Proof of some conjectures involving quadratic residues

TL;DR

This paper proves several conjectures by Sun related to quadratic residues modulo odd primes, involving products, residue comparisons, and properties of quadratic fields, advancing understanding of quadratic residue patterns and their algebraic connections.

Contribution

It provides the first proofs of Sun's conjectures on quadratic residues, connecting residue counts with class numbers and fundamental units of quadratic fields.

Findings

Confirmed conjectures involving quadratic residue products and residue comparisons.

Established relationships between residue counts and class numbers of quadratic fields.

Extended results to primes congruent to 3 mod 4 involving triangular numbers.

Abstract

We confirm several conjectures of Sun involving quadratic residues modulo odd primes. For any prime $p\equiv 1\pmod 4$ and integer $a\not\equiv0\pmod p$, we prove that \begin{align*}&(-1)^{|\{1\le k<\frac p4:\ (\frac kp)=-1\}|}\prod_{1\le j<k\le(p-1)/2}(e^{2\pi iaj^2/p}+e^{2\pi iak^2/p}) \\=&\begin{cases}1&\text{if}\ p\equiv1\pmod 8,\\\left(\frac ap\right)\varepsilon_p^{-(\frac ap)h(p)}&\text{if}\ p\equiv5\pmod8,\end{cases} \end{align*} and that \begin{align*}&\left|\left\{(j,k):\ 1\le j<k\le\frac{p-1}2\ \&\ \{aj^2\}_p>\{ak^2\}_p\right\}\right| \\&+\left|\left\{(j,k):\ 1\le j<k\le\frac{p-1}2\ \&\ \{ak^2-aj^2\}_p>\frac p2\right\}\right| \\\equiv&\left|\left\{1\le k<\frac p4:\ \left(\frac kp\right)=\left(\frac ap\right)\right\}\right|\pmod2. \end{align*} where $(\frac{a}p)$ is the Legendre symbol, $\varepsilon_p$ and $h(p)$ are the fundamental unit and the class number of the real quadratic field $\mathbb Q(\sqrt p)$ respectively, and $\{x\}_p$ is the least nonnegative residue of an integer $x$ modulo $p$. Also, for any prime $p\equiv3\pmod4$ and $\delta=1,2$, we determine $$(-1)^{\left|\left\{(j,k): \ 1\le j<k\le(p-1)/2\ \text{and}\ \{\delta T_j\}_p>\{\delta T_k\}_p\right\}\right|},$$ where $T_m$ denotes the triangular number $m(m+1)/2$.

Equations

$$(-1{)}^{|\{1⩽k<\frac{p}{4}:(\frac{k}{p})=-1\}|}\prod _{1⩽j<k⩽(p-1)/2}(e^{2\pi ia{j}^2/p}+e^{2\pi ia{k}^2/p})$$

$$=$$

$$\left|\left\{(j,k):1⩽j<k⩽\frac{p-1}{2}\&\{a{j}^2{\}}_p>\{a{k}^2{\}}_p\right\}\right|$$

$$+\left|\left\{(j,k):1⩽j<k⩽\frac{p-1}{2}\&\{a{k}^2-a{j}^2{\}}_p>\frac{p}{2}\right\}\right|$$

$$\equiv$$

$$(-1{)}^{|\left\{(j,k):1⩽j<k⩽(p-1)/2\text{and}\{\delta T_j{\}}_p>\{\delta T_k{\}}_p\right\}|},$$

$$1^2,2^2,\dots ,{\left(\frac{p-1}{2}\right)}^2$$

$$\prod _{1⩽j<k⩽(p-1)/2}({\zeta }^{a{j}^2}-{\zeta }^{a{k}^2})$$

$$=$$

$$(-1{)}^{a\frac{p+1}{2}\lfloor \frac{p-1}{4}\rfloor }{2}^{(p-1)(p-3)/8}\prod _{1⩽j<k⩽(p-1)/2}\cos \pi \frac{a(k^2-j^2)}{p}$$

$$=$$

$$\prod _{1⩽j<k⩽(p-1)/2}({\zeta }^{a{j}^2}+{\zeta }^{a{k}^2})=(-1{)}^{|\{1⩽k<\frac{p}{4}:(\frac{k}{p})=-1\}|}.$$

$$(-1{)}^{|\{1⩽k<\frac{p}{4}:(\frac{k}{p})=-1\}|}\prod _{1⩽j<k⩽(p-1)/2}({\zeta }^{a{j}^2}+{\zeta }^{a{k}^2})=\left(\frac{a}{p}\right){\epsilon }_p^{-(\frac{a}{p})h(p)}.$$

$$\left|\left\{1⩽k<\frac{p}{4}:\left(\frac{k}{p}\right)=1\right\}\right|\equiv 0(\mathrm{mod}2)⟺y\equiv \left(\frac{2}{p}\right)-1(\mathrm{mod}8).$$

$$(-1{)}^{|\{1⩽k<\frac{p}{4}:(\frac{k}{p})=-1\}|}=(-1{)}^{\frac{p-1}{4}}(-1{)}^{\frac{1}{4}(y-(\frac{2}{p})+1)}=(-1{)}^{\lfloor \frac{y}{4}\rfloor }.$$

$$s(p):=\left|\left\{(j,k):1⩽j<k⩽\frac{p-1}{2}\&\{j^2{\}}_p>\{k^2{\}}_p\right\}\right|$$

$$t(p):=\left|\left\{(j,k):1⩽j<k⩽\frac{p-1}{2}\&\{k^2-j^2{\}}_p>\frac{p}{2}\right\}\right|$$

$$(-1{)}^{s(p)}=(-1{)}^{t(p)}=\{\begin{array}{ll}1& \text{if}p\equiv 3(\mathrm{mod}8),\\ (-1{)}^{(h(-p)+1)/2}& \text{if}p\equiv 7(\mathrm{mod}8).\end{array}$$

$$s(p)+t(p)\equiv \left|\left\{1⩽k<\frac{p}{4}:\left(\frac{k}{p}\right)=1\right\}\right|(\mathrm{mod}2).$$

$$\left|\left\{(j,k):1⩽j<k⩽\frac{p-1}{2}\&\{a{j}^2{\}}_p>\{a{k}^2{\}}_p\right\}\right|$$

$$+\left|\left\{(j,k):1⩽j<k⩽\frac{p-1}{2}\&\{a{k}^2-a{j}^2{\}}_p>\frac{p}{2}\right\}\right|$$

$$\equiv$$

$$(-1{)}^{|\left\{(j,k):1⩽j<k⩽(p-1)/2\text{and}\{j(j+1){\}}_p>\{k(k+1){\}}_p\right\}|}=(-1{)}^{\lfloor (p+1)/8\rfloor }.$$

$$(-1{)}^{|\{(j,k):1⩽j<k⩽(p-1)/2\&\{T_j{\}}_p>\{T_k{\}}_p\}|}=(-1{)}^{\frac{h(-p)+1}{2}+|\{1⩽k⩽\lfloor \frac{p+1}{8}\rfloor :(\frac{k}{p})=1\}|}.$$

$$\mathrm{sign}({\pi }_c)={\left(\frac{c}{n}\right)}^{(n+1)/2},$$

$$\prod _{1⩽j<k⩽(p-1)/2}\frac{{\zeta }^{2a{j}^2}-{\zeta }^{2a{k}^2}}{{\zeta }^{a{j}^2}-{\zeta }^{a{k}^2}}=\prod _{1⩽j<k⩽(p-1)/2}\frac{{\zeta }^{a{\pi }_c(j{)}^2}-{\zeta }^{a{\pi }_c(k{)}^2}}{{\zeta }^{a{j}^2}-{\zeta }^{a{k}^2}}$$

$$=$$

$$\left(\frac{c}{p}\right)\equiv c^{(p-1)/2}=(c^2{)}^{(p-1)/4}\equiv 2^{(p-1)/4}\equiv (-1{)}^{|\{0<k<\frac{p}{4}:(\frac{k}{p})=-1\}|}(\mathrm{mod}p).$$

$$\left\{\{a{j}^2{\}}_p:1⩽j⩽\frac{p-1}{2}\right\}=\left\{\{k^2{\}}_p:1⩽k⩽\frac{p-1}{2}\right\}$$

$$(-1{)}^{|\{0<k<\frac{p}{4}:(\frac{k}{p})=-1\}|}\prod _{1⩽j<k⩽(p-1)/2}({\zeta }^{j^2}+{\zeta }^{k^2})>0.$$

$$({\zeta }^{j^2}+{\zeta }^{k^2})({\zeta }^{j_{\ast }^2}+{\zeta }^{k_{\ast }^2})=({\zeta }^{j^2}+{\zeta }^{k^2})({\zeta }^{-j^2}+{\zeta }^{-k^2})=|{\zeta }^{j^2}+{\zeta }^{k^2}{|}^2>0;$$

$$\{j,k\}=\{j_{\ast },k_{\ast }\}⟺j_{\ast }=k\text{and}{k}_{\ast }=j⟺j_{\ast }=k.$$

$${\zeta }^{j^2}+{\zeta }^{j_{\ast }^2}={\zeta }^{j^2}+{\zeta }^{-j^2}=2\cos 2\pi \frac{j^2}{p}=2\cos 2\pi \frac{j_{\ast }^2}{p}$$

$${\zeta }^{j^2}+{\zeta }^{j_{\ast }^2}>0⟺\cos 2\pi \frac{j^2}{p}>0⟺\{j^2{\}}_p<\frac{p}{4}\text{or}\{j_{\ast }^2{\}}_p<\frac{p}{4}.$$

Full text

[TABLE]

and that

[TABLE]

where $(\frac{a}{p})$ is the Legendre symbol, $\varepsilon_{p}$ and $h(p)$ are the fundamental unit and the class number of the real quadratic field $\mathbb{Q}(\sqrt{p})$ respectively, and $\{x\}_{p}$ is the least nonnegative residue of an integer $x$ modulo $p$. Also, for any prime $p\equiv 3\ ({\rm{mod}}\ 4)$ and $\delta=1,2$, we determine

[TABLE]

where $T_{m}$ denotes the triangular number $m(m+1)/2$.

2020 Mathematics Subject Classification. Primary 11A15, 05A05; Secondary 11R11, 33B10.

The work is supported by the NSFC-RFBR Cooperation and Exchange Program (grants NSFC 11811530072 and RFBR 18-51-53020-GFEN-a). The second author is also supported by the Natural Science Foundation of China (grant 11971222).

1. Introduction

Let $p$ be an odd prime. It is well known that the numbers

[TABLE]

are pairwise incongruent modulo $p$, and they give all the $(p-1)/2$ quadratic residues modulo $p$. Recently Z.-W. Sun [7] initiated the study of permutations related to quadratic residues modulo $p$ as well as evaluations of related products involving $p$th roots of unity; many of his results in [7] are related to the class number of the quadratic field $\mathbb{Q}(\sqrt{(-1)^{(p-1)/2}p})$. In this paper, we confirm some conjectures of Sun [7] in this new direction.

Let $p>3$ be a prime and let $\zeta=e^{2\pi i/p}$. Let $a\in\mathbb{Z}$ with $p\nmid a$. Recently, Z.-W. Sun [7, Theorem 1.3(ii)] showed that

[TABLE]

where $(\frac{a}{p})$ is the Legendre symbol, $\varepsilon_{p}$ with $p\equiv 1\ ({\rm{mod}}\ 4)$ is the fundamental unit of the real quadratic field $\mathbb{Q}(\sqrt{p})$, and $h(d)$ with $d\equiv 0,1\ ({\rm{mod}}\ 4)$ not a square is the class number of the quadratic field with discriminant $d$. Sun [7, Theorem 1.5] also proved that

[TABLE]

Our first theorem confirms [7, Conjecture 6.7].

Theorem 1.1**.**

Let $p$ be a prime with $p\equiv 1\ ({\rm{mod}}\ 4)$, and let $\zeta=e^{2\pi i/p}$. Let $a$ be an integer not divisible by $p$.

(i)* If $p\equiv 1\ ({\rm{mod}}\ 8)$, then*

[TABLE]

(ii)* When $p\equiv 5\ ({\rm{mod}}\ 8)$, we have*

[TABLE]

Remark 1.1*.*

Let $p$ be a prime with $p\equiv 1\ ({\rm{mod}}\ 4)$. Then $(\frac{p-1}{2}!)^{2}\equiv-1\ ({\rm{mod}}\ p)$ by Wilson’s theorem. We may write $p=x^{2}+y^{2}$ with $x,y\in\mathbb{Z}$, $x\equiv 1\ ({\rm{mod}}\ 4)$ and $y\equiv\frac{p-1}{2}!x\ ({\rm{mod}}\ p)$. As $y^{2}\equiv p-1\ ({\rm{mod}}\ 8)$, we see that $y\equiv(\frac{2}{p})-1\ ({\rm{mod}}\ 4)$. By a result of K. Burde [3], we have

[TABLE]

Thus

[TABLE]

Let $p$ be an odd prime. For each $a\in\mathbb{Z}$ we let $\{a\}_{p}$ denote the least nonnegative residue of $a$ modulo $p$. Define

[TABLE]

and

[TABLE]

as in [7], where $(j,k)$ is an ordered pair. Sun [7, Theorem 1.4(i)] showed that

[TABLE]

He also conjectured that (cf. [7, Conjecture 6.1]) if $p\equiv 1\ ({\rm{mod}}\ 4)$ then

[TABLE]

Our second theorem in the case $a=1$ confirms this conjecture.

Theorem 1.2**.**

Let $p$ be a prime with $p\equiv 1\ ({\rm{mod}}\ 4)$, and let $a\in\mathbb{Z}$ with $p\nmid a$. Then

[TABLE]

Our third theorem was first conjectured by Sun (cf. [7, Conjectures 6.3 and 6.4]).

Theorem 1.3**.**

Let $p$ be a prime with $p\equiv 3\ ({\rm{mod}}\ 4)$.

(i)* We have*

[TABLE]

(ii)* Suppose $p>3$ and write $T_{m}=m(m+1)/2$ for $m\in\mathbb{N}$. Then*

[TABLE]

We will prove Theorems 1.1-1.2 in Section 2. Based on an auxiliary theorem given in Section 3, we are going to prove Theorem 1.3 in Section 4.

2. Proofs of Theorems 1.1-1.2

In 2006, H. Pan [6] obtained the following lemma.

Lemma 2.1**.**

(H. Pan [6])* Let $n>1$ be an odd integer and let $c$ be any integer relatively prime to $n$. For each $j=1,\ldots,(n-1)/2$ let $\pi_{c}(j)$ be the unique $r\in\{1,\ldots,(n-1)/2\}$ with $cj$ congruent to $r$ or $-r$ modulo $n$. For the permutation $\pi_{c}$ on $\{1,\ldots,(n-1)/2\}$, its sign is given by*

[TABLE]

where $(\frac{c}{n})$ is the Jacobi symbol.

Proof of the First Part of Theorem 1.1. As $p\equiv 1\ ({\rm{mod}}\ 8)$, there is an integer $c$ with $c^{2}\equiv 2\ ({\rm{mod}}\ p)$. For $j=1,\ldots,(p-1)/2$ let $\pi_{c}(j)$ be the unique $r\in\{1,\ldots,(p-1)/2\}$ with $cj$ congruent to $r$ or $-r$ modulo $p$. Then $\pi_{c}$ is a permutation on $\{1,\ldots,(p-1)/2\}$, and

[TABLE]

with the aid of Lemma 2.1. In view of K. S. Williams and J. D. Currie [8, (1.4)], we have

[TABLE]

Therefore (1.2) holds in the case $p\equiv 1\ ({\rm{mod}}\ 8)$. ∎

Remark 2.1*.*

Our method to prove part (i) of Theorem 1.1 does not work for part (ii) of Theorem 1.1.

Proof of the Second Part of Theorem 1.1. We distinguish two cases.

Case 1. $(\frac{a}{p})=1$.

In this case,

[TABLE]

So it suffices to show (1.3) for $a=1$. In view of (1.1) with $a=1$, we only need to prove that

[TABLE]

As $(\frac{-1}{p})=1$, for each $1\leqslant j\leqslant(p-1)/2$ there is a unique integer $j_{*}\in\{1,\ldots,(p-1)/2\}$ such that $p-j^{2}\equiv j_{*}^{2}\ ({\rm{mod}}\ p)$. As $(\frac{2}{p})=-1$, we have $j\not=j_{*}$. For any distinct $j,k\in\{1,\ldots,(p-1)/2\}$, we have $\zeta^{j^{2}}+\zeta^{k^{2}}\not=0$ (since $\zeta^{2j^{2}}\not=\zeta^{2k^{2}}$) and

[TABLE]

also,

[TABLE]

For $1\leqslant j\leqslant(p-1)/2$, clearly

[TABLE]

and hence

[TABLE]

Thus the sign of the product

[TABLE]

is

[TABLE]

This proves (2.1).

Case 2. $(\frac{a}{p})=-1$.

By the discussion in Case 1, we have

[TABLE]

Let $\varphi_{a}$ be the element of the Galois group $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ with $\varphi_{a}(\zeta)=\zeta^{a}$. Then

[TABLE]

by the evaluation of quadratic Gauss sums (cf. [5, pp. 70-75]). Hence

[TABLE]

where $N(\varepsilon_{p})$ is the norm of $\varepsilon_{p}$ with respect to the field extension $\mathbb{Q}(\zeta_{p})/\mathbb{Q}$, and we have used the known results $N(\varepsilon_{p})=-1$ and $2\nmid h(p)$ (cf. [4, p. 185 and p. 187]). Thus, by applying the automorphism $\varphi_{a}$ to the identity (2.2), we get

[TABLE]

In view of the above, we have proven Theorem 1.1(ii). ∎

Lemma 2.2**.**

Let $p$ be an odd prime, and let $a\in\mathbb{Z}$ with $p\nmid a$. Then

[TABLE]

Proof. This can be easily checked by distinguishing the cases $\{aj^{2}\}_{p}<\{ak^{2}\}_{p}$ and $\{aj^{2}\}_{p}>\{ak^{2}\}_{p}$ for $1\leqslant j<k\leqslant(p-1)/2$. ∎

Proof of Theorem 1.2. In view of Lemma 2.2, it suffices to show that

[TABLE]

As $(\frac{-1}{p})=1$, for each $1\leqslant j\leqslant(p-1)/2$ there is a unique integer $j_{*}\in\{1,\ldots,(p-1)/2\}$ such that $-j^{2}\equiv j_{*}^{2}\ ({\rm{mod}}\ p)$ and hence $-aj^{2}\equiv aj_{*}^{2}\ ({\rm{mod}}\ p)$. Clearly, $|\{1\leqslant j\leqslant(p-1)/2:\ j=j_{*}\}|\leqslant 1$. Note that

[TABLE]

If $j$ and $k$ are distinct elements of $\{1,\ldots,(p-1)/2\}$, then $\{j,k\}=\{j_{*},k_{*}\}$ if and only if $j_{*}=k$ and $k_{*}=j$. Thus

[TABLE]

This proves the desired (2.4). ∎

3. An Auxiliary Theorem

We first need a result of Sun [7].

Lemma 3.1**.**

Let $p=2n+1$ be a prime with $n$ odd, and let $a\in\mathbb{Z}$ with $p\nmid a$. Then

[TABLE]

Proof.

By Sun [7, Theorem 1.4(ii)],

[TABLE]

This implies (3.1) since for any $1\leqslant j<k\leqslant(p-1)/2$ we have

[TABLE]

We are done. ∎

Theorem 3.2**.**

Let $p=2n+1$ be a prime with $n$ odd, and let $a,b\in\{1,\ldots,p-1\}$. Then

[TABLE]

Proof.

Let $0\leqslant t<s\leqslant n$. By comparing $\{as^{2}\}_{p}$ and $\{at^{2}\}_{p}$ with $b$, we verify case by case that

[TABLE]

is odd if and only if

[TABLE]

where for an assertion $A$ we define

[TABLE]

Note that

[TABLE]

and

[TABLE]

with $\binom{n+1}{2}\equiv\frac{p+1}{4}\ ({\rm{mod}}\ 2)$. Combining the above with (3.1), we finally obtain (3.2). ∎

4. Proof of Theorem 1.3

Lemma 4.1**.**

Let $p$ be a prime with $p\equiv 3\ ({\rm{mod}}\ 4)$.

(i) (Dirichlet (cf. [5, p. 238]))* If $p>3$ then*

[TABLE]

(ii) (B. C. Berndt and S. Chowla [1])* If $p\equiv 3\ ({\rm{mod}}\ 8)$, then $\sum_{0<k<p/4}(\frac{k}{p})=0.$ If $p\equiv 7\ ({\rm{mod}}\ 8)$, then $\sum_{p/4<k<p/2}(\frac{k}{p})=0.$*

Proof of Theorem 1.3. We just prove the second part in details since the first part can be proved similarly.

Write $n=(p-1)/2$, and set

[TABLE]

For any $r\in\mathbb{Z}$, we have

[TABLE]

Thus

[TABLE]

Note that $(\frac{a}{p})=(\frac{2}{p})$. Set

[TABLE]

Applying Theorem 3.2, from the above we obtain

[TABLE]

When $p\equiv 7\ ({\rm{mod}}\ 8)$, we have

[TABLE]

and hence (1.8) follows from (4.2).

Below we handle the case $p\equiv 3\ ({\rm{mod}}\ 8)$. Observe that

[TABLE]

Applying Lemma 4.1, we obtain

[TABLE]

So, in this case, (1.8) also follows from (4.2). ∎

Acknowledgment. We would like to thank the referee for helpful comments.