Proof of some conjectures involving quadratic residues

1.   Introduction

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

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 ${\Bbb Q}(\sqrt{(-1)^{(p-1)/2}p})$. 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 $a\in{\Bbb Z}$ with $p\nmid a$. Recently, Z.-W. Sun [7,Theorem 1.3(ⅱ)] showed that

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 ${\Bbb Q}(\sqrt{p})$, 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

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 $p$.

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

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

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 $x, y\in{\Bbb Z}$, $x \equiv 1\ ({\rm{mod}}\ 4)$ and $y \equiv \frac{p-1}2!x\ ({\rm{mod}}\ p)$. As $y^2 \equiv p-1\ ({\rm{mod}}\ 8)$, we see that $y \equiv (\frac 2p)-1\ ({\rm{mod}}\ 4)$. By a result of K. Burde [3], we have

Thus

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

and

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

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

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 $a\in{\Bbb Z}$ with $p\nmid a$. Then

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 $p \equiv 3\ ({\rm{mod}}\ 4)$.

$\rm (i)$ We have

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

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

where $(\frac cn)$ 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

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

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. $(\frac ap) = 1$.

In this case,

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

As $(\frac{-1}p) = 1$, for each $1 \leq j \leq (p-1)/2$ there is a unique integer $j_*\in\{1, \ldots, (p-1)/2\}$ such that $p-j^2 \equiv j_*^2\ ({\rm{mod}}\ p)$. As $(\frac 2p) = -1$, 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

also,

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

and hence

Thus the sign of the product

is

So (2.1) holds and (1.3) follows.

Case 2. $(\frac ap) = -1$.

By the discussion in Case 1, we have

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

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

where $N({\varepsilon}_p)$ is the norm of ${\varepsilon}_p$ with respect to the field extension ${\Bbb Q}(\zeta)/{\Bbb Q}$, 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

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

Lemma 2.2. Let $p$ be an odd prime, and let $a\in{\Bbb Z}$ with $p\nmid a$. Then

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

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

As $(\frac{-1}p) = 1$, for each $1 \leq j \leq (p-1)/2$ 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, $|\{1 \leq j \leq (p-1)/2:\ j = j_*\}| \leq 1$. Note that

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

This proves the desired (2.4). So (1.6) holds.

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 $a\in{\Bbb Z}$ with $p\nmid a$. Then

Proof. By Sun [7,Theorem 1.4(ⅱ)],

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

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

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

is odd if and only if

where for an assertion $A$ we define

Note that

and

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

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

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

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

Thus

Note that $(\frac ap) = (\frac 2p)$. Set

Applying Theorem 3.2, from the above we obtain

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

and hence (1.8) follows from (4.2).

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

Applying Lemma 4.1, we obtain

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

Acknowledgments

We would like to thank the referee for helpful comments.