Completely $p$-primitive binary quadratic forms

TL;DR

This paper investigates the properties of completely p-primitive binary quadratic forms, providing a necessary and sufficient condition for definite forms to possess this property, which relates to solutions of certain Diophantine equations.

Contribution

It introduces the concept of completely p-primitive forms and characterizes when a definite binary quadratic form is completely p-primitive.

Findings

Characterization of completely p-primitive forms for definite cases

Necessary and sufficient conditions established

Insights into solutions of Diophantine equations related to these forms

Abstract

Let $f(x,y)=ax^2+bxy+cy^2$ be a binary quadratic form with integer coefficients. For a prime $p$ not dividing the discriminant of $f$, we say $f$ is completely $p$-primitive if for any non-zero integer $N$, the diophantine equation $f(x,y)=N$ has always an integer solution $(x,y)=(m,n)$ with $(m,n,p)=1$ whenever it has an integer solution. In this article, we study various properties of completely $p$-primitive binary quadratic forms. In particular, we give a necessary and sufficient condition for a definite binary quadratic form $f$ to be completely $p$-primitive.

Equations

$$f(x,y)=a{x}^2+bxy+c{y}^2(a,b,c\in \mathbb{Z})$$

$$Q^{\ast }(f)\subset Q_p^{\ast }(f)\subset Q(f)$$

$$f\simeq g⟺\text{there is a proper class C∈GD such that f,g∈C.}$$

$$h(x,y)=\{\begin{array}{ll}{x}^2-\frac{D}{4}{y}^2& \text{if D is even,}\\ x^2+xy+\frac{1-D}{4}{y}^2& \text{otherwise}.\end{array}$$

$$[a,b,c]\simeq [a,2a({\alpha }_{1}t+x_0)+b,f({\alpha }_{1}t+x_0,1)]$$

$$[\alpha ,\beta ,\gamma ]\simeq [\alpha ,2\alpha (a_{1}t+z_0)+\beta ,g(a_{1}t+z_0,1)].$$

$$af({\alpha }_{1}t+x_0,1)=\alpha g(a_{1}t+z_0,1).$$

$$a{x}_0^2+b{x}_0+c={\alpha }_{1}a({\alpha }_1{t}^2+2t{x}_0)+(b{\alpha }_1)t-f({\alpha }_{1}t+x_0,1)\equiv 0\text{(mod α1)},$$

$$[a,b,c][\alpha ,\beta ,\gamma ]\simeq [a,2a({\alpha }_1{t}_0+x_0)+b,\alpha \eta ][\alpha ,2\alpha (a_1{t}_0+z_0)+\beta ,\ast ]\simeq [a\alpha ,\ast ,\ast ]\in \mathfrak{A}\cdot \mathfrak{Z}.$$

$$\frac{\beta -b}{2}\equiv 0(\mathrm{mod}{d}_1)\text{and}\frac{\beta +b}{2}\equiv 0(\mathrm{mod}{d}_2).$$

$$[d_1^{\prime },b,d_2^{\prime }\tilde{a}c]\in {\mathfrak{A}}_{d_1^{\prime }},[d_2^{\prime },b,d_1^{\prime }\tilde{a}c]\in {\mathfrak{A}}_{d_2^{\prime }}\text{and}[\tilde{a},b,d_1^{\prime }{d}_2^{\prime }c]\in {\mathfrak{A}}_{\tilde{a}}.$$

$$\mathfrak{A}={\mathfrak{A}}_{d_1^{\prime }}\cdot {\mathfrak{A}}_{d_2^{\prime }}\cdot {\mathfrak{A}}_{\tilde{a}}\text{and}\mathfrak{Z}={\mathfrak{Z}}_{d_1^{\prime \prime }}\cdot {\mathfrak{Z}}_{d_2^{\prime \prime }}\cdot {\mathfrak{Z}}_{\tilde{\alpha }}.$$

$$d_2^{\prime }\tilde{a}\cdot d_2^{\prime \prime }\tilde{\alpha }\in Q({\mathfrak{A}}_{d_2^{\prime }}\cdot {\mathfrak{A}}_{\tilde{a}}\cdot {\mathfrak{Z}}_{d_2^{\prime \prime }}\cdot {\mathfrak{Z}}_{\tilde{\alpha }})\text{and}(d_1^{\prime }{d}_1^{\prime \prime },d_2^{\prime }\tilde{a}\cdot d_2^{\prime \prime }\tilde{\alpha })=(d_2^{\prime }\tilde{a}\cdot d_2^{\prime \prime }\tilde{\alpha },p)=1,$$

$$Q(x)=a,2B(x,y)=b\text{and}Q(y)=c.$$

$$\begin{diagram}$$

$$\sigma (px)=\frac{(m+bn)}{2}x+(-an)y\text{and}\sigma (py)=cnx+\frac{(m-bn)}{2}y.$$

$$\sigma (px)=ux+vy\text{and}\sigma (py)=rx+sy.$$

$$p^{2}a=a{u}^2+buv+c{v}^2,$$

$$p^{2}c=a{r}^2+brs+c{s}^2,$$

$$p^{2}b=2ura+(us+vr)b+2vsc,$$

$$us-vr=p^2.$$

$$r=-cv\mu \text{and}s=(au+bv)\mu .$$

$$4p^2=(2u+bt{)}^2+(-D)t^2.$$

$${\text{ord}}_q(Q(x-\sigma (x)))\equiv {\text{ord}}_q(Q(x+\sigma (x)))\equiv 0(\text{mod 2}),$$

$$Q(x-\sigma (x))=(2a+bt{)}^{2}Q(z)\text{and}Q(x+\sigma (x))=b^2(4Q(w)-Q(z)t^2).$$

$$Q(x)=a^{2}Q(z)+2abB(z,w)+b^{2}Q(w)=\frac{b^2}{4}(4Q(w)-Q(z)t^2).$$

$$a{p}^2\in Q_p^{\ast }(\mathfrak{A}\cdot {\mathfrak{I}}_D)\cup Q_p^{\ast }(\mathfrak{A}\cdot {\mathfrak{I}}_D^{-1})=Q_p^{\ast }(\mathfrak{A}).$$

$$p^{2}a=Q(ux+vy)=a{u}^2+buv+c{v}^2\text{and}(p,u,v)=1.$$

$$\sigma (px)=ux+vy\text{and}\sigma (py)=\left(\frac{bu+cv}{a}\right)x-uy.$$

$$r(n,[a,b,c])=r(n,\mathfrak{C})=|\{(x,y)\in {\mathbb{Z}}^2:a{x}^2+bxy+c{y}^2=n\}|,$$

$$r_p^{\ast }(n,[a,b,c])=r_p^{\ast }(n,\mathfrak{C})=|\{(x,y)\in {\mathbb{Z}}^2:a{x}^2+bxy+c{y}^2=n,(x,y,p)=1\}|.$$

$$R(n,{\mathfrak{G}}_D):=\sum _{\mathfrak{C}\in {\mathfrak{G}}_D}r(n,\mathfrak{C})=\omega \sum _{k\mid n}\left(\frac{D}{k}\right),$$

Full text

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

1. Introduction

A two variable homogeneous quadratic polynomial with integer coefficients

[TABLE]

is called a binary quadratic form if the discriminant $D=b^{2}-4ac$ is a non-square integer. We always assume that $f$ is primitive, that is, $(a,b,c)=1$, unless stated otherwise. An integer $N$ is said to be (primitively) represented by $f$ if $f(x,y)=N$ has an integer solution $x,y$ ($x,y$ such that $(x,y)=1$, respectively). The set of all non-zero integers that are (primitively) represented by $f$ is denoted by $Q(f)$ ($Q^{*}(f)$, respectively). As one of diophantine equations, it is quite a difficult problem to decide $Q(f)$ for an arbitrary binary quadratic form $f$.

In 1928, in his unpublished thesis, B. W. Jones proved that if an odd prime $p$ is represented by $x^{2}+ky^{2}$, where $k$ is a positive integer relatively prime to $p$, then the diophantine equation $x^{2}+ky^{2}=N\ (N>0)$ has always an integer solution $x,y$ with $(x,y,p)=1$, if it has an integer solution. This lemma was used by many authors to solve some problems on representations of positive ternary quadratic forms (see, for example, [4], [6], [7], [8] and [10]).

The aim of this article is to fully generalize Jones’ lemma stated above. To be more precise, let $f$ be a binary quadratic form with discriminant $D$ and let $p$ be a prime not dividing $D$. An integer $a$ is said to be $p$-primitively represented by $f$ if $f(x,y)=a$ has an integer solution $(x,y)=(m,n)$ with $(m,n,p)=1$. The set of all integers that are $p$-primitively represented by $f$ is denoted by $Q_{p}^{*}(f)$. Then, clearly

[TABLE]

for any prime $p$. Note that $0\not\in Q(f)$ from the definition. A binary form $f$ is called completely $p$-primitive if $Q(f)=Q_{p}^{*}(f)$. Jones’ lemma stated above says that $x^{2}+ky^{2}$ is completely $p$-primitive for any odd prime $p\in Q(x^{2}+ky^{2})$ relatively prime to $k$.

Let $D$ be a non-square integer congruent to [math] or $1$ modulo $4$. Let $\mathfrak{G}_{D}$ be the set of all proper classes of primitive binary quadratic forms with discriminant $D$. Then it is well known that $\mathfrak{G}_{D}$ forms an abelian group with the composition law (for details, see [2]). Let $\mathfrak{I}_{D}$ be the identity class in $\mathfrak{G}_{D}$. For any proper class $\mathfrak{A}\in\mathfrak{G}_{D}$, the set $Q(\mathfrak{A})$ denotes $Q(f)$ for any binary quadratic form $f\in\mathfrak{A}$. Similarly, we also define $Q^{*}(\mathfrak{A})$ and $Q_{p}^{*}(\mathfrak{A})$. For any two binary forms $f$ and $g$ having same discriminant, we write

[TABLE]

A proper class $\mathfrak{D}\in\mathfrak{G}_{D}$ is called an ambiguous class if $\mathfrak{D}^{2}=\mathfrak{I}_{D}$. For a prime $p$, a proper class $\mathfrak{A}\in\mathfrak{G}_{D}$ is called completely $p$-primitive if any binary form $f\in\mathfrak{A}$ is completely $p$-primitive, or equivalently, $Q(\mathfrak{A})=Q_{p}^{*}(\mathfrak{A})$.

In this article, we study various properties of completely $p$-primitive binary forms for any prime $p$. If $\left(\frac{D}{p}\right)=-1$, where $\left(\frac{\cdot}{\cdot}\right)$ is the Kronecker’s symbol, then one may easily check that no integers divisible by $p$ are $p$-primitively represented by $f$. Hence we always assume that $\left(\frac{D}{p}\right)=1$. We prove that if $p^{2}\in Q_{p}^{*}(\mathfrak{I}_{D})$, then any proper class in $\mathfrak{G}_{D}$ is completely $p$-primitive. Conversely, if an ambiguous class $\mathfrak{D}\in\mathfrak{G}_{D}$ is completely $p$-primitive, then $p^{2}\in Q_{p}^{*}(\mathfrak{I}_{D})$. Finally, for any prime $p$ such that $p^{2}\not\in Q_{p}^{*}(\mathfrak{I}_{D})$, we prove that a proper class $\mathfrak{A}\in\mathfrak{G}_{D}$ is completely $p$-primitive if and only if the order of $\mathfrak{A}$ in the group $\mathfrak{G}_{D}$ is $4$ and $p^{2}\in Q_{p}^{*}(\mathfrak{A}^{2})$, under the assumption that $D<0$.

Some basic notations and terminologies on binary quadratic forms, especially the composition law between binary forms having same discriminant, can be found in [2]. See also [9] for some basic notations and terminologies on ${\mathbb{Z}}$-lattices. For a binary quadratic form $f(x,y)=ax^{2}+bxy+cy^{2}$, we simply write $f=[a,b,c]$.

2. $p$-primitive representations of integers by binary quadratic forms

Let $D$ be a non-square integer such that $D\equiv 0$ or $1\pmod{4}$. Let $\mathfrak{G}_{D}$ be the abelian group of proper equivalence classes of primitive binary quadratic forms with discriminant $D$. Let $\mathfrak{I}_{D}$ be the identity class in $\mathfrak{G}_{D}$, that is, $h(x,y)\in\mathfrak{I}_{D}$, where

[TABLE]

Lemma 2.1**.**

A binary quadratic form $f$ is completely $p$-primitive if and only if $ap^{2}\in Q_{p}^{*}(f)$ for any $a\in Q_{p}^{*}(f)$.

Proof.

Note that “only if” part is trivial. Assume that $ap^{2}\in Q_{p}^{*}(f)$ for any $a\in Q_{p}^{*}(f)$. For any $t\in Q(f)$, let $t=f(p^{\alpha}a,p^{\beta}b)$, where $a,b$ are integers that are either zero or not divisible by $p$. Then $\frac{t}{p^{2\min(\alpha,\beta)}}\in Q_{p}^{*}(f)$. Now, by assumption, we have $t\in Q_{p}^{*}(f)$. Therefore we have $Q(f)=Q_{p}^{*}(f)$. ∎

Let $\mathfrak{A},\mathfrak{Z}$ be proper classes of primitive binary quadratic forms with discriminant $D$ and $f\in\mathfrak{A}$ and $g\in\mathfrak{Z}$. It is well known that for any integers $a\in Q(\mathfrak{A})$ and $\alpha\in Q(\mathfrak{Z})$, $a\alpha\in Q(\mathfrak{A}\cdot\mathfrak{Z})$ (see [3]). Furthermore, if $a\in Q^{*}(\mathfrak{A}),\alpha\in Q^{*}(\mathfrak{Z})$ and $(a,\alpha)=1$, then $a\alpha\in Q^{*}(\mathfrak{A}\cdot\mathfrak{Z})$. However, if $a$ and $\alpha$ are not relatively prime, then it is no longer true. For example, if $[2,1,4]\in\mathfrak{A}$ and $[2,-1,4]\in\mathfrak{Z}$, then $7\in Q^{*}(\mathfrak{A})\cap Q^{*}(\mathfrak{Z})$. However, $7\!\cdot\!7\not\in Q^{*}(\mathfrak{A}\cdot\mathfrak{Z})=Q^{*}([1,1,8])$. Note that for any positive integer $a$, $a\in Q^{*}(\mathfrak{A})$ if and only if there are integers $b,c$ such that $[a,b,c]\in\mathfrak{A}$.

Lemma 2.2**.**

Under the notations given above, let $f=[a,b,c]\in\mathfrak{A}$ and $g=[\alpha,\beta,\gamma]\in\mathfrak{Z}$. If $(a,\alpha,D)=1$ and $2(a,\alpha)$ divides $\beta-b$, then $a\alpha\in Q^{*}(\mathfrak{A}\cdot\mathfrak{Z})$.

Proof.

Let $a=da_{1}$ and $\alpha=d\alpha_{1}$, where $d=(a,\alpha)$. Let $x_{0},z_{0}$ be integers such that $a_{1}x_{0}-\alpha_{1}z_{0}=\frac{\beta-b}{2d}$. Note that for any integer $t$,

[TABLE]

and

[TABLE]

Since $\det([a,b,c])=\det([\alpha,\beta,\gamma])=D$ and $2a(\alpha_{1}t+x_{0})+b=2\alpha(a_{1}t+z_{0})+\beta$, we have

[TABLE]

Therefore, $f(\alpha_{1}t+x_{0},1)$ is divisible by $\alpha_{1}$. Furthermore, since

[TABLE]

there is an integer $m$ such that $ax_{0}^{2}+bx_{0}+c=\alpha_{1}m$. Let $t_{0}$ be an integer such that $bt_{0}+m\equiv 0\pmod{d}$. Note that this is possible, for $(b,d)=1$. Since $f(\alpha_{1}t_{0}+x_{0},1)=\alpha\eta$ for some integer $\eta$, we have

[TABLE]

Therefore, we have $a\alpha\in Q^{*}(\mathfrak{A}\cdot\mathfrak{Z})$. ∎

Lemma 2.3**.**

Under the notations given above, if $a\in Q_{p}^{*}(\mathfrak{A})$, $\alpha\in Q(\mathfrak{Z})$, where $(a,\alpha)=(\alpha,p)=1$, then $a\alpha\in Q_{p}^{*}(\mathfrak{A}\cdot\mathfrak{Z})$.

Proof.

Assume that $a=f(dx_{0},dy_{0})$, where $(d,p)=(x_{0},y_{0})=1$. Then $\frac{a}{d^{2}}\in Q^{*}(\mathfrak{A})$. Since $\alpha\in Q(\mathfrak{Z})$ and $(\alpha,p)=1$, there is an integer $\delta$ relatively prime to $p$ such that $\frac{\alpha}{\delta^{2}}\in Q^{*}(\mathfrak{Z})$. From the assumption that $(a,\alpha)=1$, we have $\frac{a\alpha}{d^{2}\delta^{2}}\in Q^{*}(\mathfrak{A}\cdot\mathfrak{Z})$ by Lemma 2.2. Now, the lemma follows from the fact that $(p,d\delta)=1$. ∎

Proposition 2.4**.**

Under the notations given above, if $a\in Q_{p}^{*}(\mathfrak{A})$, $\alpha\in Q_{p}^{*}(\mathfrak{Z})$, where $(a,\alpha,D)=1$, then $a\alpha\in Q_{p}^{*}(\mathfrak{A}\cdot\mathfrak{Z})\cup Q_{p}^{*}(\mathfrak{A}\cdot\mathfrak{Z}^{-1})$.

Proof.

Without loss of generality, we may assume that $a\in Q^{*}(\mathfrak{A})$ and $\alpha\in Q^{*}(\mathfrak{Z})$. Let $d=(a,\alpha)$. Let $f=[a,b,c]\in\mathfrak{A}$ and $g=[\alpha,\beta,\gamma]\in\mathfrak{Z}$, for some integers $b,c$ and $\beta,\gamma$. Since $D=b^{2}-4ac=\beta^{2}-4\alpha\gamma$, we have $\beta^{2}-b^{2}\equiv 0\pmod{4d}$. Furthermore, since $\left(\frac{\beta-b}{2},\frac{\beta+b}{2},d\right)=1$, there are relatively prime integers $d_{1},d_{2}$ such that $d=d_{1}d_{2}$, and

[TABLE]

Choose integers $d_{1}^{\prime},d_{2}^{\prime},d_{1}^{\prime\prime}$ and $d_{2}^{\prime\prime}$ such that $a=d_{1}^{\prime}d_{2}^{\prime}\widetilde{a}$ and $\alpha=d_{1}^{\prime\prime}d_{2}^{\prime\prime}\widetilde{\alpha}$, where $(d_{1},d_{2}^{\prime}d_{2}^{\prime\prime}\cdot\widetilde{a}\widetilde{\alpha})=(d_{2},d_{1}^{\prime}d_{1}^{\prime\prime}\cdot\widetilde{a}\widetilde{\alpha})=1$. Note that $(d_{1}^{\prime},d_{1}^{\prime\prime})=d_{1}$ and $(d_{2}^{\prime},d_{2}^{\prime\prime})=d_{2}$. We define proper classes $\mathfrak{A}_{d_{1}^{\prime}}$, $\mathfrak{A}_{d_{2}^{\prime}}$ and $\mathfrak{A}_{\widetilde{a}}$ satisfying

[TABLE]

Similarly, we also define proper classes $\mathfrak{Z}_{d_{1}^{\prime\prime}},\mathfrak{Z}_{d_{2}^{\prime\prime}}$ and $\mathfrak{Z}_{\widetilde{\alpha}}$. Then, clearly we have

[TABLE]

If $p$ does not divide $d$, then the proof is almost trivial. Hence we assume that $p$ divides $d$. First, assume that $p$ divides $d_{1}$. Then by Lemma 2.2, we have $d_{1}^{\prime}d_{1}^{\prime\prime}\in Q_{p}^{*}(\mathfrak{A}_{d_{1}^{\prime}}\cdot\mathfrak{Z}_{d_{1}^{\prime\prime}})$. Since

[TABLE]

we have $a\alpha\in Q_{p}^{*}(\mathfrak{A}\cdot\mathfrak{Z})$ by Lemma 2.3. Now, assume that $p$ divides $d_{2}$. In this case, we have $d_{2}^{\prime}d_{2}^{\prime\prime}\in Q_{p}^{*}(\mathfrak{A}_{d_{2}^{\prime}}\cdot\mathfrak{Z}_{d_{2}^{\prime\prime}}^{-1})$ by Lemma 2.2. Since $d_{1}^{\prime}\widetilde{a}\cdot d_{1}^{\prime\prime}\widetilde{\alpha}\in Q(\mathfrak{A}_{d_{1}^{\prime}}\cdot\mathfrak{A}_{\widetilde{a}}\cdot\mathfrak{Z}_{d_{1}^{\prime\prime}}^{-1}\cdot\mathfrak{Z}_{\widetilde{\alpha}}^{-1})$, we may conclude that $a\alpha\in Q_{p}^{*}(\mathfrak{A}\cdot\mathfrak{Z}^{-1})$ by Lemma 2.3. ∎

3. $p$-primitive representations of binary ${\mathbb{Z}}$-lattices

Let $L$ and $M$ be non-classic integral ${\mathbb{Z}}$-lattices in a quadratic space $V$. Here a ${\mathbb{Z}}$-lattice is non-classic integral if its norm ideal $\mathfrak{n}(L)$ is ${\mathbb{Z}}$. Let $p$ be a prime. A representation $\sigma:M\to L$ is called $p$-primitive if $\sigma(M)\not\subset pL$. We say $M$ is $p$-primitively represented by $L$ if there is a $p$-primitive representation from $M$ to $L$. Note that if $\sigma(M_{p})$ is a primitive submodule of $L_{p}$, that is, $\sigma:M_{p}\to L_{p}$ is a primitive representation, then $\sigma$ is $p$-primitive. If $\text{rank}(M)=1$, then the converse is also true. However, for the higher rank case, the converse is not true in general. A $p$-primitive representation $\sigma:pL\to L$ is called essential if $\sigma\in O({\mathbb{Q}}\otimes L)$ has an infinite order. Recall that an isometry $\sigma\in O(V)$ is called proper if $\det(\sigma)=1$, and improper otherwise.

Let $D$ be a non-square integer congruent to [math] or $1$ modulo $4$. Let $\mathfrak{A}$ be a proper class with discriminant $D$ and let $f=[a,b,c]\in\mathfrak{A}$ be a binary quadratic form. The binary ${\mathbb{Z}}$-lattice corresponding to $f$ (or $\mathfrak{A}$) is defined by $L_{f}={\mathbb{Z}}\mathbb{x}+{\mathbb{Z}}\mathbb{y}$ such that

[TABLE]

Note that the binary lattice corresponding to $\mathfrak{A}$ is isometric to the binary lattice corresponding to $\mathfrak{A}^{-1}$. Conversely, for a binary ${\mathbb{Z}}$-lattice $L={\mathbb{Z}}\mathbb{x}+{\mathbb{Z}}\mathbb{y}$, the binary quadratic form $f_{L}$ corresponding to $L$ is defined by $f_{L}=[Q(\mathbb{x}),2B(\mathbb{x},\mathbb{y}),Q(\mathbb{y})]$. Note that the discriminant $D_{L}$ of the binary ${\mathbb{Z}}$-lattice $L$ is defined by $D_{L}=Q(\mathbb{x})Q(\mathbb{y})-B(\mathbb{x},\mathbb{y})^{2}$, as usual.

Theorem 3.1**.**

Let $D$ be a non-square integer congruent to [math] or $1$ modulo $4$ and let $p$ be a prime such that $\left(\frac{D}{p}\right)=1$. Then the followings are all equivalent:

• (i)

$4p^{2}\in Q_{p}^{*}([1,0,-D])$;

• (ii)

$p^{2}\in Q_{p}^{*}(\mathfrak{I}_{D})$;

• (iii)

there is a binary ${\mathbb{Z}}$-lattice $L={\mathbb{Z}}\mathbb{x}+{\mathbb{Z}}\mathbb{y}$ with discriminant $-D/4$ and a proper $p$-primitive representation $\sigma:pL\to L$;

• (iv)

for any binary ${\mathbb{Z}}$-lattice $L={\mathbb{Z}}\mathbb{x}+{\mathbb{Z}}\mathbb{y}$ with discriminant $-D/4$, there is a proper $p$-primitive representation $\sigma:pL\to L$;

• (v)

there is a binary ${\mathbb{Z}}$-lattice $L={\mathbb{Z}}\mathbb{x}+{\mathbb{Z}}\mathbb{y}$ with discriminant $-D/4$ and a proper $p$-primitive essential representation $\sigma:pL\to L$;

• (vi)

for any binary ${\mathbb{Z}}$-lattice $L={\mathbb{Z}}\mathbb{x}+{\mathbb{Z}}\mathbb{y}$ with discriminant $-D/4$, there is a proper $p$-primitive essential representation $\sigma:pL\to L$.

Proof.

Note that (i) $\iff$ (ii) is trivial. We will prove that

[TABLE]

Since proofs of all arrows placed in the middle are trivial, it suffices to show that (i) $\Rightarrow$ (vi) and (iii) $\Rightarrow$ (i).

(i) $\Rightarrow$ (vi). Let $m,n$ be integers such that $4p^{2}=m^{2}+(-D)n^{2}$ with $(m,n,p)=1$. Let $L={\mathbb{Z}}\mathbb{x}+{\mathbb{Z}}\mathbb{y}$ be any ${\mathbb{Z}}$-lattice such that $f_{L}=[a,b,c]$, where $b^{2}-4ac=D$. Define $\sigma:pL\to L$ such that

[TABLE]

Then, one may easily check that $\sigma$ is, in fact, a proper isometry of ${\mathbb{Q}}\otimes L$ such that $\sigma(pL)\not\subset pL$. The characteristic polynomial $f_{\sigma}(x)$ of $\sigma$ is $x^{2}-\frac{m}{p}x+1$. Since all roots of $f_{\sigma}$ are not roots of unity, the order of $\sigma$ is not finite.

(iii) $\Rightarrow$ (i). Assume that the corresponding binary quadratic form to $L$ is $f_{L}=[a,b,c]$, where $b^{2}-4ac=D$. Since $D$ is a non-square integer, we have $ac\neq 0$. Assume that $\sigma:pL\to L$ is a proper $p$-primitive representation such that

[TABLE]

Then we have

[TABLE]

By (3.3) and (3.4), we have $(au+bv)r+cvs=0$. Since at least one of $au+bv$ and $cv$ is non-zero, there is a rational number $\mu$ such that

[TABLE]

Now, by (3.2) and (3.5), $a\mu=\pm 1$. Since $a$ divides both $cv$ and $bv$ and $(a,(b,c))=1$, $a$ divides $v$. By letting $v=at$ in (3.1) for some integer $t$, we have

[TABLE]

Assume that $p$ divides $t$. Then $p$ divides $u$ by (3.6), and $p$ also divides both $r$ and $s$ by (3.5). This is a contradiction to the assumption that $\sigma$ is $p$-primitive. Therefore, $p$ does not divide $t$ and hence $4p^{2}\in Q_{p}^{*}([1,0,-D])$. This completes the proof. ∎

The following lemma will be used in the proof of the main theorem.

Lemma 3.2**.**

Let $L$ be a binary ${\mathbb{Z}}$-lattice with $dL=-D/4$ whose isometry group $O(L)$ contains an improper isometry $\sigma$. Then, for any vector $\mathbb{x}\in L$,

[TABLE]

for any prime $q$ not dividing $D$. In particular, if $\mathbb{x}\not\in pL$ and $Q(\mathbb{x})\equiv 0\ (\text{mod }p)$, where $p$ is a prime not dividing $D$, then $\sigma(\mathbb{x})\neq\pm\mathbb{x}$.

Proof.

Since $\sigma$ is an improper isometry, there is a primitive vector $\mathbb{z}\in L$ such that $\sigma=\tau_{\mathbb{z}}$ by Theorem 43.3 of [9]. Let $L={\mathbb{Z}}\mathbb{z}+{\mathbb{Z}}\mathbb{w}$. Since $\tau_{\mathbb{z}}(\mathbb{w})=\mathbb{w}-\frac{2B(\mathbb{z},\mathbb{w})}{Q(\mathbb{z})}\mathbb{z}\in L$, there is an integer $t$ such that $2B(\mathbb{z},\mathbb{w})=Q(\mathbb{z})t$. If $\mathbb{x}=a\mathbb{z}+b\mathbb{w}$, then $\sigma(\mathbb{x})=-a\mathbb{z}+b(\mathbb{w}-t\mathbb{z})$. Hence

[TABLE]

The lemma follows directly from the fact that $D=Q(\mathbb{z})(Q(\mathbb{z})t^{2}-4Q(\mathbb{w}))$.

Now, assume that $\mathbb{x}\in L-pL$ such that $Q(\mathbb{x})\equiv 0\ (\text{mod }p)$. If $\sigma(\mathbb{x})=-\mathbb{x}$, then $b=0$ and hence $\mathbb{x}=a\mathbb{z}$. This is a contradiction. Suppose that $\sigma(\mathbb{x})=\mathbb{x}$. Then $2a=-bt$ and

[TABLE]

Since $D$ is not divisible by $p$ by assumption, both $b$ and $a$ are divisible by $p$, which is also a contradiction. ∎

4. Classification of completely $p$-primitive binary quadratic forms

Let $D$ be a non-square integer congruent to [math] or $1$ modulo $4$, and let $\mathfrak{G}_{D}$ be a group of proper classes of primitive binary quadratic forms with discriminant $D$. Let $p$ be a prime satisfying $\left(\frac{D}{p}\right)=1$.

Theorem 4.1**.**

If $p^{2}\in Q_{p}^{*}(\mathfrak{I}_{D})$, where $\mathfrak{I}_{D}\in\mathfrak{G}_{D}$ is the identity class, then every proper class in $\mathfrak{G}_{D}$ is completely $p$-primitive. Conversely, if an ambiguous class in $\mathfrak{G}_{D}$ is completely $p$-primitive, then $p^{2}\in Q_{p}^{*}(\mathfrak{I}_{D})$.

Proof.

First, assume that $p^{2}\in Q_{p}^{*}(\mathfrak{I}_{D})$. Let $\mathfrak{A}$ be any proper class in $\mathfrak{G}_{D}$. Assume that $a\in Q_{p}^{*}(\mathfrak{A})$. Then by Proposition 2.4, we have

[TABLE]

Therefore $\mathfrak{A}$ is completely $p$-primitive by Lemma 2.1.

Conversely, assume that an ambiguous class $\mathfrak{B}\in\mathfrak{G}_{D}$ is completely $p$-primitive. Let $L={\mathbb{Z}}\mathbb{x}+{\mathbb{Z}}\mathbb{y}$ be a binary ${\mathbb{Z}}$-lattice such that $f_{L}=[a,b,c]\in\mathfrak{B}$. Since $L$ represents at least one prime primitively by [11], we may assume that $a$ is a prime. Since $p^{2}a\in Q(L)=Q_{p}^{*}(L)$, there are integers $u,v$ such that

[TABLE]

Since $a$ divides $(bu+cv)v$, $a$ divides $v$ or $bu+cv$. First, assume that $a$ divides $v$. If $v=at$ for some integer $t$, then $4p^{2}=(2u+bt)^{2}+(-D)t^{2}$. One may easily check that $t$ is not divisible by $p$ from the assumption that $(p,u,v)=1$. Therefore, we have $p^{2}\in Q_{p}^{*}(\mathfrak{I}_{D})$ by Theorem 3.1. Finally, assume that $a$ divides $bu+cv$. Define $\sigma:pL\to L$ such that

[TABLE]

Then, one may easily show that $\sigma$ is an improper $p$-primitive representation. Since $\mathfrak{B}$ is an ambiguous class and $f_{L}=[a,b,c]\in\mathfrak{B}$, there is an improper isometry $\tau\in O(L)$. Then $\tau\sigma:pL\to L$ is a proper $p$-primitive representation. The theorem follows from Theorem 3.1. ∎

From now on, we always assume that $D<0$. For a positive definite binary quadratic form $[a,b,c]\in\mathfrak{C}\in\mathfrak{G}_{D}$ and a positive integer $n$, we define

[TABLE]

and

[TABLE]

We also define $r_{p}^{\flat}(n,[a,b,c]):=r(n,[a,b,c])-r_{p}^{*}(n,[a,b,c])$. It is well known that

[TABLE]

where $\left(\frac{\cdot}{\cdot}\right)$ is the Kronecker’s symbol and

[TABLE]

Similarly, we define $R_{p}^{*}(n,\mathfrak{G}_{D}):=\sum_{\mathfrak{C}\in\mathfrak{G}_{d}}r_{p}^{*}(n,\mathfrak{C})$.

Theorem 4.2**.**

Assume that $p^{2}\not\in Q_{p}^{*}(\mathfrak{I}_{D})$. A proper class $\mathfrak{A}\in\mathfrak{G}_{D}$ is completely $p$-primitive if and only if $p^{2}\in Q_{p}^{*}(\mathfrak{A}^{2})$ and the order of $\mathfrak{A}$ in $\mathfrak{G}_{D}$ is $4$.

Proof.

Note that $p^{2}\in Q_{p}^{*}([1,1,1])$ for any prime $p\equiv 1\ (\text{mod }3)$ and $q^{2}\in Q_{q}^{*}([1,0,1])$ for any prime $q\equiv 1\ (\text{mod }4)$. Hence we may assume that $D<-4$ and the order of isometry group of $f$ is $2$ or $4$ for any binary quadratic form $f$ with discriminant $D$.

First, we prove “if” part. Assume that $\mathfrak{A}^{4}=\mathfrak{I}_{D}$ in $\mathfrak{G}_{D}$ and $p^{2}\in Q_{p}^{*}(\mathfrak{A}^{2})$. For any integer $a\in Q_{p}^{*}(\mathfrak{A})$, we have

[TABLE]

by Proposition 2.4. Therefore, $\mathfrak{A}$ is completely $p$-primitive by Lemma 2.1.

To prove “only if” part, assume that $\mathfrak{A}$ is completely $p$-primitive. Note that $\mathfrak{A}$ is not an ambiguous class by theorem 4.1. Let $q\in Q_{p}^{*}(\mathfrak{A})$ be an odd prime not dividing $p\cdot D$. Let $\mathfrak{D}\in\mathfrak{G}_{D}$ be a non-identity proper class such that $p^{2}\in Q_{p}^{*}(\mathfrak{D})$. First, we will show that $p^{2}\in Q_{p}^{*}(\mathfrak{A}^{2})$ by proving that $\mathfrak{D}=\mathfrak{A}^{2}$ or $\mathfrak{D}=\mathfrak{A}^{-2}$.

Assume that $\mathfrak{D}$ is not an ambiguous class. By Lemma 2.3, we have

[TABLE]

Since $\mathfrak{A}$ is not an ambiguous class, any of three classes among $\mathfrak{A}\cdot\mathfrak{D},\ \mathfrak{A}^{-1}\cdot\mathfrak{D}^{-1},\ \mathfrak{A}^{-1}\cdot\mathfrak{D}$ and $\mathfrak{A}\cdot\mathfrak{D}^{-1}$ cannot be the same simultaneously. Suppose that all four proper classes are different with each other in $\mathfrak{G}_{D}$. Note that $r_{p}^{\flat}(qp^{2},\mathfrak{A})=r_{p}^{\flat}(qp^{2},\mathfrak{A}^{-1})\geq 2$. Since $R(qp^{2},\mathfrak{G}_{D})=12$, we have

[TABLE]

Since $qp^{2}\in Q_{p}^{*}(\mathfrak{A})$ by Lemma 2.1, we have $\mathfrak{D}=\mathfrak{A}^{2}$ or $\mathfrak{A}^{-2}$. Assume that $\mathfrak{A}\cdot\mathfrak{D}=\mathfrak{A}^{-1}\cdot\mathfrak{D}^{-1}$ and $\mathfrak{A}\cdot\mathfrak{D}^{-1}\neq\mathfrak{A}^{-1}\cdot\mathfrak{D}$. Then, the binary ${\mathbb{Z}}$-lattice corresponding to a binary form in $\mathfrak{A}\cdot\mathfrak{D}$ has an improper isometry. Therefore we have $r_{p}^{*}(qp^{2},\mathfrak{A}\cdot\mathfrak{D})\geq 4$ by Lemma 3.2. Furthermore, one may easily show that $r_{p}^{*}(qp^{2},\mathfrak{A}\cdot\mathfrak{D})=4$ and

[TABLE]

Since $qp^{2}\in Q_{p}^{*}(\mathfrak{A})$, we have $\mathfrak{D}=\mathfrak{A}^{2}$. The proof of the case when $\mathfrak{A}\cdot\mathfrak{D}\neq\mathfrak{A}^{-1}\cdot\mathfrak{D}^{-1}$ and $\mathfrak{A}\cdot\mathfrak{D}^{-1}=\mathfrak{A}^{-1}\cdot\mathfrak{D}$ is quite similar to the above. Finally, if $\mathfrak{A}\cdot\mathfrak{D}=\mathfrak{A}^{-1}\cdot\mathfrak{D}^{-1}$ and $\mathfrak{A}\cdot\mathfrak{D}^{-1}=\mathfrak{A}^{-1}\cdot\mathfrak{D}$, then one may easily show that $\mathfrak{D}=\mathfrak{I}_{D}$, which is a contradiction.

Now, assume that $\mathfrak{D}$ is an ambiguous class. Choose integers $b,c$ suitably so that

[TABLE]

Since $r_{p}^{*}(p^{2},\mathfrak{D})\geq 4$ by Lemma 3.2, there is a vector $(x,y)\in{\mathbb{Z}}^{2}-\{\pm(1,0)\}$ such that

[TABLE]

Note that $[qp^{2},b,c]\in\mathfrak{A}\cdot\mathfrak{D}$. Since $qp^{2}x^{2}+bx(qy)+c(qy)^{2}=qp^{2}$, $(x,qy)\neq(1,0)$ and $(x,qy,p)=1$, we have $r_{p}^{*}(qp^{2},\mathfrak{A}\cdot\mathfrak{D})\geq 4$. Similarly, $r_{p}^{*}(qp^{2},\mathfrak{A}^{-1}\cdot\mathfrak{D})\geq 4$. Therefore, we have

[TABLE]

Since $qp^{2}\in Q_{p}^{*}(\mathfrak{A})$, we have $\mathfrak{D}=\mathfrak{A}^{2}$. Therefore, $p^{2}\in Q_{p}^{*}(\mathfrak{A}^{2})$ in any cases.

Suppose that $\mathfrak{A}^{4}\neq\mathfrak{I}_{D}$ and $\mathfrak{A}^{6}\neq\mathfrak{I}_{D}$. Note that $p^{2}\in Q_{p}^{*}(\mathfrak{A}^{2})$ and $p^{4}\in Q_{p}^{*}(\mathfrak{A}^{4})$ by Lemma 2.2. Hence we have

[TABLE]

Note that $R_{p}^{*}(qp^{4},\mathfrak{G}_{D})=20-R(qp^{2},\mathfrak{G}_{D})=20-12=8$. Assume that $\mathfrak{A}^{10}\neq\mathfrak{I}_{D}$ and $\mathfrak{A}^{8}\neq\mathfrak{I}_{D}$. Then by Lemma 3.2, we have

[TABLE]

This is a contradiction, for $qp^{4}\in Q_{p}^{*}(\mathfrak{A})$ by assumption. Assume that $\mathfrak{A}^{10}=\mathfrak{I}_{D}$. Then, we have

[TABLE]

Since $qp^{4}\in Q_{p}^{*}(\mathfrak{A})$, the order of $\mathfrak{A}$ is $4$, which is a contradiction. Now, assume that $\mathfrak{A}^{8}=\mathfrak{I}_{D}$. Since $\mathfrak{A}^{4}$ is an ambiguous class, $r_{p}^{*}(qp^{4},\mathfrak{A}^{5})=r_{p}^{*}(qp^{4},\mathfrak{A}^{-5})=4$ by a similar reasoning given above. Since $qp^{4}\in Q_{p}^{*}(\mathfrak{A})$, $\mathfrak{A}^{4}=\mathfrak{I}_{D}$ or $\mathfrak{A}^{6}=\mathfrak{I}_{D}$, which is a contradiction.

In the remaining, we prove that the order of $\mathfrak{A}$ is $4$ by showing $\mathfrak{A}^{6}\neq\mathfrak{I}_{D}$. Suppose, on the contrary, that $\mathfrak{A}^{6}=\mathfrak{I}_{D}$. Assume that $[p^{2},b,c]\in\mathfrak{A}^{2}$. Let $\mathfrak{F}$ be the proper class in $\mathfrak{G}_{D}$ such that $[p,b,pc]\in\mathfrak{F}$. Then clearly $\mathfrak{F}^{2}=\mathfrak{A}^{2}$. Suppose that $\mathfrak{F}=\mathfrak{A}$ or $\mathfrak{A}^{-1}$, that is, $p$ is represented by $\mathfrak{A}$. Since $\mathfrak{A}^{3}=\mathfrak{A}^{-3}$ by assmption, we have

[TABLE]

Since $R(p^{3},\mathfrak{G}_{D})=8$, we have $p^{3}\not\in Q_{p}^{*}(\mathfrak{A})$, which is a contradiction by Lemma 2.1. From now on, we assume that $\mathfrak{F}\neq\mathfrak{A}$ and $\mathfrak{F}\neq\mathfrak{A}^{-1}$. Choose an odd prime $r\in Q_{p}^{*}(\mathfrak{F}^{-1}\cdot\mathfrak{A})$ not dividing $p\cdot D$. Assume that $pr\in Q_{p}^{*}(\mathfrak{X})$ for some $\mathfrak{X}\in\mathfrak{G}_{D}$. Then by Lemma 2.2 of [3], we have

[TABLE]

Since $\mathfrak{F}$ and $\mathfrak{F}^{-1}$ are the only proper classes containing a binary quadratic form representing $p$, we have $\mathfrak{X}=\mathfrak{A}$ or $\mathfrak{A}^{-1}$. Furthermore, since $R(pr,\mathfrak{G}_{D})=8$, we have

[TABLE]

Now, let $f=[r,B,p^{3}C]\in\mathfrak{F}^{-1}\cdot\mathfrak{A}$ and $g=[p^{3},B,rC]\in\mathfrak{F}^{3}$ be binary quadratic forms. Note that $h=[p^{3}r,B,C]\in\mathfrak{F}^{-1}\cdot\mathfrak{A}\cdot\mathfrak{F}^{3}=\mathfrak{A}^{3}$. Since $\mathfrak{F}^{3}=\mathfrak{F}^{-3}$, there is a vector $(u,v)\in{\mathbb{Z}}^{2}-\{\pm(1,0)\}$ such that

[TABLE]

Note that $\text{ord}_{r}(g(u\pm 1,v))=\text{ord}_{r}(p^{3}(2\pm 2u)\pm Bv)\equiv 0\pmod{2}$ by Lemma 3.2. Since

[TABLE]

the binary ${\mathbb{Z}}$-lattice $L_{h}={\mathbb{Z}}\mathbb{x}+{\mathbb{Z}}\mathbb{y}$ corresponding to $h$ has at least $4$ primitive vectors $\pm\mathbb{x}$ and $\pm(u\mathbb{x}+rv\mathbb{y})$, all of whose norms are $p^{3}r$. Since the proper class $\mathfrak{A}^{3}$ containing $h$ is an ambiguous class, there is an improper isometry $\sigma\in O(L_{h})$. Since

[TABLE]

$\sigma(u\mathbb{x}+rv\mathbb{y})\neq\pm\mathbb{x}$ by Lemma 3.2. Therefore

[TABLE]

are all different primitive vectors in $L_{h}$ whose norms are $p^{3}r$. Furthermore, since

[TABLE]

we have $r_{p}^{*}(p^{3}r,\mathfrak{A}^{3})=8$. This implies that $p^{3}r$ is not primitively represented by $\mathfrak{A}$, though $pr$ is primitively represented by $\mathfrak{A}$. This is a contradiction. Therefore the order of $\mathfrak{A}$ in $\mathfrak{G}_{D}$ is $4$. This completes the proof. ∎

Remark 4.3*.*

(i) Note that $\mathfrak{G}_{-56}\simeq{\mathbb{Z}}/4{\mathbb{Z}}$ and the proper class $\mathfrak{A}\in\mathfrak{G}_{-56}$ containing $[3,2,5]$ is of order $4$. Furthermore, $36\in Q_{3}^{*}(\mathfrak{A}^{2})$, where $[2,0,7]\in\mathfrak{A}^{2}$. Therefore, for any positive integer $a$ such that the diophantine equation $3y^{2}+2yz+5z^{2}=a$ has an integer solution, it has an integer solution $(y,z)=(b,c)$ such that $(b,c,3)=1$ by Theorem 4.2.

(ii) For a positive integer $m\not\equiv 0\pmod{3}$, let $f_{m}(x,y,z)=m^{2}x^{2}+3y^{2}+2yz+5z^{2}$ be a ternary quadratic form. Note that $f_{m}(x,y,z)\equiv 1\pmod{3}$ if and only if

[TABLE]

Assume that for an integer $N\equiv 1\pmod{3}$, $f_{m}(x,y,z)=N$ has an integer solution. Then there is an integer solution $f_{m}(a,b,c)=N$ with $a-b+c\equiv 0\pmod{3}$, except the case when $N=f(x,0,0)=m^{2}x^{2}$ by Theorem 4.2. Therefore, if we define $\widetilde{f_{m}}(x,y,z)=m^{2}(3x+y-z)^{2}+3y^{2}+2yz+5z^{2}$, then

[TABLE]

where $S_{3,1}$ denotes the set of positive integers that are congruent to $1$ modulo $3$. In particular, assume that $m=1$. Then one may easily show that $\widetilde{f_{1}}=4x^{2}+6y^{2}+7z^{2}+6yz+2zx$. In fact, the class number of $\widetilde{f_{1}}$ is three and the other two ternary quadratic forms in the genus of $\widetilde{f_{1}}$ are

[TABLE]

One may easily check that $d\widetilde{f_{1}}=2\cdot 3^{2}\cdot 7$, and $2^{2},3^{2}$, $7^{2}$ are all represented by $\widetilde{f_{1}}$. For any prime $p$ not dividing $2\cdot 3\cdot 7$, the isometry class $[\widetilde{f_{1}}]$ is connected to $[g]$ or $[h]$ in the graph $\mathfrak{G}_{f,p}(0)$ defined in [5] (see also [1]). Furthermore, since $1\in Q(g)\cap Q(h)$, $p^{2}$ is represented by $\widetilde{f_{1}}$. Therefore, every square of an integer except $1$ is represented by $\widetilde{f_{1}}$. Consequently, we have

[TABLE]