NONCONGRUENT NUMBERS OF THE FORM

QIANQIAN CAI

doi:10.1017/S0004972726100999

NONCONGRUENT NUMBERS OF THE FORM $2pq$ WITH $pq\equiv 5\ (\textbf {mod } 8)$

Part of: Forms and linear algebraic groups Arithmetic algebraic geometry

Published online by Cambridge University Press: 23 February 2026

QIANQIAN CAI

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QIANQIAN CAI*: Affiliation:
School of Mathematics, Nanjing University , Nanjing, PR China
*: e-mail: qqcai@smail.nju.edu.cn

Article contents

Abstract
Introduction
Class numbers of imaginary quadratic fields
Proof of the main theorem
Examples
Footnotes
References

Rights & Permissions

Abstract

Let p and q be two primes with $(p,q)\equiv (1,5)$ or $(7,3) \pmod 8$. Lagrange [‘Nombres congruents et courbes elliptiques’, Séminaire Delange-Pisot-Poitou. Théorie des Nombres 16(1) (1974–1975), Article no. 16] and Qin [‘Congruent numbers, quadratic forms and $K_2$’, Math. Ann. 383(3–4) (2022), 1647–1686] showed that if $(\frac {q}{p})=-1$, then $2pq$ is not a congruent number. By using Qin’s method, we prove that if $(p,q)\equiv (1,5) \pmod 8$ and $(\frac {q}{p})=1$ with $h(-pq)\not \equiv p-1 \pmod {16}$, then $2pq$ is not a congruent number; if $(p,q)\equiv (7,3) \pmod 8$ and $(\frac {q}{p})=1$ with $h(-2pq)\not \equiv p+1 \pmod {16}$, then $2pq$ is not a congruent number. Here, $h(-d)$ denotes the class number of the imaginary quadratic field $\mathbb {Q}(\sqrt {-d})$.

Keywords

noncongruent numbers quadratic forms class numbers elliptic curves

MSC classification

Primary: 11E20: General ternary and quaternary quadratic forms; forms of more than two variables

Secondary: 11G05: Elliptic curves over global fields

Information

Type: Research Article
Information: Bulletin of the Australian Mathematical Society , Volume 114 , Issue 2 , October 2026 , pp. 256 - 265

DOI: https://doi.org/10.1017/S0004972726100999 [Opens in a new window]
Copyright: © The Author(s), 2026. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc.

1 Introduction

A positive integer n is called a congruent number if it is the area of a right triangle with rational sides. Otherwise, n is called a noncongruent number. We only need to consider the case for n being square-free. It is well known that n is a congruent number if and only if the elliptic curve $E_n: y^2 = x^3 - n^2x$ has positive Mordell–Weil rank (see [Reference Koblitz18, Ch. 1, Proposition 18]).

Let p and q be two different primes. Genocchi [Reference Genocchi11] showed that $p\equiv 3 \pmod 8$ is not congruent and that if $p\equiv 5 \pmod 8$ , then $2p$ is not congruent. Bastien [Reference Bastein1] proved that if $p=a^2+4b^2\equiv 1 \pmod 8$ with $(\frac{a+2b}{p})=-1$ , then p is not congruent and that if $p\equiv 9 \pmod {16}$ , then $2p$ is not congruent. Heegner [Reference Heegner14] and Birch [Reference Birch and Abhyankar3] showed that if $p\equiv 3 \pmod 8$ , then $2p$ is a congruent number. Stephens [Reference Stephens25] proved that ${p\equiv 5,7 \pmod 8}$ are congruent numbers. Tunnell [Reference Tunnell27] showed that if $p\equiv 3 \pmod 8$ , then p is not a congruent number; if $p=a^2+b^2\equiv 1 \pmod 8$ with $16\nmid p-1+4b$ , then p is not congruent; if $p\equiv q\equiv 5 \pmod 8$ , then $2pq$ is not congruent. Das and Saikia [Reference Das and Saikia6] showed that if $p\equiv 1 \pmod 8$ is a congruent number, then $4\mid h(2p)$ . Here and later, $h(d)$ denotes the class number of $\mathbb {Q}(\sqrt {d})$ .

Qin [Reference Qin24] established an explicit correspondence for the Shimura lift between an eigenform of weight $3/2$ , if this form arises from ternary quadratic forms, and a corresponding eigenform of weight 2. Using this, he gave a more powerful result for determining congruent numbers than Tunnell’s theorem [Reference Tunnell27].

Theorem 1.1 [Reference Qin24, Theorems 1.3 and 1.4]

Let n be a square-free positive integer. If n is an odd congruent number, then

$$ \begin{align*} \#\{(x,y,z)\in\mathbb{Z}^3 \mid n=x^2+2y^2+32z^2\} =\#\{(x,y,z)\in\mathbb{Z}^3 \mid n=2x^2+4y^2+9z^2-4yz\} \end{align*} $$

and if n is an even congruent number, then

(1.1) $$ \begin{align} \#&\{(x,y,z)\in\mathbb{Z}^3\mid n/2=x^2+4y^2+32z^2\} \nonumber\\ &\,=\#\{(x,y,z)\in\mathbb{Z}^3 \mid n/2=4x^2+4y^2+9z^2-4yz\}. \end{align} $$

Conversely, if we assume that the weak Birch–Swinnerton-Dyer conjecture is true for the elliptic curves $E_n:y^2=x^3-n^2x$ , then these equalities imply that n is a congruent number.

Using this theorem and the Minkowski–Siegel formula, Qin [Reference Qin24] showed that for a prime p, if $p\equiv 1 \pmod 8$ is a congruent number, then the 8-rank of $K_2O_{\mathbb {Q}(\sqrt {p})}$ equals one; if $h(-p)\not \equiv h(-2p) \pmod {16}$ , then $2p$ is not congruent.

For $n=2pq$ with $p,q$ being different primes, considering the congruence $(p,q)\equiv (i,j) \pmod 8$ , there are the following cases:

(1,1) noncongruent if $h(-2pq)\not \equiv h(-pq) \pmod {16}$ (proved by Qin [Reference Qin24]);
(1,3) congruent if $(\frac {q}{p})=-1$ (proved by Monsky [Reference Heath-Brown13]);
(1,5) noncongruent if $(\frac {q}{p})=-1$ (proved by Lagrange [Reference Lagrange19] and Qin [Reference Qin24]);
(1,7) congruent if $(\frac {q}{p})=-1$ (proved by Monsky [Reference Heath-Brown13]);
(3,3) noncongruent (proved by Qin [Reference Qin24]);
(3,5) congruent (proved by Monsky [Reference Heath-Brown13]);
(3,7) noncongruent if $(\frac {q}{p})=1$ (proved by Lagrange [Reference Lagrange19] and Qin [Reference Qin24]);
(5,5) noncongruent (proved by Tunnell [Reference Tunnell27] and Qin [Reference Qin24]);
(5,7) congruent (proved by Monsky [Reference Heath-Brown13]);
(7,7) noncongruent if $h(-2pq)\not \equiv h(-pq) \pmod {16}$ (proved by Qin [Reference Qin24]).

In this paper, by using Qin’s method [Reference Qin24], we consider the case for $(p,q)\equiv (1,5)$ or $(7,3)\ (\bmod \:8)$ with $(\frac {q}{p})=1$ . In [Reference Cai4, Lemma 3.1], by using the Minkowski–Siegel formula [Reference Kitaoka17, Section 6.8], we gave a relation between the number of representations of ternary quadratic forms and the class number of the relative imaginary quadratic field. Using it and [Reference Yang29], we see that

(1.2) $$ \begin{align} \#&\{(x,y,z)\in\mathbb{Z}^3 \mid pq=x^2+4y^2+32z^2\} \nonumber\\ & \, +\#\{(x,y,z)\in\mathbb{Z}^3 \mid pq=4x^2+4y^2+9z^2-4yz\}=2h(-2pq). \end{align} $$

Thus, if $2pq$ is a congruent number, then by (1.1),

(1.3) $$ \begin{align} \#&\{(x,y,z)\in\mathbb{Z}^3 \mid pq=x^2+4y^2+32z^2\} \nonumber\\ &\,=\#\{(x,y,z)\in\mathbb{Z}^3 \mid pq=4x^2+4y^2+9z^2-4yz\}=h(-2pq). \end{align} $$

Using the equality (1.3), we shall prove the following result.

Theorem 1.2. Let p and q be two primes. Then, $2pq$ is not a congruent number in the following cases:

(1) $p\equiv 1 \pmod 8$ , $q\equiv 5 \pmod 8$ , $(\frac {q}{p})=1$ and $h(-pq)\not \equiv p-1 \pmod {16}$ ;
(2) $p\equiv 7 \pmod 8$ , $q\equiv 3 \pmod 8$ , $(\frac {q}{p})=1$ and
(1.4) $$ \begin{align} h(-2pq)\not\equiv p+1 \pmod {16}. \end{align} $$

Remark 1.3. By [Reference Pizer23], if $(p,q)\equiv (7,3) \pmod 8$ with $(\frac {q}{p})=1$ , then $h(-pq)+h(-2pq)\equiv 0 \pmod {16}$ . Since $8\mid h(-pq)$ and $8\mid h(-2pq)$ (using (2.1) and (2.2)), it follows that (1.4) is equivalent to $h(-pq)\not \equiv p+1 \pmod {16}$ .

The 2-primary parts of the Shafarevich–Tate groups of the noncongruent numbers in the theorem are isomorphic to $(\mathbb {Z}/2\mathbb {Z})^2$ (see Remark 3.3).

For other results on noncongruent numbers, see [Reference Feng7–Reference Feng, Xue, Kanemitsu and Liu10, Reference Li and Qin20–Reference Ouyang and Zhang22, Reference Tian, Yuan and Zhang26, Reference Wang28].

The paper is organised as follows. In Section 2, we recall some results on the 2-primary part of class numbers of imaginary quadratic fields and give a lemma used in the proof of the main result. In Section 3, we give the proof of Theorem 1.2 as well as some criteria on the 2-primary part of the Shafarevich–Tate groups and the Mordell–Weil rank of the corresponding elliptic curves $E_{2pq}$ . In Section 4, we give some examples of noncongruent numbers of the form $2pq$ with $(p,q)\equiv \pm(1,5) \pmod 8$ and $(\frac {q}{p})=1$ .

2 Class numbers of imaginary quadratic fields

In this section, we recall some results on the 2-primary part of the class numbers of imaginary quadratic fields and show a lemma used in the proof of the main result.

We first introduce some notation. Define $[{q}/{p}]\in \mathbb {F}_2$ via $(\frac {q}{p})=(-1)^{[{q}/{p}]}$ . Let ${n=2p_1\cdots p_t}$ , where the $p_i$ are distinct odd primes. Define the $t\times t$ matrices $A=(a_{ij})_{t\times t}$ by $a_{i j}:= [{p_j}/{p_i}]$ if $i\neq j$ and $a_{i i}=\sum _{1 \leqslant j \leqslant t,\, j \neq i} a_{i j}$ , and $D_2=\operatorname {diag}\{[{2}/{p_i}]\}$ . Define the $1\times t$ matrix $\mathbf {b}=([{2}/{p_1}],\ldots ,[{2}/{p_t}])^T$ .

Let $r_{2^i} (-n)$ be the $2^i$ -rank of the class group of $F=\mathbb {Q}(\sqrt {-n})$ and $\omega (\operatorname {disc}(F))$ be the number of different prime factors of the discriminant $\operatorname {disc}(F)$ . By Gauss’s genus theory and Rédei’s criterion,

(2.1) $$ \begin{align} r_2(-n)=\omega(\operatorname{disc}(F))-1 \end{align} $$

and

(2.2) $$ \begin{align} r_4(-n)= \begin{cases} \omega(\operatorname{disc}(F))-1-\operatorname{rank}(A, \mathbf{b}) & \text { if } 4 \parallel \operatorname{disc}(F), \\ \omega(\operatorname{disc}(F))-1-\operatorname{rank}(A+D_2, \mathbf{b}) & \text { if } 8 \parallel \operatorname{disc}(F) .\end{cases} \end{align} $$

StartLayout 1st Row r 4 left parenthesis negative n right parenthesis equals StartLayout Enlarged left brace 1st Row 1st Column omega left parenthesis disc left parenthesis upper F right parenthesis right parenthesis minus 1 minus rank left parenthesis upper A comma bold b right parenthesis 2nd Column if 4 parallel to disc left parenthesis upper F right parenthesis comma 2nd Row 1st Column omega left parenthesis disc left parenthesis upper F right parenthesis right parenthesis minus 1 minus rank left parenthesis upper A plus upper D 2 comma bold b right parenthesis 2nd Column if 8 parallel to disc left parenthesis upper F right parenthesis period EndLayout EndLayout

For the computation of $r_2(-n)$ and $r_4(-n)$ , the reader can consult [Reference Wang28].

Let $D(-n)$ be the set of all positive square-free divisors of the discriminant $\operatorname {disc}(F)$ and $D^*(-n) = D(-n)\cap N_{F/\mathbb {Q}}(F^*)$ . Let $({\cdot }/{\cdot })_4$ denote the quartic residue symbol. The following two lemmas concern the 8-rank of ideal class groups.

Lemma 2.1 [Reference Jung and Yue16, Theorem 3.3]

Let $F=\mathbb {Q}(\sqrt {-d})$ , where $d=p_1 \cdots p_t$ with distinct primes $p_i \equiv 1 \:(\bmod \:4)$ , be an imaginary quadratic field and $r_4(-d)=1$ . Suppose $D^*(-d)=\{1, q_1, q_2, d\}$ , where $q_1=p_1 \cdots p_r \equiv 1 \pmod 8$ and $q_2=p_{r+1} \cdots p_t \equiv 5 \pmod 8$ . Then, $r_8(-d)=1$ if and only if $({q_2}/{q_1})_4=1$ .

Lemma 2.2 [Reference Jung and Yue16, Theorem 3.8]

Let $F=\mathbb {Q}(\sqrt {-2 d})$ be an imaginary quadratic field with $d= p_1 \cdots p_t$ and distinct primes $p_i \equiv 1 \:(\bmod \:4)$ . Suppose $r_4(-2d)=1$ , $D^*(-2d)=\{1, q_1, 2 q_2, 2 d\}$ , $q_1=p_1 \cdots p_r \equiv 1 \pmod 8, q_2=p_{r+1} \cdots p_t$ and $1 \leqslant r<t$ . Then, $r_8(-2d)=1$ if and only if $({2 q_2}/{q_1})_4=1$ .

We also need the following lemma.

Lemma 2.3 (Legendre’s theorem; [Reference Ireland and Rosen15, Proposition 17.3.1])

Let $a, b, c$ be nonzero square-free integers, pairwise relatively prime and neither all positive nor all negative. Then, $a x^2+b y^2+c z^2=0$ has a nontrivial integral solution if and only if the following conditions are satisfied:

(1) $({-a b}/{p})=1$ for every odd prime $p \mid c$ ;
(2) $({-a c}/{p})=1$ for every odd prime $p \mid b$ ;
(3) $({-b c}/{p})=1$ for every odd prime $p \mid a$ .

The following lemma is needed to prove the main theorem.

Lemma 2.4. Let $p\equiv 1 \pmod 8$ and $q\equiv 5 \pmod 8$ be two primes with $(\frac {q}{p})=1$ . Then, $h(-pq)\equiv h(-2pq) \pmod {16}$ if and only if $({2}/{p})_4=1$ .

Proof. Since $(\frac {q}{p})=1$ , we have $r_2(-pq)=r_2(-2pq)=2$ and $r_4(-pq)= r_4(-2pq)=1$ by (2.1) and (2.2). Thus, $8\mid h(-pq)$ and $8\mid h(-2pq)$ . By Lemma 2.3, it is easy to see that $D^*(-pq)=\{1,p,q,pq\}$ and $D^*(-2pq)=\{1,p,2q,2pq\}$ . Thus, by Lemma 2.1,

(2.3) $$ \begin{align} 16\mid h(-pq)\quad\mbox{if and only if } r_8(-pq)=1\quad\mbox{if and only if } \bigg(\frac{q}{p}\bigg)_4=1, \end{align} $$

and by Lemma 2.2,

(2.4) $$ \begin{align} 16\mid h(-2pq)\quad\mbox{if and only if } r_8(-2pq)=1\quad\mbox{if and only if } \bigg(\frac{2q}{p}\bigg)_4=1. \end{align} $$ 16 vertical bar h left parenthesis minus 2 p q right parenthesis if and only if r 8 left parenthesis minus 2 p q right parenthesis equals 1 if and only if left parenthesis StartFraction 2 q Over p EndFraction right parenthesis Subscript 4 Baseline equals 1 period

Combining (2.3) and (2.4), we see that $h(-pq)\equiv h(-2pq) \pmod {16}$ if and only if

$$ \begin{align*} \bigg(\frac{q}{p}\bigg)_4=\bigg(\frac{2q}{p}\bigg)_4 \quad\mbox{if and only if}\quad \bigg(\frac{2}{p}\bigg)_4=1.\\[-45pt] \end{align*} $$ left parenthesis StartFraction q Over p EndFraction right parenthesis Subscript 4 Baseline equals left parenthesis StartFraction 2 q Over p EndFraction right parenthesis Subscript 4 Baseline if and only if left parenthesis StartFraction 2 Over p EndFraction right parenthesis Subscript 4 Baseline equals 1 period

3 Proof of the main theorem

In this section, we prove Theorem 1.2 via (1.3). To estimate the number of representations of $pq$ by the ternary quadratic form $x^2+4y^2+32z^2$ , we consider the quaternary quadratic form $x^2+y^2+16z^2+32w^2$ . It leads to the following lemma.

Lemma 3.1. Let p and q be two primes. Then,

$$ \begin{align*} \begin{aligned} \#&\{(x,y,z,w)\in\mathbb{Z}^4 \mid pq=x^2+y^2+16z^2+32w^2\}\\ & \equiv\begin{cases} 2(p+7+h(-2pq)-h(-pq)) \pmod{32}&\text{if }(p,q)\equiv(1,5) \pmod 8,\\ &\text{with }\bigg(\dfrac{q}{p}\bigg)=1, \\ 2(p+1) \pmod{64}&\text{if }(p,q)\equiv(7,3) \pmod 8. \end{cases} \end{aligned} \end{align*} $$

StartLayout 1st Row StartLayout 1st Row 1st Column number sign 2nd Column StartSet left parenthesis x comma y comma z comma w right parenthesis element of double struck upper Z Superscript 4 Baseline vertical bar p q equals x squared plus y squared plus 16 z squared plus 32 w squared EndSet 2nd Row 1st Column Blank 2nd Column identical to StartLayout Enlarged left brace 1st Row 1st Column 2 left parenthesis p plus 7 plus h left parenthesis minus 2 p q right parenthesis minus h left parenthesis minus p q right parenthesis right parenthesis left parenthesis mod 32 right parenthesis 2nd Column if left parenthesis p comma q right parenthesis identical to left parenthesis 1 comma 5 right parenthesis left parenthesis mod 8 right parenthesis comma 2nd Row 1st Column Blank 2nd Column with left parenthesis StartFraction q Over p EndFraction right parenthesis equals 1 comma 3rd Row 1st Column 2 left parenthesis p plus 1 right parenthesis left parenthesis mod 64 right parenthesis 2nd Column if left parenthesis p comma q right parenthesis identical to left parenthesis 7 comma 3 right parenthesis left parenthesis mod 8 right parenthesis period EndLayout EndLayout EndLayout

Proof. For a positive integer $n\equiv 5 \pmod 8$ , from [Reference Gogišvili12],

(3.1) $$ \begin{align} \begin{aligned} \#&\{(x,y,z,w)\in\mathbb{Z}^4 \mid n=x^2+y^2+16z^2+32w^2\}\\ &\, =\sum_{d_1d_2=n}\bigg(\frac{2}{d_1}\bigg)d_2+4\sum_{\substack{n=x^2+4y^2\\ x,y>0}}\bigg(\frac{2}{x}\bigg)\bigg(\frac{-1}{y}\bigg)y. \end{aligned} \end{align} $$

StartLayout 1st Row 1st Column number sign 2nd Column StartSet left parenthesis x comma y comma z comma w right parenthesis element of double struck upper Z Superscript 4 Baseline vertical bar n equals x squared plus y squared plus 16 z squared plus 32 w squared EndSet 2nd Row 1st Column Blank 2nd Column equals sigma summation Underscript d 1 d 2 equals n Endscripts left parenthesis StartFraction 2 Over d 1 EndFraction right parenthesis d 2 plus 4 sigma summation Underscript StartLayout 1st Row n equals x squared plus 4 y squared 2nd Row x comma y greater than 0 EndLayout Endscripts left parenthesis StartFraction 2 Over x EndFraction right parenthesis left parenthesis StartFraction negative 1 Over y EndFraction right parenthesis y period EndLayout

We first show that

(3.2) $$ \begin{align} \sum_{d_1d_2=pq}\bigg(\frac{2}{d_1}\bigg)d_2 \equiv\begin{cases} 2(q-1) \pmod{32}&\text{if }(p,q)\equiv(1,5) \pmod 8, \\ 2(p+1) \pmod{64}&\text{if }(p,q)\equiv(7,3) \pmod 8. \end{cases} \end{align} $$

sigma summation Underscript d 1 d 2 equals p q Endscripts left parenthesis StartFraction 2 Over d 1 EndFraction right parenthesis d 2 identical to StartLayout Enlarged left brace 1st Row 1st Column 2 left parenthesis q minus 1 right parenthesis left parenthesis mod 32 right parenthesis 2nd Column if left parenthesis p comma q right parenthesis identical to left parenthesis 1 comma 5 right parenthesis left parenthesis mod 8 right parenthesis comma 2nd Row 1st Column 2 left parenthesis p plus 1 right parenthesis left parenthesis mod 64 right parenthesis 2nd Column if left parenthesis p comma q right parenthesis identical to left parenthesis 7 comma 3 right parenthesis left parenthesis mod 8 right parenthesis period EndLayout

In fact, the sum equals $q-p+pq-1$ . If $(p,q)\equiv (1,5) \pmod 8$ , then $8\mid p-1$ and $4\mid q-1$ ; thus, $32\mid (p-1)(q-1)$ . This implies that $p(q-1)\equiv q-1 \pmod {32}$ , which is equivalent to $q-p+pq-1\equiv 2(q-1) \pmod {32}$ . If $(p,q)\equiv (7,3) \pmod 8$ , then $8\mid q-3$ and $8\mid p+1$ ; thus, $64\mid (q-3)(p+1)$ . This implies that $q(p+1)\equiv 3(p+1) \pmod {64}$ , which is equivalent to $q-p+pq-1\equiv 2(p+1) \pmod {64}$ .

As for the second sum in (3.1), it can be nonzero only when $(p,q)\equiv (1,5) \pmod 8$ . Suppose $pq=x^2+4y^2$ . Note that $x\equiv \pm 1 \pmod 8$ if $pq\equiv 5 \pmod {16}$ and $x\equiv \pm 3 \pmod 8$ if $pq\equiv 13 \pmod {16}$ . Thus,

$$ \begin{align*} \sum_{\substack{pq=x^2+4y^2\\ x,y>0}}\bigg(\frac{2}{x}\bigg)\bigg(\frac{-1}{y}\bigg)y=(-1)^{{(pq-5)}/{8}}\sum_{\substack{pq=x^2+4y^2\\ x>0,\,y\equiv1\pmod4}}y. \end{align*} $$

sigma summation Underscript StartLayout 1st Row p q equals x squared plus 4 y squared 2nd Row x comma y greater than 0 EndLayout Endscripts left parenthesis StartFraction 2 Over x EndFraction right parenthesis left parenthesis StartFraction negative 1 Over y EndFraction right parenthesis y equals left parenthesis negative 1 right parenthesis Superscript left parenthesis p q minus 5 right parenthesis divided by 8 Baseline sigma summation Underscript StartLayout 1st Row p q equals x squared plus 4 y squared 2nd Row x greater than 0 comma y identical to 1 left parenthesis mod 4 right parenthesis EndLayout Endscripts y period

Write $p=a^2+4b^2$ and $q=c^2+4d^2$ for some $a,b,c,d\in \mathbb {Z}$ . Then, $pq=(ac\mp 4bd)^2+4(ad\pm bc)^2$ . Since $(p,q)\equiv (1,5) \pmod 8$ , we have $2\nmid acd$ and $2\mid b$ . Without loss of generality, we can assume that $a\equiv c\equiv d\equiv 1\:(\bmod \:4)$ .

Case (i): $b\equiv 2\:(\bmod \:4)$ . Then, $ad\pm bc\equiv -1\:(\bmod \:4)$ . Thus,

$$ \begin{align*}\sum_{\substack{pq=x^2+4y^2\\ x>0,\,y\equiv1 \pmod4}}y=-(ad+bc)-(ad-bc)=-2ad\equiv-2 \pmod 8.\end{align*} $$

sigma summation Underscript StartLayout 1st Row p q equals x squared plus 4 y squared 2nd Row x greater than 0 comma y identical to 1 left parenthesis mod 4 right parenthesis EndLayout Endscripts y equals minus left parenthesis a d plus b c right parenthesis minus left parenthesis a d minus b c right parenthesis equals minus 2 a d identical to negative 2 left parenthesis mod 8 right parenthesis period

Case (ii): $b\equiv 0\:(\bmod \:4)$ . Then, $ad\pm bc\equiv 1\:(\bmod \:4)$ . Thus,

$$ \begin{align*}\sum_{\substack{pq=x^2+4y^2\\ x>0,\,y\equiv1 \pmod4}}y=(ad+bc)+(ad-bc)=2ad\equiv2 \pmod 8.\end{align*} $$ sigma summation Underscript StartLayout 1st Row p q equals x squared plus 4 y squared 2nd Row x greater than 0 comma y identical to 1 left parenthesis mod 4 right parenthesis EndLayout Endscripts y equals left parenthesis a d plus b c right parenthesis plus left parenthesis a d minus b c right parenthesis equals 2 a d identical to 2 left parenthesis mod 8 right parenthesis period

Using the fact that $4\mid b$ if and only if $({2}/{p})_4=1$ (see [Reference Cox5, Theorem 4.23]) and Lemma 2.4, we see that $4\mid b$ if and only if $h(-pq)\equiv h(-2pq) \pmod {16}$ . Hence,

$$ \begin{align*}\begin{aligned} 4&\sum_{\substack{pq=x^2+4y^2\\ x,\,y>0}}\bigg(\frac{2}{x}\bigg)\bigg(\frac{-1}{y}\bigg)y\\ & \, \equiv\begin{cases} -8(-1)^{{(pq-5)}/{8}} \pmod{32}&\text{if }h(-pq)\not\equiv h(-2pq) \pmod {16}, \\ 8(-1)^{{(pq-5})/{8}} \pmod{32}&\text{if }h(-pq)\equiv h(-2pq) \pmod {16}. \end{cases} \end{aligned}\end{align*} $$

StartLayout 1st Row StartLayout 1st Row 1st Column 4 2nd Column sigma summation Underscript StartLayout 1st Row p q equals x squared plus 4 y squared 2nd Row x comma y greater than 0 EndLayout Endscripts left parenthesis StartFraction 2 Over x EndFraction right parenthesis left parenthesis StartFraction negative 1 Over y EndFraction right parenthesis y 2nd Row 1st Column Blank 2nd Column identical to StartLayout Enlarged left brace 1st Row 1st Column minus 8 left parenthesis negative 1 right parenthesis Superscript left parenthesis p q minus 5 right parenthesis divided by 8 Baseline left parenthesis mod 32 right parenthesis 2nd Column if h left parenthesis minus p q right parenthesis not identical to h left parenthesis minus 2 p q right parenthesis left parenthesis mod 16 right parenthesis comma 2nd Row 1st Column 8 left parenthesis negative 1 right parenthesis Superscript left parenthesis p q minus 5 right parenthesis slash 8 Baseline left parenthesis mod 32 right parenthesis 2nd Column if h left parenthesis minus p q right parenthesis identical to h left parenthesis minus 2 p q right parenthesis left parenthesis mod 16 right parenthesis period EndLayout EndLayout EndLayout

Since $8\mid h(-pq)$ and $8\mid h(-2pq)$ ,

(3.3) $$ \begin{align} \nonumber4\sum_{\substack{pq=x^2+4y^2\\ x,y>0}}\bigg(\frac{2}{x}\bigg)\bigg(\frac{-1}{y}\bigg)y & \equiv(-1)^{{(pq-5)}/{8}}[2(h(-2pq)-h(-pq))+8]\\ & \equiv2(h(-2pq)-h(-pq))+(-1)^{{(pq-5)}/{8}}8 \pmod{32}. \end{align} $$

StartLayout 1st Row 1st Column 4 sigma summation Underscript StartLayout 1st Row p q equals x squared plus 4 y squared 2nd Row x comma y greater than 0 EndLayout Endscripts left parenthesis StartFraction 2 Over x EndFraction right parenthesis left parenthesis StartFraction negative 1 Over y EndFraction right parenthesis y 2nd Column identical to left parenthesis negative 1 right parenthesis Superscript left parenthesis p q minus 5 right parenthesis divided by 8 Baseline left bracket 2 left parenthesis h left parenthesis minus 2 p q right parenthesis minus h left parenthesis minus p q right parenthesis right parenthesis plus 8 right bracket 2nd Row 1st Column Blank 2nd Column identical to 2 left parenthesis h left parenthesis minus 2 p q right parenthesis minus h left parenthesis minus p q right parenthesis right parenthesis plus left parenthesis negative 1 right parenthesis Superscript left parenthesis p q minus 5 right parenthesis divided by 8 Baseline 8 left parenthesis mod 32 right parenthesis period EndLayout

Combining (3.1), (3.2) and (3.3), if $(p,q)\equiv (7,3) \pmod 8$ ,

$$ \begin{align*}\#\{(x,y,z,w)\in\mathbb{Z}^4 \mid pq=x^2+y^2+16z^2+32w^2\} \equiv2(p+1) \pmod{64}.\end{align*} $$

If $(p,q)\equiv (1,5) \pmod 8$ with $(\frac {q}{p})=1$ , then

$$ \begin{align*}\begin{aligned} \#&\{(x,y,z,w)\in\mathbb{Z}^4 \mid pq=x^2+y^2+16z^2+32w^2\}\\ & \equiv2(q-1+h(-2pq)-h(-pq)+(-1)^{{(pq-5)}/{8}}4) \pmod{32}. \end{aligned}\end{align*} $$

Since $q-1+(-1)^{{(pq-5)}/{8}}4\equiv 8 \pmod {16}$ if $(p,q)\equiv (1,5),(1,13) \pmod {16}$ and $q-1+(-1)^{{(pq-5)}/{8}}4\equiv 0 \pmod {16}$ if $(p,q)\equiv (9,5),(9,13) \pmod {16}$ , we find $q-1+(-1)^{{(pq-5)}/{8}}4\equiv p+7 \pmod {16}$ , which completes the proof.

Now, we can prove Theorem 1.2.

Proof of Theorem 1.2

Since $pq\equiv 5 \pmod 8$ ,

(3.4) $$ \begin{align} \nonumber\#&\{(x,y,z,w)\in\mathbb{Z}^4 \mid pq=x^2+y^2+16z^2+32w^2\}\\ \nonumber & =\#\{(x,y,z,w)\in\mathbb{Z}^4 \mid pq=x^2+y^2+16z^2+32w^2,xyzw\neq0\}\\ \nonumber &\quad +\#\{(x,y,w)\in\mathbb{Z}^3 \mid pq=x^2+y^2+32w^2\}\\ \nonumber &\quad +\#\{(x,y,z)\in\mathbb{Z}^3 \mid pq=x^2+y^2+16z^2\}\\ \nonumber &\quad -\#\{(x,y)\in\mathbb{Z}^2 \mid pq=x^2+y^2\}\\ & \equiv 2\#\{(x,y,w)\in\mathbb{Z}^3 \mid pq=x^2+4y^2+32w^2\}\\ \nonumber &\quad +\#\{(x,y,z)\in\mathbb{Z}^3|\ \mid pq=x^2+y^2+16z^2\}\\ \nonumber &\quad -\#\{(x,y)\in\mathbb{Z}^2|\ \mid pq=x^2+y^2\} \pmod{32}. \end{align} $$

It is easy to see that

$$ \begin{align*}\begin{aligned} \#\{(x,y,z)\in\mathbb{Z}^3 \mid pq=x^2+y^2+16z^2\} & =2\#\{(x,y,z)\in\mathbb{Z}^3 \mid pq=x^2+4y^2+16z^2\}\\ & =\#\{(x,y,z)\in\mathbb{Z}^3 \mid pq=x^2+y^2+4z^2,2\nmid z\}. \end{aligned}\end{align*} $$

Thus,

$$ \begin{align*} \#&\{(x,y,z)\in\mathbb{Z}^3 \mid pq=x^2+y^2+16z^2\} \\ & =\tfrac{1}{2} (\#\{(x,y,z)\in\mathbb{Z}^3 \mid pq=x^2+y^2+16z^2\} \\ &\quad +\#\{(x,y,z)\in\mathbb{Z}^3 \mid pq=x^2+y^2+4z^2,2\nmid z\})\\ & =\tfrac{1}{2}\#\{(x,y,z)\in\mathbb{Z}^3 \mid pq=x^2+y^2+4z^2\} =4h(-pq) \end{align*} $$

StartLayout 1st Row 1st Column number sign 2nd Column StartSet left parenthesis x comma y comma z right parenthesis element of double struck upper Z cubed vertical bar p q equals x squared plus y squared plus 16 z squared EndSet 2nd Row 1st Column Blank 2nd Column equals one half left parenthesis number sign StartSet left parenthesis x comma y comma z right parenthesis element of double struck upper Z cubed vertical bar p q equals x squared plus y squared plus 16 z squared EndSet 3rd Row 1st Column Blank 2nd Column plus number sign left brace left parenthesis x comma y comma z right parenthesis element of double struck upper Z cubed vertical bar p q equals x squared plus y squared plus 4 z squared comma 2 does not divide z right brace right parenthesis 4th Row 1st Column Blank 2nd Column equals one half number sign StartSet left parenthesis x comma y comma z right parenthesis element of double struck upper Z cubed vertical bar p q equals x squared plus y squared plus 4 z squared EndSet equals 4 h left parenthesis minus p q right parenthesis EndLayout

by a well-known result of Gauss. Since $8\mid h(-pq)$ ,

(3.5) $$ \begin{align} \#\{(x,y,z)\in\mathbb{Z}^3 \mid pq=x^2+y^2+16z^2\}\equiv0 \pmod{32}. \end{align} $$

By [Reference Berndt2, Theorem 3.2.1],

(3.6) $$ \begin{align} \#\{(x,y)\in\mathbb{Z}^2 \mid pq=x^2+y^2\} =\begin{cases} 16&\text{if }(p,q)\equiv(1,5) \pmod 8, \\ 0&\text{if }(p,q)\equiv(7,3) \pmod 8. \end{cases} \end{align} $$

number sign StartSet left parenthesis x comma y right parenthesis element of double struck upper Z squared vertical bar p q equals x squared plus y squared EndSet equals StartLayout Enlarged left brace 1st Row 1st Column 16 2nd Column if left parenthesis p comma q right parenthesis identical to left parenthesis 1 comma 5 right parenthesis left parenthesis mod 8 right parenthesis comma 2nd Row 1st Column 0 2nd Column if left parenthesis p comma q right parenthesis identical to left parenthesis 7 comma 3 right parenthesis left parenthesis mod 8 right parenthesis period EndLayout

Combining (3.4), (3.5) and (3.6), and using Lemma 3.1,

$$ \begin{align*} \#&\{(x,y,w)\in\mathbb{Z}^3 \mid pq=x^2+4y^2+32w^2\}\\ & \equiv\tfrac{1}{2}\#\{(x,y,z,w)\in\mathbb{Z}^4 \mid\! pq=x^2+y^2+16z^2+32w^2\} +\tfrac{1}{2}\#\{(x,y)\in\mathbb{Z}^2 \mid\! pq=x^2+y^2\}\\ & \equiv\begin{cases} p-1+h(-2pq)-h(-pq) \pmod {16}&\text{if }(p,q)\equiv(1,5) \pmod 8,\\ p+1 \pmod {16}&\text{if }(p,q)\equiv(7,3) \pmod 8. \end{cases} \end{align*} $$

StartLayout 1st Row 1st Column number sign 2nd Column StartSet left parenthesis x comma y comma w right parenthesis element of double struck upper Z cubed vertical bar p q equals x squared plus 4 y squared plus 32 w squared EndSet 2nd Row 1st Column Blank 2nd Column identical to one half number sign StartSet left parenthesis x comma y comma z comma w right parenthesis element of double struck upper Z Superscript 4 Baseline vertical bar p q equals x squared plus y squared plus 16 z squared plus 32 w squared EndSet plus one half number sign StartSet left parenthesis x comma y right parenthesis element of double struck upper Z squared vertical bar p q equals x squared plus y squared EndSet 3rd Row 1st Column Blank 2nd Column identical to StartLayout Enlarged left brace 1st Row 1st Column p minus 1 plus h left parenthesis minus 2 p q right parenthesis minus h left parenthesis minus p q right parenthesis left parenthesis mod 16 right parenthesis 2nd Column if left parenthesis p comma q right parenthesis identical to left parenthesis 1 comma 5 right parenthesis left parenthesis mod 8 right parenthesis comma 2nd Row 1st Column p plus 1 left parenthesis mod 16 right parenthesis 2nd Column if left parenthesis p comma q right parenthesis identical to left parenthesis 7 comma 3 right parenthesis left parenthesis mod 8 right parenthesis period EndLayout EndLayout

Now, suppose $2pq$ is a congruent number. By (1.3),

$$ \begin{align*}h(-2pq)=\#\{(x,y,w)\in\mathbb{Z}^3 \mid pq=x^2+4y^2+32w^2\}.\end{align*} $$

Thus, if $(p,q)\equiv (7,3) \pmod 8$ , then $h(-2pq)\equiv p+1 \pmod {16}$ ; if $(p,q)\equiv (1,5) \pmod 8$ , then $h(-2pq)\equiv p-1+h(-2pq)-h(-pq) \pmod {16}$ , that is, $h(-pq)\equiv p-1 \pmod {16}$ . This completes the proof.

Remark 3.2. The method of using (3.1) to estimate

$$ \begin{align*}\#\{(x,y,w)\in\mathbb{Z}^3 \mid pq=x^2+4y^2+32w^2\}\end{align*} $$

was given in [Reference Feng7].

Remark 3.3. Let n be a square-free even positive integer. Assume that the Birch–Swinnerton-Dyer conjecture is true for the elliptic curves $E_n$ . By [Reference Qin24, Reference Tunnell27], if n is not congruent, then the order of the Shafarevich–Tate group is given by

where

$$ \begin{align*} r_f(n/2) & =\#\{(x,y,z)\in\mathbb{Z}^3 \mid n/2=x^2+4y^2+32z^2\}, \\ r_g(n/2) & =\#\{(x,y,z)\in\mathbb{Z}^3 \mid n/2=4x^2+4y^2+9z^2-4yz\} \end{align*} $$

and $\sigma _0(n/2)$ is the number of positive divisors of $n/2$ . Suppose $n=2pq$ satisfies the conditions in Theorem 1.2. By (1.2), $r_f(pq)+r_g(pq)=2h(-2pq)$ . By the proof of Theorem 1.2, $r_f(pq)\not \equiv h(-2pq) \pmod {16}$ . Obviously, $8\mid r_f(pq)$ and $8\mid h(-2pq)$ ; thus, $r_f(pq)- h(-2pq)\equiv 8 \pmod {16}$ . Hence,

is divisible by 4 exactly, which is equivalent to .

4 Examples

With the help of a computer, we calculated some noncongruent numbers which can be found by Theorem 1.2. All the noncongruent numbers $n=2pq$ in Table 1 are found by checking the congruence conditions $h(-pq)\not \equiv p-1 \pmod {16}$ and those in Table 2 by checking $h(-2pq)\not \equiv p+1 \pmod {16}$ , as given in Theorem 1.2.

Note that the converse of Theorem 1.2 is not true. For example, if $p=17\equiv 1 \pmod 8$ and $q=157\equiv 5 \pmod 8$ , then $(\frac {q}{p})=1$ and $h(-pq)=64\equiv p-1 \pmod {16}$ , but $n=2pq$ is not congruent. If $p=7\equiv 7 \pmod 8$ and $q=347\equiv 3 \pmod 8$ , then $(\frac {q}{p})=1$ and $h(-2pq)=40\equiv p+1 \pmod {16}$ , but $n=2pq$ is not congruent.

Table 1

$h(-pq)\not \equiv p-1 \pmod {16}$ .

A data table showing values for variables p, q, n equals 2 p q, and h of minus p q across ten rows of numerical data. See long description.

Table 1 Long description

The table consists of four columns and ten data rows.

Column headers from left to right are:

1. p

2. q

3. n = 2 p q

4. h(-p q)

The data rows are as follows:

* Row 1: p = 17, q = 53, n = 1802, h(-p q) = 24

* Row 2: p = 241, q = 5, n = 2410, h(-p q) = 40

* Row 3: p = 113, q = 13, n = 2938, h(-p q) = 56

* Row 4: p = 41, q = 37, n = 3034, h(-p q) = 48

* Row 5: p = 409, q = 5, n = 4090, h(-p q) = 48

* Row 6: p = 521, q = 5, n = 5210, h(-p q) = 16

* Row 7: p = 73, q = 37, n = 5402, h(-p q) = 48

* Row 8: p = 569, q = 5, n = 5690, h(-p q) = 32

* Row 9: p = 641, q = 5, n = 6410, h(-p q) = 24

* Row 10: p = 257, q = 13, n = 6682, h(-p q) = 72

Table 2

$h(-2pq)\not \equiv p+1 \pmod {16}$ .

A data table presenting values for variables p, q, n, and h(-2pq) across ten distinct rows. See long description.

Table 2 Long description

The table consists of four columns and ten rows of data. The columns are labeled p, q, n equals 2pq, and h(-2pq).

* Row 1: p is 47, q is 3, n is 282, h(-2pq) is 8.

* Row 2: p is 7, q is 67, n is 938, h(-2pq) is 16.

* Row 3: p is 167, q is 3, n is 1002, h(-2pq) is 16.

* Row 4: p is 7, q is 107, n is 1498, h(-2pq) is 16.

* Row 5: p is 263, q is 3, n is 1578, h(-2pq) is 16.

* Row 6: p is 311, q is 3, n is 1866, h(-2pq) is 32.

* Row 7: p is 7, q is 179, n is 2506, h(-2pq) is 32.

* Row 8: p is 431, q is 3, n is 2586, h(-2pq) is 56.

* Row 9: p is 23, q is 59, n is 2714, h(-2pq) is 80.

* Row 10: p is 127, q is 11, n is 2794, h(-2pq) is 24.

Footnotes

The author was supported by the NSFC (Grant Nos. 12231009, 11971224).

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Table 1 h(−pq)≢p−1(mod16)$h(-pq)\not \equiv p-1 \pmod {16}$.Table 1 Long description.

Table 2 h(−2pq)≢p+1(mod16)$h(-2pq)\not \equiv p+1 \pmod {16}$.Table 2 Long description.