd prime and let $a,b,c$ be positive integers which are not divisible by $p$. Let $r,s$ be positive integers. If the ternary triangular form ${\Delta}(a,p^{r}b,p^{s}c)$ is regular, then so is $\lambda_{p}({\Delta}(a,p^{r}b,p^{s}c))$.

Proof.

See [7, Lemma 3.3]. ∎

Though the proof of the next lemma is quite similar to the proof of Lemma 2.1, we provide the proof for completeness.

Lemma 2.2**.**

Let $p$ be an odd prime and let $s$ be a positive integer. Let $a,b$, and $c$ be positive integers such that $(p,abc)=1$ and $\left(\frac{-ab}{p}\right)=-1$, where $\left(\frac{\cdot}{p}\right)$ is the Legendre symbol modulo $p$. If the ternary triangular form $\Delta(a,b,p^{s}c)$ is regular, then so is $\lambda_{p}(\Delta(a,b,p^{s}c))$.

Proof.

It is enough to show that $\Delta(p^{2}a,p^{2}b,p^{s}c)$ is regular. Let $n$ be a positive integer such that the equation

[TABLE]

is soluble in ${\mathbb{Z}}_{p}$ for any prime $p$. Then

[TABLE]

Thus

[TABLE]

Since $\Delta(a,b,p^{s}c)$ is regular, there is a vector $(x,y,z)\in{\mathbb{Z}}^{3}$ with $xyz\equiv 1\ (\mathrm{mod}\ 2)$ such that $ax^{2}+by^{2}+p^{s}cz^{2}=8n+p^{2}a+p^{2}b+p^{s}c$. Since $n$ is divisible by $p$, we have $ax^{2}+by^{2}\equiv 0\ (\mathrm{mod}\ p)$. From the assumption $\left(\displaystyle\frac{-ab}{p}\right)=-1$, we have $x\equiv y\equiv 0\ (\mathrm{mod}\ p)$. So

[TABLE]

with $\frac{x}{p}\cdot\frac{y}{p}\cdot z\equiv 1\ (\mathrm{mod}\ 2)$. Thus Equation $\eqref{eq1}$ is soluble in ${\mathbb{Z}}$. This completes the proof. ∎

For an odd prime $p$ and a ternary ${\mathbb{Z}}$-lattice $L$, we say that $L$ is $p$-stable if

[TABLE]

for some $\epsilon_{p}\in{\mathbb{Z}}_{p}^{\times}$. We say that $L$ is stable if $L$ is $p$-stable for every odd prime $p$. A ternary triangular form is called $p$-stable (stable) if the corresponding quadratic form is $p$-stable (stable, respectively). Let ${\Delta}(a,b,c)$ be a regular ternary triangular form. Then by taking $\lambda_{q}$-transformations to ${\Delta}(a,b,c)$ repeatedly, if possible, for any odd prime $q$ dividing the discriminant, we may obtain a stable regular ternary triangular form ${\Delta}(a^{\prime},b^{\prime},c^{\prime})$ by Lemmas 2.1 and 2.2. Note that the corresponding quadratic form $\langle a^{\prime},b^{\prime},c^{\prime}\rangle$ has a smaller discriminant and a simpler local structure than $\langle a,b,c\rangle$.

3. Stable regular ternary triangular forms

In this section, we prove that there are exactly 17 stable regular ternary triangular forms. Throughout this section, $r_{k}$ denotes the $k$-th odd prime so that $\{r_{1}=3<r_{2}=5<r_{3}=7<\cdots\}$ is the set of all odd primes. Let ${\Delta}(a,b,c)$ be a stable regular ternary triangular form. We always assume that $0<a\leq b\leq c$.

Lemma 3.1**.**

For an integer $s$ greater than $1$, let $p_{1}<p_{2}<\cdots<p_{s}$ be odd primes. Let $u$ be an integer with $(u,p_{1}p_{2}\cdots p_{s})=1$ and let $v$ be an arbitrary integer. Then there is an integer $n$ with $0\leq n<(s+2)2^{s-1}$ such that $(un+v,p_{1}p_{2}\cdots p_{s})=1$.

Proof.

See [17, Lemma 3]. ∎

Though Lemma 3.1 gives, in general, a nice upper bound of the longitude of arithmetic progression satisfying the assumption, there is a shaper bound in some restricted situation.

Lemma 3.2**.**

Under the same notations given in Lemma 3.1, if $s<p_{1}$, then there is an integer $n$ with $0\leq n\leq s$ such that $(un+v,p_{1}p_{2}\cdots p_{s})=1$.

Proof.

Trivial. ∎

Lemma 3.3**.**

Let $p\geq 5$ be a prime and let $d$ be a positive integer with $(d,p)=1$. Let $L=\langle a,b,c\rangle$ be a $p$-stable ${\mathbb{Z}}$-lattice that is anisotropic over ${\mathbb{Z}}_{p}$. Then there is an integer $g$ such that

• (i)

$0<g<p^{2}$;

• (ii)

$dg+a+b\nrightarrow\langle a,b\rangle$* over ${\mathbb{Z}}_{p}$;*

• (iii)

$dg+a+b+c\longrightarrow\langle a,b,c\rangle$* over ${\mathbb{Z}}_{p}$;*

• (iv)

$\text{max}\left\{\text{ord}_{p}(dg+a+b),\text{ord}_{p}(dg+a+b+c)\right\}\leq 1$.

Proof.

Since $L$ is $p$-stable and is anisotropic over ${\mathbb{Z}}_{p}$ by assumption, we have

[TABLE]

for some $\epsilon_{p}\in{\mathbb{Z}}_{p}^{\times}$. First, we assume that $p$ divides $c$. Since $\langle a,b\rangle\simeq\langle 1,-\Delta_{p}\rangle$, it does not represent $\gamma\in{\mathbb{Z}}_{p}$ satisfying $\text{ord}_{p}\gamma\equiv 1\ (\mathrm{mod}\ 2)$. Since $p\geq 5$, there exists a positive integer $g_{1}$ with $g_{1}<p^{2}$ such that

[TABLE]

Then one may easily check that $g_{1}$ satisfies all conditions given above. Now, assume that $p$ divides $ab$. Without loss of generality, we may assume that $p$ divides $b$. Since $p\geq 5$, there exists an integer $a^{\prime}$ with $(p,a^{\prime})=1$ such that $aa^{\prime}$ is not a square modulo $p$ and $a^{\prime}\not\equiv-c\ (\mathrm{mod}\ p)$. We take a positive integer $g_{2}$ with $g_{2}<p$ such that $dg_{2}+a+b\equiv a^{\prime}\ (\mathrm{mod}\ p)$. One may easily show that $g_{2}$ satisfies all conditions given above, which completes the proof. ∎

Let $T$ be the set of odd primes $p$ such that the diagonal ternary quadratic form $\langle a,b,c\rangle$ is anisotropic over ${\mathbb{Z}}_{p}$. Since such primes are only finitely many, we let

[TABLE]

Let

[TABLE]

Note that $t^{\prime}=t$ if $3\not\in T$, and $t^{\prime}=t-1$ otherwise.

Lemma 3.4**.**

Under the assumptions given above, we have $t^{\prime}\leq 17$.

Proof.

Note that $\langle a,b,c\rangle$ represents every integer of the form $24n+a+b+c$ over ${\mathbb{Z}}_{3}$. Let $g$ be a positive integer satisfying Lemma 3.3 in the case when $p=q_{1}$ and $d=24$.

By Lemma 3.1, there is an integer $h$ with $0\leq h<(t^{\prime}+1)2^{t^{\prime}-2}$ such that $(24q_{1}^{2}h+24g+a+b+c,q_{2}\cdots q_{t^{\prime}})=1$. If we let $k=q_{1}^{2}h+g$, then one may easily show that

[TABLE]

and

[TABLE]

Since $\Delta(a,b,c)$ is regular, there is a vector $(x,y,z)\in{\mathbb{Z}}^{3}$ with $xyz\equiv 1\ (\mathrm{mod}\ 2)$ such that $ax^{2}+by^{2}+cz^{2}=24k+a+b+c$. From Equation (3.1), we have $z^{2}\geq 9$. So $a+b+9c\leq 24k+a+b+c$ and we have $c\leq 3k$. Now

[TABLE]

Assume to the contrary that $t^{\prime}\geq 18$. Then one may easily show that

[TABLE]

Since $q_{i}\geq r_{i+1}$ for any $i$, we have

[TABLE]

which is a contradiction. Therefore we have $t^{\prime}\leq 17$. This completes the proof. ∎

If we are able to use Lemma 3.2 instead of Lemma 3.1, then we may have more effective upper bound of $t^{\prime}$ than the previous lemma.

Lemma 3.5**.**

Under the same notations given above, if $0<t^{\prime}-j<q_{j+1}$ for some $j$ such that $1\leq j\leq t^{\prime}-1$, then we have

[TABLE]

Proof.

Note that $\langle a,b,c\rangle$ represents every integer of the form $24n+a+b+c$ over ${\mathbb{Z}}_{3}$. Let $g$ be a positive integer satisfying Lemma 3.3 in the case when $p=q_{1}$ and $d=24$. Let

[TABLE]

where for each $i$, $\epsilon_{i}$ is suitably chosen in $\{0,1\}$ so that

[TABLE]

for any $j\geq 2$. Note that $g_{1}=g<q_{1}^{2}$ and $g_{j}<q_{1}^{2}q_{2}\cdots q_{j}$ for any $j\geq 2$. Since $0<t^{\prime}-j<q_{j+1}$ by assumption, we apply Lemma 3.2 with odd primes $q_{j+1}<q_{j+2}<\cdots<q_{t^{\prime}}$, $u=24q_{1}^{2}q_{2}\cdots q_{j}$ and $v=24g_{j}+a+b+c$ so that we may conclude that there is an integer $s$ with $0\leq s\leq t^{\prime}-j$ such that

[TABLE]

Therefore, by a similar reasoning to Lemma 3.4, we have $c\leq 3q_{1}^{2}q_{2}\cdots q_{j}(t^{\prime}-j+1)$. The lemma follows directly from this. ∎

Lemma 3.6**.**

Under the assumptions given above, we have $t\leq 10$.

Proof.

By Lemma 3.4, we may assume that $t^{\prime}\leq 17$. First, assume that $q_{1}\geq 13$. Since $t^{\prime}-1<17\leq q_{2}$, we may apply Lemma 3.5 so that

[TABLE]

From this, one may easily show that $t^{\prime}\leq 8$.

Now, assume that $q_{1}=11$. Since $t^{\prime}-2<17\leq q_{3}$, we may apply Lemma 3.5 so that we may conclude that

[TABLE]

Suppose that $t^{\prime}\geq 11$. Since $r_{8}=23,r_{9}=29,r_{10}=31,\dots$, one may directly show that

[TABLE]

Since $q_{i}\geq r_{i+3}$ for any $i$, we have

[TABLE]

which is a contradiction. Therefore we have $t^{\prime}\leq 10$. Now, since $t^{\prime}-1<13\leq q_{2}$, we deduce, similarly to the above, that

[TABLE]

and thus $t^{\prime}\leq 7$.

Assume that $q_{1}=7$. Since $t^{\prime}-3<17\leq q_{4}$ in this case, one may deduce that

[TABLE]

and thus we have $t^{\prime}\leq 12$. Now, since $t^{\prime}-2<13\leq q_{3}$, we may have

[TABLE]

and hence $t^{\prime}\leq 9$. Since $t^{\prime}-1<11\leq q_{2}$,

[TABLE]

Therefore, we have $t^{\prime}\leq 7$.

Finally, assume that $q_{1}=5$. Since $t^{\prime}-4<17\leq q_{5}$, we have

[TABLE]

Now, since $t^{\prime}-3<13\leq q_{4}$, we have

[TABLE]

Then, since $t^{\prime}-2<11\leq q_{3}$, we have

[TABLE]

The lemma follows directly from this. ∎

Recall that we are assuming that $\Delta(a,b,c)$ is stable. Hence for any odd prime $p$,

[TABLE]

for some $\epsilon_{p}\in{\mathbb{Z}}_{p}^{\times}$. In the former case, every element in ${\mathbb{Z}}_{p}$ is represented by $\langle a,b,c\rangle$ over ${\mathbb{Z}}_{p}$. In the latter case,

[TABLE]

Recall that $r_{j}$ is the $j$-th odd prime. Let $u$ be a positive integer not divisible by $r_{j}$ and let $v$ be an integer. Let $\eta_{r_{j}}\in\{1,\Delta_{r_{j}}\}$. For a positive integer $i$, we define

[TABLE]

We also define

[TABLE]

Let $i={b_{e-1}b_{e-2}\dots b_{0}}_{(r_{j})}$ be the base-$r_{j}$ representation of $i$, that is,

[TABLE]

with $0\leq b_{\nu}<r_{j}$ for $\nu=1,2,\dots,e-1$ and $b_{e-1}>0$. We define

[TABLE]

We also define

[TABLE]

Lemma 3.7**.**

Under the notations and assumptions given above, we have

[TABLE]

Proof.

Since both cases can be done in a similar manner, we only provide the proof of the case when $e=2\delta$ for some positive integer $\delta$. Without loss of generality, we may assume that $u=1$. We have to show that the number of integers of the form $r_{j}^{2k-1}\eta_{r_{j}}$ ($r_{j}^{2k-1}\eta_{r_{j}}^{\prime}$) in the set $\{1+v,2+v,\dots,i+v\}$ is less than or equal to the right hand side, where $\eta_{r_{j}}$ ($\eta_{r_{j}}^{\prime}$) is a square (nonsquare, respectively) in ${\mathbb{Z}}_{r_{j}}^{\times}$.

For any integer $k$ such that $1\leq k\leq\delta$, let

[TABLE]

Let $r_{j}^{2k-1}(x+1)$ be the smallest integer greater than $v$ that is divisible by $r_{j}^{2k-1}$. Then any integer in the set $\{r_{j}^{2k-1}(x+s):1\leq s\leq r_{j}\alpha_{k}+b_{2k-1}\}$ is less than or equal to $i+v$. Note that there is at most one more integer other than these integers that is divisible by $r_{j}^{2k-1}$, and that is less than or equal to $i+v$. Note that such an integer exists only when $\epsilon_{i,j}(k)\neq 0$ (or $\beta_{k}\neq 0$). Furthermore, if such an integer exists, then it must be $r_{j}^{2k-1}(x+r_{j}\alpha_{k}+b_{2k-1}+1)$. Note that there are exactly $\frac{r_{j}-1}{2}$ quadratic residues and $\frac{r_{j}-1}{2}$ quadratic non-residues in the consecutive $r_{j}$ integers. Therefore there are exactly $\frac{r_{j}-1}{2}\alpha_{k}$ quadratic residues and $\frac{r_{j}-1}{2}\alpha_{k}$ quadratic non-residues in

[TABLE]

Note that $\alpha_{k}=\left[\frac{i}{r_{j}^{2k}}\right]$ for any $1\leq k\leq\delta$. The remaining multiples of $r_{j}^{2k-1}$ are contained in

[TABLE]

Among them, there are at most $\psi_{i,j}(k)$ quadratic residues, and at most $\psi_{i,j}(k)$ quadratic non-residues. Note that there is at most one multiple of $r_{j}^{2\delta+1}$ in $\{1+v,2+v,\dots,i+v\}$ which is, if exists, contained in the set

[TABLE]

Note that there are at most $\psi_{i,j}(\delta)$ quadratic residues or a multiple of $r_{j}$, and at most $\psi_{i,j}(\delta)$ quadratic non-residues or a multiple of $r_{j}$ in the set $\{x+1,x+2,\dots,x+b_{2\delta-1}+\epsilon_{i,j}(\delta)\}$. The lemma follows from this. ∎

For the sake of brevity, we let

[TABLE]

for positive integers $i$ and $j$.

Remark 3.8*.*

One may easily show that $a_{ij}\leq\left\lceil\displaystyle\frac{i}{r_{j}}\right\rceil$ for any positive integers $i$ and $j$, where $\left\lceil\cdot\right\rceil$ is the ceiling function. It is a little bit complicate to compute an upper bound of $\Psi_{u,v}(i,j)$ by using Lemma 3.7. Instead of that, one may easily show that

[TABLE]

Recall that $T$ is the set of all odd primes at which $\langle a,b,c\rangle$ is anisotropic, and $|T|=t\leq 10$ by Lemma 3.6.

Lemma 3.9**.**

Let $i$ be a positive integer. For any integer $s>t$, we define $b_{ij}(s)=\text{max}\left(a_{ij},\left\lceil\displaystyle\frac{i}{r_{s}}\right\rceil\right)$ for $j=1,2,\dots,s-1$. Then we have

[TABLE]

Proof.

Let $s$ be any integer greater than $t$ and let $J=\{j\in{\mathbb{N}}:r_{j}\in T\}$. We also let $J_{1}=\{j\in J:j\leq s-1\}$, $J_{2}=J-J_{1}$, and $J_{3}=\{1,2,\dots,s-1\}-J_{1}$. Note that $|J_{2}|\leq|J_{3}|$ and for any $j\in J_{3}$, $\displaystyle\left\lceil\frac{i}{r_{s}}\right\rceil\leq b_{ij}(s)$ by assumption. From Remark 3.8, for any $j\in J_{2}$, we have $a_{ij}\leq\left\lceil\displaystyle\frac{i}{r_{j}}\right\rceil\leq\left\lceil\displaystyle\frac{i}{r_{s}}\right\rceil$. Thus we have

[TABLE]

Since ${\Delta}(a,b,c)$ is stable regular, we have

[TABLE]

This completes the proof. ∎

Remark 3.10*.*

In the remaining of this section, we need the exact values of $a_{ij}$’s for some integers $i$ and $j$. We provide some of these values in Table 1 below.

Lemma 3.11**.**

Under the assumptions given above, we have $t\leq 7$.

Proof.

By Lemma 3.9 with $i=25$ and $s=11$, one may easily show that $8n_{1}+a+b+c\overset{2}{\longrightarrow}\langle a,b,c\rangle$ for some $1\leq n_{1}\leq 25$. From our assumption of $a\leq b\leq c$, we have $9a+b+c\leq 8n_{1}+a+b+c$, and thus we have $a\leq 25$. To prove the lemma, we will use Lemma 3.5 repeatedly.

First, assume that $q_{1}\geq 7$. Since $t^{\prime}-1<11\leq q_{2}$, we may apply Lemma 3.5 so that

[TABLE]

This is possible only when $t^{\prime}\leq 6$. Now, assume that $q_{1}=5$. Since $t^{\prime}-2<11\leq q_{3}$, one may deduce that

[TABLE]

and thus $t^{\prime}\leq 7$. Finally, since $t^{\prime}-1<7\leq q_{2}$, we have

[TABLE]

and thus $t^{\prime}\leq 6$. This completes the proof. ∎

Lemma 3.12**.**

For any stable regular ternary triangular form $\Delta(a,b,c)$ with $0<a\leq b\leq c$, we have $a=1$ or $2$.

Proof.

For any positive integer $n$, we define $s_{n}=8n+a+b+c$. Since

[TABLE]

we have

[TABLE]

On the other hand, by Lemma 3.9 with $i=32$ and $s=8$, one may check that

[TABLE]

By comparing these two inequalities, we have $a\leq 10$.

Now, we will show that if $3\leq a\leq 10$, then $c$ is bounded. For each positive odd integer $k$, we let

[TABLE]

and we also let $u_{k}=|U_{k}|$ and $v_{k}=|V_{k}|$. Note that $V_{k}$ does not depend on $c$. For each integer $a$ with $3\leq a\leq 10$, we will choose an integer $k$ suitably so that $v_{k}<u_{k}$. Note that if this inequality holds, then $a+b+9c\leq 8(\frac{k^{2}-1}{8}a-1)+a+b+c$ and therefore, we have

[TABLE]

In fact, we choose

[TABLE]

Now, by using Lemma 3.9 with $i=\displaystyle\frac{k^{2}-1}{8}a-1$ and $s=8$, one may easily compute the lower bound of $u_{k}$:

To compute an upper bound of $v_{k}$, note that

[TABLE]

Hence one may easily show that

[TABLE]

By comparing the lower bound for $u_{k}$ and the upper bound for $v_{k}$, we have an upper bound of $c$ for each $a=3,4,\cdots,10$, as follows:

Now, by using MAPLE program, one may check that there is no stable regular ternary triangular form ${\Delta}(a,b,c)$ for $3\leq a\leq 10$. Therefore, we have $a\leq 2$. ∎

Lemma 3.13**.**

Under the assumptions given above, we have $t\leq 5$.

Proof.

By the proof of Lemma 3.11, we have $t^{\prime}\leq 6$. First, assume that $a=2$. By Lemma 3.9 with $i=29$ and $s=8$, one may easily show, by using Table 1, that

[TABLE]

On the other hand,

[TABLE]

Thus we have $2+9b+c\leq 8\cdot 29+2+b+c$ and $b\leq 29$. Let $g$ be a positive integer satisfying Lemma 3.3 in the case when $p=q_{1}$ and $d=24$. Note that

[TABLE]

for any integer $n$. For any positive integer $r$, define

[TABLE]

Clearly $h(r)$ is represented by $\langle 2,b,c\rangle$ over ${\mathbb{Z}}_{q}$ for any $q\in\{2,3,q_{1}\}$. Note that

[TABLE]

where $b_{ij}(s)$ is an integer defined in Lemma 3.9. From this, similarly with the proof of Lemma 3.9, one may easily show that there exists a positive integer $r$ with $1\leq r\leq 7$ such that $h(r)$ is represented by $\langle 2,b,c\rangle$ over ${\mathbb{Z}}_{q_{i}}$ for any $i=2,3,\dots,t^{\prime}$. Therefore, we have

[TABLE]

Furthermore, since ${\Delta}(2,b,c)$ is regular, we have

[TABLE]

From our choices of $g$ and $r$, we have $h(r)-c\nrightarrow\langle 2,b\rangle$. Thus, $2+b+9c\leq h(r)$, which implies that $c\leq 21q_{1}^{2}$. Therefore we have

[TABLE]

This implies that $t^{\prime}\leq 4$.

Now, assume that $a=1$. By Lemma 3.9 with $i=35$ and $s=8$, one may easily show that

[TABLE]

On the other hand,

[TABLE]

Thus we have $1+9b+c\leq 8\cdot 35+1+b+c$ and $b\leq 35$. Similarly to the case when $a=2$, one may deduce that $c\leq 21q_{1}^{2}$. Therefore, we have

[TABLE]

which implies that $t^{\prime}\leq 4$. This completes the proof. ∎

Lemma 3.14**.**

For any stable regular ternary triangular form $\Delta(a,b,c)$ with $0<a\leq b\leq c$, we have $a+b\leq 21$.

Proof.

Note that $a=1$ or $2$ by Lemma 3.12. First, assume that $a=2$. By Lemma 3.9 with $i=19$ and $s=6$, one may easily show that

[TABLE]

On the other hand,

[TABLE]

Thus we have $2+9b+c\leq 8\cdot 19+2+b+c$, and $b\leq 19$. Now, assume that $a=1$. By Lemma 3.9 with $i=20$ and $s=6$, one may check that

[TABLE]

On the other hand,

[TABLE]

Thus we have $1+9b+c\leq 8\cdot 20+1+b+c$, and $b\leq 20$. ∎

Now, we are ready to classify all stable regular ternary triangular forms. The following lemma is very useful to prove the regularity.

Lemma 3.15**.**

Let $m$ be a positive integer congruent to 4 modulo 8. Then

[TABLE]

Proof.

See [18, Lemma 3.1(iii)]. ∎

Theorem 3.16**.**

There are exactly $17$ stable regular ternary triangular forms.

[TABLE]

Proof.

By Lemmas 3.12, 3.13 and 3.14, we have

[TABLE]

First, we want to find an upper bound for $c$ for each possible pair $(a,b)$. Since all the other cases can be done in a similar manner, we only consider 3 representative cases here.

(i)** $(a,b)=(2,2)$**. Let $E_{1}=\{4\!\cdot\!3,4\!\cdot\!7,4\!\cdot\!11,4\!\cdot\!19,4\!\cdot\!23,4\!\cdot\!31\}$. Suppose that $c\geq 16$. For any $e_{1}\in E_{1}$, $e_{1}$ is not represented by $\langle 2,2\rangle$. Furthermore, since $e_{1}+c<4+9c$ by assumption, $e_{1}+c\overset{2}{\nrightarrow}\langle 2,2,c\rangle$. Since $\Delta(2,2,c)$ is stable regular, there is an odd prime divisor $p$ of $e_{1}+c$ such that $\langle 2,2,c\rangle$ is anisotropic over ${\mathbb{Z}}_{p}$. Therefore, $p$ divides $c$ and also divides $e_{1}$. Furthermore, since $|E_{1}|=6$, there are at least six such odd primes. This is a contradiction to the fact that $t\leq 5$. Thus, we have $c\leq 15$ if $(a,b)=(2,2)$.

(ii)** $(a,b)=(2,3)$**. Let $E_{2}=\{69,117,141,213,285,333\}$. Suppose that $c\geq 42$. Since we are assuming that $\Delta(2,3,c)$ is 3-stable, $c$ is not divisible by 3. Any element of $E_{2}$ is of the from $8n+2+3$ for some positive integer $n$, and the elements of $E_{2}$ share no odd prime divisors other than $3$. Let $e_{2}\in E_{2}$. From the assumption that $c\geq 42$, one may easily check that $e_{2}+c\overset{2}{\nrightarrow}\langle 2,3,c\rangle$. Since ${\Delta}(2,3,c)$ is stable regular, there is an odd prime $p$ dividing $e_{2}+c$ and $\langle 2,3,c\rangle$ is anisotropic over ${\mathbb{Z}}_{p}$. Hence $p$ is greater than $3$ and divides $e_{2}$. Thus there are at least six such odd primes. This is a contradiction, and we have $c\leq 41$.

(iii)** $(a,b)=(2,6)$**. Since ${\Delta}(2,6,c)$ is 3-stable, $c$ is not a multiple of $3$. Note that $48+c=8\cdot 5+2+6+c\overset{2}{\nrightarrow}\langle 2,6,c\rangle$. Thus there is an odd prime $p>3$ dividing $48+c$ and $\langle 2,6,c\rangle$ is anisotropic over ${\mathbb{Z}}_{p}$. Therefore, $48$ is divisible by $p$, which is a contradiction. Therefore, the pair $(a,b)=(2,6)$ is impossible.

All the other cases can be done in a similar manner to one of the above three cases, and one may obtain an upper bound for $c$ in each case. After that, with the help of MAPLE program, one may show that there are $17$ candidates of stable regular ternary triangular forms given above.

For each $i=1,2,\cdots,17$, we write $\Delta_{i}=\Delta(a_{i},b_{i},c_{i})$ and $L_{i}=\langle a_{i},b_{i},c_{i}\rangle$. For any $i\in U=\{1,2,4,5,6,8,9\}$, it is well known that $\Delta_{i}$ is universal (see [10, p.23]). Hence we may assume that $i\not\in U$. Let $n_{i}$ be any positive integer such that

[TABLE]

Note that $L_{i}$ has class number 1 for any $1\leq i\leq 17$ and thus $\widetilde{n_{i}}\longrightarrow L_{i}$.

For $i\in\{11,13,14,15,16\}$, one may easily check that

[TABLE]

that is, if $a_{i}x^{2}+b_{i}y^{2}+c_{i}z^{2}=\widetilde{n_{i}}$, then $xyz\equiv 1\ (\mathrm{mod}\ 2)$. Assume that $i\in\{7,10\}$. Since the class number of $L_{i}$ is 1 and it primitively represents $\widetilde{n_{i}}$ over ${\mathbb{Z}}_{2}$, there is a vector $(x,y,z)\in R(\widetilde{n_{i}},L_{i})$ with $(x,y,z,2)=1$. One may easily check that $(x,y,z,2)=1$ implies $xyz\equiv 1\ (\mathrm{mod}\ 2)$ in this case. If $i=12$, then one may easily show that

[TABLE]

Similarly to the previous case, the existence of a vector $(x,y,z)\in R(\widetilde{n_{i}},L_{i})$ with $(x,y,z,2)=1$ implies that

[TABLE]

By Lemma 3.15,

[TABLE]

Therefore we have $r_{(1,1,1)}(\widetilde{n_{i}},x^{2}+3y^{2}+4z^{2})=2r_{(0,0,1)}(\widetilde{n_{i}},x^{2}+3y^{2}+4z^{2})>0$. If $i=3$, then one may easily check that

[TABLE]

By Lemma 3.15, we have

[TABLE]

Thus we have $r_{(1,1,1)}(\widetilde{n_{i}},x^{2}+y^{2}+3z^{2})=\frac{1}{2}r(\widetilde{n_{i}},x^{2}+y^{2}+3z^{2})>0$. Finally, assume that $i=17$. Note that if $x^{2}+3y^{2}+10z^{2}=8n+14$, then $x\equiv y\ (\mathrm{mod}\ 2)$ and $z\equiv 1\ (\mathrm{mod}\ 2)$. By Lemma 3.15 again, we have

[TABLE]

This completes the proof. ∎

4. Regular ternary triangular forms

In this section, we prove that there are exactly 49 regular ternary triangular forms. Let ${\Delta}(a^{\prime},b^{\prime},c^{\prime})$ be a regular ternary triangular form and let ${\Delta}(a,b,c)$ be the stable regular ternary triangular form obtained from it by taking $\lambda$-transformations, if necessary, repeatedly. Here, we are not assuming that $a\leq b\leq c$. It might happen that there is an odd prime $l$ dividing $a^{\prime}b^{\prime}c^{\prime}$ such that $(abc,l)=1$. We call such a prime $l$ a missing prime. Note that $\lambda_{p}\circ\lambda_{q}=\lambda_{q}\circ\lambda_{p}$ for any odd primes $p$ and $q$. Thus if $l$ is a missing prime, then one of the followings holds:

1. (i)

$\Delta(a,l^{2}b,l^{2}c)$ is regular. 2. (ii)

$\Delta(a,b,l^{2}c)$ is regular and $\left(\displaystyle\frac{-ab}{l}\right)=-1$.

Lemma 4.1**.**

There is no missing prime $l$ greater than $7$.

Proof.

Let $l$ be a missing prime. Then there is a stable regular ternary triangular form $\Delta(a,b,c)$ such that $(abc,l)=1$, and (i) or (ii) given above holds.

Assume that the case (i) holds, that is, $\Delta(a,l^{2}b,l^{2}c)$ is regular. We let

[TABLE]

First, we prove that $l\leq 131$. Assume to the contrary that $l\geq 137$. One may easily check that if

[TABLE]

with odd integers $\alpha,\beta$ and $\gamma$, then $\beta^{2}=\gamma^{2}=1$. Thus we have

[TABLE]

On the other hand, by Theorem 3.16, the set of odd primes at which $\langle a,b,c\rangle$ is anisotropic is

[TABLE]

From Remark 3.8, we have

[TABLE]

From the assumption that $l\geq 137$, we have $\displaystyle\frac{3}{25}l+3\geq\frac{4}{49}l+4$. Since

[TABLE]

we must have

[TABLE]

However, one may directly show that if $l\geq 137$, then $\left\lceil\displaystyle\frac{71}{450}l-\frac{11}{2}\right\rceil>\left[\sqrt{2l+\frac{1}{4}}\right]$. This is a contradiction and hence we have $l\leq 131$. Now, by a direct calculation with the help of MAPLE, one may check that for any prime $11\leq q\leq 131$ and any stable regular ternary triangular form $\Delta(a,b,c)$, all of the triangular forms $\Delta(a,q^{2}b,q^{2}c)$ are not regular.

Now, assume that $\Delta(a,b,l^{2}c)\ (a\leq b)$ is regular with $\left(\displaystyle\frac{-ab}{l}\right)=-1$. By Theorem 3.16, $(a,b)$ is one of the following pairs:

[TABLE]

First, suppose that $l\geq 29$. Since all the other cases can be done in a similar manner, we only consider the cases when $(a,b)=(1,1)$ or $(1,5)$. Assume that $(a,b)=(1,1)$. Since

[TABLE]

and $\Delta(a,b,l^{2}c)$ is regular, there is a vector $(x,y,z)\in z^{3}$ with $xyz\equiv 1\ (\mathrm{mod}\ 2)$ such that $x^{2}+y^{2}+l^{2}cz^{2}=418+l^{2}c$. From the assumption that $l\geq 29$, we have $z^{2}=1$. This is a contradiction, for $418$ is not a sum of two integer squares. Next, assume that $(a,b)=(1,5)$. Note that

[TABLE]

Since we are assuming that ${\Delta}(1,5,l^{2}c)$ is regular, there is a vector $(x_{1},y_{1},z_{1})\in{\mathbb{Z}}^{3}$ with $x_{1}y_{1}z_{1}\equiv 1\ (\mathrm{mod}\ 2)$ such that $x_{1}^{2}+5y_{1}^{2}+l^{2}cz_{1}^{2}=110+l^{2}c$. Since $l\geq 29$, we have $z_{1}^{2}=1$. This is a contradiction, for $110$ is not represented by $\langle 1,5\rangle$. Therefore, we have $l\leq 23$. Now, by a direct calculation with the help of MAPLE, one may check that for any prime $11\leq l\leq 23$ and any stable regular ternary triangular form $\Delta(a,b,c)$, all of the forms $\Delta(a,b,l^{2}c)$ are not regular. This completes the proof. ∎

Remark 4.2*.*

By Theorem 3.16 and Lemma 4.1, any prime divisor of the discriminant of a regular ternary triangular form is less than or equal to $7$.

Let $\Delta(a^{\prime},b^{\prime},c^{\prime})$ be a regular ternary triangular form. Then there are nonnegative integers $e_{3},e_{5}$ and $e_{7}$ such that

[TABLE]

is stable regular. Hence, to find all regular ternary triangular forms, it suffices to find all regular ternary triangular forms in the inverse image of the $\lambda_{p}$-transformation of each regular triangular form for each $p\in\{3,5,7\}$. Note that any triangular form in the inverse image $\lambda_{p}^{-1}(\Delta(a,p^{r}b,p^{s}c))$, for $abc\not\equiv 0\ (\mathrm{mod}\ p)$ and $0\leq r\leq s$, is given in Table 2.

First, we find all regular triangular forms in the inverse images of stable regular ternary triangular forms via $\lambda_{p}$-transformation for each $p\in\{3,5,7\}$, and then we repeat this process again until any inverse image does not contain a regular triangular form. As a sample, ternary triangular forms lying over $\Delta(1,1,1)$ are given in Table 3. In that table, if the triangular form is not regular, then the smallest positive integer which is represented locally, but not globally by the triangular form is given in parentheses.

Finally, one may have a list of $49$ candidates for the regular ternary triangular forms including 17 stable regular forms, which is given in Table 4. The regularities of $32$ forms except $17$ stable regular forms will be proved here. Before doing that, we need some lemmas.

Let $p$ be an odd prime and let $k$ be a positive integer relatively prime to $p$. Assume that $p$ is represented by the binary quadratic form $x^{2}+ky^{2}$. In 1928, B. W. Jones proved in his unpublished thesis that if the Diophantine equation $x^{2}+ky^{2}=N(N>0)$ has an integral solution, then it also has an integral solution $x,y$ with $(x,y,p)=1$. The following lemma follows immediately from this.

Lemma 4.3**.**

Let $N$ be a positive integer. If $x^{2}+2y^{2}=N$ for some $(x,y)\in{\mathbb{Z}}^{2}$, then there is a vector $(\tilde{x},\tilde{y})\in{\mathbb{Z}}^{2}$ such that

[TABLE]

We also need the following lemma which appeared in the middle of the proof of [21, Theorem 3.1].

Lemma 4.4**.**

Let $S\in M_{3}({\mathbb{Z}})$ be a positive-definite symmetric matrix and let $T\in M_{3}({\mathbb{Q}})$ such that ${}^{t}TST=S$. Let $(u,v,w)\in{\mathbb{Z}}^{3}$ and define

[TABLE]

Assume that

1. (i)

$T$* has an infinite order.* 2. (ii)

$(u_{n},v_{n},w_{n})\in{\mathbb{Z}}^{3}$* for any $n$.*

Then $(u,v,w)\in{\rm ker}\left(T-{\rm det}(T)I\right)$ and ${\rm dim}_{\mathbb{R}}{\rm ker}\left((T-{\rm det}(T)I)\right)=1$.

In the following 5 consecutive propositions, we prove the regularities of 5 candidates, all of whose corresponding quadratic forms are not regular(see [16]).

Proposition 4.5**.**

The ternary triangular form ${\Delta}(1,4,9)$ is regular.

Proof.

Let $L=\langle 1,4,9\rangle$ be a ternary quadratic form and let $\ell=8n+14$ be an integer such that $\ell\longrightarrow\text{gen}(L)$. One may easily check that $R(\ell,L)=R_{(1,1,1)}(\ell,L)$. Thus it suffices to show that $\ell\longrightarrow L$. Since

[TABLE]

we may assume that $\ell\longrightarrow K$.

First, assume that $\ell\equiv 0,1\ (\mathrm{mod}\ 3)$. Since $\ell\longrightarrow K$, there is a vector $(x,y,z)\in{\mathbb{Z}}^{3}$ such that $x^{2}+y^{2}+36z^{2}=\ell$. We have $x\equiv 0\ (\mathrm{mod}\ 3)$ or $y\equiv 0\ (\mathrm{mod}\ 3)$ and thus $\ell\longrightarrow\langle 1,9,36\rangle\longrightarrow L$.

Now, assume that $\ell\equiv 2\ (\mathrm{mod}\ 3)$. We assert that there is a vector $(x_{1},y_{1},z_{1})\in R(\ell,K)$ such that $x_{1}\not\equiv\pm y_{1}\ (\mathrm{mod}\ 9)$ or $z_{1}\not\equiv 0\ (\mathrm{mod}\ 3)$. Assume to the contrary that there is no such vector. Then, we may assume that there is a vector $(u,v,w)\in R(\ell,K)$ such that $u\equiv v\ (\mathrm{mod}\ 9)$ and $w\equiv 0\ (\mathrm{mod}\ 3)$. Let

[TABLE]

Note that

[TABLE]

If we let

[TABLE]

then one may check that $(u_{1},v_{1},w_{1})\in{\mathbb{Z}}^{3}$ and thus $(u_{1},v_{1},w_{1})\in R(\ell,K)$. Thus $u_{1}\equiv\pm v_{1}\ (\mathrm{mod}\ 9)$ and $w_{1}\equiv 0\ (\mathrm{mod}\ 3)$ by assumption. Since

[TABLE]

we have $u_{1}\equiv v_{1}\ (\mathrm{mod}\ 9)$. From this, one may easily check that $T$ satisfies all conditions given in Lemma 4.4 with $S=M_{K}$, and thus we have $(u,v,w)\in\text{ker}(T-I)$. Since $\text{ker}(T-I)=\langle(1,1,0)\rangle$, we have $(u,v,w)=k(1,1,0)$ for some integer $k$ and $u^{2}+v^{2}+36w^{2}=2k^{2}$. This is a contradiction to the fact that $\ell\equiv 6\ (\mathrm{mod}\ 8)$, and we may conclude that there is a vector $(x_{2},y_{2},z_{2})\in R(\ell,K)$ such that

[TABLE]

By changing signs of $x_{2},y_{2},z_{2}$ and by interchanging the role of $x_{2}$ and $y_{2}$, if necessary, we may assume that there is a vector $(x_{3},y_{3},z_{3})\in R(\ell,K)$ such that $2x_{3}+y_{3}+12z_{3}\equiv 0\ (\mathrm{mod}\ 9)$. If we let

[TABLE]

then one may easily show that $(x_{4},y_{4},z_{4})\in R\left(\ell,L\right)$. This completes the proof. ∎

Proposition 4.6**.**

The ternary triangular form ${\Delta}(1,3,27)$ is regular.

Proof.

Let $L=\langle 1,3,27\rangle$ be a ternary quadratic form and let $\ell=8n+31$ be an integer such that $\ell\longrightarrow\text{gen}(L)$. Note that

[TABLE]

By [21, Theorem 2.3] one may show that any integer congruent to 7 modulo 8 that is represented by $K$ is also represented by $L$. Therefore, $\ell$ is represented by $L$. Note that if $x^{2}+3y^{2}+27z^{2}=\ell$, then

[TABLE]

Therefore, if there is a vector $(x,y,z)\in R(\ell,L)$ with $x\equiv y\ (\mathrm{mod}\ 2)$, then we are done by Lemma 3.15. Thus we may assume that for any $(x,y,z)\in R(\ell,L)$,

[TABLE]

Suppose that $xy\not\equiv 0\ (\mathrm{mod}\ 3)$ for any $(x,y,z)\in R\left(\ell,L\right)$. Let $(u,v,w)\in R\left(\ell,L\right)$ with $u\equiv v\ (\mathrm{mod}\ 3)$. For a rational isometry

[TABLE]

of $M_{L}$, we apply Lemma 4.4. Then we have $(u,v,w)\in\text{ker}(T+I)$. Since $\text{ker}(T+I)=\langle(2,-1,0)\rangle$, we have $(u,v,w)=k(2,-1,0)$ for some integer $k$. One may easily check that $|k|>1$ and $(k,6)=1$. Hence there is a prime $q\geq 5$ such that $k=qs$ and $s\in{\mathbb{Z}}$. Then

[TABLE]

On the other hand,

[TABLE]

If we let $M_{1}=\langle 27\rangle\perp\begin{pmatrix}4&2\\ 2&4\end{pmatrix}$, then

[TABLE]

Note that $7\longrightarrow M_{2}$. By [1, Proposition 1], we have $7q^{2}\longrightarrow M_{1}$ and thus $\ell=7q^{2}s^{2}\longrightarrow M_{1}$. Thus $r_{(1,1,1)}(\ell,L)>0$ and we are done with this case.

Now, suppose that there is a vector $(x_{1},y_{1},z_{1})\in R\left(\ell,L\right)$ such that $x_{1}y_{1}\equiv 0\ (\mathrm{mod}\ 3)$. We define

[TABLE]

Then, one may easily check that $(x_{2},y_{2},z_{2})\in R_{(1,1,1)}\left(\ell,L\right)$. ∎

Proposition 4.7**.**

The ternary triangular form ${\Delta}(1,6,27)$ is regular.

Proof.

Let $L=\langle 1,6,27\rangle$ be a ternary quadratic form and let $\ell=8n+34$ be an integer such that $\ell\longrightarrow\text{gen}(L)$. Note that

[TABLE]

Since $\lambda_{2}(L)\simeq\langle 3\rangle\perp\begin{pmatrix}2&1\\ 1&14\end{pmatrix}\simeq\lambda_{2}(K)$, we have

[TABLE]

and thus $\ell\longrightarrow L$. If $(x,y,z)\in R\left(\ell,L\right)$, then

[TABLE]

Thus we may assume that for any $(x,y,z)\in R\left(\ell,L\right)$,

[TABLE]

First, suppose that there is a vector $(x_{1},y_{1},z_{1})\in R(\ell,L)$ with $x_{1}\equiv 0\ (\mathrm{mod}\ 3)$. If we let

[TABLE]

then one may easily check that $(x_{2},y_{2},z_{2})\in R_{(1,1,1)}(\ell,L)$. Hence we may further assume that for any $(x,y,z)\in R(\ell,L)$, $x\not\equiv 0\ (\mathrm{mod}\ 3)$.

Now, suppose that there is a vector $(x_{3},y_{3},z_{3})\in R(\ell,L)$ with $y_{3}\equiv 0\ (\mathrm{mod}\ 3)$. Let $y_{3}=3y_{3}^{\prime}$. Then we have $x_{3}^{2}+27(2y_{3}^{\prime 2}+z_{3}^{2})=\ell$. Since $y_{3}^{\prime}\equiv 1\ (\mathrm{mod}\ 2)$, we have $2y_{3}^{\prime 2}+z_{3}^{2}\neq 0$. By Lemma 4.3, there is a vector $(x_{4},y_{4},z_{4})\in{\mathbb{Z}}^{3}$ with $y_{4}\not\equiv z_{4}\ (\mathrm{mod}\ 3)$ such that $x_{4}^{2}+27(2y_{4}^{2}+z_{4}^{2})=\ell$. Thus $(x_{4},3y_{4},z_{4})\in R(\ell,L)$ such that $y_{4}\not\equiv 0\ (\mathrm{mod}\ 3)$ or $z_{4}\not\equiv 0\ (\mathrm{mod}\ 3)$. By changing signs of $x_{4},y_{4},z_{4}$, if necessary, we may assume that $x_{4}\equiv y_{4}+z_{4}\ (\mathrm{mod}\ 3)$. If we let

[TABLE]

then one may easily check that $(x_{5},y_{5},z_{5})\in R_{(1,1,1)}(\ell,L)$. Therefore, we further assume that for any $(x,y,z)\in R(\ell,L)$, $xy\not\equiv 0\ (\mathrm{mod}\ 3)$.

Suppose that there is a vector $(x_{6},y_{6},z_{6})\in R(\ell,L)$ such that $y_{6}\not\equiv\pm 4x_{6}\ (\mathrm{mod}\ 9)$ or $z_{6}\not\equiv 0\ (\mathrm{mod}\ 3)$. Then one may check that by changing signs of $x_{6},y_{6},z_{6}$, if necessary, we may assume that

[TABLE]

If $x_{6}+y_{6}-3z_{6}\equiv 0\ (\mathrm{mod}\ 9)$, then we define

[TABLE]

If $x_{6}-4y_{6}-3z_{6}\equiv 0\ (\mathrm{mod}\ 9)$, then we define

[TABLE]

Then one may easily check that $(x_{7},y_{7},z_{7})\in R_{(1,1,1)}(\ell,L)$ in each case. Now, we further assume that for any $(x,y,z)\in R(\ell,L)$,

[TABLE]

Suppose that there is a vector $(x_{8},y_{8},z_{8})\in R(\ell,L)$ such that $z_{8}\not\equiv 0\ (\mathrm{mod}\ 9)$. By changing signs of $y_{8}$ and $z_{8}$, if necessary, we may assume that $y_{8}\equiv 4x_{8}\ (\mathrm{mod}\ 9)$ and $\displaystyle\frac{x_{8}-y_{8}}{3}+z_{8}\not\equiv\pm 4x_{8}\ (\mathrm{mod}\ 9)$. If we let

[TABLE]

then $(x_{9},y_{9},z_{9})\in R(\ell,L)$ and $y_{9}\not\equiv\pm 4x_{9}\ (\mathrm{mod}\ 9)$. This contradicts to our assumption (4.2). Therefore, we further assume that for any $(x,y,z)\in R(\ell,L)$,

[TABLE]

Take a vector $(u,v,w)\in R(\ell,L)$ with $u\equiv v\ (\mathrm{mod}\ 3)$ so that $u+2v+6w\equiv 0\ (\mathrm{mod}\ 9)$. If we let

[TABLE]

then one may easily check that

[TABLE]

If we let

[TABLE]

then clearly, $(u_{1},v_{1},w_{1})\in{\mathbb{Z}}^{3}$, and thus $(u_{1},v_{1},w_{1})\in R(\ell,L)$. Note that $u_{1}-v_{1}\equiv 0\ (\mathrm{mod}\ 3)$. From this, one may show that $T$ satisfies all conditions given in Lemma 4.4 with $S=M_{L}$, and thus we have $(u,v,w)\in\text{ker}(T+I)$. Since $\text{ker}(T+I)=\langle(2,-1,0)\rangle$, we have $(u,v,w)=k(2,-1,0)$ for some integer $k$ with $|k|>1$ and $(k,6)=1$. Thus there is a prime divisor $q\geq 5$ of $k$. Now $\ell=10q^{2}s^{2}$ for some odd integer $s$. Note that

[TABLE]

Let $M_{1}=\langle 6\rangle\perp\begin{pmatrix}16&4\\ 4&28\end{pmatrix}$. Then

[TABLE]

Note that $10\longrightarrow M_{2}$. By [1, Proposition 1], we have $r(10q^{2}s^{2},M_{1})>0$, and this completes the proof. ∎

Proposition 4.8**.**

The ternary triangular form ${\Delta}(1,9,18)$ is regular.

Proof.

Let $L=\langle 1,9,18\rangle$ be a ternary quadratic form and let $\ell=8n+28$ be an integer such that $\ell\longrightarrow\text{gen}(L)$. Note that

[TABLE]

Since $\lambda_{2}(L)\simeq\langle 9\rangle\perp\begin{pmatrix}2&1\\ 1&5\end{pmatrix}\simeq\lambda_{2}(K)$, we have $\ell\longrightarrow L$. Let $(x,y,z)\in R(\ell,L)$. We may assume that $x\equiv y\equiv z\equiv 0\ (\mathrm{mod}\ 2)$. Then $x\not\equiv y\ (\mathrm{mod}\ 4)$.

First, assume that $x\not\equiv 0\ (\mathrm{mod}\ 3)$ and $y^{2}+2z^{2}>0$. Then by Lemma 4.3, there is a vector $(y_{1},z_{1})\in{\mathbb{Z}}^{2}$ with $y_{1}\not\equiv z_{1}\ (\mathrm{mod}\ 3)$, $y_{1}\equiv y\ (\mathrm{mod}\ 4)$ and $z_{1}\equiv z\ (\mathrm{mod}\ 2)$ such that $y_{1}^{2}+2z_{1}^{2}=y^{2}+2z^{2}$. So $x^{2}+9y_{1}^{2}+18z_{1}^{2}=\ell$. By replacing $x$ by $-x$, if necessary, we may assume $x+y_{1}-z_{1}\equiv 0\ (\mathrm{mod}\ 3)$. If we let

[TABLE]

then one may easily check that $(x_{2},y_{2},z_{2})\in R_{(1,1,1)}\left(\ell,L\right)$.

Now, assume that $x\not\equiv 0\ (\mathrm{mod}\ 3)$ and $y=z=0$. Note that

[TABLE]

If we let $M_{1}=\langle 9\rangle\perp\begin{pmatrix}5&2\\ 2&8\end{pmatrix}$, then

[TABLE]

Then by [1, Proposition 1], $2p^{2}\longrightarrow M_{1}$ for any prime $p\geq 5$. Note that

[TABLE]

So there is a prime divisor $q$ of $\displaystyle\frac{x}{2}$ with $q\geq 5$. Thus we have $r\left(\displaystyle\frac{\ell}{2},M_{1}\right)>0$.

Finally, assume that $x\equiv 0\ (\mathrm{mod}\ 3)$. If we let

[TABLE]

then one may easily check that $(x_{3},y_{3},z_{3})\in R_{(1,1,1)}(\ell,L)$. ∎

Proposition 4.9**.**

The ternary triangular form ${\Delta}(1,1,18)$ is regular.

Proof.

Let $L=\langle 1,1,18\rangle$ be a ternary quadratic form and let $\ell=8n+20$ be an integer such that $\ell\longrightarrow\text{gen}(L)$. Note that

[TABLE]

Since $\lambda_{2}(L)\simeq\langle 1,1,9\rangle\simeq\lambda_{2}(K)$, we have $\ell\longrightarrow L$. Let $(x,y,z)\in R(\ell,L)$.

First, assume that $\ell\equiv 0\ (\mathrm{mod}\ 3)$. Then $x\equiv y\equiv 0\ (\mathrm{mod}\ 3)$ and thus $\ell\equiv 0\ (\mathrm{mod}\ 9)$. So

[TABLE]

Note that $\displaystyle\frac{\ell}{9}\geq 4$ and $\displaystyle\frac{\ell}{9}\equiv 4\ (\mathrm{mod}\ 8)$. Since the triangular form $\Delta(1,1,2)$ is universal, there is a vector $(x_{1},y_{1},z_{1})\in R_{(1,1,1)}\left(\displaystyle\frac{\ell}{9},\langle 1,1,2\rangle\right)$ and thus $(3x_{1},3y_{1},z_{1})\in R_{(1,1,1)}\left(\ell,L\right)$.

Now, assume $\ell\equiv 1\ (\mathrm{mod}\ 3)$. Note that $xy\equiv 0\ (\mathrm{mod}\ 3)$. Without loss of generality, we may assume that $y\equiv 0\ (\mathrm{mod}\ 3)$. Then

[TABLE]

Note that $\ell\geq 28$, $\ell\equiv 4\ (\mathrm{mod}\ 8)$. Since $\Delta(1,9,18)$ is regular by Proposition 4.8, there is a vector $(x_{2},y_{2},z_{2})\in R_{(1,1,1)}\left(\ell,\langle 1,9,18\rangle\right)$ and thus $(x_{2},3y_{2},z_{2})\in R_{(1,1,1)}\left(\ell,\langle 1,1,18\rangle\right)$.

Finally, assume that $\ell\equiv 2\ (\mathrm{mod}\ 3)$. Since $x^{2}+y^{2}+18z^{2}\equiv 4\ (\mathrm{mod}\ 8)$, we may assume that $x\equiv 0\ (\mathrm{mod}\ 4)$, $y\equiv 2\ (\mathrm{mod}\ 4)$ and $z\equiv 0\ (\mathrm{mod}\ 2)$. Since $xy\not\equiv 0\ (\mathrm{mod}\ 3)$, we may further assume that $x\equiv y\ (\mathrm{mod}\ 3)$. If we let

[TABLE]

then one may easily check that $(x_{3},y_{3},z_{3})\in R_{(1,1,1)}\left(\ell,L\right)$. ∎

Theorem 4.10**.**

There are exactly $49$ regular ternary triangular forms, which are listed in Table 4.

Proof.

For $1\leq i\leq 49$, we write $\Delta_{i}=\Delta(a_{i},b_{i},c_{i})$. Let $L_{i}=\langle a_{i},b_{i},c_{i}\rangle$ be a ternary quadratic form and let $\ell_{i}(n)=8n+a_{i}+b_{i}+c_{i}$ be any integer such that $\ell_{i}(n)\longrightarrow\text{gen}(L_{i})$. In Theorem 3.16 and Propositions 4.5$\sim$4.9, we have already proved the regularity of each $\Delta_{i}$ when

[TABLE]

Hence we may assume that $i$ is not contained in the above set. Note that for any integer $i$ which is not contained in $\{16,26,35,44,45\}$, which we alreay considered in Propositions 4.5$\sim$4.9, the corresponding quadratic form $L_{i}$ has class number $1$ and thus $\ell_{i}(n)\longrightarrow L_{i}$. If $i\in\{10,36,39,40,41,49\}$, then one may easily show that $R(\ell_{i}(n),L_{i})=R_{(1,1,1)}(\ell_{i}(n),L_{i})$. Hence $\ell_{i}(n)\overset{2}{\longrightarrow}L_{i}$ in this case.

Now, we consider the case when $i=30$. Note that if $x^{2}+3y^{2}+18z^{2}=8n+22$, then we have $z\equiv 1\ (\mathrm{mod}\ 2)$ and $x\equiv y\ (\mathrm{mod}\ 2)$. By Lemma 3.15, we have

[TABLE]

Since the proof of the case when $i=48$ is quite similar to this, we omit the proof.

Assume that $i=31$. Since the quadratic form $\langle 1,6,9\rangle$ has class number $1$ and it primitively represents $8n+16$ over ${\mathbb{Z}}_{2}$, there is a vector

[TABLE]

Since $x^{2}+6y^{2}+9z^{2}\equiv 0\ (\mathrm{mod}\ 8)$, we have $xyz\equiv 1\ (\mathrm{mod}\ 2)$.

For the remaining $i$, that is,

[TABLE]

one may check that $\Delta(a_{i},b_{i},c_{i})$ can be obtained from a ternary triangular form whose regularity is already proved by taking $\lambda_{p}$-transformations several times for some $p\in\{3,5,7\}$. Furthermore, one may easily check that the regularity is preserved during taking the $\lambda_{p}$-transformation. This completes the proof. ∎