# On the Hybrid Mean Value Involving Kloosterman Sums and Sums Analogous to Dedekind Sums

Authors: Di Han; Wenpeng Zhang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The main purpose of this paper is using the properties of Gauss sums and the mean value theorem of DirichletL-functions to study one kind of hybrid mean value problems involving Kloosterman sums and sums analogous to Dedekind sums and give two exact computational formulae for them.


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## Body

## 1. Introduction

Letc be a natural number and let d be an integer prime to c. The classical Dedekind sums
(1)S(d,c)=∑j=1c‍((jc))((djc)),
where
(2)((x))={x-[x]-12,ifxisnotaninteger;0,ifxisaninteger,describe the behaviour of the logarithm of the eta-function (see [1, 2]) under modular transformations. Gandhi [3] also introduced another sum analogous to Dedekind sums S(h,k) as follows:
(3)S2(h,k)=∑j=1k‍(-1)j((jk))((hjk)),
where k denotes any positive even number and h denotes any integer with (h,k)=1.About the arithmetical properties ofS2(h,k) and related sums, many authors had studied them and obtained a series of interesting results; see [1–9]. For example, the second author [7] proved the following conclusion.Letk=2βM be a positive integer with β≥1 and (M,2)=1. Then we have the asymptotic formula
(4)∑‍′h=1k|S2(h,k)|2=5112kϕ(k)(35-222β)×∏pα∥M‍[(1+(1/p))2-(1/p3α+1)](1+(1/p)+(1/p2))+O(k·exp(4lnlnklnk)),
where ∑‍h=1′k denotes the summation over all integers 1≤h≤k such that (h,k)=1, ∏pα∥M‍ denotes the product over all prime divisors of M such that pα∣M and pα+1†M, ϕ(k) is the Euler function, and exp(y)=ey.The sumS2(h,k) is important, because it has close relations with the classical Dedekind sums S(h,k). But unfortunately, so far, we knew that all results of S2(h,k) are the properties of their own, or the relationships between S2(h,k) and S(h,k), and had nothing to do with the other arithmetic functions. If we can find some relations between S2(h,k) and other arithmetic function, that will be very useful for further study of the properties of S2(h,k).On the other hand, we introduce the classical Kloosterman sumsK(n,q), which are defined as follows. For any positive integer q>1 and integer n,
(5)K(n,q)=∑‍′b=1qe(nb+b-q),
where b¯ denotes the solution of the congruence x·b≡1modq and e(x)=e2πix.Some elementary properties ofK(n,q) can be found in [10, 11].The main purpose of this paper is using the properties of the Gauss sums and the mean square value theorem of DirichletL-functions to study a hybrid mean value problem involving S2(h,k) and Kloosterman sums and give two exact computational formulae for them. That is, we will prove the following.Theorem 1.
Letp be an odd prime. Then one has the identity
(6)∑m=1p‍(2m-1,p)=1∑n=1p‍(2n-1,p)=1K(2m-1,p)·K(2n-1,p)dddddddddddddd·S2((2m-1)·2n-1¯,2p)=-p(p-1)4,
where (2n-1)·(2n-1)¯≡1mod2p.Theorem 2.
Letp be an odd prime; then one has the identity
(7)∑m=1p‍(2m-1,p)=1∑n=1p‍(2n-1,p)=1|K(2m-1,p)|2·|K(2n-1,p)|2ddddddddddddddd·S2((2m-1)·(2n-1)¯,p)={-14p2(p-1)+3·p2·hp2,ifp≡3mod8;-14p2(p-1)-p2·hp2,ifp≡7mod8;-14p2(p-1),ifp≡1mod4,
where hp denotes the class number of the quadratic field Q(-p).


## 2. Several Lemmas

In this section, we will give several lemmas, which are necessary in the proof of our theorems. Hereinafter, we will use many properties of Gauss sums, all of which can be found in [12], so they will not be repeated here. First we have the following.Lemma 3.
Letp be an odd prime; then one has the identity
(8)∑n=1p‍χ(2n-1)·|K(2n-1,p)|2=χ¯(-1)·τ3(χ)·τ(χ¯2)τ(χ¯).Proof.
It is clear that ifn pass through a complete residue system modp, then 2n-1 also pass through a complete residue system modp. So for any nonprincipal character χmodp, from the properties of Gauss sums τ(χ) (see Theorem 8.9 of [12])
(9)χ(a)=1τ(χ¯)∑b=1p-1‍χ¯(b)e(bap),
we have the identity
(10)∑n=1p‍χ(2n-1)·|K(2n-1,p)|2=∑n=1p‍χ(n)|K(n,p)|2=∑a=1p-1‍∑b=1p-1‍∑n=1p-1‍χ(n)e(n(a-b)+(a¯-b¯)p)=∑a=1p-1‍∑b=1p-1‍∑n=1p-1‍χ(n)e(nb(a-1)+b¯(a¯-1)p)=τ(χ)·∑a=1p-1‍∑b=1p-1‍χ¯(b(a-1))e(b¯(a¯-1)p)=τ2(χ)·∑a=1p-1‍χ¯(a-1)χ¯(a¯-1)=τ2(χ)·∑a=1p-1‍χ(a)χ¯(-(a-1)2)=χ¯(-1)·τ2(χ)·∑a=1p-2‍χ(a+1)χ¯(a2)=χ¯(-1)·τ2(χ)·∑a=1p-2‍χ(a¯+a¯2)=χ¯(-1)·τ2(χ)·∑a=1p-1‍χ(a2+a)=χ¯(-1)·τ2(χ)·1τ(χ¯)∑a=1p-1‍∑b=1p-1‍χ¯(b)e(b(a2+a)p)=χ¯(-1)·τ2(χ)·1τ(χ¯)·∑b=1p-1‍χ¯(b)e(bp)∑a=1p-1‍χ(a)e(bap)=χ¯(-1)·τ3(χ)·1τ(χ¯)·∑b=1p-1‍χ¯2(b)e(bp)=χ¯(-1)·τ3(χ)·τ(χ¯2)τ(χ¯).
This proves Lemma 3.Lemma 4.
Letq>2 be an integer; then for any integer a with (a,q)=1, one has the identity
(11)S(a,q)=1π2q∑d|q‍d2ϕ(d)∑χmodd‍χ(-1)=-1χ(a)|L(1,χ)|2,
where L(1,χ) denotes the Dirichlet L-function corresponding to character χmodd.Proof.
See Lemma 2 of [8].Lemma 5.
Letp be an odd prime. Then for any odd number h with (h,p)=1, one has the identity
(12)S2(h,2p)=-S(h,p)+S(2h,p)+S(2¯h,p),
where 2¯ satisfies the congruence 2·2¯≡1modp.Proof.
Note that the divisors of2p are 1,2,p, and 2p. So from Lemma 4 and the definition of S2(h,2p) and S(h,k) we have
(13)S2(h,2p)=∑j=12p‍(-1)j((j2p))((hj2p))=2∑j=1p‍((jp))((hjp))-∑j=12p‍((j2p))((hj2p))=2S(h,p)-S(h,2p)=2S(h,p)-12π2p∑d|2p‍d2ϕ(d)∑χmodd‍χ(-1)=-1χ(h)|L(1,χ)|2=2S(h,p)-p2π2(p-1)∑χmodp‍χ(-1)=-1χ(h)|L(1,χ)|2(14)-2pπ2(p-1)∑χmod2p‍χ(-1)=-1χ(h)|L(1,χ)|2=2S(h,p)-p2π2(p-1)∑χmodp‍χ(-1)=-1χ(h)|L(1,χ)|2-2pπ2(p-1)∑χmodp‍χ(-1)=-1χ(h)λ(h)|L(1,χλ)|2,
where λ denotes the principal character mod2.
From the Euler infinite product formula (see Theorem 11.6 of [12]) we have,
(15)|L(1,χλ)|2=∏p1‍|1-χ(p1)λ(p1)p1|-2=∏p1>2‍|1-χ(p1)p1|-2=|1-χ(2)2|2·∏p1‍|1-χ(p1)p1|-2=(54-χ(2)2-χ¯(2)2)·|L(1,χ)|2,
where ∏p‍ denotes the product over all primes p.
From Lemma4 we also have the identity
(16)S(h,p)=1π2·pp-1∑χmodp‍χ(-1)=-1χ(h)|L(1,χ)|2.
Note that h is an odd number; combining (14), (15), and (16) we have the identity
(17)S2(h,2p)=2S(h,p)-12S(h,p)-2pπ2(p-1)·∑χmodp‍χ(-1)=-1χ(h)|L(1,χλ)|2=32S(h,p)-2pπ2(p-1)·∑χmodp‍χ(-1)=-1χ(h)(54-χ(2)2-χ¯(2)2)·|L(1,χ)|2=-S(h,p)+S(2h,p)+S(2¯h,p).
This proves Lemma 5.Lemma 6.
Letp be an odd prime. Then one has the identities(A)
(18)∑χmodpχ(-1)=-1|L(1,χ)|2=π212·(p-1)2·(p-2)p2;(B)
(19)∑χmodpχ(-1)=-1χ(2)·|L(1,χ)|2=π224·(p-1)2·(p-5)p2.Proof.
From the definition of Dedekind sums we have(20)S(1,c)=∑a=1c-1‍(ac-12)2=(c-1)(c-2)12c.
Ifp≡1modc, then, from (20) and noting that the reciprocity theorem of Dedekind sums (see [5]), we have the computational formula
(21)S(c,p)=p2+c2+112pc-14-S(p,c)=p2+c2+112pc-14-S(1,c)=p2+c2+112pc-14-(c-1)(c-2)12c=(p-1)(p-1-c2)12pc.
Now taking c=1 in (21), from (16) we may immediately deduce the identity
(22)∑χmodp‍χ(-1)=-1|L(1,χ)|2=π212·(p-1)2·(p-2)p2.
Taking c=2 in (21), from (16) we can also deduce the identity
(23)∑χmodp‍χ(-1)=-1χ(2)·|L(1,χ)|2=π224·(p-1)2·(p-5)p2.
Now Lemma 6 follows from (22) and (23).


## 3. Proof of the Theorems

In this section, we will complete the proof of our theorems. First we prove Theorem1. Note that if χ is a nonprincipal character modp, then |τ(χ)|=p and
(24)|∑‍m=1pχ(2m-1)K(2m-1,p)|=|∑a=1p-1‍∑m=1p‍χ(m)e(ma+a¯p)|=|τ2(χ)|=p.
From (24) and Lemmas 4, 5, and 6 we have
(25)∑m=1p‍(2m-1,p)=1∑n=1p‍(2n-1,p)=1K(2m-1,p)·K(2n-1,p)dddddddddddddd·S2((2m-1)·2n-1¯,2p)=-pπ2(p-1)∑χmodp‍χ(-1)=-1|∑n=1p‍χ(2n-1)·K(2n-1,p)|2dddddddddddddddsdds·|L(1,χ)|2+pπ2(p-1)×∑χmodp‍χ(-1)=-1χ(2)|∑n=1p‍χ(2n-1)·K(2n-1,p)|2dddddddddsdds·|L(1,χ)|2+pπ2(p-1)×∑χmodp‍χ(-1)=-1χ¯(2)|∑n=1p‍χ(2n-1)·K(2n-1,p)|2dddddddddsdds·|L(1,χ)|2=-p3π2(p-1)∑χmodp‍χ(-1)=-1|L(1,χ)|2+p3π2(p-1)×∑χmodp‍χ(-1)=-1χ(2)|L(1,χ)|2+p3π2(p-1)×∑χmodp‍χ(-1)=-1χ¯(2)·|L(1,χ)|2=-p(p-1)(p-2)12+p(p-1)(p-5)12=-p(p-1)4.
This proves Theorem 1.Now we prove Theorem2. If p≡1mod4, then from Lemmas 3, 5, and 6 we have
(26)∑m=1p‍(2m-1,p)=1∑n=1p‍(2n-1,p)=1|K(2m-1,p)|2·|K(2n-1,p)|2ddddddddddddddd·S2((2m-1)·(2n-1)¯,p)=-pπ2(p-1)∑χmodp‍χ(-1)=-1|∑n=1p‍χ(2n-1)·|K(2n-1,p)|2|2dddddddddddddddsdds·|L(1,χ)|2+pπ2(p-1)×∑χmodp‍χ(-1)=-1χ(2)|∑n=1p‍χ(2n-1)·|K(2n-1,p)|2|2dddddddddsdds·|L(1,χ)|2+pπ2(p-1)×∑χmodp‍χ(-1)=-1χ¯(2)|∑n=1p‍χ(2n-1)·|K(2n-1,p)|2|2dddsddddsdds·|L(1,χ)|2=-p4π2(p-1)∑χmodp‍χ(-1)=-1|L(1,χ)|2+p4π2(p-1)×∑χmodp‍χ(-1)=-1χ(2)|L(1,χ)|2+p4π2(p-1)×∑χmodp‍χ(-1)=-1χ¯(2)·|L(1,χ)|2=-112p2(p-1)(p-2)+112p2(p-1)(p-5)=-14p2(p-1).
If p≡3mod4, then note that the Legendre symbol (-1/p)=χ2(-1)=-1, L(1,χ2)=π·hp/p (see Dirichlet's class number formula, Chapter 6 of [13]), and
(27)τ(χ22)=∑a=1p-1‍(ap)2e(ap)=∑a=1p-1‍e(ap)=-1,
so from Lemmas 3, 5, and 6 we have
(28)∑m=1p‍(2m-1,p)=1∑n=1p‍(2n-1,p)=1|K(2m-1,p)|2·|K(2n-1,p)|2ddddddddddddddd·S2((2m-1)·(2n-1)¯,p)=-pπ2(p-1)∑χmodp‍χ(-1)=-1|∑n=1p‍χ(2n-1)·|K(2n-1,p)|2|2dddddddddddddddsdds·|L(1,χ)|2+pπ2(p-1)×∑χmodp‍χ(-1)=-1χ(2)|∑n=1p‍χ(2n-1)·|K(2n-1,p)|2|2dddddddddsdds·|L(1,χ)|2+pπ2(p-1)×∑χmodp‍χ(-1)=-1χ¯(2)|∑n=1p‍χ(2n-1)·|K(2n-1,p)|2|2dddddddddsdds·|L(1,χ)|2=-p4π2(p-1)∑χmodp‍χ(-1)=-1|L(1,χ)|2+p4π2(p-1)×∑χmodp‍χ(-1)=-1χ(2)|L(1,χ)|2+p4π2(p-1)×∑χmodp‍χ(-1)=-1χ¯(2)·|L(1,χ)|2+p3π2·|L(1,χ2)|2-p3π2·(2p)·|L(1,χ2)|2-p3π2·(2p)·|L(1,χ2)|2=-14p2(p-1)+p2·hp2-2(2p)·p2·hp2.
Note that (2/p)=(-1)(p2-1)/8=-1 if p≡3mod8; and (2/p)=1 if p≡7mod8, from (28) we may immediately deduce
(29)∑m=1p‍(2m-1,p)=1∑n=1p‍(2n-1,p)=1|K(2m-1,p)|2·|K(2n-1,p)|2ddddddddddddddd·S2((2m-1)·(2n-1)¯,p)={-14p2(p-1)+3·p2·hp2,ifp≡3mod8;-14p2(p-1)-p2·hp2,ifp≡7mod8.
Now Theorem 2 follows from (26) and (29).This completes the proofs of all results.



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*Source: 760505-2013-10-31.xml*