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On the quadratic residues and their distribution properties

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Open Mathematics
From the journal Open Mathematics Volume 21 Issue 1

Abstract

The main purpose of this article is to use elementary methods and properties of classical Gauss sums to determine identities for the number of residue systems of a mod p such that a , a + a ¯ , and a − a ¯ are all quadratic residues, improving on results of Wang and Lv.

Keywords: quadratic residues; quadratic non-residues; the classical Gauss sums; counting functions; identity
MSC 2010: 11A15; 11L40

1 Introduction

As usual, let p be an odd prime. For any integer a with ( a , p ) = 1 , if there exists an integer x such that the congruence x 2 ≡ a mod p holds, then we call a a quadratic residue modulo p . Otherwise, we call a a quadratic non-residue modulo p . The study of the properties of quadratic residues modulo p is an important content in elementary number theory. Because of this, many great mathematicians have engaged in research in this field, such as Gauss, Legendre and so on. In fact, Legendre first introduced the characteristic function of the quadratic residues modulo p . That is, for any integer a , the Legendre’s symbol * p is defined as

a p = 1 , if  a  is a quadratic residue  mod p ; − 1 , if  a  is a quadratic non-residue  mod p ; 0 , if  p ∣ a .

It is the introduction of this function that greatly promotes the study of quadratic residues theory, and achieved many important research results. Perhaps the most representative results are the following two conclusions:

Let p and q be two distinct odd primes. Then one has the quadratic reciprocity formula (see [1, Theorem 9.8] or [2, Theorems 4–6])

p q ⋅ q p = ( − 1 ) ( p − 1 ) ( q − 1 ) 4 .

For any odd prime p with p ≡ 1 mod 4 , there must exist two non-zero integers α ( p ) and β ( p ) such that

(1) p = α 2 ( p ) + β 2 ( p ) ,

where α ( p ) and β ( p ) in (1) can be represented by the Legendre’s symbol modulo p (see [2, Theorems 4–11]), i.e.,

(2) α ( p ) = 1 2 ∑ a = 1 p − 1 a + a ¯ p and β ( p ) = 1 2 ∑ a = 1 p − 1 a + r a ¯ p ,

where r is any quadratic non-residue modulo p , a ¯ is the inverse of a modulo p . That is, a ¯ satisfies the equation x ⋅ a ≡ 1 mod p .

In addition, there are many papers related to quadratic residues and non-residues modulo p . Two famous results about quadratic residues are

a a p = a + 1 p = 1 , a ∈ R ( p ) = p − 3 4 ,

and

a a p = a + 1 p = − 1 , a ∈ R ( p ) = p − 1 4 ,

where p is an odd prime, R ( p ) is a complete residue system of modulo p , and [ * ] is the greatest integer function. Other scholars have also studied the distribution of quadratic residues, see [3–14]. For example, Sun [3] figured up all the values of a ( a ∈ R ( p ) ) such that a p = a + 1 p or a − 1 p = a p = a + 1 p . In recent years, Wang and Lv [4] studied the distribution properties of certain quadratic residues and non-residues modulo p , and obtained the following two conclusions:

Theorem A

For any prime p with p ≡ 3 mod 4 , one has the identities

N ( p , 1 ) = 1 8 ( p − 3 ) , if p ≡ 3 mod 8 ; 1 8 ( p − 7 ) , if p ≡ 7 mod 8

and

N ( p , − 1 ) = 1 8 ( p − 3 ) , if p ≡ 3 mod 8 ; 1 8 ( p + 1 ) , if p ≡ 7 mod 8 ,

where N ( p , 1 ) denotes the number of all integers 1 ≤ a ≤ p − 1 such that a , a + a ¯ , and a − a ¯ are all quadratic residues modulo p, N ( p , − 1 ) denotes the number of all integers 1 ≤ a ≤ p − 1 such that a is a quadratic non-residue modulo p, and both a + a ¯ and a − a ¯ are quadratic residues modulo p.

Theorem B

For any prime p with p ≡ 1 mod 4 , one has the asymptotic formulae

N ( p , 1 ) = 1 8 ( p − 3 ) + E ( p , 1 ) , if p ≡ 5 mod 8 ; 1 8 ( p − 17 ) + E 1 ( p , 1 ) , if p ≡ 1 mod 8

and

N ( p , − 1 ) = 1 8 ( p + 3 ) + E ( p , − 1 ) , if p ≡ 5 mod 8 ; 1 8 ( p + 3 ) + E 1 ( p , − 1 ) , if p ≡ 1 mod 8 ,

where we have the estimates ∣ E ( p , 1 ) ∣ ≤ 3 4 ⋅ p , ∣ E 1 ( p , 1 ) ∣ ≤ 5 4 ⋅ p , ∣ E ( p , − 1 ) ∣ ≤ 3 4 ⋅ p , and ∣ E 1 ( p , − 1 ) ∣ ≤ 5 4 ⋅ p .

Obviously, Theorem A looks very beautiful, but Theorem B does not. This leads to our curiosity, which naturally asks: If p ≡ 1 mod 4 , are there exact computational formulae for N ( p , 1 ) and N ( p , − 1 ) ?

In this article, we utilize some results of classical Gauss sums to obtain better estimates of character sums, and give more accurate expressions for N ( p , 1 ) and N ( p , − 1 ) in the cases p ≡ 1 mod 4 . That is, we prove the following:

Theorem 1

For any prime p with p ≡ 5 mod 8 , we have the identities

N ( p , 1 ) = N ( p , − 1 ) = 1 8 ( p − 7 − 2 α ( p ) ) ,

where the constant α ( p ) is the same as defined in (2).

Theorem 2

For any prime p with p ≡ 1 mod 8 , we have the identities

N ( p , 1 ) = 1 8 ⋅ ( p − 23 + 6 ⋅ α ( p ) ) + p 4 ⋅ τ ( χ 8 ) τ ( χ 8 5 ) + τ ( χ 8 5 ) τ ( χ 8 )

and

N ( p , − 1 ) = 1 8 ⋅ ( p + 1 + 6 ⋅ α ( p ) ) − p 4 ⋅ τ ( χ 8 ) τ ( χ 8 5 ) + τ ( χ 8 5 ) τ ( χ 8 ) ,

where χ 8 denotes any eighth-order primitive character modulo p , and τ ( χ ) is the classical Gauss sums defined by

(3) τ ( χ ) = ∑ a = 1 p − 1 χ ( a ) e a p , e ( y ) = e 2 π i y .

Some notes: It is clear that our theorems improve the corresponding results in [4]. For any prime p with p ≡ 1 mod 8 and integer k, let

S k ( p ) = τ k ( χ 8 ) τ k ( χ 8 5 ) + τ k ( χ 8 5 ) τ k ( χ 8 ) .

It is clear that for any integer k , we have the estimate

∣ S k ( p ) ∣ = τ k ( χ 8 ) τ k ( χ 8 5 ) + τ k ( χ 8 5 ) τ k ( χ 8 ) ≤ 2 .

But whether there exists an exact value is still an open problem. We only know that S 1 ( p ) is an irrational number. Besides, for any integer k ≥ 1 ,

S 1 ( p ) ⋅ S k ( p ) = τ ( χ 8 ) τ ( χ 8 5 ) + τ ( χ 8 5 ) τ ( χ 8 ) ⋅ τ k ( χ 8 ) τ k ( χ 8 5 ) + τ k ( χ 8 5 ) τ k ( χ 8 ) = τ k + 1 ( χ 8 ) τ k + 1 ( χ 8 5 ) + τ k + 1 ( χ 8 5 ) τ k + 1 ( χ 8 ) + τ k − 1 ( χ 8 ) τ k − 1 ( χ 8 5 ) + τ k − 1 ( χ 8 5 ) τ k − 1 ( χ 8 ) = S k + 1 ( p ) + S k − 1 ( p ) .

Thus, S k ( p ) satisfies the second-order linear recurrence formula

S k + 1 ( p ) = S 1 ( p ) ⋅ S k ( p ) − S k − 1 ( p ) , k = 1 , 2 , 3 , … .

Noting that S 0 ( p ) = 2 , so if we knew the exact value of S 1 ( p ) , then we can obtain all values of S k ( p ) for any integer k .

From our theorems we can also deduce the following two corollaries:

Corollary 1

For any prime p ≡ 5 mod 8 , we have the asymptotic formulae

N ( p , 1 ) = N ( p , − 1 ) = 1 8 ( p − 7 ) + E ( p ) , ∣ E ( p ) ∣ ≤ 1 4 ⋅ p .

Corollary 2

For any prime p ≡ 1 mod 8 , we have the asymptotic formulae

N ( p , 1 ) = 1 8 ⋅ ( p − 23 + 6 ⋅ α ( p ) ) + E 1 ( p ) , ∣ E 1 ( p ) ∣ ≤ 1 2 ⋅ p ; N ( p , − 1 ) = 1 8 ⋅ ( p + 1 + 6 ⋅ α ( p ) ) + E 2 ( p ) , ∣ E 2 ( p ) ∣ ≤ 1 2 ⋅ p .

2 Several lemmas

To complete the proofs of our main results, we need the following five simple lemmas. For convenience, we always write a p = χ 2 ( a ) . The first lemma is a result of Chen and Zhang [15].

Lemma 1

Let p be a prime with p ≡ 1 mod 4 . Then for any fourth-order character χ 4 modulo p, we have the identity

τ 2 ( χ 4 ) + τ 2 ( χ ¯ 4 ) = 2 p ⋅ α ( p ) ,

where τ ( χ ) is the same as defined in (3), and α ( p ) is the same as defined in (2).

Lemma 2

Let p be a prime with p ≡ 1 mod 4 . Then we have the identity

∑ a = 1 p − 1 a p a 2 − 1 p = 2 χ 4 ( − 1 ) ⋅ α ( p ) ,

where χ 4 is any fourth-order primitive character modulo p.

Proof

Noting that χ 4 2 = χ ¯ 4 2 = χ 2 , and for any integer a with ( a , p ) = 1 ,

1 + χ 2 ( a ) = 2 , if  a  is a quadratic residue modulo  p ; 0 , if  a  is a quadratic non-residue modulo  p .

Thus,

(4) ∑ a = 1 p − 1 a p a 2 − 1 p = ∑ a = 1 p − 1 χ 4 2 ( a ) χ 2 ( a 2 − 1 ) = ∑ a = 1 p − 1 χ 4 ( a 2 ) χ 2 ( a 2 − 1 ) = 2 ⋅ ∑ a = 1 p − 1 a is a quadratic residue modulo p χ 4 ( a ) χ 2 ( a − 1 ) = ∑ a = 1 p − 1 ( 1 + χ 2 ( a ) ) χ 4 ( a ) χ 2 ( a − 1 ) = ∑ a = 1 p − 1 χ 4 ( a ) χ 2 ( a − 1 ) + ∑ a = 1 p − 1 χ ¯ 4 ( a ) χ 2 ( a − 1 ) .

Noting that χ 4 ( − 1 ) = χ ¯ 4 ( − 1 ) and when p ≡ 1 mod 4 , τ ( χ 2 ) = p . We have

∑ a = 1 p − 1 χ 4 ( a ) χ 2 ( a − 1 ) = τ ( χ 2 ) p ∑ a = 1 p − 1 χ 4 ( a ) χ 2 ( a − 1 ) = 1 p ∑ b = 1 p − 1 χ 2 ( b ) e b p ∑ a = 1 p − 1 χ 4 ( a ) χ 2 ( a − 1 ) .

When p ∣ a − 1 , ∑ a = 1 p − 1 χ 4 ( a ) χ 2 ( a − 1 ) = 0 . When ( a − 1 , p ) = 1 , b passes through a reduced residue system modulo p is equivalent to ( a − 1 ) ⋅ b passes through a reduced residue system modulo p . Thus,

(5) ∑ a = 1 p − 1 χ 4 ( a ) χ 2 ( a − 1 ) = 1 p ∑ a = 1 p − 1 ∑ b = 1 p − 1 χ 2 ( b ( a − 1 ) ) e b ( a − 1 ) p χ 4 ( a ) χ 2 ( a − 1 ) = 1 p ∑ a = 1 p − 1 ∑ b = 1 p − 1 χ 2 ( b ) χ 4 ( a ) e b ( a − 1 ) p = 1 p ∑ b = 1 p − 1 χ 2 ( b ) e − b p ∑ a = 1 p − 1 χ 4 ( a ) e a b p = τ ( χ 4 ) p ∑ b = 1 p − 1 χ 2 ( b ) χ 4 ¯ ( b ) e − b p = τ ( χ 4 ) p ∑ b = 1 p − 1 χ 4 ( b ) e − b p = χ 4 ( − 1 ) ⋅ τ 2 ( χ 4 ) p .

Similarly, we have

(6) ∑ a = 1 p − 1 χ ¯ 4 ( a ) χ 2 ( a − 1 ) = χ 4 ( − 1 ) ⋅ τ 2 ( χ ¯ 4 ) p .

Applying (4), (5), (6), and Lemma 1 we can deduce that

∑ a = 1 p − 1 a p a 2 − 1 p = χ 4 ( − 1 ) p ⋅ ( τ 2 ( χ 4 ) + τ 2 ( χ ¯ 4 ) ) = 2 χ 4 ( − 1 ) ⋅ α ( p ) .

This proves Lemma 2.□

Lemma 3

Let p be a prime with p ≡ 1 mod 4 . Then we have

∑ a = 1 p − 1 a 4 − 1 p = − 2 + 2 χ 4 ( − 1 ) ⋅ α ( p ) .

Proof

Let χ 4 be any fourth-order primitive character modulo p . Then from Lemma 1 and the properties of the reduced residue system modulo p we have

∑ a = 1 p − 1 a 4 − 1 p = ∑ a = 1 p − 1 ( 1 + χ 4 ( a ) + χ 2 ( a ) + χ ¯ 4 ( a ) ) ⋅ χ 2 ( a − 1 ) = ∑ a = 1 p − 1 χ 2 ( a − 1 ) + ∑ a = 1 p − 1 χ 2 ( 1 − a ¯ ) + ∑ a = 1 p − 1 χ 4 ( a ) χ 2 ( a − 1 ) + ∑ a = 1 p − 1 χ ¯ 4 ( a ) χ 2 ( a − 1 ) = − 2 + χ 4 ( − 1 ) ⋅ τ 2 ( χ 4 ) p + χ ¯ 4 ( − 1 ) ⋅ τ 2 ( χ ¯ 4 ) p = − 2 + χ 4 ( − 1 ) p ⋅ ( τ 2 ( χ 4 ) + τ 2 ( χ ¯ 4 ) ) = − 2 + 2 χ 4 ( − 1 ) ⋅ α ( p ) .

This proves Lemma 3.□

Lemma 4

Let p be a prime with p ≡ 5 mod 8 . Then we have the identity

∑ a = 1 p − 1 a p a 4 − 1 p = 0 .

Proof

Since p ≡ 5 mod 8 , we have χ 4 ( − 1 ) = χ ¯ 4 ( − 1 ) = − 1 , from the method of proving Lemma 3 we have

(7) ∑ a = 1 p − 1 a p a 4 − 1 p = ∑ a = 1 p − 1 χ 4 2 ( a ) χ 2 ( a 4 − 1 ) = ∑ a = 1 p − 1 ( 1 + χ 2 ( a ) ) χ 4 ( a ) χ 2 ( a 2 − 1 ) = ∑ a = 1 p − 1 χ 4 ( a ) χ 2 ( a 2 − 1 ) + ∑ a = 1 p − 1 χ ¯ 4 ( a ) χ 2 ( a 2 − 1 ) = ∑ a = 1 p − 1 χ 4 ( − a ) χ 2 ( ( − a ) 2 − 1 ) + ∑ a = 1 p − 1 χ ¯ 4 ( − a ) χ 2 ( ( − a ) 2 − 1 ) = − ∑ a = 1 p − 1 χ 4 ( a ) χ 2 ( a 2 − 1 ) − ∑ a = 1 p − 1 χ ¯ 4 ( a ) χ 2 ( a 2 − 1 ) .

From (7) we may immediately deduce the identity

∑ a = 1 p − 1 a p a 4 − 1 p = 0 .

This proves Lemma 4.□

Lemma 5

Let p be a prime with p ≡ 1 mod 8 . Then we have the identity

∑ a = 1 p − 1 a p a 4 − 1 p = 2 ⋅ p ⋅ τ ( χ 8 ) τ ( χ 8 5 ) + τ ( χ 8 5 ) τ ( χ 8 ) ,

where χ 8 denotes any eighth-order primitive character modulo p .

Proof

Since p ≡ 1 mod 8 , there must exist four eighth-order primitive characters modulo p . Let χ 8 be any one of the eighth-order primitive characters modulo p and χ 4 = χ 8 2 . Then from the properties of the fourth-order primitive characters and Gauss sums modulo p , we have the identity

(8) ∑ a = 1 p − 1 a p a 4 − 1 p = ∑ a = 1 p − 1 χ 8 ( a 4 ) χ 2 ( a 4 − 1 ) = ∑ a = 1 p − 1 ( 1 + χ 4 ( a ) + χ 2 ( a ) + χ ¯ 4 ( a ) ) χ 8 ( a ) χ 2 ( a − 1 ) = ∑ a = 1 p − 1 χ 8 ( a ) χ 2 ( a − 1 ) + ∑ a = 1 p − 1 χ 8 3 ( a ) χ 2 ( a − 1 ) + ∑ a = 1 p − 1 χ 8 5 ( a ) χ 2 ( a − 1 ) + ∑ a = 1 p − 1 χ ¯ 8 ( a ) χ 2 ( a − 1 ) = 1 p ∑ a = 1 p − 1 χ 8 ( a ) ∑ b = 1 p − 1 χ 2 ( b ) e b ( a − 1 ) p + 1 p ∑ a = 1 p − 1 χ 8 3 ( a ) ∑ b = 1 p − 1 χ 2 ( b ) e b ( a − 1 ) p + 1 p ∑ a = 1 p − 1 χ 8 5 ( a ) ∑ b = 1 p − 1 χ 2 ( b ) e b ( a − 1 ) p + 1 p ∑ a = 1 p − 1 χ ¯ 8 ( a ) ∑ b = 1 p − 1 χ 2 ( b ) e b ( a − 1 ) p = 2 ⋅ χ 8 ( − 1 ) ⋅ τ ( χ 8 ) ⋅ τ ( χ 8 3 ) p + 2 ⋅ χ ¯ 8 ( − 1 ) ⋅ τ ( χ ¯ 8 ) ⋅ τ ( χ ¯ 8 3 ) p .

If p ≡ 1 mod 8 , then noting that τ ( χ 8 ) ⋅ τ ( χ ¯ 8 ) = χ ¯ 8 ( − 1 ) ⋅ p , τ ( χ 8 3 ) ⋅ τ ( χ 8 5 ) = χ ¯ 8 ( − 1 ) ⋅ p and τ ( χ ¯ 8 3 ) = τ ( χ 8 5 ) . From (8) we have the identity

∑ a = 1 p − 1 a p a 4 − 1 p = 2 ⋅ p ⋅ τ ( χ 8 ) τ ( χ 8 5 ) + τ ( χ 8 5 ) τ ( χ 8 ) .

This proves Lemma 5.□

3 Proofs of the theorems

In this section, we complete the proofs of our main results. For any integer a with ( a , p ) = 1 , we have

1 k ⋅ ( 1 + χ ( a ) + χ 2 ( a ) + ⋅ ⋅ ⋅ + χ k − 1 ( a ) ) = 1 , if  a  is a  k th residue modulo  p ; 0 , if  a  is a  k th non-residue modulo  p ,

where χ is a k -th-order character modulo p . In addition, for any prime p with p ≡ 1 mod 4 , noting that − 1 p = 1 and a + a ¯ ≡ 0 mod p has two solutions λ and − λ with λ ≠ ± 1 and λ ¯ = − λ . So from the definition of N ( p , 1 ) and the properties of the Legendre’s symbol mod p , we have

(9) N ( p , 1 ) = 1 8 ∑ a = 1 ( a ⋅ ( a + a ¯ ) ⋅ ( a − a ¯ ) , p ) = 1 p − 1 1 + a p 1 + a + a ¯ p 1 + a − a ¯ p = 1 8 ∑ a = 2 ( a + a ¯ , p ) = 1 p − 2 1 + a p 1 + a + a ¯ p 1 + a − a ¯ p = 1 8 ∑ a = 1 p − 1 1 + a p 1 + a + a ¯ p 1 + a − a ¯ p − 1 8 ⋅ 1 + 1 p ⋅ 1 + 1 + 1 ¯ p ⋅ 1 + 1 − 1 ¯ p − 1 8 ⋅ 1 + − 1 p ⋅ 1 + ( − 1 ) + ( − 1 ) ¯ p ⋅ 1 + ( − 1 ) − ( − 1 ) ¯ p − 1 8 ⋅ 1 + λ p ⋅ 1 + λ + λ ¯ p ⋅ 1 + λ − λ ¯ p − 1 8 ⋅ 1 + − λ p ⋅ 1 + ( − λ ) + ( − λ ) ¯ p ⋅ 1 + ( − λ ) − ( − λ ) ¯ p = p − 1 8 + 1 8 ∑ a = 1 p − 1 a p + 1 8 ∑ a = 1 p − 1 a + a ¯ p + 1 8 ∑ a = 1 p − 1 a − a ¯ p + 1 8 ∑ a = 1 p − 1 a 2 + 1 p + 1 8 ∑ a = 1 p − 1 a 2 − 1 p + 1 8 ∑ a = 1 p − 1 a p a 2 − a ¯ 2 p + 1 8 ∑ a = 1 p − 1 a 2 − a ¯ 2 p − 1 4 1 + λ p + 2 λ p + 2 p − 1 2 1 + 2 p = p − 1 8 + 1 8 ⋅ 0 + 1 8 ∑ a = 1 p − 1 a + a ¯ p + 1 8 ∑ a = 1 p − 1 a − a ¯ p ⋅ a 2 p + 1 8 ∑ a = 1 p − 1 a 2 + 1 p + 1 8 ∑ a = 1 p − 1 a 2 − 1 p + 1 8 ∑ a = 1 p − 1 a p a 2 − a ¯ 2 p ⋅ a 2 p + 1 8 ∑ a = 1 p − 1 a 2 − a ¯ 2 p a 2 p − 1 4 1 + λ p + 2 λ p + 2 p − 1 2 1 + 2 p = p − 7 8 + 1 8 ∑ a = 1 p − 1 a + a ¯ p + 1 8 ∑ a = 1 p − 1 a p a 2 − 1 p + 1 8 ∑ a = 1 p − 1 a 2 + 1 p + 1 8 ∑ a = 1 p − 1 a 2 − 1 p + 1 8 ∑ a = 1 p − 1 a p a 4 − 1 p + 1 8 ∑ a = 1 p − 1 a 4 − 1 p − 1 4 1 + 2 p λ p − 3 4 2 p .

If p = 8 k + 5 , we have 2 p = λ p = − 1 . In fact in this case, noting that λ + λ ¯ ≡ 0 mod p , we have λ 2 ≡ − 1 mod p and λ p − 1 2 ≡ ( − 1 ) p − 1 4 ≡ − 1 mod p . From Euler’s criterion (see [1, Theorem 9.2]) we have λ p = − 1 and 2 p = ( − 1 ) p 2 − 1 8 = − 1 . We also have

(10) ∑ a = 1 p − 1 a 2 ± 1 p = ∑ a = 1 p − 1 a ± 1 p + ∑ a = 1 p − 1 a p a ± 1 p = − 1 + ∑ a = 1 p − 1 1 ± a ¯ p = − 1 + ∑ a = 1 p − 1 1 ± a p = − 2 .

From (9), (10), and Lemmas 2–4, we have

(11) N ( p , 1 ) = p − 7 8 + α ( p ) 4 − α ( p ) 4 − 1 2 − 1 + α ( p ) 4 + 3 4 = 1 8 ( p − 7 − 2 α ( p ) ) .

On the other hand, noting that the identity

(12) N ( p , − 1 ) = 1 8 ∑ a = 2 ( a + a ¯ , p ) = 1 p − 2 1 − a p 1 + a + a ¯ p 1 + a − a ¯ p = 1 4 ∑ a = 2 ( a + a ¯ , p ) = 1 p − 2 1 + a + a ¯ p 1 + a − a ¯ p − N ( p , 1 ) = 1 4 ∑ a = 1 p − 1 1 + a + a ¯ p 1 + a − a ¯ p − 1 4 ⋅ 1 + 1 + 1 ¯ p ⋅ 1 + 1 − 1 ¯ p − 1 4 ⋅ 1 + ( − 1 ) + ( − 1 ) ¯ p ⋅ 1 + ( − 1 ) − ( − 1 ) ¯ p − 1 4 ⋅ 1 + λ + λ ¯ p ⋅ 1 + λ − λ ¯ p − 1 4 ⋅ 1 + − λ + ( − λ ) ¯ p ⋅ 1 + − λ − ( − λ ) ¯ p − N ( p , 1 ) = p − 1 4 + 1 4 ∑ a = 1 p − 1 a + a ¯ p + 1 4 ∑ a = 1 p − 1 a − a ¯ p + 1 4 ∑ a = 1 p − 1 a 4 − 1 p − 1 2 1 + 2 p − 1 2 1 + 2 λ p − N ( p , 1 ) .

From (11), (12), Lemmas 2, 3, and the definition of α ( p ) we have

(13) N ( p , − 1 ) = p − 1 4 + α ( p ) 2 − α ( p ) 2 − 1 + α ( p ) 2 − 1 − 1 8 ( p − 7 − 2 α ( p ) ) = 1 8 ( p − 7 − 2 α ( p ) ) .

Now Theorem 1 follows from (11) and (13).

If p ≡ 1 mod 8 , then 2 p = λ p = 1 and χ 4 ( − 1 ) = 1 , from (9), (10), Lemmas 2, 3, and 5, we have

(14) N ( p , 1 ) = p − 7 8 + α ( p ) 4 + α ( p ) 4 − 1 2 + − 1 + α ( p ) 4 − 5 4 + p 4 ⋅ τ ( χ 8 ) τ ( χ 8 5 ) + τ ( χ 8 5 ) τ ( χ 8 ) = 1 8 ⋅ ( p − 23 + 6 ⋅ α ( p ) ) + p 4 ⋅ τ ( χ 8 ) τ ( χ 8 5 ) + τ ( χ 8 5 ) τ ( χ 8 ) .

From (10), (12), (14), Lemmas 2, 3, and 5 we also have

(15) N ( p , − 1 ) = p − 1 4 + α ( p ) 2 + α ( p ) 2 + − 1 + α ( p ) 2 − 1 − 1 − N ( p , 1 ) = 1 8 ⋅ ( p + 1 + 6 ⋅ α ( p ) ) − p 4 ⋅ τ ( χ 8 ) τ ( χ 8 5 ) + τ ( χ 8 5 ) τ ( χ 8 ) .

From (14) and (15) we may immediately complete the proof of Theorem 2.

Acknowledgments

The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this article.

  1. Funding information: This work was supported by the N. S. F. (12126357) of P. R. China.

  2. Author contributions: All authors have equally contributed to this work. All authors read and approved the final manuscript.

  3. Conflict of interest: The authors declare that there are no conflicts of interest regarding the publication of this article.

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Received: 2023-03-23
Revised: 2023-08-07
Accepted: 2023-08-27
Published Online: 2023-09-26

© 2023 the author(s), published by De Gruyter

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  26. A derivative-Hilbert operator acting on Dirichlet spaces
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  58. Uniqueness theorems of the Hahn difference operator of entire function with a Picard exceptional value
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  62. A multigrid discretization scheme based on the shifted inverse iteration for the Steklov eigenvalue problem in inverse scattering
  63. Existence and nonexistence of solutions for elliptic problems with multiple critical exponents
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  122. Approximate solvability method for nonlocal impulsive evolution equation
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https://doi.org/10.1515/math-2023-0123
Keywords for this article
quadratic residues; quadratic non-residues; the classical Gauss sums; counting functions; identity
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  77. Multiplicity of solutions for a class of critical Schrödinger-Poisson systems on the Heisenberg group
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  80. Compression with wildcards: All exact or all minimal hitting sets
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  82. Geometric classifications of k-almost Ricci solitons admitting paracontact metrices
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  85. Existence of solutions for semilinear retarded equations with non-instantaneous impulses, non-local conditions, and infinite delay
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  124. Global well-posedness of initial-boundary value problem of fifth-order KdV equation posed on finite interval
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  134. The structure fault tolerance of burnt pancake networks
  135. Average value of the divisor class numbers of real cubic function fields
  136. Uniqueness of exponential polynomials
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