On the Mean Values of Certain Character Sums

Abstract

Let 𝑞≥5 be an odd number. In this paper, we study the fourth power mean of certain character sums ∑^∗_{𝜒mod𝑞,(𝜒−1)=−1}|∑_{1≤𝑎≤𝑞/4}𝑎𝜒(𝑎)|⁴ and ∑^∗_{𝜒mod𝑞,(𝜒−1)=1}|∑_{1≤𝑎≤𝑞/4}𝑎𝜒(𝑎)|⁴, where ∑^∗ denotes the summation over primitive characters modulo 𝑞, and give some asymptotic formulae.

1. Introduction

The sum 𝑆_𝜒(𝑛)=1𝑞^𝑛𝑞_𝑎=1𝑎^𝑛𝜒(𝑎) (1) appears frequently in number theory, where 𝜒 is a nonprincipal primitive character modulo 𝑞, and has been studied by several experts. For example, for 𝑞≡3mod4 being a prime 𝑝 and 𝜒 being the Legendre symbol, Ayoub et al. [1] have proved that 𝑆_𝜒(𝑛)<0 for 𝑛=1,2 and for 𝑛≥𝑝−2. Fine [2] has showed that for 𝑛>2, there exist infinitely many primes 𝑝≡3mod4 with 𝑆_𝜒(𝑛)>0 and infinitely many with 𝑆_𝜒(𝑛)<0.

Williams [3] proved that 𝑆_𝜒(𝑛)=𝑂𝑝^1/2log𝑝 (2) for 𝜒 being the Legendre symbol modulo 𝑝. For primitive character 𝜒 modulo 𝑞, Toyoizumi [4] used the generalized Bernoulli numbers to express 𝑆_𝜒(𝑛) in terms of Gauss sums and Dirichlet 𝐿-functions as follows: ^𝑞_𝑎=1𝑎^𝑛𝜒(𝑎)=2𝑞^𝑛𝜏𝜒×∑_{1≤𝑚≤𝑛/2}(^𝑛_2𝑚−1)(2𝑚−1)!𝐿2𝑚,𝜒(−1)^𝑚+1(2𝜋)^2𝑚,if𝜒(−1)=1,2𝑞^𝑛𝜏𝜒×∑_{0≤𝑚≤(𝑛−1)/2}(^𝑛_2𝑚)(2𝑚)!𝐿2𝑚+1,𝜒(−1)^𝑚+1(2𝜋)^2𝑚+1𝑖,if𝜒(−1)=−1, (3) where 𝜏(𝜒)=∑^𝑞_𝑎=1𝜒(𝑎)𝑒(𝑎/𝑞) is the Gauss sum, 𝑒(𝑦)=𝑒^{2𝜋𝑖𝑦}, 𝐿(𝑠,𝜒) is the Dirichlet 𝐿-function corresponding to 𝜒, and (^𝑛_𝑚) denotes the binomial coefficient.

Toyoizumi [4] also gave explicit bounds for 𝑆_𝜒(𝑛).

Proposition 1. (a) Assume that 𝜒(−1)=1 and 𝑛≥2. Then for any primitive character 𝜒mod𝑞, one has 𝑆_𝜒(𝑛)<𝐶₁(𝑛)𝑞^1/2, (4) where 𝐶₁(𝑛)=2𝜁(2)𝑛!(2𝜋)^𝑛+1_{1≤𝑚≤𝑛/2}(2𝜋)^{𝑛+1−2𝑚}()!,𝑛+1−2𝑚 (5) and 𝜁(𝑠) is the Riemann zeta function.
(b) Assume that 𝜒(−1)=−1 and 𝑛≥3. Then for any primitive character 𝜒mod𝑞, one has 𝑆_𝜒(𝑛)<𝐶₂(𝑛)+𝐿1,𝜒𝜋𝑞^1/2, (6) where 𝐶₂(𝑛)=2𝜁(3)𝑛!(2𝜋)^𝑛+1_{1≤𝑚≤(𝑛−1)/2}(2𝜋)^𝑛−2𝑚()!.𝑛−2𝑚 (7)

In [5], Peral used the Gauss sums and adequate Fourier expansion to greatly improve the result in Proposition 1.

Proposition 2. (a) Assume that 𝜒(−1)=1 is a primitive nonprincipal character modulo 𝑞, and then 𝑆_𝜒(𝑛)≤𝑞^1/2𝑛−12().𝑛+1 (8)
(b) Assume that 𝜒(−1)=−1 is a primitive character modulo 𝑞; then, 𝑆_𝜒(𝑛)+𝜏𝜒𝐿1,𝜒𝜋𝑖≤𝑞^1/2𝑛𝜋¹₀1ln()𝑡2sin𝜋𝑡^𝑛−1d𝑡. (9)

Furthermore, Liu and Zhang [6] gave an upper bound for 𝑆_𝜒(𝑛) when 𝜒 is a nonprincipal character modulo 𝑞.

It may be interesting to consider the mean value of certain character sums. For example, Burgess [7] proved that ^∗𝑞_{𝜒mod𝑞𝑛=1}^ℎ_𝑚=1𝜒(𝑛+𝑚)⁴≤8𝑑²𝑞𝑞²ℎ², (10) where ∑^∗ denotes the summation over primitive characters modulo 𝑞, and 𝑑(𝑞) is the Dirichlet divisor function. Xu and Zhang studied the power mean ^∗_{𝜒mod𝑞𝜒(−1)=±11≤𝑎≤𝑞/4}𝜒(𝑎)⁴ (11) in [8, 9] and obtained some sharper results.

In this paper, we study the fourth power mean of certain character sums 𝑎𝜒(^∗_{𝜒mod𝑞𝜒(−1)=−11≤𝑎≤𝑞/4}𝑎)⁴,_{∗𝜒mod𝑞𝜒(−1)=11≤𝑎≤𝑞/4}𝑎𝜒(𝑎)⁴ (12) and give a few asymptotic formulae.

Theorem 3. Let 𝑞≥5 be an odd number. Then one has 𝑎𝜒(^∗_{𝜒mod𝑞𝜒(−1)=−11≤𝑎≤𝑞/4}𝑎)⁴=72¹⁷⋅3²𝑞⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝²³1+1/𝑝²+352¹⁶⋅3²𝑞⋅17⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝⁴³1+1/𝑝⁴−𝐿²3,𝜒₄2¹⁴⋅3²⋅𝜋⋅𝐿5,𝜒₄×𝑞⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝²²1−𝜒₄𝑝/𝑝³²1−𝜒₄𝑝/𝑝⁵+𝐿3,𝜒₄2⁸⋅5⋅3²⋅𝜋²𝑞⁶𝐽𝑞×_𝑝∣𝑞1−𝜒₄𝑝/𝑝³1−1/𝑝⁴×_𝑝∤𝑞11+𝑝²1−𝜒₄𝑝𝑞/𝑝+𝑂^6+𝜖, (13) where 𝐽(𝑞) is the number of primitive characters modulo 𝑞, 𝜒₄ is the nonprincipal character modulo 4, and 𝜖 is any fixed positive real number.

Theorem 4. Let 𝑞≥5 be an odd number. Then one has 𝑎𝜒(^∗_{𝜒mod𝑞𝜒(−1)=11≤𝑎≤𝑞/4}𝑎)⁴=3852²⁰⋅51⋅𝑞⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝⁴³1+1/𝑝⁴+15𝐿⁴3,𝜒₄𝜋¹²⋅𝑞⁶𝐽𝑞_𝑝∣𝑞1−𝜒₄𝑝/𝑝³³1+𝜒₄𝑝/𝑝³−1423𝜋𝐿²3,𝜒₄2¹⁶⋅5²⋅3⁴⋅𝐿7,𝜒₄⋅𝑞⁶𝐽𝑞_𝑝∣𝑞1−𝜒₄𝑝/𝑝³²1−1/𝑝⁴²1−𝜒₄𝑝/𝑝⁷+32¹⁶⋅𝑞⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝²³1+1/𝑝²−𝐿²3,𝜒₄2¹²⋅𝜋⋅𝐿5,𝜒₄⋅𝑞⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝²²1−𝜒₄𝑝/𝑝³²1−𝜒₄𝑝/𝑝⁵+𝐿3,𝜒₄2¹⁰⋅3⋅𝜋²⋅𝑞⁶𝐽𝑞×_𝑝∣𝑞1−𝜒₄𝑝/𝑝³1−1/𝑝⁴×_𝑝∤𝑞11+𝑝²1−𝜒₄𝑝𝑞/𝑝+𝑂^6+𝜖. (14)

From Theorems 3 and 4, we immediately get the following corollaries.

Corollary 5. Let 𝑝≥5 be a prime. Then one has 𝑎𝜒(_{𝜒mod𝑝𝜒(−1)=−11≤𝑎≤𝑝/4}𝑎)⁴=212¹⁷𝑝⋅17⁷−𝐿²3,𝜒₄2¹⁴⋅3²⋅𝜋⋅𝐿5,𝜒₄𝑝⁷+𝐿3,𝜒₄2⁸⋅5⋅3²⋅𝜋²𝑝⁷_𝑝111+𝑝²₁1−𝜒₄𝑝₁/𝑝₁+𝑂𝑝^6+𝜖, (15) where ∏_𝑝1 denotes the product over all primes.

Corollary 6. Let 𝑝≥5 be a prime. Then 𝑎𝜒(_{𝜒mod𝑝𝜒(−1)=1𝜒≠𝜒01≤𝑎≤𝑝/4}𝑎)⁴=28332²⁰𝑝⋅51⁷+15𝐿⁴3,𝜒₄𝜋¹²𝑝⁷−1423𝜋𝐿²3,𝜒₄2¹⁶⋅5²⋅3⁴⋅𝐿7,𝜒₄𝑝⁷−𝐿²3,𝜒₄2¹²⋅𝜋⋅𝐿5,𝜒₄𝑝⁷+𝐿3,𝜒₄2¹⁰⋅3⋅𝜋²𝑝⁷_𝑝111+𝑝²₁1−𝜒₄𝑝₁/𝑝₁+𝑂𝑝^6+𝜖. (16)

Remark 7. It seems that the contributions of odd and even primitive characters to the fourth power moment of character sums over [1,𝑞/4] are very different.

2. Express the Character Sum in terms of Gauss Sums and 𝐿-Functions (I)

Let 𝜒 be an odd primitive character modulo 𝑞. In this section, we will express ∑_{1≤𝑎≤𝑞/4}𝑎𝜒(𝑎) in terms of Gauss sums and Dirichlet 𝐿-functions. We need the following lemmas.

Lemma 8. Suppose that 𝑞≥5 is an odd number, and 𝜒 is an odd character modulo 𝑞.(i) For 𝑞≡1(mod4), one has 𝑎𝜒(_{1≤𝑎≤(3𝑞−3)/4}𝑎)=_{1≤𝑎≤𝑞}𝑎𝜒(𝑎)+𝑞_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎)−_{1≤𝑎≤(𝑞−1)/4}𝑎𝜒_{1≤𝑎≤(3𝑞−3)/4}(𝑎),𝜒(𝑎)=_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎). (17)(ii) For 𝑞≡3(mod4), one has 𝑎𝜒(_{1≤𝑎≤(3𝑞−1)/4}𝑎)=_{1≤𝑎≤𝑞}𝑎𝜒(𝑎)+𝑞_{1≤𝑎≤(𝑞−3)/4}𝜒(𝑎)−_{1≤𝑎≤(𝑞−3)/4}𝑎𝜒_{1≤𝑎≤(3𝑞−1)/4}(𝑎),𝜒(𝑎)=_{1≤𝑎≤(𝑞−3)/4}𝜒(𝑎). (18)

Proof. It is easy to show that 𝑎𝜒(_{1≤𝑎≤(3𝑞−3)/4}𝑎)=_{1≤𝑎≤𝑞}𝑎𝜒(𝑎)−_{(3𝑞+1)/4≤𝑎≤𝑞}(𝑎)=𝑎𝜒_{1≤𝑎≤𝑞}𝑎𝜒(𝑎)−_{1≤𝑏≤(𝑞−1)/4}𝑞−𝑏𝜒𝑞−𝑏=_{1≤𝑎≤𝑞}𝑎𝜒(𝑎)+𝑞_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎)−_{1≤𝑎≤(𝑞−1)/4}𝑎𝜒_{1≤𝑎≤(3𝑞−3)/4}(𝑎),𝜒(𝑎)=_{1≤𝑎≤𝑞}𝜒(𝑎)−_{(3𝑞+1)/4≤𝑎≤𝑞}𝜒(𝑎)=_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎). (19) This proves (i). Similarly, we can deduce (ii).

Lemma 9. Suppose that 𝑞≥5 is an odd number, and 𝜒 is an odd character modulo 𝑞. Let 𝜒₄ be the nonprincipal character modulo 4. For 𝑞≡1(mod4), one has ^4𝑞_𝑎=1𝑎²𝜒(𝑎)𝜒₄(𝑎)=16𝜒(4)𝑞^𝑞_𝑎=1𝑎𝜒(𝑎)+16𝜒(4)𝑞²_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎)−64𝜒(4)𝑞_{1≤𝑎≤(𝑞−1)/4}𝑎𝜒(𝑎). (20)

Proof. Note that 𝜒₄(1)=1 and 𝜒₄(3)=−1, and we get ^4𝑞_𝑎=1𝑎²𝜒(𝑎)𝜒₄(𝑎)=^𝑞−1_𝑎=0(4𝑎+1)²𝜒()−4𝑎+1_𝑞−1𝑎∑₌₀(4𝑎+3)²𝜒(4𝑎+3). (21)
First we have ^𝑞−1_𝑎=0(4𝑎+1)²𝜒(4𝑎+1)=16^𝑞−1_𝑎=0𝑎²𝜒(4𝑎+1)+8^𝑞−1_𝑎=0()+𝑎𝜒4𝑎+1_{𝑞−1𝑎=0}𝜒()(4)4𝑎+1=16𝜒^𝑞−1_𝑎=0𝑎²𝜒𝑎+4+8𝜒(4)^𝑞−1_𝑎=0𝑎𝜒𝑎+4=16𝜒(4)^𝑞−1_𝑎=0𝑎+4²𝜒𝑎+4+8𝜒(4)1−4⋅4^𝑞−1_𝑎=0𝑎+4𝜒𝑎+4, (22) where 4 is the inverse of 4 modulo 𝑞 with 4⋅4≡1(mod𝑞) and 1≤4≤𝑞. Since 𝑞≡1(mod4), we get 4=(3𝑞+1)/4. Then from Lemma 8, we have ^𝑞−1_𝑎=0𝑎+4²𝜒𝑎+4=_{0≤𝑎≤(𝑞−1)/4}𝑎+4²𝜒𝑎+4+_{(𝑞−1)/4<𝑎≤𝑞−1}𝑎+4²𝜒𝑎+4=_{0≤𝑎≤(𝑞−1)/4}𝑎+4²𝜒𝑎+4+_{(𝑞−1)/4<𝑎≤𝑞−1}𝑎+4−𝑞²𝜒𝑎+4−𝑞+2𝑞_{(𝑞−1)/4<𝑎≤𝑞−1}𝑎+4−𝑞𝜒𝑎+4−𝑞+𝑞²_{(𝑞−1)/4<𝑎≤𝑞−1}𝜒𝑎+=4−𝑞_𝑞𝑎=1𝑎²𝜒(𝑎)+2𝑞_{1≤𝑎≤(3𝑞−3)/4}𝑎𝜒(𝑎)+𝑞²_{1≤𝑎≤(3𝑞−3)/4}𝜒(𝑎)=^𝑞_𝑎=1𝑎²𝜒(𝑎)+2𝑞^𝑞_𝑎=1𝑎𝜒(𝑎)+3𝑞²_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎)−2𝑞_{1≤𝑎≤(𝑞−1)/4}𝑎𝜒_{𝑞−1𝑎=0}(𝑎),𝑎+4𝜒𝑎+4=_{0≤𝑎≤(𝑞−1)/4}𝑎+4𝜒𝑎+4+_{(𝑞−1)/4<𝑎≤𝑞−1}𝑎+4𝜒𝑎+4=_{0≤𝑎≤(𝑞−1)/4}𝑎+4𝜒𝑎+4+_{(𝑞−1)/4<𝑎≤𝑞−1}𝑎+4−𝑞𝜒𝑎+4−𝑞+𝑞_{(𝑞−1)/4<𝑎≤𝑞−1}𝜒𝑎+=4−𝑞_𝑞𝑎=1𝑎𝜒(𝑎)+𝑞_{(𝑞−1)/4≤𝑎≤𝑞−1}𝜒𝑎+=4−𝑞_𝑞𝑎=1𝑎𝜒(𝑎)+𝑞_{1≤𝑎≤(3𝑞−3)/4}𝜒(𝑎)=^𝑞_𝑎=1𝑎𝜒(𝑎)+𝑞_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎). (23) Therefore ^𝑞−1_𝑎=0(4𝑎+1)²𝜒()(4)4𝑎+1=16𝜒^𝑞−1_𝑎=0𝑎+4²𝜒𝑎+4+8𝜒(4)1−4⋅4^𝑞−1_𝑎=0𝑎+4𝜒𝑎+4=16𝜒(4)^𝑞_𝑎=1𝑎²𝜒(𝑎)+8𝜒(4)𝑞^𝑞_𝑎=1(𝑎)(4)𝑞𝑎𝜒+24𝜒²_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎)−32𝜒(4)𝑞_{1≤𝑎≤(𝑞−1)/4}𝑎𝜒(𝑎). (24)
On the other hand, we get ^𝑞−1_𝑎=0(4𝑎+3)²𝜒(4𝑎+3)=16^𝑞−1_𝑎=0𝑎²𝜒(4𝑎+3)+24^𝑞−1_𝑎=0𝑎𝜒(4𝑎+3)+9^𝑞−1_𝑎=0𝜒()(4)4𝑎+3=16𝜒^𝑞−1_𝑎=0𝑎²𝜒𝑎+3⋅4+24𝜒(4)^𝑞−1_𝑎=0𝑎𝜒𝑎+3⋅4. (25)
Since 4=(3𝑞+1)/4 and 3⋅4=2𝑞+(𝑞+3)/4, we have ^𝑞−1_𝑎=0(4𝑎+3)²𝜒()(4)4𝑎+3=16𝜒^𝑞−1_𝑎=0𝑎²𝜒𝑎+𝑞+34+24𝜒(4)^𝑞−1_𝑎=0𝑎𝜒𝑎+𝑞+34=16𝜒(4)^𝑞−1_𝑎=0𝑎+𝑞+34²𝜒𝑎+𝑞+34+241−𝑞+33𝜒(4)×^𝑞−1_𝑎=0𝑎+𝑞+34𝜒𝑎+𝑞+34. (26)
Note that ^𝑞−1_𝑎=0𝑎+𝑞+34²𝜒𝑎+𝑞+34=_{0≤𝑎≤(3𝑞−3)/4}𝑎+𝑞+34²𝜒𝑎+𝑞+34+_{(3𝑞+1)/4≤𝑎≤𝑞−1}𝑎+𝑞+34²𝜒𝑎+𝑞+34=_{0≤𝑎≤(3𝑞−3)/4}𝑎+𝑞+34²𝜒𝑎+𝑞+34+_{(3𝑞+1)/4≤𝑎≤𝑞−1}𝑎+𝑞+34−𝑞²𝜒𝑎+𝑞+34−𝑞+2𝑞_{(3𝑞+1)/4≤𝑎≤𝑞−1}𝑎+𝑞+34−𝑞𝜒𝑎+𝑞+34−𝑞+𝑞²_{(3𝑞+1)/4≤𝑎≤𝑞−1}𝜒𝑎+𝑞+34=−𝑞_𝑞𝑎=1𝑎²𝜒(𝑎)+𝑞²_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎)+2𝑞_{1≤𝑎≤(𝑞−1)/4}𝑎𝜒_{𝑞−1𝑎=0}(𝑎),𝑎+𝑞+34𝜒𝑎+𝑞+34=_{0≤𝑎≤(3𝑞−3)/4}𝑎+𝑞+34𝜒𝑎+𝑞+34+_{(3𝑞+1)/4≤𝑎≤𝑞−1}𝑎+𝑞+34𝜒𝑎+𝑞+34=_{0≤𝑎≤(3𝑞−3)/4}𝑎+𝑞+34𝜒𝑎+𝑞+34+_{(3𝑞+1)/4≤𝑎≤𝑞−1}𝑎+𝑞+34−𝑞𝜒𝑎+𝑞+34−𝑞+𝑞_{(3𝑞+1)/4≤𝑎≤𝑞−1}𝜒𝑎+𝑞+34=−𝑞_𝑞𝑎=1𝑎𝜒(𝑎)+𝑞_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎), (27) so we get ^𝑞−1_𝑎=0(4𝑎+3)²𝜒()(4)4𝑎+3=16𝜒^𝑞−1_𝑎=0𝑎+𝑞+34²𝜒𝑎+𝑞+34+241−𝑞+33𝜒(4)×^𝑞−1_𝑎=0𝑎+𝑞+34𝜒𝑎+𝑞+34=16𝜒(4)^𝑞_𝑎=1𝑎²𝜒(𝑎)−8𝜒(4)𝑞^𝑞_𝑎=1(𝑎)(4)𝑞𝑎𝜒+8𝜒²_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎)+32𝜒(4)𝑞_{1≤𝑎≤(𝑞−1)/4}𝑎𝜒(𝑎). (28)
Now combine (21)–(28); we have ^4𝑞_𝑎=1𝑎²𝜒(𝑎)𝜒₄(𝑎)=^𝑞−1_𝑎=0(4𝑎+1)²𝜒()−4𝑎+1_{𝑞−1𝑎=0}(4𝑎+3)²𝜒()(4)𝑞4𝑎+3=16𝜒^𝑞_𝑎=1(𝑎)(4)𝑞𝑎𝜒+16𝜒²_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎)−64𝜒(4)𝑞_{1≤𝑎≤(𝑞−1)/4}𝑎𝜒(𝑎). (29)

Lemma 10. Suppose that 𝑞≥5 is an odd number, and 𝜒 is an odd character modulo 𝑞. Let 𝜒₄ be the nonprincipal character modulo 4. For 𝑞≡3(mod4), one has ^4𝑞_𝑎=1𝑎²𝜒(𝑎)𝜒₄(𝑎)=−16𝜒(4)𝑞^𝑞_𝑎=1(𝑎)(4)𝑞𝑎𝜒−16𝜒²_{1≤𝑎≤(𝑞−1)/4}𝜒(𝑎)+64𝜒(4)𝑞_{1≤𝑎≤(𝑞−1)/4}𝑎𝜒(𝑎). (30)

Proof. For 𝑞≡3(mod4), we get 4=(𝑞+1)/4 and 3⋅4=(3𝑞+3)/4. Using the methods of proving Lemma 9, we have ^4𝑞_𝑎=1𝑎²𝜒(𝑎)𝜒₄(𝑎)=^𝑞−1_𝑎=0(4𝑎+1)²𝜒(4𝑎+1)−^𝑞−1_𝑎=0(4𝑎+3)²𝜒(4𝑎+3_{𝑞−1𝑎=0}),(4𝑎+1)²𝜒()(4)4𝑎+1=16𝜒^𝑞−1_𝑎=0𝑎+4²𝜒𝑎+4+8𝜒(4)1−4⋅4^𝑞−1_𝑎=0𝑎+4𝜒𝑎+4,_{𝑞−1𝑎=0}(4𝑎+3)²𝜒(4𝑎+3)=16^𝑞−1_𝑎=0𝑎²𝜒(4𝑎+3)+24^𝑞−1_𝑎=0𝑎𝜒(4𝑎+3)+9^𝑞−1_𝑎=0𝜒()(4)4𝑎+3=16𝜒^𝑞−1_𝑎=0𝑎²𝜒𝑎+3⋅4+24𝜒(4)^𝑞−1_𝑎=0𝑎𝜒𝑎+3⋅4=16𝜒(4)^𝑞−1_𝑎=0𝑎+3𝑞+34²𝜒𝑎+3𝑞+34−24𝜒(4)𝑞^𝑞−1_𝑎=0𝑎+3𝑞+34𝜒𝑎+3𝑞+34. (31)(32)(33)
It is not hard to show that ^𝑞−1_𝑎=0𝑎+4²𝜒𝑎+4=_{0≤𝑎≤(3𝑞−1)/4}𝑎+4²𝜒𝑎+4+_{(3𝑞−1)/4<𝑎≤𝑞−1}𝑎+4²𝜒𝑎+4=_{0≤𝑎≤(3𝑞−1)/4}𝑎+4²𝜒𝑎+4+_{(3𝑞−1)/4<𝑎≤𝑞−1}𝑎+4−𝑞²𝜒𝑎+4−𝑞+2𝑞_{(3𝑞−1)/4<𝑎≤𝑞−1}𝑎+4−𝑞𝜒𝑎+4−𝑞+𝑞²_{(3𝑞−1)/4<𝑎≤𝑞−1}𝜒𝑎+=4−𝑞_𝑞𝑎=1𝑎²𝜒(𝑎)+𝑞²_{1≤𝑎≤(𝑞−3)/4}𝜒(𝑎)+2𝑞_{1≤𝑎≤(𝑞−3)/4}𝑎𝜒_{𝑞−1𝑎=0}(𝑎),𝑎+4𝜒𝑎+4=_{0≤𝑎≤(3𝑞−1)/4}𝑎+4𝜒𝑎+4+_{(3𝑞−1)/4<𝑎≤𝑞−1}𝑎+4𝜒𝑎+4=_{0≤𝑎≤(3𝑞−1)/4}𝑎+4𝜒𝑎+4+_{(3𝑞−1)/4<𝑎≤𝑞−1}𝑎+4−𝑞𝜒𝑎+4−𝑞+𝑞_{(3𝑞−1)/4<𝑎≤𝑞−1}𝜒𝑎+=4−𝑞_𝑞𝑎=1𝑎𝜒(𝑎)+𝑞_{1≤𝑎≤(𝑞−3)/4}𝜒(𝑎). (34) Then by (32), we have ^𝑞−1_𝑎=0(4𝑎+1)²𝜒()(4)4𝑎+1=16𝜒^𝑞−1_𝑎=0𝑎+4²𝜒𝑎+4+8𝜒(4)1−4×4^𝑞−1_𝑎=0𝑎+4𝜒𝑎+4=16𝜒(4)^𝑞_𝑎=1𝑎²𝜒(𝑎)−8𝜒(4)𝑞^𝑞_𝑎=1𝑎𝜒(𝑎)+8𝜒(4)𝑞²_{1≤𝑎≤(𝑞−3)/4}𝜒(𝑎)+32𝜒(4)𝑞_{1≤𝑎≤(𝑞−3)/4}𝑎𝜒(𝑎). (35)
On the other hand, by Lemma 8, we get ^𝑞−1_𝑎=0𝑎+3𝑞+34²𝜒𝑎+3𝑞+34=_{0≤𝑎≤(𝑞−3)/4}𝑎+3𝑞+34²𝜒𝑎+3𝑞+34+_{(𝑞+1)/4≤𝑎≤𝑞−1}𝑎+3𝑞+34²𝜒𝑎+3𝑞+34=_{0≤𝑎≤(𝑞−3)/4}𝑎+3𝑞+34²𝜒𝑎+3𝑞+34+_{(𝑞+1)/4≤𝑎≤𝑞−1}𝑎+3𝑞+34−𝑞²𝜒𝑎+3𝑞+34−𝑞+2𝑞_{(𝑞+1)/4≤𝑎≤𝑞−1}𝑎+3𝑞+34−𝑞𝜒𝑎+3𝑞+34−𝑞+𝑞²_{(𝑞+1)/4≤𝑎≤𝑞−1}𝜒𝑎+3𝑞+34=−𝑞_𝑞𝑎=1𝑎²𝜒(𝑎)+2𝑞_{1≤𝑎≤(3𝑞−1)/4}𝑎𝜒(𝑎)+𝑞²_{1≤𝑎≤(3𝑞−1)/4}𝜒(𝑎)=^𝑞_𝑎=1𝑎²𝜒(𝑎)+2𝑞^𝑞_𝑎=1𝑎𝜒(𝑎)+3𝑞²_{1≤𝑎≤(𝑞−3)/4}𝜒(𝑎)−2𝑞_{1≤𝑎≤(𝑞−3)/4}𝑎𝜒_{𝑞−1𝑎=0}(𝑎),𝑎+3𝑞+34𝜒𝑎+3𝑞+34=_{0≤𝑎≤(𝑞−3)/4}𝑎+3𝑞+34𝜒𝑎+3𝑞+34+_{(𝑞+1)/4≤𝑎≤𝑞−1}𝑎+3𝑞+34𝜒𝑎+3𝑞+34=_{0≤𝑎≤(𝑞−3)/4}𝑎+3𝑞+34𝜒𝑎+3𝑞+34+_{(𝑞+1)/4≤𝑎≤𝑞−1}𝑎+3𝑞+34−𝑞𝜒𝑎+3𝑞+34−𝑞+𝑞_{(𝑞+1)/4≤𝑎≤𝑞−1}𝜒𝑎+3𝑞+34=−𝑞_𝑞𝑎=1𝑎𝜒(𝑎)+𝑞_{1≤𝑎≤(3𝑞−1)/4}𝜒(𝑎)=^𝑞_𝑎=1𝑎𝜒(𝑎)+𝑞_{1≤𝑎≤(𝑞−3)/4}𝜒(𝑎). (36) Then from (33), we have ^𝑞−1_𝑎=0(4𝑎+3)²𝜒()(4)4𝑎+3=16𝜒^𝑞−1_𝑎=0𝑎+3𝑞+34²𝜒𝑎+3𝑞+34−24𝜒(4)𝑞^𝑞−1_𝑎=0𝑎+3𝑞+34𝜒𝑎+3𝑞+34=16𝜒(4)^𝑞_𝑎=1𝑎²𝜒(𝑎)+8𝜒(4)𝑞^𝑞_𝑎=1(𝑎)(4)𝑞𝑎𝜒+24𝜒²_{1≤𝑎≤(𝑞−3)/4}𝜒(𝑎)−32𝜒(4)𝑞_{1≤𝑎≤(𝑞−3)/4}𝑎𝜒(𝑎). (37)
Combining (31), (35), and (37), we have ^4𝑞_𝑎=1𝑎²𝜒(𝑎)𝜒₄(𝑎)=^𝑞−1_𝑎=0(4𝑎+1)²𝜒()−4𝑎+1_{𝑞−1𝑎=0}(4𝑎+3)²𝜒()(4)𝑞4𝑎+3=−16𝜒^𝑞_𝑎=1(𝑎)(4)𝑞𝑎𝜒−16𝜒²_{1≤𝑎≤(𝑞−3)/4}𝜒(𝑎)+64𝜒(4)𝑞_{1≤𝑎≤(𝑞−3)/4}𝑎𝜒(𝑎). (38)

Now we can express ∑_{1≤𝑎≤𝑞/4}𝑎𝜒(𝑎) in terms of Gauss sums and Dirichlet 𝐿-functions.

Theorem 11. Let 𝜒 be an odd primitive character modulo odd integer 𝑞≥5, and let 𝜒₄ be the nonprincipal character modulo 4. Then one has _{1≤𝑎≤𝑞/4}(𝑎)=𝑞𝑎𝜒𝜏𝜒8𝜋𝑖𝜒(2)𝐿1,𝜒−𝜒(4)𝐿1,𝜒+4𝜋𝐿2,𝜒𝜒₄. (39)

Proof. By Lemmas 9 and 10, we get 16𝜒(4)𝑞^𝑞_𝑎=1𝑎𝜒(𝑎)+16𝜒(4)𝑞²_{1≤𝑎≤𝑞/4}𝜒(𝑎)−64𝜒(4)𝑞_{1≤𝑎≤𝑞/4}(𝑎)=𝑎𝜒_4𝑞𝑎∑₌₁𝑎²𝜒(𝑎)𝜒₄(𝑎)()−,if𝑞≡1mod4_4𝑞𝑎∑₌₁𝑎²𝜒(𝑎)𝜒₄(𝑎),if𝑞≡3(mod4)=𝜒₄𝑞^4𝑞_𝑎=1𝑎²𝜒𝜒₄(𝑎). (40) From the Fourier expansion for primitive character sums (see [10] or [11]) _{𝑎<𝜆𝑞}𝜒(𝑎)=𝜏𝜒𝜋^+∞_𝑛∑₌₁𝜒(𝑛)sin(2𝜋𝑛𝜆)𝑛()𝜏𝜒𝐿,if𝜒−1=1,1,𝜒−𝜏𝜒𝜋𝑖×𝜋𝑖_+∞𝑛∑₌₁𝜒(𝑛)cos(2𝜋𝑛𝜆)𝑛,if𝜒(−1)=−1, (41) we easily have _{1≤𝑎≤𝑞/4}𝜒(𝑎)=2+𝜒(2)−𝜒(4)𝜏𝜒𝐿2𝜋𝑖1,𝜒. (42)
Note that 𝜒𝜒₄ is a primitive character modulo 4𝑞 satisfying 𝜒𝜒₄(−1)=1, and 𝜏𝜒𝜒₄=^4𝑞_𝑎=1𝜒𝜒₄(𝑎)e𝑎=4𝑞_{4𝑞𝑎=1𝑏=1}𝜒4𝑏+𝑞𝑎𝜒₄4𝑏+𝑞𝑎e4𝑏+𝑞𝑎=4𝑞_{4𝑞𝑎=1𝑏=1}𝜒(4𝑏)𝜒₄e𝑏𝑞𝑎𝑞+𝑎4=𝜒(4)𝜒₄𝑞⁴_𝑎=1𝜒₄(𝑎)e𝑎4×^𝑞_𝑏=1𝜒(𝑏)e𝑏𝑞=2𝑖𝜒(4)𝜒₄𝑞𝜏𝜒, (43) then from (3) we have ^𝑞_𝑎=1(𝑎)𝑞𝑎𝜒=−𝜏𝜒𝐿𝜋𝑖1,𝜒,_4𝑞𝑎=1𝑎²𝜒𝜒₄(𝑎)=16𝑞²𝜋²𝜏𝜒𝜒₄𝐿2,𝜒𝜒₄=32𝑖𝜒(4)𝜒₄𝑞𝜋²𝑞²𝜏𝜒𝐿2,𝜒𝜒₄. (44) Therefore −16𝜒(4)𝑞𝜋𝑖²𝜏𝜒𝐿1,𝜒+8𝜒(4)2+𝜒(2)−𝜒(4)𝑞𝜋𝑖²𝜏𝜒𝐿1,𝜒−64𝜒(4)𝑞_{1≤𝑎≤𝑞/4}(𝑎)=(4)𝑎𝜒32𝑖𝜒𝜋²𝑞²𝜏𝜒𝐿2,𝜒𝜒₄. (45) Then we have _{1≤𝑎≤𝑞/4}(𝑎)=𝑞𝑎𝜒𝜏𝜒8𝜋𝑖𝜒(2)𝐿1,𝜒−𝜒(4)𝐿1,𝜒+4𝜋𝐿2,𝜒𝜒₄. (46)

3. Express the Character Sum in terms of Gauss Sums and 𝐿-Functions (II)

Let 𝜒 be an even primitive character modulo 𝑞. In this section, we express ∑_{1≤𝑎≤𝑞/4}𝑎𝜒(𝑎) in terms of Gauss sums and Dirichlet 𝐿-functions.

Lemma 12. Let 𝑞>2 be an odd number, and let 𝜒 be a nonprincipal character modulo 𝑞. Then 4𝜒(2)𝑞^(𝑞−1)/2_𝑎=1(𝑎)=(2)𝑞𝑎𝜒2𝜒+1^𝑞_𝑎=1𝑎𝜒(𝑎)−4𝜒(2)−1^𝑞_𝑎=1𝑎²𝜒(𝑎),4𝜒(2)𝑞^𝑞_{𝑎=(𝑞+1)/2}(𝑎)=(2)𝑞𝑎𝜒2𝜒−1^𝑞_𝑎=1𝑎𝜒(𝑎)+4𝜒(2)−1^𝑞_𝑎=1𝑎²𝜒(𝑎). (47)

Proof. We have ^𝑞_𝑎=1(2𝑎)²𝜒()=2𝑎_{(𝑞−1)/2𝑎=1}(2𝑎)²𝜒(2𝑎)+^𝑞_{𝑎=(𝑞+1)/2}(2𝑎)²𝜒()=2𝑎_{(𝑞−1)/2𝑎=1}(2𝑎)²𝜒()+2𝑎_{(𝑞+1)/2𝑏=1}2𝑏+𝑞−1²𝜒=2𝑏+𝑞−1_{(𝑞−1)/2𝑎=1}(2𝑎)²𝜒()+2𝑎_{(𝑞+1)/2𝑏=1}(2𝑏−1)²𝜒(2𝑏−1)+2𝑞^(𝑞+1)/2_𝑏=1(2𝑏−1)𝜒(2𝑏−1)+𝑞^2(𝑞+1)/2_𝑏=1𝜒()=2𝑏−1_𝑞𝑎=1𝑎²𝜒(𝑎)+2𝑞^(𝑞+1)/2_𝑎=1(2𝑎−1)𝜒(2𝑎−1)+𝑞^2(𝑞+1)/2_𝑎=1𝜒(2𝑎−1). (48)
Since ^(𝑞+1)/2_𝑎=1𝜒(2𝑎−1)+^(𝑞−1)/2_𝑎=1𝜒(2𝑎)=^𝑞_𝑎=1𝜒(𝑎)=0,_{(𝑞+1)/2𝑎=1}(2𝑎−1)𝜒(2𝑎−1)+^(𝑞−1)/2_𝑎=12𝑎𝜒(2𝑎)=^𝑞_𝑎=1𝑎𝜒(𝑎), (49) we have ^𝑞_𝑎=1(2𝑎)²𝜒(2𝑎)=^𝑞_𝑎=1𝑎²𝜒(𝑎)+2𝑞^𝑞_𝑎=1𝑎𝜒(𝑎)−4𝜒(2)𝑞^(𝑞−1)/2_𝑎=1(𝑎)(2)𝑞𝑎𝜒−𝜒^{2(𝑞−1)/2}_𝑎=1𝜒(𝑎). (50)
It is not hard to show that 𝜒(2)𝑞^(𝑞−1)/2_𝑎=1𝜒(𝑎)=1−2𝜒(2)^𝑞_𝑎=1(𝑎),𝜒(2)𝑞𝑎𝜒_{𝑞𝑎=(𝑞+1)/2}𝜒(𝑎)=2𝜒(2)−1^𝑞_𝑎=1𝑎𝜒(𝑎). (51) Therefore 4𝜒(2)𝑞^(𝑞−1)/2_𝑎=1(𝑎)=(2)𝑞×𝑎𝜒2𝜒+1_𝑞𝑎=1𝑎𝜒(𝑎)−4𝜒(2)−1^𝑞_𝑎=1𝑎²𝜒(𝑎). (52) Note that 𝑎𝜒(^(𝑞−1)/2_𝑎=1𝑎)+^𝑞_{𝑎=(𝑞+1)/2}𝑎𝜒(𝑎)=^𝑞_𝑎=1𝑎𝜒(𝑎), (53) we have 4𝜒(2)𝑞^𝑞_{𝑎=(𝑞+1)/2}(𝑎)=(2)𝑞×𝑎𝜒2𝜒−1_𝑞𝑎=1𝑎𝜒(𝑎)+4𝜒(2)−1^𝑞_𝑎=1𝑎²𝜒(𝑎). (54)

Lemma 13. Let 𝑞≥5 be an odd number, and let 𝜒 be an nonprincipal even character modulo 𝑞. If 𝑞≡1(mod4), then 16𝜒(4)^𝑞−1_𝑎=1𝑎²𝜒(𝑎)=(4)(2)×8𝜒−2𝜒+1_𝑞𝑎=1𝑎²𝜒(𝑎)(4)×−16𝑞𝜒_{(𝑞−1)/4𝑎=1}𝑎𝜒(𝑎)+4𝑞²𝜒(4)^(𝑞−1)/4_𝑎=1𝜒(𝑎). (55) While if 𝑞≡3(mod4), we have 16𝜒(4)^𝑞−1_𝑎=1𝑎²𝜒(𝑎)=8𝜒(4)−2𝜒(2)+1^𝑞_𝑎=1𝑎²𝜒(𝑎)(4)×−16𝑞𝜒_{(𝑞−3)/4𝑎=1}𝑎𝜒(𝑎)+4𝑞²𝜒(4)^(𝑞−3)/4_𝑎=1𝜒(𝑎). (56)

Proof. First suppose that 𝑞≡1(mod4). Then 4=(3𝑞+1)/4. We have 16𝜒(4)^𝑞−1_𝑎=1𝑎²𝜒(𝑎)=^𝑞−1_𝑎=1(4𝑎)²𝜒(4𝑎)=^(𝑞−1)/4_𝑎=1(4𝑎)²𝜒()+4𝑎_{(2𝑞−2)/4𝑎=𝑞+3/4}(4𝑎)²𝜒(4𝑎)+^{(3𝑞−3)/4}_{𝑎=(2𝑞+2)/4}(4𝑎)²𝜒()+4𝑎_{𝑞−1𝑎=(3𝑞+1)/4}(4𝑎)²𝜒()=4𝑎_{(𝑞−1)/4𝑎=1}(4𝑎)²𝜒()+4𝑎_{(𝑞−1)/4𝑎=1}4𝑎+𝑞−1²𝜒()+4𝑎−1_{(𝑞−1)/4𝑎=1}4𝑎+2𝑞−2²𝜒()+4𝑎−2_{(𝑞−1)/4𝑎=1}4𝑎+3𝑞−3²𝜒()=4𝑎−3_{(𝑞−1)/4𝑎=1}(4𝑎)²𝜒()+4𝑎_{(𝑞−1)/4𝑎=1}(4𝑎−1)²𝜒()+4𝑎−1_{(𝑞−1)/4𝑎=1}(4𝑎−2)²𝜒()+4𝑎−2_{(𝑞−1)/4𝑎=1}(4𝑎−3)²𝜒(4𝑎−3)+2𝑞^(𝑞−1)/4_𝑎=1(4𝑎−1)𝜒(4𝑎−1)+4𝑞^(𝑞−1)/4_𝑎=1(4𝑎−2)𝜒(4𝑎−2)+6𝑞^(𝑞−1)/4_𝑎=1(4𝑎−3)𝜒(4𝑎−3)+𝑞^{2(𝑞−1)/4}_𝑎=1𝜒(4𝑎−1)+4𝑞^{2(𝑞−1)/4}_𝑎=1𝜒(4𝑎−2)+9𝑞^{2(𝑞−1)/4}_𝑎=1𝜒()=4𝑎−3_{𝑞−1𝑎=1}𝑎²𝜒(𝑎)+8𝑞𝜒(4)^(𝑞−1)/4_𝑎=1𝑎+𝑞−14𝜒𝑎+𝑞−14−2𝑞²𝜒(4)^(𝑞−1)/4_𝑎=1𝜒𝑎+𝑞−14+16𝑞𝜒(4)^(𝑞−1)/4_𝑎=1𝑎+𝑞−12𝜒𝑎+𝑞−12−8𝑞²𝜒(4)^(𝑞−1)/4_𝑎=1𝜒𝑎+𝑞−12+24𝑞𝜒(4)^(𝑞−1)/4_𝑎=1𝑎+3𝑞−34𝜒𝑎+3𝑞−34−18𝑞²𝜒(4)^(𝑞−1)/4_𝑎=1𝜒𝑎+3𝑞−34+𝑞²𝜒(4)^(𝑞−1)/4_𝑎=1𝜒𝑎+𝑞−14+4𝑞²𝜒(4)^(𝑞−1)/4_𝑎=1𝜒𝑎+𝑞−12+9𝑞²𝜒(4)^(𝑞−1)/4_𝑎=1𝜒𝑎+3𝑞−34=^𝑞−1_𝑎=1𝑎²𝜒(𝑎)+8𝑞𝜒(4)^(𝑞−1)/2_{𝑎=𝑞+3/4}(𝑎)(4)𝑎𝜒+16𝑞𝜒^{(3𝑞−3)/4}_{𝑎=(𝑞+1)/2}(𝑎)(4)𝑎𝜒+24𝑞𝜒^𝑞−1_{𝑎=(3𝑞+1)/4}𝑎𝜒(𝑎)−𝑞²𝜒(4)^(𝑞−1)/2_{𝑎=(𝑞+3)/4}𝜒(𝑎)−4𝑞²𝜒(4)^{(3𝑞−3)/4}_{𝑎=(𝑞+1)/2}𝜒(𝑎)−9𝑞²𝜒(4)^𝑞−1_{𝑎=(3𝑞+1)/4}𝜒(𝑎). (57)
Note that ∑_{1≤𝑎≤(𝑞−1)/2}𝜒(𝑎)=0 and ∑^𝑞_𝑎=1𝑎𝜒(𝑎)=0 for even character 𝜒. By Lemma 12 we have 16𝜒(4)^𝑞−1_𝑎=1𝑎²𝜒(𝑎)=^𝑞_𝑎=1𝑎²𝜒(𝑎)(4)×−16𝑞𝜒_{(𝑞−1)/4𝑎=1}𝑎𝜒(𝑎)−8𝑞𝜒(4)^(𝑞−1)/2_𝑎=1𝑎𝜒(𝑎)+4𝑞²𝜒(4)^(𝑞−1)/4_𝑎=1𝜒(𝑎)=(4)(2)×8𝜒−2𝜒+1_𝑞𝑎=1𝑎²𝜒(𝑎)−16𝑞𝜒(4)^(𝑞−1)/4_𝑎=1𝑎𝜒(𝑎)+4𝑞²𝜒(4)^(𝑞−1)/4_𝑎=1𝜒(𝑎). (58)
Now assume that 𝑞≡3(mod4). Then 4=(𝑞+1)/4. We have 16𝜒(4)^𝑞−1_𝑎=1𝑎²𝜒(𝑎)=^𝑞−1_𝑎=1(4𝑎)²𝜒()=4𝑎_{(𝑞−3)/4𝑎=1}(4𝑎)²𝜒(4𝑎)+^{(2𝑞−2)/4}_{𝑎=(𝑞+1)/4}(4𝑎)²𝜒()+4𝑎_{(3𝑞−1)/4𝑎=(2𝑞+2)/4}(4𝑎)²𝜒(4𝑎)+^𝑞−1_{𝑎=(3𝑞+3)/4}(4𝑎)²𝜒()=4𝑎_{(𝑞−3)/4𝑎=1}(4𝑎)²𝜒()+4𝑎_{(𝑞+1)/4𝑎=1}4𝑎+𝑞−3²𝜒()+4𝑎−3_{(𝑞+1)/4𝑎=1}4𝑎+2𝑞−2²𝜒()+4𝑎−2_{(𝑞−3)/4𝑎=1}4𝑎+3𝑞−1²𝜒()=4𝑎−1_{(𝑞−3)/4𝑎=1}(4𝑎)²𝜒(4𝑎)+^(𝑞−3)/4_𝑎=1(4𝑎−1)²𝜒()+4𝑎−1_{(𝑞+1)/4𝑎=1}(4𝑎−2)²𝜒()+4𝑎−2_{(𝑞+1)/4𝑎=1}(4𝑎−3)²𝜒(4𝑎−3)+6𝑞^(𝑞−3)/4_𝑎=1(4𝑎−1)𝜒(4𝑎−1)+4𝑞^(𝑞+1)/4_𝑎=1(4𝑎−2)𝜒(4𝑎−2)+2𝑞^(𝑞+1)/4_𝑎=1(4𝑎−3)𝜒(4𝑎−3)+9𝑞^{2(𝑞−3)/4}_𝑎=1𝜒(4𝑎−1)+4𝑞^2(𝑞+1)/4_𝑎=1𝜒(4𝑎−2)+𝑞^2(𝑞+1)/4_𝑎=1𝜒()=4𝑎−3_{𝑞−1𝑎=1}𝑎²𝜒(𝑎)+24𝑞𝜒(4)^(𝑞−3)/4_𝑎=1𝑎+3𝑞−14𝜒𝑎+3𝑞−14−18𝑞²𝜒(4)^(𝑞−3)/4_𝑎=1𝜒𝑎+3𝑞−14+16𝑞𝜒(4)^(𝑞+1)/4_𝑎=1𝑎+𝑞−12𝜒𝑎+𝑞−12−8𝑞²𝜒(4)^(𝑞+1)/4_𝑎=1𝜒𝑎+𝑞−12+8𝑞𝜒(4)^(𝑞+1)/4_𝑎=1𝑎+𝑞−34𝜒𝑎+𝑞−34−2𝑞²𝜒(4)^(𝑞+1)/4_𝑎=1𝜒𝑎+𝑞−34+9𝑞²𝜒(4)^(𝑞−3)/4_𝑎=1𝜒𝑎+3𝑞−14+4𝑞²𝜒(4)^(𝑞+1)/4_𝑎=1𝜒𝑎+𝑞−12+𝑞²𝜒(4)^(𝑞+1)/4_𝑎=1𝜒𝑎+𝑞−34=^𝑞−1_𝑎=1𝑎²𝜒(𝑎)+24𝑞𝜒(4)^𝑞−1_{𝑎=(3𝑞+3)/4}(𝑎)(4)𝑎𝜒+16𝑞𝜒^{(3𝑞−1)/4}_{𝑎=(𝑞+1)/2}(𝑎)(4)𝑎𝜒+8𝑞𝜒^(𝑞−1)/2_{𝑎=(𝑞+1)/4}𝑎𝜒(𝑎)−9𝑞²𝜒(4)^𝑞−1_{𝑎=(3𝑞+3)/4}𝜒(𝑎)−4𝑞²𝜒(4)^{(3𝑞−1)/4}_{𝑎=(𝑞+1)/2}𝜒(𝑎)−𝑞²𝜒(4)^(𝑞−1)/2_{𝑎=(𝑞+1)/4}𝜒(𝑎). (59)
Note that ∑_{1≤𝑎≤(𝑞−1)/2}𝜒(𝑎)=0 and ∑^𝑞_𝑎=1𝑎𝜒(𝑎)=0 for even character 𝜒. By Lemma 12 we have 16𝜒(4)^𝑞−1_𝑎=1𝑎²𝜒(𝑎)=^𝑞_𝑎=1𝑎²𝜒(𝑎)−16𝑞𝜒(4)^(𝑞−3)/4_𝑎=1(𝑎)(4)𝑎𝜒−8𝑞𝜒^(𝑞−1)/2_𝑎=1𝑎𝜒(𝑎)+4𝑞²𝜒(4)^(𝑞−3)/4_𝑎=1𝜒(𝑎)=8𝜒(4)−2𝜒(2)+1^𝑞_𝑎=1𝑎²𝜒(𝑎)−16𝑞𝜒(4)^(𝑞−3)/4_𝑎=1𝑎𝜒(𝑎)+4𝑞²𝜒(4)^(𝑞−3)/4_𝑎=1𝜒(𝑎). (60)

Now we express ∑_{1≤𝑎≤𝑞/4}𝑎𝜒(𝑎) in terms of Gauss sums and Dirichlet 𝐿-functions.

Theorem 14. Let 𝜒 be an even primitive character modulo odd integer 𝑞≥5. Then one has _{1≤𝑎≤𝑞/4}(𝑎)=𝑞𝑎𝜒16𝜋²𝜒(4)−2𝜒(2)−8𝜏𝜒𝐿2,𝜒+𝑞𝜏𝜒𝐿4𝜋1,𝜒𝜒₄. (61)

Proof. By Lemma 13, (3), and (37), we have 16𝑞𝜒(4)_{1≤𝑎≤𝑞/4}(𝑎)=(2)(4)𝑎𝜒1−2𝜒−8𝜒^𝑞_𝑎=1𝑎²𝜒(𝑎)+4𝑞²𝜒(4)_{1≤𝑎≤𝑞/4}𝜒(𝑎)=𝑞²𝜋²1−2𝜒(2)−8𝜒(4)𝜏𝜒𝐿2,𝜒+4𝑞²𝜋𝜒(4)𝜏𝜒𝐿1,𝜒𝜒₄. (62) Therefore _{1≤𝑎≤𝑞/4}(𝑎)=𝑞𝑎𝜒16𝜋²𝜒(4)−2𝜒(2)𝜒𝐿−8×𝜏2,𝜒+𝑞𝜏𝜒𝐿4𝜋1,𝜒𝜒₄. (63)

4. Mean Values of Dirichlet 𝐿-Functions

In this section, we will study the mean values of Dirichlet 𝐿-functions, which will be used to prove Theorems 3 and 4.

Lemma 15. Let 𝑞 and 𝑟 be integers with 𝑞≥2 and (𝑟,𝑞)=1. Then one has the identities ^∗_𝜒mod𝑞𝜒(𝑟)=_{𝑑∣(𝑞,𝑟−1)}𝜇𝑞𝑑𝜙(𝑑),𝐽𝑞=_𝑑∣𝑞𝜇𝑞𝑑𝜙(𝑑), (64) where ∑^∗_𝜒mod𝑞 denotes the summation over all primitive characters modulo 𝑞, and 𝐽(𝑞) is the number of primitive characters modulo 𝑞.

Proof. This is Lemma 3 of [12].

Lemma 16. Let 𝑞≥2 be an odd number, and let 𝑘≥0 be an integer. Then one has ^∞_{𝑛=1(𝑛,𝑞)=1}𝑑(𝑛)𝑑2^𝑘𝑛𝑛²=5+3𝑘5⋅𝜁⁴(2)𝜁(4)_𝑝∣𝑞1−1/𝑝²³1+1/𝑝²_{∞𝑛=1(𝑛,𝑞)=1},𝑑(𝑛)𝑑2^𝑘𝑛𝑛⁴=15𝑘+17⋅𝜁17⁴(4)𝜁(8)_𝑝∣𝑞1−1/𝑝⁴³1+1/𝑝⁴_{∞𝑛=1(𝑛,2𝑞)=1},𝑑²(𝑛)𝑛²=27⋅𝜁80⁴(2)𝜁(4)_𝑝∣𝑞1−1/𝑝²³1+1/𝑝²_{∞𝑛=1(𝑛,2𝑞)=1},𝑑²(𝑛)𝑛⁴=3375⋅𝜁4352⁴(4)𝜁(8)_𝑝∣𝑞1−1/𝑝⁴³1+1/𝑝⁴_{∞𝑛=1(𝑛,2𝑞)=1},𝑑²(𝑛)𝜒₄(𝑛)𝑛³=11−1/2⁶⋅𝐿⁴3,𝜒₄𝜁(6)_𝑝∣𝑞1−𝜒₄𝑝/𝑝³³1+𝜒₄𝑝/𝑝³_{∞𝑛=1(𝑛,𝑞)=1},𝑑(𝑛)𝜏₁2^𝑘𝑛𝑛²=𝜁²(2)𝐿²3,𝜒₄𝐿5,𝜒₄×_𝑝∣𝑞1−1/𝑝²²1−𝜒₄𝑝/𝑝³²1−𝜒₄𝑝/𝑝⁵_{∞𝑛=1(𝑛,𝑞)=1},𝜏₁(𝑛)𝜏₁2^𝑘𝑛𝑛²=𝜁(4)𝐿3,𝜒_4𝑝∣𝑞𝜒1−⁴𝑝𝑝³11−𝑝⁴×_𝑝∤𝑞11+𝑝²1−𝜒₄𝑝,/𝑝_{∞𝑛=1(𝑛,𝑞)=1}𝑑2^𝑘𝑛𝜏₂(𝑛)𝑛³=15⋅𝜁16𝑘+1²(4)𝐿²3,𝜒₄𝐿7,𝜒₄×_𝑝∣𝑞1−𝜒₄𝑝/𝑝³²1−1/𝑝⁴²1−𝜒₄𝑝/𝑝⁷_{∞𝑛=1(𝑛,𝑞)=1},𝑑(𝑛)𝜏₂2^𝑘𝑛𝑛³=12^𝑘⋅𝜁²(4)𝐿²3,𝜒₄𝐿7,𝜒₄×_𝑝∣𝑞1−𝜒₄𝑝/𝑝³²1−1/𝑝⁴²1−𝜒₄𝑝/𝑝⁷_{∞𝑛=1(𝑛,2𝑞)=1},𝜒₄(𝑛)𝑑(𝑛)𝜏₂(𝑛)𝑛²=9⋅𝜁16²(2)𝐿²3,𝜒₄𝐿5,𝜒₄×_𝑝∣𝑞1−1/𝑝²²1−𝜒₄𝑝/𝑝³²1−𝜒₄𝑝/𝑝⁵_{∞𝑛=1(𝑛,𝑞)=1},𝜏₂(𝑛)𝜏₂2^𝑘𝑛𝑛²=42^𝑘5⋅𝜁(4)𝐿3,𝜒₄×_𝑝∣𝑞𝜒1−⁴𝑝𝑝³11−𝑝⁴×_𝑝∤𝑞11+𝑝²1−𝜒₄𝑝,/𝑝 (65) where 𝑑(𝑛)=∑_𝑑∣𝑛1, 𝜏₁(𝑛)=∑_𝑑∣𝑛(𝜒₄(𝑑)/𝑑), and 𝜏₂(𝑛)=∑_𝑑∣𝑛(𝜒₄(𝑛/𝑑)/𝑑).

Proof. By the Euler product, we have ^∞_{𝑛=1𝑛,𝑞=1}𝑑(𝑛)𝑑2^𝑘𝑛𝑛²=^∞∞_{𝑖=0𝑚=1𝑚,2𝑞=1}𝑑2^𝑖𝑚𝑑2^𝑘+𝑖𝑚2^𝑖𝑚²=^∞_𝑖=0𝑑2^𝑖𝑑2^𝑘+𝑖2^∞2𝑖_{𝑚=1𝑚,2𝑞=1}𝑑²(𝑚)𝑚²=^∞_𝑖=0(𝑖+1)(𝑘+𝑖+1)2^2𝑖_𝑝∤2𝑞^∞_𝑗=0𝑗+1²𝑝^2𝑗=5/4+(3/4)𝑘1−1/2²³_𝑝∤2𝑞1+1/𝑝²1−1/𝑝²³=5+3𝑘5_𝑝1+1/𝑝²1−1/𝑝²³_𝑝∣𝑞1−1/𝑝²³1+1/𝑝²=5+3𝑘5⋅𝜁⁴(2)𝜁(4)_𝑝∣𝑞1−1/𝑝²³1+1/𝑝². (66) Similarly, we can deduce the other identities.

Lemma 17. Let 𝑞≥2 be an odd number. For integers 𝑘≥0 and 𝑙≥1, one has ^∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒2^𝑘𝐿1,𝜒⁴=5+3𝑘2^𝑘+1⋅𝜁⋅5⁴(2)𝜁(4)𝐽𝑞_𝑝∤𝑞1−1/𝑝²³1+1/𝑝²+𝑂𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=−1}𝐿2,𝜒𝜒₄⁴=3375⋅𝜁8704⁴(4)𝜁(8)𝐽𝑞_𝑝∣𝑞1−1/𝑝⁴³1+1/𝑝⁴+𝑂𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=−1}𝜒2^𝑘𝐿²1,𝜒𝐿1,𝜒𝐿2,𝜒𝜒₄=𝜁²(2)𝐿²3,𝜒₄2^𝑘+1𝐿5,𝜒₄𝐽𝑞×_𝑝∣𝑞1−1/𝑝²²1−𝜒₄𝑝/𝑝³²1−𝜒₄𝑝/𝑝⁵+𝑂𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=−1}𝜒2^𝑙𝐿²1,𝜒𝐿²2,𝜒𝜒₄≪𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=−1}𝜒2^𝑙𝐿1,𝜒𝐿2,𝜒𝜒₄𝐿²2,𝜒𝜒₄≪𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=−1}𝜒2^𝑘𝐿1,𝜒²𝐿2,𝜒𝜒₄²=𝜁(4)𝐿3,𝜒₄2^𝑘+1𝐽𝑞_𝑝∣𝑞𝜒1−⁴𝑝𝑝³11−𝑝⁴×_𝑝∤𝑞11+𝑝²1−𝜒₄𝑝𝑞/𝑝+𝑂^𝜖. (67)

Proof. We only prove the first formula since, similarly, we can get the others. Let 𝑑(𝑛)=∑_𝑑∣𝑛1 be the divisor function. For 𝑁≥𝑞², by Abel’s identity, we get 𝐿²1,𝜒=^∞_𝑛=1𝜒(𝑛)𝑛²=_{1≤𝑛≤𝑁}𝜒(𝑛)𝑑(𝑛)𝑛𝑞+𝑂^1/2log𝑞log𝑁𝑁^1/2. (68)
For (𝑟,𝑞)=1, from Lemma 15, we have ^∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(𝑟)=12^∗_𝜒mod𝑞()𝜒(𝑟)=11−𝜒−12^∗_𝜒mod𝑞𝜒(𝑟)−12^∗_𝜒mod𝑞𝜒()=1−𝑟2_{𝑑∣(𝑞,𝑟−1)}𝜇𝑞𝑑𝜙(𝑑)−12_{𝑑∣(𝑞,𝑟+1)}𝜇𝑞𝑑𝜙(𝑑). (70) Then from Lemma 16 we get^∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒2^𝑘𝐿1,𝜒⁴=^∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒2^𝑘_{1≤𝑛≤𝑁}𝜒(𝑛)𝑑(𝑛)𝑛𝑞+𝑂^1/2log𝑞log𝑁𝑁^1/22=_{1≤𝑛≤𝑁1≤𝑚≤𝑁}𝑑(𝑛)𝑑(𝑚)𝑛𝑚^∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒2^𝑘𝑛𝜒(𝑚)𝑞+𝑂^3/2log𝑞log³𝑁𝑁^1/2=12_{1≤𝑛≤𝑁(𝑛,𝑞)=11≤𝑚≤𝑁(𝑚,𝑞)=1}𝑑(𝑛)𝑑(𝑚)𝑛𝑚_{𝑑|(𝑞,2𝑘𝑛−𝑚)}𝜇𝑞𝑑𝜙(𝑑)−12_{1≤𝑛≤𝑁(𝑛,𝑞)=11≤𝑚≤𝑁(𝑚,𝑞)=1}𝑑(𝑛)𝑑(𝑚)𝑛𝑚_{𝑑∣(𝑞,2𝑘𝑛+𝑚)}𝜇𝑞𝑑𝜙(𝑑)𝑞+𝑂^3/2log𝑞log³𝑁𝑁^1/2=𝐽𝑞2^𝑘+1_{1≤𝑛≤𝑁/2𝑘(𝑛,𝑞)=1}𝑑(𝑛)𝑑2^𝑘𝑛𝑛²+12_𝑑∣𝑞𝜇𝑞𝑑𝜙(𝑑)_{1≤𝑛≤𝑁(𝑛,𝑞)=121≤𝑚≤𝑁(𝑚,𝑞)=1𝑘2𝑛≡𝑚(mod𝑑)𝑘𝑛≠𝑚}𝑑(𝑛)𝑑(𝑚)−1𝑛𝑚2_𝑑|𝑞𝜇𝑞𝑑𝜙(𝑑)_{1≤𝑛≤𝑁(𝑛,𝑞)=121≤𝑚≤𝑁(𝑚,𝑞)=1𝑘𝑛≡−𝑚(mod𝑑)}𝑑(𝑛)𝑑(𝑚)𝑞𝑛𝑚+𝑂^3/2log𝑞log³𝑁𝑁^1/2=𝐽𝑞2^∞𝑘+1_{𝑛=1(𝑛,𝑞)=1}𝑑(𝑛)𝑑2^𝑘𝑛𝑛²+𝑂_𝑑∣𝑞𝜙(𝑑)_{1≤𝑛≤𝑁(1−2𝑘𝑛)/𝑑≤𝑙≤(𝑁−2𝑘𝑛)/𝑑}𝑑(𝑛)𝑑𝑙𝑑+2^𝑘𝑛𝑛𝑙𝑑+2^𝑘𝑛+𝑂_𝑑∣𝑞𝜙(𝑑)_{1≤𝑛≤𝑁(1+2𝑘𝑛)/𝑑≤𝑙≤(𝑁+2𝑘𝑛)/𝑑}𝑑(𝑛)𝑑𝑙𝑑−2^𝑘𝑛𝑛𝑙𝑑−2^𝑘𝑛𝑞+𝑂^3/2log𝑞log³𝑁𝑁^1/2=𝐽𝑞2^∞𝑘+1_{𝑛=1(𝑛,𝑞)=1}𝑑(𝑛)𝑑2^𝑘𝑛𝑛²+𝑂𝑁^𝜖𝑞+𝑂^3/2log𝑞log³𝑁𝑁^1/2=5+3𝑘2^𝑘+1⋅𝜁⋅5⁴(2)𝜁(4)𝐽𝑞_𝑝∣𝑞1−1/𝑝²³1+1/𝑝²+𝑂𝑁^𝜖𝑞+𝑂^3/2log𝑞log³𝑁𝑁^1/2. (71)Now taking 𝑁=𝑞⁴, we immediately get ^∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒2^𝑘𝐿1,𝜒⁴=5+3𝑘2^𝑘+1⋅𝜁⋅5⁴(2)𝜁(4)𝐽𝑞×_𝑝∣𝑞1−1/𝑝²³1+1/𝑝²+𝑂𝑞^𝜖. (72)

Lemma 18. Let 𝑞≥2 be an odd number. For integers 𝑘≥0 and 𝑙≥1, one has ^∗_{𝜒()𝜒mod𝑞−1=1}𝜒2^𝑘𝐿2,𝜒⁴=15𝑘+172^2𝑘+1⋅𝜁⋅17⁴(4)𝜁(8)⋅𝐽𝑞_𝑝∣𝑞1−1/𝑝⁴³1+1/𝑝⁴+𝑂𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=1}𝐿²2,𝜒𝐿²1,𝜒𝜒₄=121−1/2⁶⋅𝐿⁴3,𝜒₄𝜁(6)⋅𝐽𝑞_𝑝∣𝑞1−𝜒₄𝑝/𝑝³³1+𝜒₄𝑝/𝑝³+𝑂𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=1}𝜒2^𝑙𝐿²2,𝜒𝐿²1,𝜒𝜒₄≪𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=1}𝜒2^𝑘𝐿²2,𝜒𝐿1,𝜒𝜒₄𝐿=))2,𝜒((15/16𝑘+12^2𝑘+1⋅𝜁²(4)𝐿²3,𝜒₄𝐿7,𝜒₄⋅𝐽𝑞_𝑝∣𝑞1−𝜒₄𝑝/𝑝³²1−1/𝑝⁴²1−𝜒₄𝑝/𝑝⁷+𝑂𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=1}𝜒2^𝑘𝐿²2,𝜒𝐿1,𝜒𝜒₄𝐿=12,𝜒2^𝑘+1⋅𝜁²(4)𝐿²3,𝜒₄𝐿7,𝜒₄⋅𝐽𝑞_𝑝∣𝑞1−𝜒₄(𝑝)/𝑝³²1−1/𝑝⁴²1−𝜒₄𝑝/𝑝⁷+𝑂𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=1}𝐿(1,𝜒𝜒₄)⁴=27⋅𝜁160⁴(2)𝜁(4)⋅𝐽𝑞_𝑝∣𝑞1−1/𝑝²³1+1/𝑝²+𝑂𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=1}𝜒2^𝑙𝐿²1,𝜒𝜒₄𝐿1,𝜒𝜒₄𝐿2,𝜒≪𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=1}𝐿²1,𝜒𝜒₄𝐿1,𝜒𝜒₄𝐿=92,𝜒⋅𝜁32²(2)𝐿²3,𝜒₄𝐿5,𝜒₄⋅𝐽𝑞_𝑝∣𝑞1−1/𝑝²²1−𝜒₄𝑝/𝑝³²1−𝜒₄𝑝/𝑝⁵+𝑂𝑞^𝜖,_{∗𝜒mod𝑞𝜒(−1)=1}𝜒2^𝑘𝐿1,𝜒𝜒₄²𝐿2,𝜒²=22^2𝑘5⋅𝜁(4)𝐿3,𝜒₄⋅𝐽𝑞_𝑝∣𝑞1−𝜒₄𝑝/𝑝³1−1/𝑝⁴×_𝑝∤𝑞11+𝑝²1−𝜒₄𝑝𝑞/𝑝+𝑂^𝜖. (73)

Proof. By Lemma 17 and the methods proving Lemma 18, we can get this lemma.

5. Proof of Theorems 3 and 4

First we prove Theorem 3. By Theorem 11 and Lemma 18, we have 𝑎𝜒(^∗_{𝜒mod𝑞𝜒(−1)=−11≤𝑎≤𝑞/4}𝑎)⁴=𝑞⁶8⁴𝜋^4∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(2)𝐿1,𝜒−𝜒(4)𝐿1,𝜒+4𝜋𝐿2,𝜒𝜒₄⁴=𝑞⁶8⁴𝜋^4∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(4)𝐿²1,𝜒+𝜒(16)𝐿²1,𝜒+16𝜋²𝐿²2,𝜒𝜒₄−2𝜒(8)𝐿²1,𝜒+8𝜋𝜒(2)𝐿1,𝜒𝐿2,𝜒𝜒₄−8𝜋𝜒(4)𝐿1,𝜒𝐿2,𝜒𝜒₄²=32048𝜋⁴𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝐿1,𝜒⁴−1512𝜋⁴𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(2)𝐿1,𝜒⁴+12048𝜋⁴𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(4)𝐿1,𝜒⁴+116𝜋⁸𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝐿2,𝜒𝜒₄⁴−1256𝜋⁵𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝐿²1,𝜒𝐿1,𝜒𝐿2,𝜒𝜒₄+3256𝜋⁵𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(2)𝐿²1,𝜒𝐿1,𝜒𝐿2,𝜒𝜒₄−3256𝜋⁵𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(4)𝐿²1,𝜒𝐿1,𝜒𝐿2,𝜒𝜒₄+1256𝜋⁵𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(8)𝐿²1,𝜒𝐿1,𝜒𝐿2,𝜒𝜒₄+1128𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(4)𝐿²1,𝜒𝐿²2,𝜒𝜒₄−164𝜋⁶𝑞^6∗_{𝜒()𝜒mod𝑞−1=−1}𝜒(8)𝐿²1,𝜒𝐿²2,𝜒𝜒₄+1128𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(16)𝐿²1,𝜒𝐿²2,𝜒𝜒₄+116𝜋⁷𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(2)𝐿1,𝜒𝐿2,𝜒𝜒₄𝐿²2,𝜒𝜒₄−116𝜋⁷𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(4)𝐿1,𝜒𝐿2,𝜒𝜒₄𝐿²2,𝜒𝜒₄+132𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝐿1,𝜒²𝐿2,𝜒𝜒₄²−132𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=−1}𝜒(2)𝐿1,𝜒²𝐿2,𝜒𝜒₄²=72¹⁷⋅3²𝑞⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝²³1+1/𝑝²+352¹⁶⋅3²𝑞⋅17⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝⁴³1+1/𝑝⁴−𝐿²3,𝜒₄2¹⁴⋅3²⋅𝜋⋅𝐿5,𝜒₄×𝑞⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝²²1−𝜒₄(𝑝)/𝑝³²1−𝜒₄𝑝/𝑝⁵+𝐿3,𝜒₄2⁸⋅5⋅3²⋅𝜋²𝑞⁶𝐽𝑞×_𝑝∣𝑞𝜒1−⁴𝑝𝑝³11−𝑝⁴×_𝑝∤𝑞11+𝑝²1−𝜒₄𝑝𝑞/𝑝+𝑂^6+𝜖. (74) This proves Theorem 3.

On the other hand, by Theorem 14 and Lemma 18, we have 𝑎𝜒(^∗_{𝜒mod𝑞𝜒(−1)=11≤𝑎≤𝑞/4}𝑎)⁴=𝑞⁶16⁴𝜋^8∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(4)−2𝜒(2)−8𝐿2,𝜒+4𝜋𝐿1,𝜒𝜒₄⁴=𝑞⁶16⁴𝜋^8∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(4)−2𝜒(2)−8²𝐿²2,𝜒+16𝜋²𝐿²1,𝜒𝜒₄+8𝜋𝜒(4)−2𝜒(2)−8×𝐿1,𝜒𝜒₄𝐿2,𝜒²=528165536𝜋⁸𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝐿2,𝜒⁴+4278192𝜋⁸𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(2)𝐿2,𝜒⁴−2278192𝜋⁸𝑞^6∗_{𝜒()𝜒mod𝑞−1=1}𝜒(4)𝐿2,𝜒⁴−71024𝜋⁸𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(8)𝐿2,𝜒⁴+1512𝜋⁸𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(16)𝐿(2,𝜒)⁴+132𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝐿²2,𝜒𝐿²1,𝜒𝜒₄+164𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(2)𝐿²2,𝜒𝐿²1,𝜒𝜒₄−3512𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(4)𝐿²2,𝜒𝐿²1,𝜒𝜒₄−1512𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(8)𝐿²2,𝜒𝐿²1,𝜒𝜒₄+12048𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(16)𝐿²2,𝜒𝐿²1,𝜒𝜒₄+164𝜋⁷𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(4)𝐿²2,𝜒𝐿1,𝜒𝜒₄𝐿−32,𝜒128𝜋⁷𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(2)𝐿²2,𝜒𝐿1,𝜒𝜒₄𝐿−2,𝜒1471024𝜋⁷𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝐿²2,𝜒𝐿1,𝜒𝜒₄𝐿−2,𝜒591024𝜋⁷𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(2)𝐿²2,𝜒𝐿1,𝜒𝜒₄𝐿+2,𝜒1054096𝜋⁷𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(4)𝐿²2,𝜒𝐿1,𝜒𝜒₄𝐿+2,𝜒152048𝜋⁷𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(8)𝐿²2,𝜒𝐿1,𝜒𝜒₄𝐿−12,𝜒512𝜋⁷𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(16)𝐿²2,𝜒𝐿1,𝜒𝜒₄𝐿+12,𝜒256𝜋⁴𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝐿1,𝜒𝜒₄⁴+1256𝜋⁵𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(4)𝐿²1,𝜒𝜒₄𝐿1,𝜒𝜒₄𝐿−12,𝜒128𝜋⁵𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(2)𝐿²1,𝜒𝜒₄𝐿1,𝜒𝜒₄𝐿−12,𝜒32𝜋⁵𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝐿²1,𝜒𝜒₄𝐿1,𝜒𝜒₄𝐿+2,𝜒691024𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝐿1,𝜒𝜒₄²𝐿2,𝜒²+7256𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(2)𝐿1,𝜒𝜒₄²𝐿2,𝜒²−164𝜋⁶𝑞^6∗_{𝜒mod𝑞𝜒(−1)=1}𝜒(4)𝐿1,𝜒𝜒₄²𝐿2,𝜒²=3852²⁰⋅51⋅𝑞⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝⁴³1+1/𝑝⁴+15𝐿⁴3,𝜒₄𝜋¹²⋅𝑞⁶𝐽𝑞_𝑝∣𝑞1−𝜒₄𝑝/𝑝³³1+𝜒₄𝑝/𝑝³−1423𝜋𝐿²3,𝜒₄2¹⁶⋅5²⋅3⁴⋅𝐿7,𝜒₄⋅𝑞⁶𝐽𝑞_𝑝∣𝑞1−𝜒₄𝑝/𝑝³²1−1/𝑝⁴²1−𝜒₄𝑝/𝑝⁷+32¹⁶⋅𝑞⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝²³1+1/𝑝²−𝐿²3,𝜒₄2¹²⋅𝜋⋅𝐿5,𝜒₄⋅𝑞⁶𝐽𝑞_𝑝∣𝑞1−1/𝑝²²1−𝜒₄𝑝/𝑝³²1−𝜒₄𝑝/𝑝⁵+𝐿3,𝜒₄2¹⁰⋅3⋅𝜋²⋅𝑞⁶𝐽𝑞_𝑝∣𝑞1−𝜒₄𝑝/𝑝³1−1/𝑝⁴×_𝑝∤𝑞11+𝑝²1−𝜒₄𝑝𝑞/𝑝+𝑂^6+𝜖. (75) This completes the proof of Theorem 4.

Acknowledgments

This work is supported by National Natural Science Foundation of China under Grant no. 11001218, the Natural Science Foundation of Shaanxi Province of China under Grant no. 2013JM1017 and 2011JQ1010, and the Natural Science Foundation of the Education Department of Shaanxi Province of China under Grant no. 2013JK0558.