# On a Kind of Dirichlet Character Sums

Authors: Rong Ma; Yulong Zhang; Guohe Zhang
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

Letp≥3 be a prime and let χ denote the Dirichlet character modulo p. For any prime q with q<p, define the set Eq,p=a∣1≤a,a-≤p,aa-≡1modpanda≡a-modq. In this paper, we study a kind of mean value of Dirichlet character sums ∑a≤pa∈Eq,pχ(a), and use the properties of the Dirichlet L-functions and generalized Kloosterman sums to obtain an interesting estimate.


---

## Body

## 1. Introduction

Letk≥3 be an integer and let χ denote the Dirichlet character modulo k, for any real number x≥1, many scholars have studied the following sums:
(1)∑n≤x‍χ(n),
where n are positive integers.Perhaps one of the most famous results is Pólya’s inequality [1]. That is, when χ is the primitive character modulo k, we have
(2)∑n≤x‍χ(n)<k1/2logk.In fact, the result can be extended to the nonprincipal characterχ modulo k [2]. Further details about the estimates of character sums can be found in the literature, for example, [3, 4].For any fixed integerH>0 and any positive integer k≥3, define the following set:
(3)L(H,k)={a∣1≤a,a¯≤k-1,(a,k)=1,aa¯≡1(modk),|a-a¯|≤H},
let χ denote the Dirichlet character modulo k, define the sums as follows:
(4)∑n≤kn∈L(H,k)χ(n).Xi and Yi [5] studied the problem for χ the nonprincipal Dirichlet character modulo k, and got
(5)∑n≤kn∈L(H,k)χ(n)≪k1/2d(k)logH,
where 0<H≤q was a constant and d(k) was the divisor function. Before this, Wenpeng [6] got an asymptotic formula for the case that χ was the principal Dirichlet character modulo k.On the other hand, for each integera with 1≤a≤k and (a,k)=1, we know that there exists one and only one b with 1≤b≤k such that ab≡1(modk). Let r2(k) be the number of solutions of the congruent equation ab≡1(modk) for 1≤a, b≤k in which a and b are of opposite parity, this can be expressed as follows:
(6)r2(k)=∑‍a=1kaa¯≡1(modk)2†(a+a¯)1.Richard [7] asks us to find r2(k) or at least to say something nontrivial about it. About this problem, a lot of scholars have studied it [8–12]. Now we let m be another integer with m<k and let rm(k) denote the number of all pairs of integers a, b satisfying ab≡1(modk), 1≤a, b≤k, and m†(a+b). Lu and Yi  [13] have obtained the asymptotic formula of generalized D. H. Lehmer problem as follows:
(7)rm(k)=∑‍a=1kaa¯≡1(modk)m†(a+a¯)1=(1-1m)ϕ(k)+O(k1/2log2k),
where the O constant only depends on m.In this paper, letp be an odd prime and let q be a fixed prime with q<p, define the set E(q,p) for a(1≤a≤p) such that aa¯≡1(modp) and a≡a¯(modq), that is,
(8)E(q,p)={a∣1≤a,a¯≤p-1,aa¯≡1(modp),a≡a¯(modq)}.As another case of (7), we will consider the mean value of Dirichlet character sums as follows:
(9)∑a≤p-1a∈E(q,p)‍χ(a),
and get an interesting estimate. That is, we will prove the following theorem.Theorem 1.
Letp be an odd prime and let q be a fixed prime with q<p, and let χ denote the Dirichlet character modulo p. Let E(q,p) denote the following set:
(10)E(q,p)={a∣1≤a≤p-1,aa¯≡1(modp),a≡a¯(modq)},
then, for any nonprincipal Dirichlet character χmodp, we have the following estimate:
(11)∑a≤p-1a∈E(q,p)‍χ(a)=O(p1/2+ϵ),
where the O constant only depends on q.
From this Theorem we can get(12)∑‍a=1p-1aa¯≡1(modp)q†(a-a¯)χ(a)=∑‍a=1p-1aa¯≡1(modp)χ(a)-∑a≤p-1a∈E(q,p)‍χ(a)=O(p1/2+ϵ).
For any integerk and fixed integer m such that (m,k)=1, whether or not there exists an estimate for
(13)∑a≤ka∈E(m,k)‍χ(a)
is still an open problem.


## 2. Some Lemmas

In this section, we will give several lemmas which are necessary in the proof of the theorem.Lemma 2.
LetQ be an integer, and let χ be a primitive character modulo Q. Then, for any real number u and v with u<v, we have
(14)∑uQ<n≤vQ‍χ(n)=τ(χ)∑0<|h|≤H‍χ¯(h)e(-hu)-e(-hv)2πih+O(1+QlogQH),
where e(x)=e2πix and τ(χ)=∑a=1Qχ(a)e(a/Q) are Gauss sums.
Especially, letu=0, we have a slight modification
(15)∑0<n≤vQ‍χ(n)={τ(χ)π∑n=1∞‍χ¯(n)sin(2πnv)n+O(1),ifχ(-1)=1,τ(χ)πi∑n=1∞‍χ¯(n)(1-cos(2πnv))n+O(1),ifχ(-1)=-1.Proof.
(See [1]).Lemma 3.
Letq be a prime, let Q be an integer with Q>q, and let χ be a primitive character modulo Q, then, we have
(16)∑n≤Q/q‍χ(n)={τ(χ)(q-1)π∑l=1q-1‍∑χ2modq‍(χ2(l)sin2πlq)×L(1,χχ¯2)+O(1),χ(-1)=1,τ(χ)(q-1)πi∑l=1q-1‍∑χ2modq‍χ2(l)(1-cos2πlq)×L(1,χχ¯2)+O(1),χ(-1)=-1,
where L(1,χ) are the Dirichlet L-functions corresponding to χ.Proof.
From Lemma2, we take u=0, and v=1/q and get
(17)∑n≤Q/q‍χ(n)=τ(χ)∑0<h≤H‍χ¯(h)1-e(-h/q)2πih+τ(χ)∑-H<h≤0‍χ¯(h)1-e(-h/q)2πih+O(1+QlogQH).
Whenχ(-1)=1, we have
(18)∑n≤Q/q‍χ(n)=τ(χ)∑0<h≤H‍χ¯(h)sin(2πh/q)πh+O(1+QlogQH)=τ(χ)π∑l=1q-1‍(sin2πlq)∑0<h≤Hh≡l(modq)‍χ¯(h)h+O(1+QlogQH)=τ(χ)(q-1)π∑l=1q-1‍(sin2πlq)∑0<h≤H‍χ¯(h)h×∑χ2modq‍χ¯2(h)χ2(l)+O(1+QlogQH)=τ(χ)(q-1)π∑l=1q-1‍∑χ2modq‍(χ2(l)sin2πlq)×∑0<h≤H‍χχ¯2(h)h+O(1+QlogQH).
LetH→∞, then, we have
(19)∑n≤Q/q‍χ(n)=τ(χ)(q-1)π∑l=1q-1‍∑χ2modq‍(χ2(l)sin2πlq)L(1,χχ¯2)+O(1).
The case ofχ(-1)=-1 can be treated in the same way. This proves Lemma 3.Lemma 4.
Letp,q be odd primes and let χ1,χ2 be the Dirichlet characters modulo p and q, respectively, such that (p,q)=1, denote χ=χ1χ2, k=pq, and χ is the Dirichlet character modulo pq, the famous Gauss sums are defined as follows:
(20)G(n,χ)=∑a=1k‍χ(a)e(nak),
where e(y)=e2πiy. Hence, we have
(21)G(n,χ)=χ1(q)χ2(p)G(n,χ1)G(n,χ2).
Whenn=1, we denote τ(χ)=G(1,χ); therefore, we have
(22)τ(χ)=χ1(q)χ2(p)τ(χ1)τ(χ2).Proof.
(See [14]).Lemma 5.
Letm, and n be integers and let q≥3 be prime, let χ denote the Dirichlet character modulo q, the generalized Kloosterman sums are defined by
(23)Sχ(m,n;q)=∑amodq‍χ(a)e(ma¯+naq),
where aa¯≡1(modq) and e(y)=e2πiy.
Then, we have the following estimate:(24)Sχ(m,n;q)≪q1/2+ϵ(m,n,q)1/2,
where (m,n,q) denotes the gcd of m,n, and q.Proof.
(See [15]).Lemma 6.
Letp≥3 be an odd prime, let χ, χ1 be a Dirichlet character modulo p, and χχ1≠χ0p. For any odd prime q with q<p, let χ2,χ3,χ4 be any Dirichlet characters with χ2modq, χ3modq and χ4modq, respectively, then, no matter χ is odd character or even character modulo p, we have
(25)∑χ1≠χp0χ1χ2(-1)=1‍χχ1(q)χ1(q)τ(χχ1)τ(χ1)∑χ1≠χp0χ1χ2(-1)=1‍×L(1,χχ1χ2χ¯3)L(1,χ¯1χ2χ¯4)≪p3/2+ϵ;(26)∑χ1≠χp0χ1χ2(-1)=-1χχ1(q)χ1(q)τ(χχ1)τ(χ1)∑χ1≠χp0χ1χ2(-1)=1‍×L(1,χχ1χ2χ¯3)L(1,χ¯1χ2χ¯4)≪p3/2+ϵ;‍(27)∑χ1≠χp0χ(-1)=1χχ1(q)χ1(q)τ(χχ1)τ(χ1)∑χ1≠χp0χ1(-1)=1‍×L(1,χχ1χ¯3)L(1,χ1χ¯4)≪p3/2+ϵ;‍(28)∑χ1≠χp0χ(-1)=-1‍χχ1(q)χ1(q)τ(χχ1)τ(χ1)∑χ1≠χp0χ1(-1)=1‍×L(1,χχ1χ¯3)L(1,χ1χ¯4)≪p3/2+ϵ,
where the ≪ constant only depends on q.Proof.
For any integern with (n,k)=1 (k≥3 is any positive integer), we have
(29)∑χmodkχ(-1)=1‍χ(n)={12ϕ(k),n≡±1(modk),0,otherwise.∑χmodkχ(-1)=-1‍χ(n)={12ϕ(k),n≡1(modk),-12ϕ(k),n≡-1(modk),0,otherwise.
Now lety>p and let A(y,χ)=∑p<n≤y‍χ(n). Then, from the Pólya-Vinogradov inequality, we obtain
(30)A(y,χ)≪plogp.
Hence, from Abel’s identity, for any Re(s)≥1, we can easily get
(31)L(s,χ)=∑n≤p‍χ(n)ns+∫p∞‍A(y,χ)ys+1dy=∑n≤p‍χ(n)ns+O(logpps-1/2)=∑n≤p‍χ(n)ns+O(logpp1/2).
We will take (25), for example, to prove this lemma. For (q,p)=1, from the definition of τ(χ) and (31), since χ1 is not the principle character modulo p and χχ1 is not the principle character modulo p, we have
(32)∑χ1≠χp0χ1χ2(-1)=1‍χχ1(q)χ1(q)τ(χχ1)τ(χ1)∑χ1≠χp0χ1χ2(-1)=1‍×L(1,χχ1χ2χ¯3)L(1,χ¯1χ2χ¯4)=∑χ1modpχ1χ2(-1)=1‍χχ1(q)χ1(q)τ(χχ1)τ(χ1)×(∑k≤p‍χχ1χ2χ¯3(k)k+O(logpq(pq)1/2))×(∑n≤p‍χ¯1χ2χ¯4(n)n+O(logpq(pq)1/2))-χ(q)τ(χ)τ(χ0p)L(1,χχ2χ¯3)L(1,χ2χ¯4)=∑χ1modpχ1χ2(-1)=1‍χχ1(q)χ1(q)τ(χχ1)τ(χ1)×∑k≤p‍χχ1χ2χ¯3(k)k∑n≤p‍χ¯1χ2χ¯4(n)n+O(p3/2log2p)=∑k≤p‍∑n≤p‍χχ1χ¯3(k)χ2χ¯4(n)χ(qa)kn×∑‍a=1p′∑‍b=1p′χ(a)e(a+bp)×∑χ1modpχ1χ2(-1)=1‍χ1(q2ab)χ¯1(kn)+O(p3/2log2p)=ϕ(p)2∑k≤q‍∑n≤p‍χχ1χ¯3(k)χ2χ¯4(n)χ(qa)kn×(∑‍a=1p′∑‍b=1p′q2ab≡kn(modp)χ(a)e(a+bp)±∑‍a=1p′∑‍b=1p′q2ab≡-kn(modp)χ(a)e(a+bp))+O(p3/2log2p)=ϕ(p)2∑k≤p‍∑n≤p‍χχ1χ¯3(k)χ2χ¯4(n)χ(qa)kn×(Sχ(1,q¯2kn;p)±Sχ(1,-q¯2kn;p))+O(p3/2log2p),
where abq2kn¯≡1(modp). From Lemma 5, we can easily obtain
(33)∑χ1≠χp0χ1χ2(-1)=1‍χχ1(q)χ1(q)τ(χχ1)τ(χ1)L(1,χχ1χ2χ¯3)∑χ1≠χp0χ1χ2(-1)=1‍×L(1,χ¯1χ2χ¯4)≪p3/2+ϵ1log2p≪p3/2+ϵ,
where the ≪ constant only depends on q. Therefore, this completes the proof of Lemma 6.Lemma 7.
Letq be a fixed odd prime and let p be a prime with p>q, let χ denote the nonprincipal Dirichlet character modulo p, and χ1 denote the Dirichlet character modulo p, then, we have
(34)∑χ1modpχ1≠χp0‍χχ1(q)χ1(q)∑χ1modpχ1≠χp0‍×∑a≤(p-1)/q‍χχ1(a)∑b≤(p-1)/q‍χ1(b)≪p3/2+ϵ,
where the ≪ constant only depends on q.Proof.
For primesp and q, χ1 and χχ1 are nonprincipal and primitive characters modulo p, hence, from Lemmas 3 and 6, when χ(-1)=1, we can get
(35)∑χ1modpχ1≠χp0‍χχ1(q)χ1(q)∑a≤(p-1)/q‍χχ1(a)∑b≤(p-1)/q‍χ1(b)=∑χ1(-1)=1χ1≠χp0‍χχ1(q)χ1(q)×({(∑u=1q-1‍)2}τ(χχ1)(q-1)π∑u=1q-1‍∑χ3modq‍(χ3(u)sin2πuq)τ(χχ1)(q-1)π∑u=1q-1‍∑χ3modq‍×L(1,χχ1χ3¯)+O(1){(∑u=1q-1‍)2})×({(∑u=1q-1‍)2}τ(χ1)(q-1)π∑v=1q-1‍∑χ4modq‍(χ4(v)sin2πvq)τ(χ1)(q-1)π∑v=1q-1‍∑χ4modq‍×L(1,χ1χ4¯)+O(1){(∑u=1q-1‍)2})+∑χ1(-1)=-1χ1≠χp0‍χχ1(q)χ1(q)×({(∑u=1q-1‍)2}τ(χχ1)(q-1)πi∑u=1q-1‍∑χ3modq‍(χ3(u)(1-cos2πuq))τ(χχ1)(q-1)πi∑χ3modq‍∑u=1q-1‍×L(1,χχ1χ3¯)+O(1){(∑u=1q-1‍)2})×({(∑u=1q-1‍)2}τ(χ1)(q-1)πi∑v=1q-1‍∑χ4modq‍(χ4(v)(1-cos2πvq))τ(χ1)(q-1)πi∑χ4modq‍∑v=1q-1‍×L(1,χ1χ4¯)+O(1){(∑u=1q-1‍)2})=1(q-1)2π2×∑u=1q-1‍∑v=1q-1‍∑χ3modq‍∑χ4modq‍χ3(u)χ4(v)sin2πuqsin2πvq×∑χ1≠χp0χ1(-1)=1‍χχ1(q)χ1(m)τ(χχ1)τ(χ1)×∑χ1≠χp0χ1(-1)=1‍×L(1,χχ1χ3¯)L(1,χ1χ4¯)+-1(q-1)2π2×∑u=1q-1‍∑v=1q-1‍∑χ3modq‍∑χ4modm‍χ3(u)χ4(v)×∑u=1q-1‍∑v=1q-1‍∑χ3modq‍∑χ4modm‍×(1-cos2πuq)(1-cos2πvq)×∑χ1≠χp0χ1(-1)=-1‍χχ1(q)χ1(q)τ(χχ1)τ(χ1)×∑χ1≠χp0χ1(-1)=-1‍×L(1,χχ1χ3¯)L(1,χ1χ4¯)+O(p1/2+ϵ)≪p3/2+ϵ,
where the ≪ constant is only concerned with q.
Whenχ(-1)=-1, by the similar method, we can also obtain
(36)∑χ1modpχ1≠χp0‍χχ1(q)χ1(q)∑a≤(p-1)/q‍χχ1(a)∑b≤(p-1)/q‍χ1(b)≪p3/2+ϵ,
where the ≪ constant is only concerned with q. Combining (35) and (36), we can obtain Lemma 7. This completes the proof of Lemma 7.Lemma 8.
Letq be a fixed odd prime and let p be a prime with p>q, let χ denote the nonprincipal Dirichlet character modulo p, let χ1,χ2 denote the Dirichlet character modulo p,q respectively, then, we have
(37)∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍∑a≤p-1q†a‍∑b≤p-1q†b‍χχ1χ2(a)χ1χ¯2(b)≪p3/2+ϵ,
where the ≪ constant is only concerned with q.Proof.
According to the properties of Dirichlet character, we can get(38)∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍∑a≤p-1q†a‍∑b≤p-1q†b‍χχ1χ2(a)χ1χ¯2(b)=∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍(∑a≤p-1‍χχ1χ2(a)-∑a≤p-1q∣a‍χχ1χ2(a))∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍×(∑b≤p-1‍χ1χ¯2(b)-∑b≤p-1q∣b‍χ1χ¯2(b))=∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍({∑χ2modqχ2≠χq0‍}∑a≤p-1‍χχ1χ2(a)-χχ1χ2(q)∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍×∑a≤(p-1)/q‍χχ1χ2(a){∑χ2modqχ2≠χq0‍})×(∑b≤p-1‍χ1χ¯2(b)-χ1χ¯2(q)∑b≤(p-1)/q‍χ1χ¯2(b))=∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍∑a≤p-1‍χχ1χ2(a)∑b≤p-1‍χ1χ¯2(b).
For primesp and q, let χ′=χχ1 be a nonprincipal and primitive character modulo p, χ2 is also a primitive character modulo q, so χχ1χ2 is a primitive character modulo pq; therefore, from Lemmas 3, 4, and 6 and from (38), it is clear that when χ(-1)=1, we can obtain
(39)∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍∑a≤p-1q†a‍∑b≤p-1q†b‍χχ1χ2(a)χ1χ¯2(b)=∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍χ1χ2(-1)=1×(τ(χχ1χ2)(q-1)π∑u=1q-1‍∑χ3modq‍(χ3(u)sin2πuq)τ(χχ1χ2)(q-1)π∑u=1q-1‍∑χ3modq‍×L(1,χχ1χ2χ3¯)+O(1){∑u=1q-1‍∑χ3modq‍})×({(∑v=1q-1‍)2}τ(χ1χ¯2)(q-1)π∑v=1q-1‍∑χ4modq‍(χ4(v)sin2πvq)τ(χ1χ¯2)(q-1)π∑v=1q-1‍∑χ4modq‍×L(1,χ¯1χ2χ¯4)+O(1){(∑v=1q-1‍)2})+∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍χ1χ2(-1)=-1×({(∑v=1q-1‍)2}τ(χχ1χ2)(q-1)πi∑u=1q-1‍∑χ3modq‍(χ3(u)(1-cos2πuq))τ(χχ1χ2)(q-1)πi∑u=1q-1‍∑χ3modq‍×L(1,χχ1χ2χ3¯)+O(1){(∑v=1q-1‍)2})×({(∑v=1q-1‍)2}τ(χ1χ¯2)(q-1)πi∑v=1q-1∑χ4modq‍(χ4(v)(1-cos2πvq))τ(χ1χ¯2)(q-1)πi∑v=1q-1‍∑χ4modq‍×L(1,χ¯1χ2χ¯4)+O(1){(∑v=1q-1‍)2})=1(q-1)2π2∑u=1q-1‍∑v=1q-1‍∑χ2modqχ2≠χq0‍∑χ3modq‍∑χ4modq‍×χ2(p)τ(χ2)χ¯2(p)τ(χ¯2)χ3(u)χ4(v)sin2πuq×sin2πvq∑χ1≠χp0χ1χ2(-1)=1‍χχ1(q)χ1(q)τ(χχ1)τ(χ1)×sin2πvq∑χ1≠χp0χ1χ2(-1)=1‍×L(1,χχ1χ2χ3¯)L(1,χ¯1χ2χ¯4)+-1(q-1)2π2∑u=1q-1‍∑v=1q-1‍∑χ2modqχ2≠χq0‍∑χ3modq‍∑χ4modq‍×χ2(p)τ(χ2)χ¯2(p)τ(χ¯2)χ3(u)χ4(v)×(1-cos2πuq)(1-cos2πvq)×∑χ1≠χp0χ1χ2(-1)=-1‍χχ1(q)χ1(q)τ(χχ1)τ(χ1)∑χ1≠χp0χ1χ2(-1)=-1‍×L(1,χχ1χ2χ3¯)L(1,χ¯1χ2χ¯4)≪p3/2+ϵ,
where the ≪ constant is only concerned with q.
Whenχ(-1)=-1, in the similar way, we can also obtain
(40)∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍∑a≤p-1q†a‍∑b≤p-1q†b‍χχ1χ2(a)χ1χ¯2(b)≪p3/2+ϵ,
where the ≪ constant is only concerned with q.
Therefore, from (39) and (40), we can easily get Lemma 8. This completes the proof of Lemma 8.


## 3. Proof of Theorem

In this section, we will complete the proof of the theorem. According to the orthogonality relation for character sums, we have(41)∑a≤pa∈E(q,p)‍χ(a)=∑‍b=1p-1∑‍a=1p-1a≡b(modq)ab≡1(modp)χ(a)=1p-1∑χ1modp‍∑‍b=1p-1∑‍a=1p-1a≡b(modq)χχ1(a)χ1(b)=1p-1∑‍a=1p-1∑‍b=1p-1a≡b(modq)χ(a)+1p-1∑χ1modpχ1≠χp0‍∑‍a=1p-1∑‍b=1p-1a≡b(modq)χχ1(a)χ1(b)=1p-1∑a=1p-1‍χ(a)(∑b=1b≡a(modq)p-1‍1)+1p-1∑χ1modpχ1≠χp0‍∑a=1q∣ap-1‍∑b=1q∣bp-1‍χχ1(a)χ1(b)+1p-1∑χ1modpχ1≠χp0‍∑‍a=1q†ap-1∑‍b=1q†bp-1a≡b(modq)χχ1(a)χ1(b)=1p-1∑a=1p-1‍χ(a)(pq+O(1))+1p-1∑χ1modpχ1≠χp0‍χχ1(q)χ1(q)×∑a≤(p-1)/q‍χχ1(a)∑b≤(p-1)/q‍χ1(b)+1(p-1)ϕ(q)×∑χ1modpχ1≠χp0‍∑χ2modq‍∑a≤p-1q†a‍∑b≤p-1q†b‍χχ1(a)χ1(b)∑χ1modpχ1≠χp0‍∑χ2modq‍∑a≤p-1q†a‍∑b≤p-1q†b‍×χ2(a)χ¯2(b)=p(p-1)q∑a=1p-1‍χ(a)+O(1)+1p-1∑χ1modpχ1≠χp0‍χχ1(q)χ1(q)×∑a≤(p-1)/q‍χχ1(a)∑b≤(p-1)/q‍χ1(b)+1(p-1)ϕ(q)×∑χ1modpχ1≠χp0‍∑a≤p-1m†a‍∑b≤p-1q†b‍χχ1(a)χ1(b)+1(p-1)ϕ(q)×∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍∑a≤p-1q†a‍∑b≤p-1q†b‍χχ1(a)χ1(b)∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍∑a≤p-1q†a‍∑b≤p-1q†b‍×χ2(a)χ¯2(b)=O(1)+1p-1∑χ1modpχ1≠χp0‍χχ1(q)χ1(q)×∑a≤(p-1)/q‍χχ1(a)∑b≤(p-1)/q‍χ1(b)+1(p-1)ϕ(q)×∑χ1modpχ1≠χp0‍∑a≤p-1q†a‍∑b≤p-1q†b‍χχ1(a)χ1(b)+1(p-1)ϕ(q)×∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍∑a≤p-1q†a‍∑b≤p-1q†b‍χχ1χ2(a)χ1χ¯2(b).Note that for any nonprincipal Dirichlet characterχ modulo k (k≥3 is an positive integer), we have ∑n=1kχ(n)=0, hence, we obtain
(42)∑a≤p-1q†a‍∑b≤p-1q†b‍χχ1(a)χ1(b)=(∑a≤p-1‍χχ1(a)-∑a≤p-1q∣a‍χχ1(a))×(∑b≤p-1‍χ1(b)-∑b≤p-1q∣b‍χ1(b))=∑a≤p-1q∣a‍χχ1(a)∑b≤p-1q∣b‍χ1(b)=χχ1(q)χ1(q)∑a≤(p-1)/q‍χχ1(a)∑b≤(p-1)/q‍χ1(b).
From (41) and (42) and Lemmas 7 and 8, we get
(43)∑a≤pa∈E(q,p)‍χ(a)=1p-1(1+1ϕ(q))×∑χ1modpχ1≠χp0‍χχ1(q)χ1(q)∑a≤(p-1)/q‍χχ1(a)∑b≤(p-1)/q‍χ1(b)+1(p-1)ϕ(q)×∑χ1modpχ1≠χp0‍∑χ2modqχ2≠χq0‍∑a≤p-1q†a‍∑b≤p-1q†b‍χχ1χ2(a)χ1χ¯2(b)+O(1)≪p1/2+ϵ,
where the ≪ constant only depends on q. This completes the proof of Theorem.



---

*Source: 750964-2013-09-15.xml*