Research article

A note on the hybrid power mean involving the cubic Gauss sums and Kloosterman sums

  • Received: 17 March 2022 Revised: 19 June 2022 Accepted: 22 June 2022 Published: 01 July 2022
  • MSC : 11L03, 11L05

  • The main purpose of this paper is to study the calculating problem of one kind hybrid power mean involving the cubic Gauss sums and Kloosterman sums, and using the elementary methods, analytic methods and the properties of the classical Gauss sums to give some interesting calculating formula for them. At the same time, the paper also provides an effective calculating method for the study of the hybrid power mean involving the k-th Gauss sums and Kloosterman sums.

    Citation: Xiaoxue Li, Wenpeng Zhang. A note on the hybrid power mean involving the cubic Gauss sums and Kloosterman sums[J]. AIMS Mathematics, 2022, 7(9): 16102-16111. doi: 10.3934/math.2022881

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  • The main purpose of this paper is to study the calculating problem of one kind hybrid power mean involving the cubic Gauss sums and Kloosterman sums, and using the elementary methods, analytic methods and the properties of the classical Gauss sums to give some interesting calculating formula for them. At the same time, the paper also provides an effective calculating method for the study of the hybrid power mean involving the k-th Gauss sums and Kloosterman sums.



    Let q be a positive integer. For any integers m and n, The famous Kloosterman sums K(m,n;q) is defined as follows:

    K(m,n;q)=q    a=1(a,q)=1e(ma+n¯aq),

    where e(y)=e2πiy, i2=1 and ¯a is defined by the congruence a¯a1modq.

    This sum plays a very important role in the study of analytic number theory, many number theory problems are closely related to it, so many scholars have studied the various properties about K(m,n;q) and obtained a series of significant research results. For example, Kloosterman's pioneering work [3] proved the identity

    pm=1|p1a=1e(ma+¯ap)|4=p(2p23p3),

    where p is an odd prime.

    Estermann [4] studied the upper bound estimation of K(m,n;q), and obtained the best estimation

    |K(m,n;q)|(m,n,q)12d(q)q12,

    where (m,n,q) denotes the greatest common divisor of m, n and q, d(q) denotes the Dirichlet divisor function.

    Zhang [5] used the elementary method to study the fourth power mean of K(m,n;q) for general modulo q, and proved the identity

    qm=1|qa=1(a,q)=1e(ma+n¯aq)|4=3ω(q)q2ϕ(q)pq(2313p43(p1)),

    where ω(q) denotes the number of all distinct prime divisors of q, ϕ(q) denotes the Euler's function, and pq denotes the product over all prime divisors of q such that pq and p2q.

    In addition, Chen and Hu [6] studied the hybrid power mean involving the cubic Gauss sums A(m) and Kloosterman sums K(m,n;p), i.e.,

    Sk(p)=p1m=1Ak(m)K2(m,1;p)=p1m=1(p1a=0e(ma3p))k|p1a=1e(ma+¯ap)|2,

    and proved that for any positive integer k1, one has the third-order linear recurrence formula

    Sk+3(p)=3pSk+1(p)+dpSk(p),

    with the first three values

    S1(p)=2p2+dpA(1)pA2(1),
    S2(p)=2p3+2(d1)p2+p(d2p)A(1)dpA2(1),
    S3(p)=(d+6)p3+3dp2A(1)3p2A2(1)dp(p+1),

    where 4p=d2+27b2, and d is uniquely determined by d1mod3.

    Some related contents can also be found in [7,8,9,10,11,12,13,14,15,16], we would not list them all here.

    Obviously, the result in [6] is meaningful, it gives a calculation method for all Sk(p). However, this result does not look pretty, because it contains parameter A(1). Therefore, we can not calculate the exact value of Sk(p).

    On the other hand, Kloosterman sums can be used to solve the Waring-Goldbach problem, to solve the problems of prime distribution over short intervals, mean estimation of the Riemann Zata function and Fourier coefficients in modular form, etc. Not only that, the properties of Kloosterman sums can also be used to determine the generalized Hamming weight or the weight distribution of some linear codes, so it has important applications in the field of communication, cryptography and coding theory. Therefore, it is necessary to further study the properties of Klooaterman sums.

    Based on the above, in this paper, as a note of [6], we modify some parameters in [6] as follows:

    Ck(h,p)=p1m=1Ak(¯m)Kh(m,p)=p1m=1(p1b=0e(¯mb3p))k(p1a=1e(ma+¯ap))h,

    then we will get some more concise and prettier results. That is, we will prove the following two conclusions:

    Theorem 1. Let p be a prime with p1mod3. Then for any integer k, we have the third-order linear recurrence formula

    Ck(1,p)=3pCk2(1,p)+dpCk3(1,p),  k3,

    with the first three values C0(1,p)=1, C1(1,p)=dp and C2(1,p)=2p(p+1), where 4p=d2+27b2, and d is uniquely determined by d1mod3.

    Theorem 2. Let p be a prime with p1mod3. Then for any integer k, we have the third-order linear recurrence formula

    Ck(2,p)=3pCk2(2,p)+dpCk3(2,p),  k3,

    with the first three values

    C0(2,p)=p2p1,C1(2,p)=p(d22p)

    and

    C2(2,p)=p(2p2+dp2p2).

    From these two theorems we may immediately deduce the following corollaries:

    Corollary 1. For any prime p with p1mod3, we have the identity

    p1m=1p1a=1e(ma+¯ap)p1b=0e(¯mb3p)=2p1d.

    Corollary 2. For any prime p with p1mod3, we have the identity

    p1m=1|p1a=1e(ma+¯ap)p1b=0e(¯mb3p)|2=3p25dp+d33p3d2.

    Corollary 3. For any prime p with p1mod3, we have the identity

    p1m=1|p1b=0e(¯mb3p)|4|p1a=1e(ma+¯ap)|2=p2(6p2+dp+d36p6).

    Some notes: In our theorems, we only consider the prime p with p1mod3. In fact if 3(p1), then for any integer m with (m,p)=1, we have A(¯m)=0 and Ck(h,p)=0 for all k1. So in this case, the results are trivial.

    Note that K(m,n;p) is a real number, so if we replace A(¯m) in Theorem 2 with A(m), then the values of Ck(h,p) are the same as in [6].

    In addition, since A(¯m)0 for all (m,p)=1, so the third-order linear recurrence formulas in Theorem 1 and Theorem 2 are also hold for all integers k<3.

    For any integer h3, whether there is an exact calculating formula for Ck(h,p) is an open problem. It remains to be further studied.

    To complete the proofs of our all results, we need three necessary lemmas. The proofs of these lemmas requires some knowledge of elementary or analytic number theory, all these can be found in references [1,2], so we do not repeat them here. First we have the following:

    Lemma 1. Let p be an odd prime with p1mod3. Then for any third-order character λ modulo p, we have the identity

    τ3(λ)+τ3(¯λ)=dp,

    where τ(χ)=p1a=1χ(a)e(ap) denotes the classical Gauss sums, 4p=d2+27b2, and d is uniquely determined by d1mod3.

    Proof. The proof of this Lemma see Zhang and Hu [17] or Berndt and Evans [18].

    Lemma 2. Let p be a prime with p1mod3, then for any three-order character λ modulo p, we have

    p1m=1(p1a=0e(¯ma3p))(p1a=1e(ma+¯ap))=dp

    and

    p1m=1(p1a=0e(¯ma3p))2(p1a=1e(ma+¯ap))=2p(p+1).

    Proof. Let p be a prime with p1mod3, let λ be any third-order character modulo p. Then for any integer m with (m,p)=1, from the properties of the third-order character modulo p we have the identity

    A(¯m)=1+p1a=1(1+λ(a)+¯λ(a))e(¯map)=λ(m)τ(λ)+¯λ(m)τ(¯λ). (2.1)

    From the properties of the classical Gauss sums we have

    p1m=1λ(m)(p1a=1e(ma+¯ap))=p1a=1e(¯ap)p1m=1λ(m)e(map)=τ(λ)p1a=1¯λ(a)e(¯ap)=τ2(λ) (2.2)

    and

    p1m=1¯λ(m)(p1a=1e(ma+¯ap))=τ2(¯λ). (2.3)

    From (2.1)–(2.3) and Lemma 1, we have

    p1m=1(p1a=0e(¯ma3p))(p1a=1e(ma+¯ap))=τ3(λ)+τ3(¯λ)=dp. (2.4)

    Note that τ(λ)τ(¯λ)=p, from (2.4) and λ2=¯λ we also have

    A2(¯m)=λ(m)τ2(¯λ)+¯λ(m)τ2(λ)+2p. (2.5)
    p1m=1(p1a=1e(ma+¯ap))=1. (2.6)

    From (2.2), (2.3), (2.5), (2.6) and Lemma 1, we have

    p1m=1(p1a=0e(¯ma3p))2(p1a=1e(ma+¯ap))=p1m=1(¯λ(m)τ2(λ)+λ(m)τ2(¯λ)+2p)(p1a=1e(ma+¯ap))=τ2(λ)τ2(¯λ)+τ2(λ)τ2(¯λ)+2p=2p(p+1). (2.7)

    Now Lemma 2 follows from (2.4) and (2.7).

    Lemma 3. Let p be a prime with p1mod3. Then we have the identities

    p1m=1A(¯m)(p1a=1e(ma+¯ap))2=p(d22p)

    and

    p1m=1A2(¯m)(p1a=1e(ma+¯ap))2=p(2p2+dp2p2).

    Proof. From the properties of the classical Gauss sums we have

    p1m=1λ(m)(p1a=1e(ma+¯ap))2=τ(λ)p1a=1p1b=1¯λ(a+b)e(¯a+¯bp)=τ(λ)p1a=1¯λ(a+1)p1b=1¯λ(b)e(¯b(¯a+1)p)=τ2(λ)p1a=1¯λ(a+1)¯λ(¯a+1)=τ2(λ)p1a=1λ(a)¯λ2(a+1)=τ2(λ)p1a=1λ(a)λ(a+1)=τ2(λ)τ(¯λ)p1b=1¯λ(b)p1a=1λ(a)e(b(a+1)p)=τ3(λ)τ(¯λ)p1b=1¯λ2(b)e(bp)=τ4(λ)τ(¯λ)=τ5(λ)p. (2.8)

    Similarly, we also have

    p1m=1¯λ(m)(p1a=1e(ma+¯ap))2=τ5(¯λ)p. (2.9)

    From (2.1), (2.8), (2.9) and Lemma 1, we have

    p1m=1A(¯m)(p1a=1e(ma+¯ap))2=p1m=1(λ(m)τ(λ)+¯λ(m)τ(¯λ))(p1a=1e(ma+¯ap))2=τ6(λ)p+τ6(¯λ)p=1p[(τ3(λ)+τ3(¯λ))22τ3(λ)τ3(¯λ)]=1p(d2p22p3)=p(d22p). (2.10)

    Note that the identity

    p1m=1(p1a=1e(ma+¯ap))2=p2p1, (2.11)

    from (2.5), (2.8), (2.9), (2.11) and Lemma 1, we have the identity

    p1m=1A2(¯m)(p1a=1e(ma+¯ap))2=p1m=1(¯λ(m)τ2(λ)+λ(m)τ2(¯λ)+2p)(p1a=1e(ma+¯ap))2=τ2(λ)τ5(¯λ)p+τ2(¯λ)τ5(λ)p+2p(p2p1)=p(τ3(λ)+τ3(¯λ))+2p(p2p1)=p(2p2+dp2p2). (2.12)

    Now Lemma 3 follows from (2.10) and (2.12).

    Applying Lemma 1 and Lemma 2 we can easily prove our theorems. First we prove Theorem 1. For any prime p with p1mod3 and integer m with (m,p)=1, from (2.1) and Lemma 1 we have the identity

    A3(¯m)=τ3(λ)+τ3(¯λ)+3pA(¯m)=dp+3pA(¯m). (3.1)

    From (3.1) we can deduce that

    Ak(¯m)=Ak3(¯m)A3(¯m)=Ak3(¯m)(dp+3pA(¯m))=3pAk2(¯m)+dpAk3(¯m). (3.2)

    From (3.2) we can deduce the third-order linear recurrence formula

    Ck(h,p)=p1m=1Ak(¯m)(p1a=1e(ma+¯ap))h=3pCk2(h,p)+dpCk3(h,p),  k3. (3.3)

    Taking h=1, from Lemma 2 we have C0(1,p)=1, C1(1,p)=dp and C2(1,p)=2p(p+1). This proves Theorem 1.

    Now we prove Theorem 2. Taking h=2 in (3.3), from (2.11) and Lemma 3 we have

    C0(2,p)=p2p1,C1(2,p)=p(d22p)

    and

    C2(2,p)=p(2p2+dp2p2).

    This proves Theorem 2.

    From (3.1) we know that A(¯m)0. So formula (3.3) also holds for all integers k<0. Taking k=2 in Theorem 1 we have

    C2(1,p)=3pC0(1,p)+dpC1(1,p)

    or

    p1m=1p1a=1e(ma+¯ap)p1b=0e(¯mb3p)=2p1d.

    This proves Corollary 1.

    Taking k=2 in Theorem 2, note that

    C2(2,p)=3pC0(2,p)+dpC1(2,p)

    and

    C1(2,p)=3pC1(2,p)+dpC2(2,p).

    Therefore, we have

    C1(2,p)=1dp[2p3+dp22p22p3p(p2p1)]=p2+dp+p+1d

    and

    C2(2,p)=1dp(C1(2,p)3pC1(2,p))=1dp[d2p2p23pp2+dp+p+1d]=3p25dp+d33p3d2.

    This completes the proofs of our all results.

    The main results of this paper is to give two theorems for the hybrid power mean involving the cubic Gauss sums and Kloosterman sums. In addition, we also obtained a third-order linear recurrence formula for Ck(h,p) with h=1 and 2. That is, for any integer k, we have the three-order linear recurrence formula

    Ck(h,p)=3pCk2(h,p)+dpCk3(h,p)

    with the exact values C0(h,p), C1(h,p) and C2(h,p), where d is uniquely determined by 4p=d2+27b2 and d1mod3.

    At the same time, our results also provides an effective method for the study of the hybrid power mean involving the k-th Gauss sums and Kloosterman sums.

    This work is supported by the N. S. F. (12126357) of China, the Doctoral Scientific Research Project of XAAI. The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper.

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



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