# On the Mean Values of Certain Character Sums

Authors: Zhefeng Xu; Huaning Liu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Letq≥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/4aχa4 and  ∑χmodq,χ-1=1*∑1≤a≤q/4aχa4, where ∑‍* denotes the summation over primitive characters modulo q, and give some asymptotic formulae.


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## Body

## 1. Introduction

The sum(1)Sχ(n)=1qn∑a=1q‍anχ(a)
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 Sχ(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 Sχ(n)>0 and infinitely many with Sχ(n)<0.Williams [3] proved that
(2)Sχ(n)=O(p1/2logp)
for χ being the Legendre symbol modulo p. For primitive character χ modulo q, Toyoizumi [4] used the generalized Bernoulli numbers to express Sχ(n) in terms of Gauss sums and Dirichlet L-functions as follows:
(3)∑a=1q‍anχ(a)={2qnτ(χ)×∑1≤m≤n/2‍(n2m-1)(2m-1)!L(2m,χ¯)(-1)m+1(2π)2m,ifχ(-1)=1,2qnτ(χ)×∑0≤m≤(n-1)/2‍(n2m)(2m)!L(2m+1,χ¯)(-1)m+1(2π)2m+1i,ifχ(-1)=-1,
where τ(χ)=∑a=1q‍χ(a)e(a/q) is the Gauss sum, e(y)=e2πiy, L(s,χ) is the Dirichlet L-function corresponding to χ, and (nm) denotes the binomial coefficient.Toyoizumi [4] also gave explicit bounds for Sχ(n).Proposition 1.
(a) Assume thatχ(-1)=1 and n≥2. Then for any primitive character χmodq, one has
(4)|Sχ(n)|<C1(n)q1/2,
where
(5)C1(n)=2ζ(2)n!(2π)n+1∑1≤m≤n/2‍(2π)n+1-2m(n+1-2m)!,
and ζ(s) is the Riemann zeta function.
(b) Assume thatχ(-1)=-1 and n≥3. Then for any primitive character χmodq, one has
(6)|Sχ(n)|<(C2(n)+|L(1,χ)|π)q1/2,
where
(7)C2(n)=2ζ(3)n!(2π)n+1∑1≤m≤(n-1)/2‍(2π)n-2m(n-2m)!.In [5], Peral used the Gauss sums and adequate Fourier expansion to greatly improve the result in Proposition 1.Proposition 2.
(a) Assume thatχ(-1)=1 is a primitive nonprincipal character modulo q, and then
(8)|Sχ(n)|≤q1/2(n-12(n+1)).
(b) Assume thatχ(-1)=-1 is a primitive character modulo q; then,
(9)|Sχ(n)+τ(χ)L(1,χ¯)πi|≤q1/2(nπ∫01‍ln12sin(πt)tn-1dt).Furthermore, Liu and Zhang [6] gave an upper bound for Sχ(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
(10)∑‍*χmodq∑n=1q‍|∑m=1h‍χ(n+m)|4≤8d2(q)q2h2,
where ∑‍* denotes the summation over primitive characters modulo q, and d(q) is the Dirichlet divisor function. Xu and Zhang studied the power mean
(11)∑‍*χmodqχ(-1)=±1|∑1≤a≤q/4‍χ(a)|4
in [8, 9] and obtained some sharper results.In this paper, we study the fourth power mean of certain character sums(12)∑‍*χmodqχ(-1)=-1|∑1≤a≤q/4‍aχ(a)|4,∑‍*χmodqχ(-1)=1|∑1≤a≤q/4‍aχ(a)|4
and give a few asymptotic formulae.Theorem 3.
Letq≥5 be an odd number. Then one has
(13)∑‍*χmodqχ(-1)=-1|∑1≤a≤q/4‍aχ(a)|4=7217·32q6J(q)∏p∣q‍(1-1/p2)31+1/p2+35216·32·17q6J(q)∏p∣q‍(1-1/p4)31+1/p4-L2(3,χ4)214·32·π·L(5,χ4)×q6J(q)∏p∣q‍(1-1/p2)2(1-χ4(p)/p3)21-χ4(p)/p5+L(3,χ4)28·5·32·π2q6J(q)×∏p∣q‍(1-χ4(p)/p3)(1-1/p4)×∏p∤q‍(1+1p2(1-χ4(p)/p))+O(q6+ϵ),
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.
Letq≥5 be an odd number. Then one has
(14)∑‍*χmodqχ(-1)=1|∑1≤a≤q/4‍aχ(a)|4=385220·51·q6J(q)∏p∣q‍(1-1/p4)31+1/p4+15L4(3,χ4)π12·q6J(q)∏p∣q‍(1-χ4(p)/p3)31+χ4(p)/p3-1423πL2(3,χ4)216·52·34·L(7,χ4)·q6J(q)∏p∣q‍(1-χ4(p)/p3)2(1-1/p4)21-χ4(p)/p7+3216·q6J(q)∏p∣q‍(1-1/p2)31+1/p2-L2(3,χ4)212·π·L(5,χ4)·q6J(q)∏p∣q‍(1-1/p2)2(1-χ4(p)/p3)21-χ4(p)/p5+L(3,χ4)210·3·π2·q6J(q)×∏p∣q‍(1-χ4(p)/p3)(1-1/p4)×∏p∤q‍(1+1p2(1-χ4(p)/p))+O(q6+ϵ).From Theorems3 and 4, we immediately get the following corollaries.Corollary 5.
Letp≥5 be a prime. Then one has
(15)∑χmodp‍χ(-1)=-1|∑1≤a≤p/4‍aχ(a)|4=21217·17p7-L2(3,χ4)214·32·π·L(5,χ4)p7+L(3,χ4)28·5·32·π2p7∏p1‍(1+1p12(1-χ4(p1)/p1))+O(p6+ϵ),
where ∏p1‍ denotes the product over all primes.Corollary 6.
Letp≥5 be a prime. Then
(16)∑χmodpχ(-1)=1χ≠χ0|∑1≤a≤p/4‍aχ(a)|4=2833220·51p7+15L4(3,χ4)π12p7-1423πL2(3,χ4)216·52·34·L(7,χ4)p7-L2(3,χ4)212·π·L(5,χ4)p7+L(3,χ4)210·3·π2p7∏p1‍(1+1p12(1-χ4(p1)/p1))+O(p6+ϵ).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 andL-Functions (I)

Letχ be an odd primitive character modulo q. In this section, we will express ∑1≤a≤q/4‍aχ(a) in terms of Gauss sums and Dirichlet L-functions. We need the following lemmas.Lemma 8.
Suppose thatq≥5 is an odd number, and χ is an odd character modulo q.(i)
Forq≡1(mod4), one has
(17)∑1≤a≤(3q-3)/4‍aχ(a)=∑1≤a≤q‍aχ(a)+q∑1≤a≤(q-1)/4‍χ(a)-∑1≤a≤(q-1)/4‍aχ(a),∑1≤a≤(3q-3)/4‍χ(a)=∑1≤a≤(q-1)/4‍χ(a).(ii)
Forq≡3(mod4), one has
(18)∑1≤a≤(3q-1)/4‍aχ(a)=∑1≤a≤q‍aχ(a)+q∑1≤a≤(q-3)/4‍χ(a)-∑1≤a≤(q-3)/4‍aχ(a),∑1≤a≤(3q-1)/4‍χ(a)=∑1≤a≤(q-3)/4‍χ(a).Proof.
It is easy to show that(19)∑1≤a≤(3q-3)/4‍aχ(a)=∑1≤a≤q‍aχ(a)-∑(3q+1)/4≤a≤q‍aχ(a)=∑1≤a≤q‍aχ(a)-∑1≤b≤(q-1)/4‍(q-b)χ(q-b)=∑1≤a≤q‍aχ(a)+q∑1≤a≤(q-1)/4‍χ(a)-∑1≤a≤(q-1)/4‍aχ(a),∑1≤a≤(3q-3)/4‍χ(a)=∑1≤a≤q‍χ(a)-∑(3q+1)/4≤a≤q‍χ(a)=∑1≤a≤(q-1)/4χ(a).
This proves (i). Similarly, we can deduce (ii).Lemma 9.
Suppose thatq≥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
(20)∑a=14q‍a2χ(a)χ4(a)=16χ(4)q∑a=1q‍aχ(a)+16χ(4)q2∑1≤a≤(q-1)/4‍χ(a)-64χ(4)q∑1≤a≤(q-1)/4‍aχ(a).Proof.
Note thatχ4(1)=1 and χ4(3)=-1, and we get
(21)∑a=14q‍a2χ(a)χ4(a)=∑a=0q-1‍(4a+1)2χ(4a+1)-∑a=0q-1‍(4a+3)2χ(4a+3).
First we have(22)∑a=0q-1‍(4a+1)2χ(4a+1)=16∑a=0q-1‍a2χ(4a+1)+8∑a=0q-1‍aχ(4a+1)+∑a=0q-1‍χ(4a+1)=16χ(4)∑a=0q-1‍a2χ(a+4¯)+8χ(4)∑a=0q-1‍aχ(a+4¯)=16χ(4)∑a=0q-1‍(a+4¯)2χ(a+4¯)+8χ(4)(1-4·4¯)∑a=0q-1‍(a+4¯)χ(a+4¯),
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
(23)∑a=0q-1‍(a+4¯)2χ(a+4¯)=∑0≤a≤(q-1)/4‍(a+4¯)2χ(a+4¯)+∑(q-1)/4<a≤q-1‍(a+4¯)2χ(a+4¯)=∑0≤a≤(q-1)/4‍(a+4¯)2χ(a+4¯)+∑(q-1)/4<a≤q-1‍(a+4¯-q)2χ(a+4¯-q)+2q∑(q-1)/4<a≤q-1‍(a+4¯-q)χ(a+4¯-q)+q2∑(q-1)/4<a≤q-1‍χ(a+4¯-q)=∑a=1q‍a2χ(a)+2q∑1≤a≤(3q-3)/4‍aχ(a)+q2∑1≤a≤(3q-3)/4‍χ(a)=∑a=1q‍a2χ(a)+2q∑a=1q‍aχ(a)+3q2∑1≤a≤(q-1)/4‍χ(a)-2q∑1≤a≤(q-1)/4‍aχ(a),∑a=0q-1‍(a+4¯)χ(a+4¯)=∑0≤a≤(q-1)/4‍(a+4¯)χ(a+4¯)+∑(q-1)/4<a≤q-1‍(a+4¯)χ(a+4¯)=∑0≤a≤(q-1)/4‍(a+4¯)χ(a+4¯)+∑(q-1)/4<a≤q-1‍(a+4¯-q)χ(a+4¯-q)+q∑(q-1)/4<a≤q-1‍χ(a+4¯-q)=∑a=1q‍aχ(a)+q∑(q-1)/4≤a≤q-1‍χ(a+4¯-q)=∑a=1q‍aχ(a)+q∑1≤a≤(3q-3)/4‍χ(a)=∑a=1q‍aχ(a)+q∑1≤a≤(q-1)/4‍χ(a).
Therefore
(24)∑a=0q-1‍(4a+1)2χ(4a+1)=16χ(4)∑a=0q-1‍(a+4¯)2χ(a+4¯)+8χ(4)(1-4·4¯)∑a=0q-1‍(a+4¯)χ(a+4¯)=16χ(4)∑a=1q‍a2χ(a)+8χ(4)q∑a=1q‍aχ(a)+24χ(4)q2∑1≤a≤(q-1)/4‍χ(a)-32χ(4)q∑1≤a≤(q-1)/4‍aχ(a).
On the other hand, we get(25)∑a=0q-1‍(4a+3)2χ(4a+3)=16∑a=0q-1‍a2χ(4a+3)+24∑a=0q-1‍aχ(4a+3)+9∑a=0q-1‍χ(4a+3)=16χ(4)∑a=0q-1‍a2χ(a+3·4¯)+24χ(4)∑a=0q-1‍aχ(a+3·4¯).
Since4¯=(3q+1)/4 and 3·4¯=2q+(q+3)/4, we have
(26)∑a=0q-1‍(4a+3)2χ(4a+3)=16χ(4)∑a=0q-1‍a2χ(a+q+34)+24χ(4)∑a=0q-1‍aχ(a+q+34)=16χ(4)∑a=0q-1‍(a+q+34)2χ(a+q+34)+24(1-q+33)χ(4)×∑a=0q-1‍(a+q+34)χ(a+q+34).
Note that(27)∑a=0q-1‍(a+q+34)2χ(a+q+34)=∑0≤a≤(3q-3)/4‍(a+q+34)2χ(a+q+34)+∑(3q+1)/4≤a≤q-1‍(a+q+34)2χ(a+q+34)=∑0≤a≤(3q-3)/4‍(a+q+34)2χ(a+q+34)+∑(3q+1)/4≤a≤q-1‍(a+q+34-q)2χ(a+q+34-q)+2q∑(3q+1)/4≤a≤q-1‍(a+q+34-q)χ(a+q+34-q)+q2∑(3q+1)/4≤a≤q-1‍χ(a+q+34-q)=∑a=1q‍a2χ(a)+q2∑1≤a≤(q-1)/4‍χ(a)+2q∑1≤a≤(q-1)/4‍aχ(a),∑a=0q-1‍(a+q+34)χ(a+q+34)=∑0≤a≤(3q-3)/4‍(a+q+34)χ(a+q+34)+∑(3q+1)/4≤a≤q-1‍(a+q+34)χ(a+q+34)=∑0≤a≤(3q-3)/4‍(a+q+34)χ(a+q+34)+∑(3q+1)/4≤a≤q-1‍(a+q+34-q)χ(a+q+34-q)+q∑(3q+1)/4≤a≤q-1‍χ(a+q+34-q)=∑a=1q‍aχ(a)+q∑1≤a≤(q-1)/4‍χ(a),
so we get
(28)∑a=0q-1‍(4a+3)2χ(4a+3)=16χ(4)∑a=0q-1‍(a+q+34)2χ(a+q+34)+24(1-q+33)χ(4)×∑a=0q-1‍(a+q+34)χ(a+q+34)=16χ(4)∑a=1q‍a2χ(a)-8χ(4)q∑a=1q‍aχ(a)+8χ(4)q2∑1≤a≤(q-1)/4‍χ(a)+32χ(4)q∑1≤a≤(q-1)/4‍aχ(a).
Now combine (21)–(28); we have
(29)∑a=14q‍a2χ(a)χ4(a)=∑a=0q-1‍(4a+1)2χ(4a+1)-∑a=0q-1‍(4a+3)2χ(4a+3)=16χ(4)q∑a=1q‍aχ(a)+16χ(4)q2∑1≤a≤(q-1)/4‍χ(a)-64χ(4)q∑1≤a≤(q-1)/4‍aχ(a).Lemma 10.
Suppose thatq≥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
(30)∑a=14q‍a2χ(a)χ4(a)=-16χ(4)q∑a=1q‍aχ(a)-16χ(4)q2∑1≤a≤(q-1)/4‍χ(a)+64χ(4)q∑1≤a≤(q-1)/4aχ(a).Proof.
Forq≡3(mod4), we get 4¯=(q+1)/4 and 3·4¯=(3q+3)/4. Using the methods of proving Lemma 9, we have
(31)∑a=14q‍a2χ(a)χ4(a)=∑a=0q-1‍(4a+1)2χ(4a+1)-∑a=0q-1‍(4a+3)2χ(4a+3),(32)∑a=0q-1‍(4a+1)2χ(4a+1)=16χ(4)∑a=0q-1‍(a+4¯)2χ(a+4¯)+8χ(4)(1-4·4¯)∑a=0q-1‍(a+4¯)χ(a+4¯),(33)∑a=0q-1‍(4a+3)2χ(4a+3)=16∑a=0q-1‍a2χ(4a+3)+24∑a=0q-1‍aχ(4a+3)+9∑a=0q-1‍χ(4a+3)=16χ(4)∑a=0q-1‍a2χ(a+3·4¯)+24χ(4)∑a=0q-1‍aχ(a+3·4¯)=16χ(4)∑a=0q-1‍(a+3q+34)2χ(a+3q+34)-24χ(4)q∑a=0q-1‍(a+3q+34)χ(a+3q+34).
It is not hard to show that(34)∑a=0q-1‍(a+4¯)2χ(a+4¯)=∑0≤a≤(3q-1)/4‍(a+4¯)2χ(a+4¯)+∑(3q-1)/4<a≤q-1‍(a+4¯)2χ(a+4¯)=∑0≤a≤(3q-1)/4‍(a+4¯)2χ(a+4¯)+∑(3q-1)/4<a≤q-1‍(a+4¯-q)2χ(a+4¯-q)+2q∑(3q-1)/4<a≤q-1‍(a+4¯-q)χ(a+4¯-q)+q2∑(3q-1)/4<a≤q-1‍χ(a+4¯-q)=∑a=1q‍a2χ(a)+q2∑1≤a≤(q-3)/4‍χ(a)+2q∑1≤a≤(q-3)/4‍aχ(a),∑a=0q-1‍(a+4¯)χ(a+4¯)=∑0≤a≤(3q-1)/4‍(a+4¯)χ(a+4¯)+∑(3q-1)/4<a≤q-1‍(a+4¯)χ(a+4¯)=∑0≤a≤(3q-1)/4‍(a+4¯)χ(a+4¯)+∑(3q-1)/4<a≤q-1‍(a+4¯-q)χ(a+4¯-q)+q∑(3q-1)/4<a≤q-1‍χ(a+4¯-q)=∑a=1q‍aχ(a)+q∑1≤a≤(q-3)/4‍χ(a).
Then by (32), we have
(35)∑a=0q-1‍(4a+1)2χ(4a+1)=16χ(4)∑a=0q-1‍(a+4¯)2χ(a+4¯)+8χ(4)(1-4×4¯)∑a=0q-1‍(a+4¯)χ(a+4¯)=16χ(4)∑a=1q‍a2χ(a)-8χ(4)q∑a=1q‍aχ(a)+8χ(4)q2∑1≤a≤(q-3)/4‍χ(a)+32χ(4)q∑1≤a≤(q-3)/4‍aχ(a).
On the other hand, by Lemma8, we get
(36)∑a=0q-1‍(a+3q+34)2χ(a+3q+34)=∑0≤a≤(q-3)/4‍(a+3q+34)2χ(a+3q+34)+∑(q+1)/4≤a≤q-1‍(a+3q+34)2χ(a+3q+34)=∑0≤a≤(q-3)/4‍(a+3q+34)2χ(a+3q+34)+∑(q+1)/4≤a≤q-1‍(a+3q+34-q)2χ(a+3q+34-q)+2q∑(q+1)/4≤a≤q-1‍(a+3q+34-q)χ(a+3q+34-q)+q2∑(q+1)/4≤a≤q-1‍χ(a+3q+34-q)=∑a=1q‍a2χ(a)+2q∑1≤a≤(3q-1)/4‍aχ(a)+q2∑1≤a≤(3q-1)/4‍χ(a)=∑a=1q‍a2χ(a)+2q∑a=1q‍aχ(a)+3q2∑1≤a≤(q-3)/4‍χ(a)-2q∑1≤a≤(q-3)/4‍aχ(a),∑a=0q-1‍(a+3q+34)χ(a+3q+34)=∑0≤a≤(q-3)/4‍(a+3q+34)χ(a+3q+34)+∑(q+1)/4≤a≤q-1‍(a+3q+34)χ(a+3q+34)=∑0≤a≤(q-3)/4‍(a+3q+34)χ(a+3q+34)+∑(q+1)/4≤a≤q-1‍(a+3q+34-q)χ(a+3q+34-q)+q∑(q+1)/4≤a≤q-1‍χ(a+3q+34-q)=∑a=1q‍aχ(a)+q∑1≤a≤(3q-1)/4‍χ(a)=∑a=1q‍aχ(a)+q∑1≤a≤(q-3)/4‍χ(a).
Then from (33), we have
(37)∑a=0q-1‍(4a+3)2χ(4a+3)=16χ(4)∑a=0q-1‍(a+3q+34)2χ(a+3q+34)-24χ(4)q∑a=0q-1‍(a+3q+34)χ(a+3q+34)=16χ(4)∑a=1q‍a2χ(a)+8χ(4)q∑a=1q‍aχ(a)+24χ(4)q2∑1≤a≤(q-3)/4‍χ(a)-32χ(4)q∑1≤a≤(q-3)/4‍aχ(a).
Combining (31), (35), and (37), we have
(38)∑a=14q‍a2χ(a)χ4(a)=∑a=0q-1‍(4a+1)2χ(4a+1)-∑a=0q-1‍(4a+3)2χ(4a+3)=-16χ(4)q∑a=1q‍aχ(a)-16χ(4)q2∑1≤a≤(q-3)/4‍χ(a)+64χ(4)q∑1≤a≤(q-3)/4‍aχ(a).Now we can express∑1≤a≤q/4‍aχ(a) in terms of Gauss sums and Dirichlet L-functions.Theorem 11.
Letχ be an odd primitive character modulo odd integer q≥5, and let χ4 be the nonprincipal character modulo 4. Then one has
(39)∑1≤a≤q/4‍aχ(a)=q8πiτ(χ)(4π4πL(2,χ¯χ4)χ¯(2)L(1,χ¯)-χ¯(4)L(1,χ¯)+4πL(2,χ¯χ4)).Proof.
By Lemmas9 and 10, we get
(40)16χ(4)q∑a=1q‍aχ(a)+16χ(4)q2∑1≤a≤q/4‍χ(a)-64χ(4)q∑1≤a≤q/4‍aχ(a)={∑a=14q‍a2χ(a)χ4(a),ifq≡1(mod4)-∑a=14q‍a2χ(a)χ4(a),ifq≡3(mod4)=χ4(q)∑a=14q‍a2χχ4(a).
From the Fourier expansion for primitive character sums (see [10] or [11])
(41)∑a<λq‍χ(a)={τ(χ)π∑n=1+∞‍χ¯(n)sin(2πnλ)n,ifχ(-1)=1,τ(χ)L(1,χ¯)πi-τ(χ)πi×∑n=1+∞‍χ¯(n)cos(2πnλ)n,ifχ(-1)=-1,
we easily have
(42)∑1≤a≤q/4‍χ(a)=2+χ¯(2)-χ¯(4)2πiτ(χ)L(1,χ¯).
Note thatχχ4 is a primitive character modulo 4q satisfying χχ4(-1)=1, and
(43)τ(χχ4)=∑a=14q‍χχ4(a)e(a4q)=∑a=14‍∑b=1q‍χ(4b+qa)χ4(4b+qa)e(4b+qa4q)=∑a=14‍∑b=1q‍χ(4b)χ4(qa)e(bq+a4)=χ(4)χ4(q)(∑a=14‍χ4(a)e(a4))×(∑b=1q‍χ(b)e(bq))=2iχ(4)χ4(q)τ(χ),
then from (3) we have
(44)∑a=1q‍aχ(a)=-qπiτ(χ)L(1,χ¯),∑a=14q‍a2χχ4(a)=16q2π2τ(χχ4)L(2,χ¯χ4)=32iχ(4)χ4(q)π2q2τ(χ)L(2,χ¯χ4).
Therefore
(45)-16χ(4)πiq2τ(χ)L(1,χ¯)+8χ(4)(2+χ¯(2)-χ¯(4))πiq2τ(χ)L(1,χ¯)-64χ(4)q∑1≤a≤q/4‍aχ(a)=32iχ(4)π2q2τ(χ)L(2,χ¯χ4).
Then we have
(46)∑1≤a≤q/4‍aχ(a)=q8πiτ(χ)(4πL(2,χ¯χ4)χ¯(2)L(1,χ¯)-χ¯(4)L(1,χ¯)+4πL(2,χ¯χ4)).


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

Letχ be an even primitive character modulo q. In this section, we express ∑1≤a≤q/4‍aχ(a) in terms of Gauss sums and Dirichlet L-functions.Lemma 12.
Letq>2 be an odd number, and let χ be a nonprincipal character modulo q. Then
(47)4χ(2)q∑a=1(q-1)/2‍aχ(a)=(2χ(2)+1)q∑a=1q‍aχ(a)-(4χ(2)-1)∑a=1q‍a2χ(a),4χ(2)q∑a=(q+1)/2q‍aχ(a)=(2χ(2)-1)q∑a=1q‍aχ(a)+(4χ(2)-1)∑a=1q‍a2χ(a).Proof.
We have(48)∑a=1q‍(2a)2χ(2a)=∑a=1(q-1)/2‍(2a)2χ(2a)+∑a=(q+1)/2q‍(2a)2χ(2a)=∑a=1(q-1)/2‍(2a)2χ(2a)+∑b=1(q+1)/2‍(2b+q-1)2χ(2b+q-1)=∑a=1(q-1)/2‍(2a)2χ(2a)+∑b=1(q+1)/2‍(2b-1)2χ(2b-1)+2q∑b=1(q+1)/2‍(2b-1)χ(2b-1)+q2∑b=1(q+1)/2‍χ(2b-1)=∑a=1q‍a2χ(a)+2q∑a=1(q+1)/2‍(2a-1)χ(2a-1)+q2∑a=1(q+1)/2‍χ(2a-1).
Since(49)∑a=1(q+1)/2‍χ(2a-1)+∑a=1(q-1)/2‍χ(2a)=∑a=1q‍χ(a)=0,∑a=1(q+1)/2‍(2a-1)χ(2a-1)+∑a=1(q-1)/2‍2aχ(2a)=∑a=1q‍aχ(a),
we have
(50)∑a=1q‍(2a)2χ(2a)=∑a=1q‍a2χ(a)+2q∑a=1q‍aχ(a)-4χ(2)q∑a=1(q-1)/2‍aχ(a)-χ(2)q2∑a=1(q-1)/2‍χ(a).
It is not hard to show that(51)χ(2)q∑a=1(q-1)/2‍χ(a)=(1-2χ(2))∑a=1q‍aχ(a),χ(2)q∑a=(q+1)/2q‍χ(a)=(2χ(2)-1)∑a=1q‍aχ(a).
Therefore
(52)4χ(2)q∑a=1(q-1)/2‍aχ(a)=(2χ(2)+1)q×∑a=1q‍aχ(a)-(4χ(2)-1)∑a=1q‍a2χ(a).
Note that
(53)∑a=1(q-1)/2‍aχ(a)+∑a=(q+1)/2q‍aχ(a)=∑a=1q‍aχ(a),
we have
(54)4χ(2)q∑a=(q+1)/2q‍aχ(a)=(2χ(2)-1)q×∑a=1q‍aχ(a)+(4χ(2)-1)∑a=1q‍a2χ(a).Lemma 13.
Letq≥5 be an odd number, and let χ be an nonprincipal even character modulo q. If q≡1(mod4), then
(55)16χ(4)∑a=1q-1‍a2χ(a)=(8χ(4)-2χ(2)+1)×∑a=1q‍a2χ(a)-16qχ(4)×∑a=1(q-1)/4‍aχ(a)+4q2χ(4)∑a=1(q-1)/4‍χ(a).
While if q≡3(mod4), we have
(56)16χ(4)∑a=1q-1‍a2χ(a)=(8χ(4)-2χ(2)+1)∑a=1q‍a2χ(a)-16qχ(4)×∑a=1(q-3)/4‍aχ(a)+4q2χ(4)∑a=1(q-3)/4‍χ(a).Proof.
First suppose thatq≡1(mod4). Then 4¯=(3q+1)/4. We have
(57)16χ(4)∑a=1q-1‍a2χ(a)=∑a=1q-1‍(4a)2χ(4a)=∑a=1(q-1)/4‍(4a)2χ(4a)+∑a=(q+3)/4(2q-2)/4‍(4a)2χ(4a)+∑a=(2q+2)/4(3q-3)/4‍(4a)2χ(4a)+∑a=(3q+1)/4q-1‍(4a)2χ(4a)=∑a=1(q-1)/4‍(4a)2χ(4a)+∑a=1(q-1)/4‍(4a+q-1)2χ(4a-1)+∑a=1(q-1)/4‍(4a+2q-2)2χ(4a-2)+∑a=1(q-1)/4‍(4a+3q-3)2χ(4a-3)=∑a=1(q-1)/4‍(4a)2χ(4a)+∑a=1(q-1)/4‍(4a-1)2χ(4a-1)+∑a=1(q-1)/4(4a-2)2‍χ(4a-2)+∑a=1(q-1)/4‍(4a-3)2χ(4a-3)+2q∑a=1(q-1)/4‍(4a-1)χ(4a-1)+4q∑a=1(q-1)/4‍(4a-2)χ(4a-2)+6q∑a=1(q-1)/4‍(4a-3)χ(4a-3)+q2∑a=1(q-1)/4‍χ(4a-1)+4q2∑a=1(q-1)/4‍χ(4a-2)+9q2∑a=1(q-1)/4‍χ(4a-3)=∑a=1q-1‍a2χ(a)+8qχ(4)∑a=1(q-1)/4‍(a+q-14)χ(a+q-14)-2q2χ(4)∑a=1(q-1)/4‍χ(a+q-14)+16qχ(4)∑a=1(q-1)/4‍(a+q-12)χ(a+q-12)-8q2χ(4)∑a=1(q-1)/4‍χ(a+q-12)+24qχ(4)∑a=1(q-1)/4‍(a+3q-34)χ(a+3q-34)-18q2χ(4)∑a=1(q-1)/4‍χ(a+3q-34)+q2χ(4)∑a=1(q-1)/4‍χ(a+q-14)+4q2χ(4)∑a=1(q-1)/4‍χ(a+q-12)+9q2χ(4)∑a=1(q-1)/4‍χ(a+3q-34)=∑a=1q-1‍a2χ(a)+8qχ(4)∑a=(q+3)/4(q-1)/2‍aχ(a)+16qχ(4)∑a=(q+1)/2(3q-3)/4‍aχ(a)+24qχ(4)∑a=(3q+1)/4q-1‍aχ(a)-q2χ(4)∑a=(q+3)/4(q-1)/2‍χ(a)-4q2χ(4)∑a=(q+1)/2(3q-3)/4‍χ(a)-9q2χ(4)∑a=(3q+1)/4q-1‍χ(a).
Note that∑1≤a≤(q-1)/2‍χ(a)=0 and ∑a=1q‍aχ(a)=0 for even character χ. By Lemma 12 we have
(58)16χ(4)∑a=1q-1‍a2χ(a)=∑a=1q‍a2χ(a)-16qχ(4)×∑a=1(q-1)/4‍aχ(a)-8qχ(4)∑a=1(q-1)/2‍aχ(a)+4q2χ(4)∑a=1(q-1)/4‍χ(a)=(8χ(4)-2χ(2)+1)×∑a=1q‍a2χ(a)-16qχ(4)∑a=1(q-1)/4‍aχ(a)+4q2χ(4)∑a=1(q-1)/4‍χ(a).
Now assume thatq≡3(mod4). Then 4¯=(q+1)/4. We have
(59)16χ(4)∑a=1q-1‍a2χ(a)=∑a=1q-1‍(4a)2χ(4a)=∑a=1(q-3)/4‍(4a)2χ(4a)+∑a=(q+1)/4(2q-2)/4‍(4a)2χ(4a)+∑a=(2q+2)/4(3q-1)/4‍(4a)2χ(4a)+∑a=(3q+3)/4q-1‍(4a)2χ(4a)=∑a=1(q-3)/4‍(4a)2χ(4a)+∑a=1(q+1)/4‍(4a+q-3)2χ(4a-3)+∑a=1(q+1)/4‍(4a+2q-2)2χ(4a-2)+∑a=1(q-3)/4‍(4a+3q-1)2χ(4a-1)=∑a=1(q-3)/4‍(4a)2χ(4a)+∑a=1(q-3)/4‍(4a-1)2χ(4a-1)+∑a=1(q+1)/4‍(4a-2)2χ(4a-2)+∑a=1(q+1)/4‍(4a-3)2χ(4a-3)+6q∑a=1(q-3)/4‍(4a-1)χ(4a-1)+4q∑a=1(q+1)/4‍(4a-2)χ(4a-2)+2q∑a=1(q+1)/4‍(4a-3)χ(4a-3)+9q2∑a=1(q-3)/4‍χ(4a-1)+4q2∑a=1(q+1)/4‍χ(4a-2)+q2∑a=1(q+1)/4‍χ(4a-3)=∑a=1q-1‍a2χ(a)+24qχ(4)∑a=1(q-3)/4‍(a+3q-14)χ(a+3q-14)-18q2χ(4)∑a=1(q-3)/4‍χ(a+3q-14)+16qχ(4)∑a=1(q+1)/4‍(a+q-12)χ(a+q-12)-8q2χ(4)∑a=1(q+1)/4‍χ(a+q-12)+8qχ(4)∑a=1(q+1)/4‍(a+q-34)χ(a+q-34)-2q2χ(4)∑a=1(q+1)/4‍χ(a+q-34)+9q2χ(4)∑a=1(q-3)/4‍χ(a+3q-14)+4q2χ(4)∑a=1(q+1)/4‍χ(a+q-12)+q2χ(4)∑a=1(q+1)/4‍χ(a+q-34)=∑a=1q-1‍a2χ(a)+24qχ(4)∑a=(3q+3)/4q-1‍aχ(a)+16qχ(4)∑a=(q+1)/2(3q-1)/4‍aχ(a)+8qχ(4)∑a=(q+1)/4(q-1)/2‍aχ(a)-9q2χ(4)∑a=(3q+3)/4q-1‍χ(a)-4q2χ(4)∑a=(q+1)/2(3q-1)/4‍χ(a)-q2χ(4)∑a=(q+1)/4(q-1)/2‍χ(a).
Note that∑1≤a≤(q-1)/2‍χ(a)=0 and ∑a=1q‍aχ(a)=0 for even character χ. By Lemma 12 we have
(60)16χ(4)∑a=1q-1‍a2χ(a)=∑a=1q‍a2χ(a)-16qχ(4)∑a=1(q-3)/4‍aχ(a)-8qχ(4)∑a=1(q-1)/2‍aχ(a)+4q2χ(4)∑a=1(q-3)/4‍χ(a)=(8χ(4)-2χ(2)+1)∑a=1q‍a2χ(a)-16qχ(4)∑a=1(q-3)/4‍aχ(a)+4q2χ(4)∑a=1(q-3)/4‍χ(a).Now we express∑1≤a≤q/4‍aχ(a) in terms of Gauss sums and Dirichlet L-functions.Theorem 14.
Letχ be an even primitive character modulo odd integer q≥5. Then one has
(61)∑1≤a≤q/4‍aχ(a)=q16π2(χ¯(4)-2χ¯(2)-8)τ(χ)L(2,χ¯)+q4πτ(χ)L(1,χ¯χ4).Proof.
By Lemma13, (3), and (37), we have
(62)16qχ(4)∑1≤a≤q/4‍aχ(a)=(1-2χ(2)-8χ(4))∑a=1q‍a2χ(a)+4q2χ(4)∑1≤a≤q/4‍χ(a)=q2π2(1-2χ(2)-8χ(4))τ(χ)L(2,χ¯)+4q2πχ(4)τ(χ)L(1,χ¯χ4).
Therefore
(63)∑1≤a≤q/4‍aχ(a)=q16π2(χ¯(4)-2χ¯(2)-8)×τ(χ)L(2,χ¯)+q4πτ(χ)L(1,χ¯χ4).


## 4. Mean Values of DirichletL-Functions

In this section, we will study the mean values of DirichletL-functions, which will be used to prove Theorems 3 and 4.Lemma 15.
Letq and r be integers with q≥2 and (r,q)=1. Then one has the identities
(64)∑‍*χmodqχ(r)=∑d∣(q,r-1)‍μ(qd)ϕ(d),J(q)=∑d∣q‍μ(qd)ϕ(d),
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.
Letq≥2 be an odd number, and let k≥0 be an integer. Then one has
(65)∑n=1∞‍(n,q)=1d(n)d(2kn)n2=5+3k5·ζ4(2)ζ(4)∏p∣q‍(1-1/p2)31+1/p2,∑n=1∞‍(n,q)=1d(n)d(2kn)n4=15k+1717·ζ4(4)ζ(8)∏p∣q‍(1-1/p4)31+1/p4,∑n=1∞‍(n,2q)=1d2(n)n2=2780·ζ4(2)ζ(4)∏p∣q‍(1-1/p2)31+1/p2,∑n=1∞‍(n,2q)=1d2(n)n4=33754352·ζ4(4)ζ(8)∏p∣q‍(1-1/p4)31+1/p4,∑n=1∞‍(n,2q)=1d2(n)χ4(n)n3=1(1-1/26)·L4(3,χ4)ζ(6)∏p∣q‍(1-χ4(p)/p3)31+χ4(p)/p3,∑n=1∞‍(n,q)=1d(n)τ1(2kn)n2=ζ2(2)L2(3,χ4)L(5,χ4)×∏p∣q‍(1-1/p2)2(1-χ4(p)/p3)21-χ4(p)/p5,∑n=1∞‍(n,q)=1τ1(n)τ1(2kn)n2=ζ(4)L(3,χ4)∏p∣q‍(1-χ4(p)p3)(1-1p4)×∏p∤q‍(1+1p2(1-χ4(p)/p)),∑n=1∞‍(n,q)=1d(2kn)τ2(n)n3=(1516k+1)·ζ2(4)L2(3,χ4)L(7,χ4)×∏p∣q‍(1-χ4(p)/p3)2(1-1/p4)21-χ4(p)/p7,∑n=1∞‍(n,q)=1d(n)τ2(2kn)n3=12k·ζ2(4)L2(3,χ4)L(7,χ4)×∏p∣q‍(1-χ4(p)/p3)2(1-1/p4)21-χ4(p)/p7,∑n=1∞‍(n,2q)=1χ4(n)d(n)τ2(n)n2=916·ζ2(2)L2(3,χ4)L(5,χ4)×∏p∣q‍(1-1/p2)2(1-χ4(p)/p3)21-χ4(p)/p5,∑n=1∞‍(n,q)=1τ2(n)τ2(2kn)n2=42k5·ζ(4)L(3,χ4)×∏p∣q‍(1-χ4(p)p3)(1-1p4)×∏p∤q‍(1+1p2(1-χ4(p)/p)),
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(66)∑n=1∞‍(n,q)=1d(n)d(2kn)n2=∑i=0∞‍∑m=1∞‍(m,2q)=1d(2im)d(2k+im)(2im)2=∑i=0∞‍d(2i)d(2k+i)22i∑m=1∞‍(m,2q)=1d2(m)m2=(∑i=0∞‍(i+1)(k+i+1)22i)∏p∤2q‍(∑j=0∞‍(j+1)2p2j)=5/4+(3/4)k(1-1/22)3∏p∤2q‍1+1/p2(1-1/p2)3=5+3k5∏p‍1+1/p2(1-1/p2)3∏p∣q‍(1-1/p2)31+1/p2=5+3k5·ζ4(2)ζ(4)∏p∣q‍(1-1/p2)31+1/p2.
Similarly, we can deduce the other identities.Lemma 17.
Letq≥2 be an odd number. For integers k≥0 and l≥1, one has
(67)∑‍*χmodqχ(-1)=-1χ(2k)|L(1,χ)|4=5+3k2k+1·5·ζ4(2)ζ(4)J(q)∏p∤q‍(1-1/p2)31+1/p2+O(qϵ),∑‍*χmodqχ(-1)=-1|L(2,χχ4)|4=33758704·ζ4(4)ζ(8)J(q)∏p∣q‍(1-1/p4)31+1/p4+O(qϵ),∑‍*χmodqχ(-1)=-1χ¯(2k)L2(1,χ¯)L(1,χ)L(2,χχ4)=ζ2(2)L2(3,χ4)2k+1L(5,χ4)J(q)×∏p∣q‍(1-1/p2)2(1-χ4(p)/p3)21-χ4(p)/p5+O(qϵ),∑‍*χmodqχ(-1)=-1χ¯(2l)L2(1,χ¯)L2(2,χχ4)≪qϵ,∑‍*χmodqχ(-1)=-1χ¯(2l)L(1,χ¯)L(2,χ¯χ4)L2(2,χχ4)≪qϵ,∑‍*χmodqχ(-1)=-1χ(2k)|L(1,χ)|2|L(2,χχ4)|2=ζ(4)L(3,χ4)2k+1J(q)∏p∣q‍(1-χ4(p)p3)(1-1p4)×∏p∤q‍(1+1p2(1-χ4(p)/p))+O(qϵ).Proof.
We only prove the first formula since, similarly, we can get the others. Letd(n)=∑d∣n‍1 be the divisor function. For N≥q2, by Abel’s identity, we get
(68)L2(1,χ)=(∑n=1∞‍χ(n)n)2=∑1≤n≤N‍χ(n)d(n)n+O(q1/2logqlogNN1/2).
For(r,q)=1, from Lemma 15, we have
(70)∑‍*χmodqχ(-1)=-1χ(r)=12∑‍*χmodq(1-χ(-1))χ(r)=12∑‍*χmodqχ(r)-12∑‍*χmodqχ(-r)=12∑d∣(q,r-1)‍μ(qd)ϕ(d)-12∑d∣(q,r+1)‍μ(qd)ϕ(d).
Then from Lemma 16 we get(71)∑‍*χmodqχ(-1)=-1χ(2k)|L(1,χ)|4=∑‍*χmodqχ(-1)=-1χ(2k)|∑1≤n≤N‍χ(n)d(n)n+O(q1/2logqlogNN1/2)|2=∑1≤n≤N‍∑1≤m≤N‍d(n)d(m)nm∑‍*χmodqχ(-1)=-1χ(2kn)χ¯(m)+O(q3/2logqlog3NN1/2)=12∑1≤n≤N‍(n,q)=1∑1≤m≤N‍(m,q)=1d(n)d(m)nm∑d|(q,2kn-m)‍μ(qd)ϕ(d)-12∑1≤n≤N‍(n,q)=1∑1≤m≤N‍(m,q)=1d(n)d(m)nm∑d∣(q,2kn+m)‍μ(qd)ϕ(d)+O(q3/2logqlog3NN1/2)=J(q)2k+1∑1≤n≤N/2k‍(n,q)=1d(n)d(2kn)n2+12∑d∣q‍μ(qd)ϕ(d)∑1≤n≤N‍(n,q)=1∑1≤m≤N‍(m,q)=12kn≡m(modd)2kn≠md(n)d(m)nm-12∑d|q‍μ(qd)ϕ(d)∑1≤n≤N‍(n,q)=1∑1≤m≤N‍(m,q)=12kn≡-m(modd)d(n)d(m)nm+O(q3/2logqlog3NN1/2)=J(q)2k+1∑n=1∞‍(n,q)=1d(n)d(2kn)n2+O(∑d∣q‍ϕ(d)∑1≤n≤N‍∑(1-2kn)/d≤l≤(N-2kn)/d‍d(n)d(ld+2kn)n(ld+2kn))+O(∑d∣q‍ϕ(d)∑1≤n≤N‍∑(1+2kn)/d≤l≤(N+2kn)/d‍d(n)d(ld-2kn)n(ld-2kn))+O(q3/2logqlog3NN1/2)=J(q)2k+1∑n=1∞‍(n,q)=1d(n)d(2kn)n2+O(Nϵ)+O(q3/2logqlog3NN1/2)=5+3k2k+1·5·ζ4(2)ζ(4)J(q)∏p∣q‍(1-1/p2)31+1/p2+O(Nϵ)+O(q3/2logqlog3NN1/2).Now taking N=q4, we immediately get
(72)∑‍*χmodqχ(-1)=-1χ(2k)|L(1,χ)|4=5+3k2k+1·5·ζ4(2)ζ(4)J(q)×∏p∣q‍(1-1/p2)31+1/p2+O(qϵ).Lemma 18.
Letq≥2 be an odd number. For integers k≥0 and l≥1, one has
(73)∑‍*χmodqχ(-1)=1χ¯(2k)|L(2,χ¯)|4=15k+1722k+1·17·ζ4(4)ζ(8)·J(q)∏p∣q‍(1-1/p4)31+1/p4+O(qϵ),∑‍*χmodqχ(-1)=1L2(2,χ¯)L2(1,χχ4)=12(1-1/26)·L4(3,χ4)ζ(6)·J(q)∏p∣q‍(1-χ4(p)/p3)31+χ4(p)/p3+O(qϵ),∑‍*χmodqχ(-1)=1χ¯(2l)L2(2,χ¯)L2(1,χχ4)≪qϵ,∑‍*χmodqχ(-1)=1χ(2k)L2(2,χ¯)L(1,χχ4)L(2,χ)=((15/16)k+1)22k+1·ζ2(4)L2(3,χ4)L(7,χ4)·J(q)∏p∣q‍(1-χ4(p)/p3)2(1-1/p4)21-χ4(p)/p7+O(qϵ),∑‍*χmodqχ(-1)=1χ¯(2k)L2(2,χ¯)L(1,χχ4)L(2,χ)=12k+1·ζ2(4)L2(3,χ4)L(7,χ4)·J(q)∏p∣q‍(1-χ4(p)/p3)2(1-1/p4)21-χ4(p)/p7+O(qϵ),∑‍*χmodqχ(-1)=1|L(1,χ¯χ4)|4=27160·ζ4(2)ζ(4)·J(q)∏p∣q‍(1-1/p2)31+1/p2+O(qϵ),∑‍*χmodqχ(-1)=1χ(2l)L2(1,χ¯χ4)L(1,χχ4)L(2,χ)≪qϵ,∑‍*χmodqχ(-1)=1L2(1,χ¯χ4)L(1,χχ4)L(2,χ)=932·ζ2(2)L2(3,χ4)L(5,χ4)·J(q)∏p∣q‍(1-1/p2)2(1-χ4(p)/p3)21-χ4(p)/p5+O(qϵ),∑‍*χmodqχ(-1)=1χ¯(2k)|L(1,χ¯χ4)|2|L(2,χ¯)|2=222k5·ζ(4)L(3,χ4)·J(q)∏p∣q‍(1-χ4(p)/p3)(1-1/p4)×∏p∤q‍(1+1p2(1-χ4(p)/p))+O(qϵ).Proof.
By Lemma17 and the methods proving Lemma 18, we can get this lemma.


## 5. Proof of Theorems3 and 4

First we prove Theorem3. By Theorem 11 and Lemma 18, we have
(74)∑‍*χmodqχ(-1)=-1|∑1≤a≤q/4‍aχ(a)|4=q684π4∑‍*χmodqχ(-1)=-1|4πL(2,χ¯χ4)χ¯(2)L(1,χ¯)-χ¯(4)L(1,χ¯)+4πL(2,χ¯χ4)|4=q684π4∑‍*χmodqχ(-1)=-1|8πχ¯(4)L(1,χ¯)L(2,χ¯χ4)χ¯(4)L2(1,χ¯)+χ¯(16)L2(1,χ¯)+16π2L2(2,χ¯χ4)-2χ¯(8)L2(1,χ¯)+8πχ¯(2)L(1,χ¯)L(2,χ¯χ4)-8πχ¯(4)L(1,χ¯)L(2,χ¯χ4)|2=32048π4q6∑‍*χmodqχ(-1)=-1|L(1,χ)|4-1512π4q6∑‍*χmodqχ(-1)=-1χ(2)|L(1,χ)|4+12048π4q6∑‍*χmodqχ(-1)=-1χ(4)|L(1,χ)|4+116π8q6∑‍*χmodqχ(-1)=-1|L(2,χχ4)|4-1256π5q6∑‍*χmodqχ(-1)=-1L2(1,χ¯)L(1,χ)L(2,χχ4)+3256π5q6∑‍*χmodqχ(-1)=-1χ¯(2)L2(1,χ¯)L(1,χ)L(2,χχ4)-3256π5q6∑‍*χmodqχ(-1)=-1χ¯(4)L2(1,χ¯)L(1,χ)L(2,χχ4)+1256π5q6∑‍*χmodqχ(-1)=-1χ¯(8)L2(1,χ¯)L(1,χ)L(2,χχ4)+1128π6q6∑‍*χmodqχ(-1)=-1χ¯(4)L2(1,χ¯)L2(2,χχ4)-164π6q6∑‍*χmodqχ(-1)=-1χ¯(8)L2(1,χ¯)L2(2,χχ4)+1128π6q6∑‍*χmodqχ(-1)=-1χ¯(16)L2(1,χ¯)L2(2,χχ4)+116π7q6∑‍*χmodqχ(-1)=-1χ¯(2)L(1,χ¯)L(2,χ¯χ4)L2(2,χχ4)-116π7q6∑‍*χmodqχ(-1)=-1χ¯(4)L(1,χ¯)L(2,χ¯χ4)L2(2,χχ4)+132π6q6∑‍*χmodqχ(-1)=-1|L(1,χ)|2|L(2,χχ4)|2-132π6q6∑‍*χmodqχ(-1)=-1χ(2)|L(1,χ)|2|L(2,χχ4)|2=7217·32q6J(q)∏p∣q‍(1-1/p2)31+1/p2+35216·32·17q6J(q)∏p∣q‍(1-1/p4)31+1/p4-L2(3,χ4)214·32·π·L(5,χ4)×q6J(q)∏p∣q‍(1-1/p2)2(1-χ4(p)/p3)21-χ4(p)/p5+L(3,χ4)28·5·32·π2q6J(q)×∏p∣q‍(1-χ4(p)p3)(1-1p4)×∏p∤q‍(1+1p2(1-χ4(p)/p))+O(q6+ϵ).
This proves Theorem 3.On the other hand, by Theorem14 and Lemma 18, we have
(75)∑‍*χmodqχ(-1)=1|∑1≤a≤q/4‍aχ(a)|4=q6164π8∑‍*χmodqχ(-1)=1|(χ¯(4)-2χ¯(2)-8)L(2,χ¯)+4πL(1,χ¯χ4)|4=q6164π8∑‍*χmodqχ(-1)=1|(χ¯(4)-2χ¯(2)-8)2L2(2,χ¯)+16π2L2(1,χ¯χ4)+8π(χ¯(4)-2χ¯(2)-8)×L(1,χ¯χ4)L(2,χ¯)(χ¯(4)-2χ¯(2)-8)2|2=528165536π8q6∑‍*χmodqχ(-1)=1|L(2,χ¯)|4+4278192π8q6∑‍*χmodqχ(-1)=1χ¯(2)|L(2,χ¯)|4-2278192π8q6∑‍*χmodqχ(-1)=1χ¯(4)|L(2,χ¯)|4-71024π8q6∑‍*χmodqχ(-1)=1χ¯(8)|L(2,χ¯)|4+1512π8q6∑‍*χmodqχ(-1)=1χ¯(16)|L(2,χ¯)|4+132π6q6∑‍*χmodqχ(-1)=1L2(2,χ¯)L2(1,χχ4)+164π6q6∑‍*χmodqχ(-1)=1χ¯(2)L2(2,χ¯)L2(1,χχ4)-3512π6q6∑‍*χmodqχ(-1)=1χ¯(4)L2(2,χ¯)L2(1,χχ4)-1512π6q6∑‍*χmodqχ(-1)=1χ¯(8)L2(2,χ¯)L2(1,χχ4)+12048π6q6∑‍*χmodqχ(-1)=1χ¯(16)L2(2,χ¯)L2(1,χχ4)+164π7q6∑‍*χmodqχ(-1)=1χ(4)L2(2,χ¯)L(1,χχ4)L(2,χ)-3128π7q6∑‍*χmodqχ(-1)=1χ(2)L2(2,χ¯)L(1,χχ4)L(2,χ)-1471024π7q6∑‍*χmodqχ(-1)=1L2(2,χ¯)L(1,χχ4)L(2,χ)-591024π7q6∑‍*χmodqχ(-1)=1χ¯(2)L2(2,χ¯)L(1,χχ4)L(2,χ)+1054096π7q6∑‍*χmodqχ(-1)=1χ¯(4)L2(2,χ¯)L(1,χχ4)L(2,χ)+152048π7q6∑‍*χmodqχ(-1)=1χ¯(8)L2(2,χ¯)L(1,χχ4)L(2,χ)-1512π7q6∑‍*χmodqχ(-1)=1χ¯(16)L2(2,χ¯)L(1,χχ4)L(2,χ)+1256π4q6∑‍*χmodqχ(-1)=1|L(1,χ¯χ4)|4+1256π5q6∑‍*χmodqχ(-1)=1χ(4)L2(1,χ¯χ4)L(1,χχ4)L(2,χ)-1128π5q6∑‍*χmodqχ(-1)=1χ(2)L2(1,χ¯χ4)L(1,χχ4)L(2,χ)-132π5q6∑‍*χmodqχ(-1)=1L2(1,χ¯χ4)L(1,χχ4)L(2,χ)+691024π6q6∑‍*χmodqχ(-1)=1|L(1,χ¯χ4)|2|L(2,χ¯)|2+7256π6q6∑‍*χmodqχ(-1)=1χ¯(2)|L(1,χ¯χ4)|2|L(2,χ¯)|2-164π6q6∑‍*χmodqχ(-1)=1χ¯(4)|L(1,χ¯χ4)|2|L(2,χ¯)|2=385220·51·q6J(q)∏p∣q‍(1-1/p4)31+1/p4+15L4(3,χ4)π12·q6J(q)∏p∣q‍(1-χ4(p)/p3)31+χ4(p)/p3-1423πL2(3,χ4)216·52·34·L(7,χ4)·q6J(q)∏p∣q‍(1-χ4(p)/p3)2(1-1/p4)21-χ4(p)/p7+3216·q6J(q)∏p∣q‍(1-1/p2)31+1/p2-L2(3,χ4)212·π·L(5,χ4)·q6J(q)∏p∣q‍(1-1/p2)2(1-χ4(p)/p3)21-χ4(p)/p5+L(3,χ4)210·3·π2·q6J(q)∏p∣q‍(1-χ4(p)/p3)(1-1/p4)×∏p∤q‍(1+1p2(1-χ4(p)/p))+O(q6+ϵ).
This completes the proof of Theorem 4.



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*Source: 326597-2013-11-27.xml*