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Research article

On the fourth power mean of one special two-term exponential sums

  • Received: 17 August 2022 Revised: 14 January 2023 Accepted: 28 January 2023 Published: 06 February 2023
  • MSC : 11L03, 11L05

  • The main purpose of this paper is using the elementary methods and the number of the solutions of some congruence equations to study the calculating problem of the fourth power mean of one special two-term exponential sums, and give an exact calculating formula for it.

    Citation: Wenpeng Zhang, Yuanyuan Meng. On the fourth power mean of one special two-term exponential sums[J]. AIMS Mathematics, 2023, 8(4): 8650-8660. doi: 10.3934/math.2023434

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  • The main purpose of this paper is using the elementary methods and the number of the solutions of some congruence equations to study the calculating problem of the fourth power mean of one special two-term exponential sums, and give an exact calculating formula for it.



    Let p be an odd prime. For any integers k>h1 and integers m and n with (n,p)=1, the two-term exponential sums S(m,n,k,h;p) is defined as

    S(m,n,k,h;p)=p1a=0e(mak+nahp),

    where e(y)=e2πiy and i2=1.

    This sum play a very important role in the study of analytic number theory, so many number theorists and scholars had studied the various properties of S(m,n,k,h;p), and obtained a series of meaningful research results. For example, H. Zhang and W. P. Zhang [1] proved that for any odd prime p, one has

    p1m=1|p1a=0e(ma3+nap)|4={2p3p2 if  3p1;2p37p2 if  3p1, (1.1)

    where n represents any integer with (n,p)=1.

    W. P. Zhang and D. Han [2] used elementary and analytic methods to obtain the identity:

    p1m=1|p1a=0e(a3+map)|6=5p48p3p2, (1.2)

    where p denotes an odd prime with 3(p1).

    Recently, W. P. Zhang and Y. Y. Meng [3] studied the sixth power mean of S(m,1,3,1;p), and proved that for any odd prime p, we have the identities

    p1m=1|p1a=0e(ma3+ap)|6={5p3(p1), if  p5mod6;p2(5p223pd2), if p1mod6, (1.3)

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

    On the other hand, L. Chen and X. Wang [4] studied the fourth power mean of S(m,1,4,1;p), and proved that the identities

    p1m=1|p1a=0e(ma4+ap)|4={2p2(p2), if  p7mod12,2p3, if p11mod12,2p(p210p2α2), if p1mod24,2p(p24p2α2), if p5mod24,2p(p26p2α2), if p13mod24,2p(p28p2α2), if p17mod24, (1.4)

    where α=α(p)=p12a=1(a+¯ap) is an integer satisfying the following identity (see Theorem 4–11 in [5]):

    p=α2+β2=(p12a=1(a+¯ap))2+(p12a=1(a+r¯ap))2,

    (p) denotes the Legendre's symbol, r is any quadratic non-residue modulo p and ¯a is the solution of the congruence equation ax1modp.

    In addition, T. T. Wang and W. P. Zhang [6] gave calculating formulae of the fourth and sixth power mean of S(m,n,2,1;p).

    From the formulae (1.1)–(1.4), it is not difficult to see that the content of all these papers only involves h=1 in S(m,n,k,h;p). Through searching literature, we have not found papers dealing with the fourth power mean of the two-term exponential sums S(m,n,k,2;p) so far. Therefore, when k>h=2, it is difficult to obtain some ideal results.

    In this paper, we use the elementary and analytic methods, and the number of the solutions of some congruence equations to study the calculating problem of the 2k-th power mean:

    S2k(p)=p1m=0|p1a=0e(ma3+a2p)|2k,

    and give an exact calculating formula for S4(p) with p3mod4.

    That is, we give the following two conclusions:

    Theorem 1.1. Let p be a prime with p11 mod 12, then we have the identity

    p1m=0|p1a=0e(ma3+a2p)|4=2p3p2(3+p1a=1(a1+¯ap)),

    where (p) denotes the Legendre's symbol modulo p, and ¯a is the solution of the congruence equation ax1 mod p.

    Theorem 1.2. Let p be a prime with p7 mod 12, then we have the identity

    p1m=0|p1a=0e(ma3+a2p)|4=2p3p2(9p1a=1(a1+¯ap)).

    To complete the proofs of our theorems, we need four simple lemmas. Of course, the proofs of these lemmas need some knowledge of elementary or analytic number theory, all these can be found in references [5] and [7,8], so we do not repeat them here. First, according to W. P. Zhang and J. Y. Hu [9] or B. C. Berndt and R. J. Evans [10], we have the following:

    Lemma 2.1. Let p be an odd prime with p1 mod 3. 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 d1 mod 3.

    Related results can also be found in [11,12,13].

    Lemma 2.2. Let p be an odd prime, then we have the identities

    p1a=0p1c=0a3+1c3modpa2+1c2modp1=2+(2p)

    and

    p1a=0p1b=0p1c=0a3+b3c3+1modpa2+b2c2+1modp1=3p5(2p)+(3p)p1a=1(a1+ˉap),

    where (p) denotes the Legendre's symbol modulo p, and a¯a1 mod p.

    Proof. We only prove the second formula in Lemma 2.2. Similarly, we can deduce the first one. It is clear that from the properties of the complete residue system and the Legendre's symbol modulo p we have

    p1a=0p1b=0p1c=0a3+b3c3+1modpa2+b2c2+1modp1=p1a=0p1b=0p1c=0a3+3a2c+3ac2+b3+3b2+3b0modpa2+2ac+b2+2b0modp1=p1a=0p1b=1p1c=0a3+3a2c+3ac2+1+3¯b+3¯b20modpa2+2ac+1+2¯b0modp1+p1a=0p1c=0a3+3a2c+3ac20modpa2+2ac0modp1=p1a=0p1b=0p1c=0a3+3a(2c+a)2+1+3(2b+1)20modpa(a+2c)+1+2b0modp1+pp1a=0p1c=0a3+3a2c+3ac2+10modpa2+2ac+10modp1=p1a=0p1b=0p1c=0a3+3ac2+1+3b20modpac+b0modp1+pp1a=0p1c=0a3+3a(2c+a)2+40modpa(2c+a)+10modp1=p1a=0p1c=0(a+1)(a2a+1+3ac2)0modp1+pp1a=1p1c=1a3+3ac2+40modpac+10modp1=2p+p2a=0p1c=0a2a+1+3ac20modp1p1a=1a3+3¯a+40modp1=2p2+p1a=1p1c=03a(a2a+1)+c20modp1p1a=1(a+1)(a3a2+a+3)0modp1=2p3+p1a=1(1+(3a(a2a+1)p))p2a=1a3a2+a+30modp1=3p4+(3p)p1a=1(a1+¯ap)p2a=1(a+1)(a22a+3)0modp1=3p4+(3p)p1a=1(a1+¯ap)p1a=0(a1)2+20modp1=3p5(2p)+(3p)p1a=1(a1+¯ap).

    This proves Lemma 2.2.

    Lemma 2.3. Let p be an odd prime, then we have the identities

    p1a=0p1c=0a3+1c3modp1={p if 3(p1);p+d2 if 3(p1),

    and

    p1a=0p1b=0p1c=0a3+b3c3+1modp1={p2 if 3(p1);p2d+6p if 3(p1),

    where d is the same as defined in Lemma 2.1.

    Proof. We only prove the second formula in Lemma 2.3. Similarly, we can deduce the first one. If (3,p1)=1, then from the properties of the complete residue system modulo p we have

    p1a=0p1b=0p1c=0a3+b3c3+1modp1=p1a=0p1b=0p1c=0a+bc+1modp1=p2. (2.1)

    If (3,p1)=3, let λ denote any third-order primitive character modulo p and

    A(m)=p1a=0e(ma3p).

    Then for any integer m with (m,p)=1, from Lemma 2.1, the definition and properties of the classical Gauss sums we have

    A(m)=1+p1a=1(1+λ(a)+¯λ(a))e(map)=¯λ(m)τ(λ)+λ(m)τ(¯λ), (2.2)

    and

    A3(m)=τ3(λ)+τ3(¯λ)+3pA(m)=dp+3pA(m). (2.3)

    Note that A(m)=A(m), then from (2.2), (2.3) and the properties of the trigonometric sums we have

    p1a=0p1b=0p1c=0a3+b3c3+1modp1=1pp1a=0p1b=0p1c=0p1m=0e(m(a3+b3c31)p)=p2+1pp1m=1e(mp)A2(m)A(m)=p2+1pp1m=1e(mp)A3(m)=p2+1pp1m=1(dp+3pA(m))e(mp)=p2d+6τ(λ)¯τ(λ)=p2d+6p. (2.4)

    Now Lemma 2.3 follows from (2.1) and (2.4).

    Lemma 2.4. Let p be an odd prime, then we have the identities

    p1a=0p1c=0a3c3modpe(a2c2p)={p, if 3(p1);p2+τ(χ2)p1a=1a31modp(a21p), if 3(p1),

    where χ2=(p) denotes the Legendre's symbol modulo p.

    Proof. For any integer n with (n,p)=1, note the identity

    p1a=0e(na2p)=1+p1a=1(1+χ2(a))e(nap)=χ2(n)τ(χ2). (2.5)

    If 3(p1), then from (2.5) and the properties of the complete residue system modulo p and Lemma 2.2, we have

    p1a=0p1c=0a3c30modpe(a2c2p)=1+p1a=1p1c=1a31modpe(c2(a21)p)=1+p1=p. (2.6)

    If 3(p1), then we have

    p1a=0p1c=0a3c30modpe(a2c2p)=1+p1a=1p1c=1a31modpe(c2(a21)p)=1+p1a=1p1c=0a31modpe(c2(a21)p)p1a=1a31modp1=1+p+τ(χ2)p1a=1a31modp(a21p)3=p2+τ(χ2)p1a=1a31modp(a21p). (2.7)

    Now Lemma 2.4 follows from (2.6) and (2.7).

    In this paper, we prove the following two conclusions:

    Theorem 3.1. Let p be a prime with p11 mod 12, then we have the identity

    p1m=0|p1a=0e(ma3+a2p)|4=2p3p2(3+p1a=1(a1+¯ap)),

    where (p) denotes the Legendre's symbol modulo p, and ¯a is the solution of the congruence equation ax1 mod p.

    Proof. Now we apply the lemmas to complete the proofs of our theorems. For any integer n, note that the trigonometrical identities

    p1a=0e(nap)={p if  pn;0 if  pn.

    From (2.5) and the properties of the reduced residue system modulo p we have,

    S4(p)=p1m=0|p1a=0e(ma3+a2p)|4=pp1a=0p1b=0p1c=0p1d=0a3+b3c3+d3modpe(a2+b2c2d2p)=pp1a=0p1b=0p1c=0p1d=1a3+b3c3+1modpe(d2(a2+b2c21)p)+pp1a=0p1b=0p1c=0a3+b3c3modpe(a2+b2c2p)=pp1a=0p1b=0p1c=0a3+b3c3+1modpp1d=0e(d2(a2+b2c21)p)+pp1a=0p1b=0p1c=0a3+b3c3modpe(a2+b2c2p)pp1a=0p1b=0p1c=0a3+b3c3+1modp1=p2p1a=0p1b=0p1c=0a3+b3c3+1modpa2+b2c2+1modp1+pτ(χ2)p1a=0p1b=0p1c=0a3+b3c3+1modp(a2+b2c21p)+pp1a=0p1b=1p1c=0a3+1c3modpe(b2(a2+1c2)p)pp1a=0p1b=0p1c=0a3+b3c3+1modp1+pp1a=0p1c=0a3c3modpe(a2c2p)=p2p1a=0p1b=0p1c=0a3+b3c3+1modpa2+b2c2+1modp1+pτ(χ2)p1a=0p1b=0p1c=0a3+b3c3+1modp(a2+b2c21p)+p2p1a=0p1c=0a3+1c3modpa2+1c2modp1+pτ(χ2)p1a=0p1c=0a3+1c3modp(a2+1c2p)pp1a=0p1c=0a3+1c3modp1pp1a=0p1b=0p1c=0a3+b3c3+1modp1+pp1a=0p1c=0a3c3modpe(a2c2p). (3.1)

    It is clear that S4(p) is a real number. If p=12k+11, then 3(p1), p3mod4, (1p)=1 and (3p)=1. Note that τ(χ2)=ip is a pure imaginary number. So from Lemma 2.2, Lemma 2.3, Lemma 2.4 and (3.1), we have

    S4(p)=p1m=0|p1a=0e(ma3+a2p)|4=p2(3p5(2p)+(3p)p1a=1(a1+¯ap))+p2(2+(2p))+pτ(χ2)p1a=0p1b=0p1c=0a3+b3c3+1modp(a2+b2c21p)+pτ(χ2)p1a=0p1c=0a3+1c3modp(a2+1c2p)p(p+p2)+p2=2p3p2(3+p1a=1(a1+¯ap))pτ(χ2)p1a=0p1b=0p1c=0a3+b3c3+1modp(a2+b2c21p)pτ(χ2)p1a=0p1c=0a3+1c3modp(a2+1c2p)=2p3p2(3+p1a=1(a1+¯ap)).

    This proves Theorem 3.1.

    Theorem 3.2. Let p be a prime with p7 mod 12, then we have the identity

    p1m=0|p1a=0e(ma3+a2p)|4=2p3p2(9p1a=1(a1+¯ap)).

    Proof. Similarly, if p=12k+7, then S4(p) is also a real number, note that 3(p1), (1p)=1, (3p)=1 and τ(χ2)=ip, so from Lemma 2.2, Lemma 2.3, Lemma 2.4 and (3.1) we have

    S4(p)=p1m=0|p1a=0e(ma3+a2p)|4=p2(3p5(2p)+(3p)p1a=1(a1+¯ap))+p2(2+(2p))p(p+d2+p2d+6p)+p(p2)=2p3p2(9p1a=1(a1+¯ap)).

    This completes the proof of Theorem 3.2.

    From A. Weil's important works [14] and [15] we have the estimate:

    |p1a=1(a1+¯ap)|=|p1a=1(a3a2+ap)|p.

    Combining this estimate and our theorems we can deduce the following:

    Corollary 3.1. Let a be an integer, p1a0. Let p be an odd prime with p3 mod 4. Then we have the asymptotic formula

    p1m=0|p1a=0e(ma3+a2p)|4=2p3+O(p52).

    Remark 3.1. In this paper, we only discussed the case p3 mod 4. If p1mod4, then we can not calculate the exact value of S4(p) yet. The reason is that we do not know the exact values of

    p1a=0p1b=0p1c=0a3+b3c3+1modp(a2+b2c21p)+p1a=0p1c=0a3+1c3modp(a2+1c2p).

    This will await our further study.

    If p3mod4, then we can easily deduce the identities

    p1a=0p1b=0p1c=0a3+b3c3+1modp(a2+b2c21p)+p1a=0p1c=0a3+1c3modp(a2+1c2p)=0.

    The main results of this paper is to give an exact calculating formula for the fourth power mean of one special two-term exponential sums. That is,

    S4(p)=p1m=0|p1a=0e(ma3+a2p)|4,

    with the case p3mod4.

    Here, we give an example to calculate the exact results of the prime number p satisfying conditions p7mod12 or p11mod12. The exact results of calculation are summarised in Table 1.

    Table 1.  The calculation of S4(p).
    p S4(p) p S4(p)
    7 S4(7)=245 11 S4(11)=2783
    19 S4(19)=11913 23 S4(23)=18515
    31 S4(31)=43245 47 S4(47)=201019
    43 S4(43)=134977 59 S4(59)=414239
    67 S4(67)=579081 71 S4(71)=741027
    79 S4(79)=979837 83 S4(83)=1095351
    103 S4(103)=1920229 107 S4(107)=2278351
    127 S4(127)=4080637 131 S4(131)=4376055
    139 S4(139)=5429201 167 S4(167)=9900595
    151 S4(151)=7045509 179 S4(179)=11759047

     | Show Table
    DownLoad: CSV

    At the same time, our results also provides some new and effective method for the calculating problem of the fourth power mean of the higher order two-term exponential sums. We have reasons to believe that these works will play a positive role in promoting the study of relevant problems. Furthermore, it is still an open problem for the case of k>h3 for S(m,n,k,h;p), interested readers can continue this research.

    The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper.

    This work is supported by the N. S.F. (12126357) of P. R. China.

    The authors declare no conflict of interest.



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