Research article

New identities involving Hardy sums S3(h,k) and general Kloosterman sums

  • Received: 26 May 2020 Accepted: 23 November 2020 Published: 25 November 2020
  • MSC : 11F20, 11L05

  • The main purpose of this paper is to obtain some exact computational formulas or upper bounds for hybrid mean value involving Hardy sums S3(h,p) and general Kloosterman sums K(r,l,λ;p). By applying the properties of Gauss sums and the mean value theorems of Dirichlet L-function, we derive some new identities. As the special cases, we also deduce some exact computational formulas for hybrid mean value involving S3(h,p) and classical Kloosterman sums K(n,p).

    Citation: Wenjia Guo, Yuankui Ma, Tianping Zhang. New identities involving Hardy sums S3(h,k) and general Kloosterman sums[J]. AIMS Mathematics, 2021, 6(2): 1596-1606. doi: 10.3934/math.2021095

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  • The main purpose of this paper is to obtain some exact computational formulas or upper bounds for hybrid mean value involving Hardy sums S3(h,p) and general Kloosterman sums K(r,l,λ;p). By applying the properties of Gauss sums and the mean value theorems of Dirichlet L-function, we derive some new identities. As the special cases, we also deduce some exact computational formulas for hybrid mean value involving S3(h,p) and classical Kloosterman sums K(n,p).


    If h and k are integers with k>0, the classical Dedekind sums S(h,k) are defined as

    S(h,k)=ka=1((ak))((ahk)),

    where

    ((x))={x[x]12,if x is not an integer; 0,if x is an integer. 

    The various properties of S(h,k) were investigated by many authors, one of which is reciprocity theorem (see Tom M. Apostol [1] or L. Carlitz [2]). That is, for all positive integers h and k with (h,k)=1, we have the identity

    S(h,k)+S(k,h)=h2+k2+112hk14.

    Conrey et al. [3] studied the mean value distribution of S(h,k) and deduced the important asymptotic formula

    kh=1|S(h,k)|2m=fm(k)(k12)2m+O((k95+k2m1+1m+1)log3k),

    where kh=1 denotes the summation over all h such that (h,k)=1 and

    n=1fm(n)ns=2ζ2(2m)ζ(4m)ζ(s+4m1)ζ2(s+2m)ζ(s).

    Moreover, X. L. He and W. P. Zhang [4] gave an interesting asymptotic formula for the Dedekind sums with a weight of Hurwitz zeta-function as follows:

    kh=1ζ2(12,hk)S2(h,k)=k3144ζ(3)p|k(11p3)+O(k52exp(3logkloglogk)).

    Other sums analogous to the Dedekind sums are the Hardy sums. Using the notation of Berndt and Goldberg [5], they defined

    S1(h,k)=k1j=1(1)j+1+[hjk],

    where h and k are integers with k>0.

    In 2014, H. Zhang and W. P. Zhang [6] obtained some beautiful identities involving S1(h,k) in the forms of

    p1m=1p1n=1K(m,p)K(n,p)S1(2m¯n,p),
    p1m=1p1n=1|K(m,p)|2|K(n,p)|2S1(2m¯n,p),

    where K(n,p) denotes the reduced form of the general Kloosterman sums attached to a Dirichlet character λ modulo k as

    K(r,l,λ;k)=ka=1λ(a)e(ra+l¯ak),

    where e(x)=e2πix, ¯a denotes the solution of the congruence xa1modk.

    Recently, H. F. Zhang and T. P. Zhang [7] extended the results in [6] to a more general situation as

    p1m=1p1n=1K(m,s,λ;p)¯K(n,t,λ;p)S1(2m¯n,p),
    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S1(2m¯n,p),

    where ¯K(n,t,λ;p) denotes complex conjugate of K(n,t,λ;p).

    Actually there are six forms of Hardy sums (see Berndt [8] and Goldberg [9]). A natural question is whether we can obtain similar results by replacing S1(h,k) with other forms of Hardy sums. Due to some technical reasons, for most of other forms of Hardy sums, the answer is no! Thanks to the important relationships among Hardy sums and Dedekind sums built by R. Sitaramachandrarao [10], we are lucky to find the only one S3(h,p) to replace, with

    S3(h,k)=kj=1(1)j((hjk)).

    Our starting point relies heavily on the following in [10] as:

    Proposition 1. Let k be an odd positive integer, h be an integer with (h,k)=1. Then

    S3(h,k)=2S(h,k)4S(2h,k).

    Then applying the properties of Gauss sums and the mean square value of Dirichlet L-functions, we have

    Theorem 1. Let p be an odd prime. Then for any Dirichlet character λmodp and any integer s, t with (s,p)=(t,p)=1, we have

    p1m=1p1n=1K(m,s,λ;p)¯K(n,t,λ;p)S3(m¯n,p)={p12,if ¯λχ=χ0;p(p1)2,if ¯λχχ0,

    where χ is an odd Dirichlet character modulo p and χ0 is the principal character modulo p.

    Theorem 2. Let p be an odd prime with p1mod4. Then for any Dirichlet character λmodp and any integer s, t with (s,p)=(t,p)=1, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p){=p2(p1)2,if ¯λχχ0, ¯λχχ0;=p(p1)2,if ¯λχχ0, ¯λχ=χ0;p92+12p44p72p3+5p52+p22p3212p,if ¯λχ=χ0, ¯λχ=χ0;p53p4+3p312p212p,if ¯λχ=χ0, ¯λχχ0.

    Theorem 3. Let p be an odd prime with p3mod8. Then for any Dirichlet character λmodp and any integer s, t with (s,p)=(t,p)=1, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p){=p2(p1)26p2h2p,if ¯λχχ0, ¯λχχ0;=p(p1)26ph2p,if ¯λχχ0, ¯λχ=χ0;p92+12p44p72p3+5p52+p22p3212p+6p2h2p,if ¯λχ=χ0, ¯λχ=χ0;p53p4+3p312p212p+6p3h2p,if ¯λχ=χ0, ¯λχχ0,

    where hp denotes the class number of the quadratic field Q(p).

    Theorem 4. Let p be an odd prime with p7mod8. Then for any Dirichlet character λmodp and any integer s, t with (s,p)=(t,p)=1, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p){=p2(p1)2+2p2h2p,if ¯λχχ0, ¯λχχ0;=p(p1)2+2ph2p,if ¯λχχ0, ¯λχ=χ0;p92+12p44p72p3+5p52+p22p3212p+2p2h2p,if ¯λχ=χ0, ¯λχ=χ0;p53p4+3p312p212p+2p3h2p,if ¯λχ=χ0, ¯λχχ0.

    Taking λ=λ0, s=t=1 in Theorems 1–4, we immediately deduce the following results.

    Corollary 1. Let p be an odd prime. Then we have the identity

    p1m=1p1n=1K(m,p)K(n,p)S3(m¯n,p)=p(p1)2.

    Corollary 2. Let p be an odd prime. Then we have

    p1m=1p1n=1|K(m,p)|2|K(n,p)|2S3(m¯n,p)={p2(p1)2,if p1mod4;p2(p1)26p2h2p,if p3mod8;p2(p1)2+2p2h2p,if p7mod8.

    To prove the Theorems, we need the following Lemmas.

    Lemma 1. Let k>2 be an integer. Then for any integer a with (a,k)=1, we have the identity

    S(a,k)=1π2kdkd2ϕ(d)χmoddχ(1)=1χ(a)|L(1,χ)|2,

    where L(1,χ) denotes the Dirichlet L-function corresponding to Dirichlet character χ mod d.

    Proof. See Lemma 2 of [11].

    Lemma 2. Let p be an odd prime, s be any integer with (s,p)=1. Then for any non-principal character χmodp and any Dirichlet character λmodp, we have

    |p1m=1χ(m)K(m,s,λ;p)|={p12,if ¯λχ=χ0;p,if ¯λχχ0.

    Proof. See Lemma 2 of reference [7].

    Lemma 3. Let p be an odd prime, s be any integer with (s,p)=1. Then for any non-principal character χmodp and any Dirichlet character λmodp, we have

    |p1m=1χ(m)|K(m,s,λ;p)|2|={p|τ(¯χ2)|,if ¯λχχ0, ¯λχχ0;p12|τ(¯χ2)|,if ¯λχχ0, ¯λχ=χ0;p|τ(¯χ2)+(p1)|,if ¯λχ=χ0, ¯λχ=χ0;p|τ(¯χ2)τ(¯λχ)+(p1)|,if ¯λχ=χ0, ¯λχχ0,

    where τ(χ)=pa=1χ(a)e(ap) denotes the classical Gauss sums.

    Proof. See Lemma 1 of reference [7].

    Lemma 4. Let p be an odd prime, then we have

    χmodpχ(1)=1|L(1,χ)|2=π212(p1)2(p2)p2,
    χmodpχ(1)=1χ(2)|L(1,χ)|2=π224(p1)2(p5)p2.

    Proof. See Lemma 5 of reference [6].

    Now we come to prove our Theorems.

    Firstly, we prove Theorem 1. Applying Proposition 1 and Lemma 1, we obtain

    p1m=1p1n=1K(m,s,λ;p)¯K(n,t,λ;p)S3(m¯n,p)=2pπ2(p1)χmodpχ(1)=1p1m=1χ(m)K(m,s,λ;p)p1n=1χ(¯n)¯K(n,t,λ;p)|L(1,χ)|24pπ2(p1)χmodpχ(1)=1χ(2)p1m=1χ(m)K(m,s,λ;p)p1n=1χ(¯n)¯K(n,t,λ;p)|L(1,χ)|2=2pπ2(p1)χmodpχ(1)=1|p1m=1χ(m)K(m,s,λ;p)|2|L(1,χ)|24pπ2(p1)χmodpχ(1)=1χ(2)|p1m=1χ(m)K(m,s,λ;p)|2|L(1,χ)|2.

    Then from Lemma 2 and Lemma 4, if ¯λχ=χ0, we have

    p1m=1p1n=1K(m,s,λ;p)¯K(n,t,λ;p)S3(m¯n,p)=2p2π2(p1)χmodpχ(1)=1|L(1,χ)|24p2π2(p1)χmodpχ(1)=1χ(2)|L(1,χ)|2=2p2π2(p1)π212(p1)2(p2)p24p2π2(p1)π224(p1)2(p5)p2=p12.

    While if ¯λχχ0, we have

    p1m=1p1n=1K(m,s,λ;p)¯K(n,t,λ;p)S3(m¯n,p)=2p3π2(p1)χmodpχ(1)=1|L(1,χ)|24p3π2(p1)χmodpχ(1)=1χ(2)|L(1,χ)|2=2p3π2(p1)π212(p1)2(p2)p24p3π2(p1)π224(p1)2(p5)p2=p(p1)2.

    This completes the proof of Theorem 1.

    Then we prove Theorem 2. From Proposition 1 and Lemma 1, we obtain

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)=2pπ2(p1)χmodpχ(1)=1|p1m=1χ(m)|K(m,s,λ;p)|2|2|L(1,χ)|24pπ2(p1)χmodpχ(1)=1χ(2)|p1m=1χ(m)|K(m,s,λ;p)|2|2|L(1,χ)|2.

    Since p1mod4, and notice that |τ(¯χ2)|=p. From Lemma 3 and Lemma 4, if ¯λχχ0, ¯λχχ0, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)=2p4π2(p1)χmodpχ(1)=1|L(1,χ)|24p4π2(p1)χmodpχ(1)=1χ(2)|L(1,χ)|2=2p4π2(p1)π212(p1)2(p2)p24p4π2(p1)π224(p1)2(p5)p2=p2(p1)2.

    Similarly, if ¯λχχ0, ¯λχ=χ0, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)=p(p1)2.

    If ¯λχ=χ0, ¯λχ=χ0, we can obtain

    |p1m=1χ(m)|K(m,s,λ;p)|2|2=p2[(Re τ(¯χ2)+(p1))2+(Im τ(¯χ2))2]=p2[p+2(p1)Re τ(¯χ2)+(p1)2].

    So we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)=2p3π2(p1)χmodpχ(1)=1[p+2(p1)Re τ(¯χ2)+(p1)2]|L(1,χ)|24p3π2(p1)χmodpχ(1)=1χ(2)[p+2(p1)Re τ(¯χ2)+(p1)2]|L(1,χ)|2=2p3[p+(p1)2]π2(p1)χmodpχ(1)=1|L(1,χ)|2+4p3π2χmodpχ(1)=1Re (τ(¯χ2))|L(1,χ)|24p3[p+(p1)2]π2(p1)χmodpχ(1)=1χ(2)|L(1,χ)|28p3π2χmodpχ(1)=1χ(2)Re (τ(¯χ2))|L(1,χ)|2=p(p1)(p2p+1)2+4p3π2χmodpχ(1)=1Re (τ(¯χ2))|L(1,χ)|28p3π2χmodpχ(1)=1χ(2)Re (τ(¯χ2))|L(1,χ)|2.

    Noting that x|x| holds for any real number x, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)|p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)|p(p1)(p2p+1)2+4p72π2χmodpχ(1)=1|L(1,χ)|2+8p72π2χmodpχ(1)=1|L(1,χ)|2=p(p1)(p2p+1)2+4p72π2π212(p1)2(p2)p2+8p72π2π212(p1)2(p2)p2=p92+12p44p72p3+5p52+p22p3212p.

    Similarly, if ¯λχ=χ0, ¯λχχ0, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)p53p4+3p312p212p.

    This completes the proof of Theorem 2.

    Next we turn to prove Theorem 3. Since p3mod4, note that (1p)=χ2(1)=1, L(1,χ2)=πhpp, and τ(¯χ22)=1. From Lemma 3 and Lemma 4, if ¯λχχ0, ¯λχχ0, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)=2p4π2(p1)χmodpχ(1)=1|L(1,χ)|22p4π2(p1)|L(1,χ2)|2+2p3π2(p1)|L(1,χ2)|24p4π2(p1)χmodpχ(1)=1χ(2)|L(1,χ)|2+4p4π2(p1)χ2(2)|L(1,χ2)|24p3π2(p1)χ2(2)|L(1,χ2)|2=p2(p1)22p3π2|L(1,χ2)|2+4p3π2χ2(2)|L(1,χ2)|2=p2(p1)22p2h2p+4p2h2p(2p).

    Similarly, if ¯λχχ0, ¯λχ=χ0, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)=p(p1)22ph2p+4ph2p(2p).

    If ¯λχ=χ0, ¯λχ=χ0, we can obtain

    p2|τ(¯χ22)+(p1)|2=p2[1+2(p1)Re τ(¯χ22)+(p1)2].

    So we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)=2p3π2(p1)χmodpχ(1)=1[p+2(p1)Re τ(¯χ2)+(p1)2]|L(1,χ)|22p3π2(p1)[p+2(p1)Re τ(¯χ22)+(p1)2]|L(1,χ2)|2+2p3π2(p1)[1+2(p1)Re τ(¯χ22)+(p1)2]|L(1,χ2)|24p3π2(p1)χmodpχ(1)=1χ(2)[p+2(p1)Re τ(¯χ2)+(p1)2]|L(1,χ)|2+4p3π2(p1)[p+2(p1)Re τ(¯χ22)+(p1)2]χ2(2)|L(1,χ2)|24p3π2(p1)[1+2(p1)Re τ(¯χ22)+(p1)2]χ2(2)|L(1,χ2)|2=p(p1)(p2p+1)2+4p3π2χmodpχ(1)=1Re (τ(¯χ2))|L(1,χ)|28p3π2χmodpχ(1)=1χ(2)Re (τ(¯χ2))|L(1,χ)|22p2h2p+4p2h2p(2p).

    Similarly, if ¯λχ=χ0, ¯λχχ0, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)=p(p1)(2p22p+1)24p3π2χmodpχ(1)=1Re (τ(¯χ2)τ(¯λχ))|L(1,χ)|2+8p3π2χmodpχ(1)=1χ(2)Re (τ(¯χ2)τ(¯λχ))|L(1,χ)|22p3h2p+4p3h2p(2p).

    Combining the fact that

    (2p)=(1)p218={1,if p±1mod8;1,if p±3mod8,

    we deduce that if p3mod8, then

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)={p2(p1)26p2h2p,if ¯λχχ0, ¯λχχ0;p(p1)26ph2p,if ¯λχχ0, ¯λχ=χ0.

    If ¯λχ=χ0, ¯λχ=χ0, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)=p(p1)(p2p+1)2+4p3π2χmodpχ(1)=1Re (τ(¯χ2))|L(1,χ)|28p3π2χmodpχ(1)=1χ(2)Re (τ(¯χ2))|L(1,χ)|26p2h2p.

    Then

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)p(p1)(p2p+1)2+4p3π2|χmodpχ(1)=1Re (τ(¯χ2))|L(1,χ)|2|+8p3π2|χmodpχ(1)=1χ(2)Re (τ(¯χ2))|L(1,χ)|2|+6p2h2p=p92+12p44p72p3+5p52+p22p3212p+6p2h2p.

    Similarly, if ¯λχ=χ0, ¯λχχ0, we have

    p1m=1p1n=1|K(m,s,λ;p)|2|K(n,t,λ;p)|2S3(m¯n,p)p53p4+3p312p212p+6p3h2p.

    This completes the proof of Theorem 3.

    Theorem 4 can be derived by the same method. This completes the proof of our Theorems.

    In this paper, we obtain some exact computational formulas or upper bounds for hybrid mean value involving Hardy sums and Kloosterman sums (both classical Kloosterman sums and general Kloosterman sums) by applying the properties of Gauss sums and the mean value of Dirichlet L-function. But in some cases, unluckily, it is difficult to get the exact formula. So how to get the exact formula in all cases remains open.

    The authors want to show their great thanks to the anonymous referee for his/her helpful comments and suggestions.

    This work is supported by the National Natural Science Foundation of China (No. 11871317, 11926325, 11926321) and the Fundamental Research Funds for the Central Universities (No. GK201802011).

    The authors declare that there are no conflicts of interest regarding the publication of this paper.



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    1. Jiankang Wang, Zhefeng Xu, Minmin Jia, Distribution of values of Hardy sums over Chebyshev polynomials, 2024, 9, 2473-6988, 3788, 10.3934/math.2024186
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