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Research article

A certain two-term exponential sum and its fourth power means

  • Received: 31 July 2020 Accepted: 18 September 2020 Published: 23 September 2020
  • MSC : 11L03, 11L05

  • The main purpose of this article is using the properties of the Legendre's symbol and the classical Gauss sums to study the calculating problem of the fourth power mean of a certain two-term exponential sums, and give an interesting calculating formula for it.

    Citation: Jin Zhang, Wenpeng Zhang. A certain two-term exponential sum and its fourth power means[J]. AIMS Mathematics, 2020, 5(6): 7500-7509. doi: 10.3934/math.2020480

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  • The main purpose of this article is using the properties of the Legendre's symbol and the classical Gauss sums to study the calculating problem of the fourth power mean of a certain two-term exponential sums, and give an interesting calculating formula for it.


    Let q3 be a fixed integer. For any integers k2 and m with (m,q)=1, we define the two-term exponential sums G(m,k;q) as

    G(m,k;q)=q1a=0e(mak+aq),

    where as usual, e(y)=e2πiy and i2=1.

    This sum play a very important role in the study of analytic number theory, many number theory problems are closely related to it. Such is the case with the famous Waring problem. Therefore, it is necessary to study the various properties of G(m,k;q), in order to promote the development of research work in related fields. The work in this area mainly includes two aspects, one is the upper bound estimate of the exponential sum, the other is the mean value of the exponential sum. In fact, a great deal of work has been done in this field and a series of meaningful research results have been obtained, we do not want to enumerate here, interested readers can refer to references [5,6,7,8,9,10,11,12,13,14,15].

    For example, H. Zhang and W. P. Zhang [5] proved that for any odd prime p, one has the identity

    p1m=1|p1a=0 e(ma3+nap)|4={2p3p2if p1$,2p37p2if |p1$,

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

    W. P. Zhang and D. Han [6] obtained the identity

    p1a=1|p1n=0e(n3+anp)|6=5p48p3p2,

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

    X. Y. Liu and W. P. Zhang [8] proved that for any prime p with 3(p1), one has the identity

    χmodpp1m=0|p1a=1χ(a)e(ma3+ap)|6=p(p1)(6p328p2+39p+5),

    where χmodp denotes the summation over all Dirichlet characters modulo p.

    In this paper, we will use the properties of Legendre's symbol and the classical Gauss sums to prove the following main result:

    Theorem. Let p>3 be a prime, then we have the identity

    p1m=1(mp)|p1a=0e(ma3+ap)|4={p2(δ3),if p1mod6;p2(δ+3),if p1mod6,

    where as usual, (p) denotes the Legendre's symbol modulo p, d¯d1modp, δ=p1d=1(d1+¯dp) is an integer which satisfies the estimate |δ|2p.

    From this theorem and H. Zhang and W. P. Zhang [5] we may immediately deduce the following three corollaries:

    Corollary 1. Let p>3 be a prime, then we have the identity

    p1m=1|p1a=0e(m2a3+ap)|4={2p3+p2(δ10),if p1mod6;2p3+p2(δ+2),if p1mod6.

    Corollary 2. Let p>3 be a prime, n be any quadratic non-residue modulo p. Then we have the identity

    p1m=1|p1a=0e(nm2a3+ap)|4=2p3p2(δ+4).

    Corollary 3. Let p>3 be a prime. Then for any integer n with (n,p)=1, we have the asymptotic formula

    p1m=1|p1a=0e(nm2a3+ap)|4=2p3+O(p52).

    Some notes: From our theorem and A. Weil's work [15] we can also deduce the following nontrivial estimate:

    |p1m=1(mp)|p1a=0e(ma3+ap)|4|=O(p52).

    Obviously, this estimate saves p12 compared to the trivial estimate p3, and may be used in some analytic number theory problems.

    In addition, the distribution of exponential sums is very irregular, so it should be meaningful to study their power means and obtain an accurate calculation formula.

    It is interesting to ask whether a formula analogous to our theorem can be obtained for the general 2h-th power means

    p1m=1(mp)|p1a=0e(ma3+ap)|2h, where h3.

    This is an open problem. We hope that interested readers will join us in considering this problem.

    In this section, we will give some basic lemmas. Of course, the proofs of these lemmas need some knowledge of elementary and analytic number theory. They can be found in references [1,2,3,4], and we do not repeat them. At the beginning, we are going to introduce the definition of the classical Gauss sums τ(χ) as follows: For any integer q>1, let χ denotes any Dirichlet character modulo q, then the classical Gauss sums τ(χ) is defined as

    τ(χ)=qa=1χ(a)e(aq),

    where e(y)=e2πiy. Based on this definition and its properties:

    qa=1χ(a)e(maq)=¯χ(m)τ(χ),

    providing (m,q)=1 or χ is a primitive character modulo q, we have the following:

    Lemma 1. If p>3 be an odd prime, then we have the identity

    p1m=1(mp)(p1a=0e(ma3+ap))2=(3p)p,

    where (p)=χ2 denotes the Legendre symbol modp.

    Proof. Note that χ32=χ2 and τ2(χ2)=χ2(1)p, from the definition and properties of the Gauss sums and reduced residue system modulo p we have

    p1m=1χ2(m)(p1a=0e(ma3+ap))2=p1m=1χ2(m)p1a=0e(ma3+ap)(1+p1b=1e(mb3+bp))=p1m=1χ2(m)p1a=0e(ma3+ap)+p1m=1χ2(m)p1a=0p1b=1e(ma3+mb3+a+bp)=τ(χ2)p1a=1χ2(a)e(ap)+τ(χ2)p1a=0p1b=1χ2(a3+b3)e(a+bp)=τ(χ2)p1a=1χ2(a)e(ap)+τ(χ2)p1a=0χ2(a3+1)p1b=1χ2(b)e(b(a+1)p)=τ2(χ2)+τ2(χ2)p1a=0χ2(a3+1)χ2(a+1)=χ2(1)p+χ2(1)pp2a=0χ2(a2a+1)=χ2(1)p+χ2(1)pp2a=0((2a1)2+3p)=χ2(1)p(3p)p+χ2(1)pp1b=0(b2+3p). (2.1)

    Note that for any odd prime p, we have the identity

    p1a=0(a2+np)=(np)+p1a=1(1+(ap))(a+np)=p1a=0(a+np)+p1a=1(a(a+n)p)=p1a=1(1+n¯a)p)=p1a=1(1+nap)={p1if pn,1if pn. (2.2)

    From (2.1) and (2.2) we have

    p1m=1(mp)(p1a=0e(ma3+ap))2=(3p)p.

    This proves Lemma 1.

    Lemma 2. Let p>3 be an odd prime, then we have the identity

    p1m=1(mp)(p1a=0e(ma3+ap))2(p1b=0e(mb3bp))=(1p)p2+(6p)p2p1a=0a3+20modp(a+2p).

    Proof. From Lemma 1 and its proving methods we have

    p1m=1(mp)(p1a=0e(ma3+ap))2(p1b=0e(mb3bp))=p1m=1(mp)(p1a=0e(ma3+ap))2(1+p1b=1e(mb3bp))=(3p)p+p1m=1(mp)(p1a=0e(ma3+ap))2(p1b=1e(mb3bp))=χ2(3)p+χ2(1)pp1a=0p1b=0(a3+b31p)(a+b1p). (2.3)

    Let c=b1, if b pass through a complete residue system modp, then c also passes a complete residue system modp, from (2.2), (2.3) and the properties of the complete residue system modp we have

    p1a=0p1b=0(a3+b31p)(a+b1p)=p1a=0p1c=0(a3+(c+1)31p)(a+cp)=p1a=0p1c=1(a3c3+c3+3c2+3cp)(ac+cp)+p1a=1(a4p)=(p1)+p1a=0(a+1p)p1c=1(a3+1+3¯c+3¯c2p)=(p1)p1a=0((a3+1)(a+1)p)+p1a=0(a+1p)p1c=0(3c2+3c+a3+1p)=(p1)p2a=0(a2a+1p)+p1a=0(a+1p)p1c=0(3(2c+1)2+4a3+1p)=p1+(3p)p1a=0((2a1)2+3p)+p1a=0(a+1p)p1c=0(3c2+4a3+1p)=p+(3p)+(3p)pp1a=04a3+10modp(a+1p)(3p)p1a=0(a+1p)=p+(3p)+(3p)pp1a=0¯2a3+10modp(¯2a+1p)=p+(3p)+(3p)pp1a=0a3+20modp(a+2p)(2p)=p+(3p)+(6p)pp1a=0a3+20modp(a+2p). (2.4)

    Combining (2.3) and (2.4) we have the identity

    p1m=1(mp)(p1a=0e(ma3+ap))2(p1b=0e(mb3bp))=(1p)p2+(6p)p2p1a=0a3+20modp(a+2p).

    This proves Lemma 2.

    Lemma 3. If p>3 is an odd prime, then we have

    p1a=0p1b=0p1c=0(a3+b3c31p)(a+bc1p)=(1p)pp1d=1(d1+¯dp)p3(3p)p(6p)pp1d=1d3+20modp(d+2p).

    Proof. Let d=a1, e=bc. Then from the properties of the complete residue system modulo p we have

    p1a=0p1b=0p1c=0(a3+b3c31p)(a+bc1p)=p1d=0p1e=0p1c=0((d+1)3+(e+c)3c31p)(d+ep)=p1d=0p1e=0p1c=0(d3+3d2+3d+e3+3e2c+3ec2p)(d+ep)=p1d=0p1e=1p1c=0(d3+3d2e+3de2+1+3ec+3e2c2p)(d+1p)+pp1d=1(d2+3d+3p). (2.5)

    If c pass through a complete residue system mod p, then 2ce+1 also passes a complete residue system modp, so from (2.2) and (2.5) we have

    p1a=0p1b=0p1c=0(a3+b3c31p)(a+bc1p)=p1d=0p1e=1p1c=0(3(2ce+1)2+1+4d3+12d2e+12de2p)(d+1p)+pp1d=0((2d+3)2+3p)(3p)p=p1d=0p1e=1p1c=0(3c2+1+4d3+12d2e+12de2p)(d+1p)+pp1d=0(d2+3p)(3p)p=(3p)pp1d=0p1e=112de2+12d2e+4d3+10modp(d+1p)(3p)p1d=0p1e=1(d+1p)p(3p)p=(3p)pp1d=1p1e=13d2(2e+d)2+d4+d0modp(d+1p)p(3p)p=(3p)pp1d=1p1e=03e2+d4+d0modp(d+1p)(3p)pp1d=18d3+20modp(d+1p)p(3p)p=(3p)pp1d=1(1+(3(d4+d)p))(d+1p)p(3p)p(3p)pp1d=1(2d)3+20modp(2d+2p)(2p)=pp1d=1(d4dp)(d+1p)p2(3p)p(6p)pp1d=1d3+20modp(d+2p)=(1p)pp2d=1(dp)(d2d+1p)p2(3p)p(6p)pp1d=1d3+20modp(d+2p)=(1p)pp1d=1(d1+¯dp)p3(3p)p(6p)pp1d=1d3+20modp(d+2p).

    This proves Lemma 3.

    Applying the basic lemmas of the previous section, we can easily complete the proof of our theorem. Our proof idea is to decompose the proof process of the theorem into the form of lemmas, and then apply Lemma 2 and Lemma 3 directly. In fact for any prime p>3, note that τ2(χ2)=χ2(1)p and the identity

    p1m=1(mp)(p1a=0e(ma3+ap))2(p1b=0e(mb3bp))(p1c=1e(mc3cp))=p1a=0p1b=0p1c=0p1d=1p1m=1(mp)e(m(a3+b3c3d3)+a+bcdp)=τ(χ2)p1a=0p1b=0p1c=0p1d=1(a3+b3c3d3p)e(a+bcdp)=τ(χ2)p1a=0p1b=0p1c=0(a3+b3c31p)p1d=1(dp)e(d(a+bc1)p)=τ2(χ2)p1a=0p1b=0p1c=0(a3+b3c31p)(a+bc1p)=(1p)pp1a=0p1b=0p1c=0(a3+b3c31p)(a+bc1p), (3.1)

    from (3.1), Lemma 2 and Lemma 3 we have

    p1m=1(mp)|p1a=0e(ma3+ap)|4=p1m=1(mp)(p1a=0e(ma3+ap))2(p1b=0e(mb3bp))(1+p1b=1e(mb3bp))=p1m=1(mp)(p1a=0e(ma3+ap))2(p1b=0e(mb3bp))+p1m=1(mp)(p1a=0e(ma3+ap))2(p1b=0e(mb3bp))(p1b=1e(mb3bp))=p1m=1(mp)(p1a=0e(ma3+ap))2(p1b=0e(mb3bp))+p(1p)p1a=0p1b=0p1c=0(a3+b3c31p)(a+bc1p)=(1p)p2+(6p)p2p1a=0a3+20modp(a+2p)+p2p1d=1(d1+¯dp)(1p)p23(3p)p2(6p)p2p1d=1d3+20modp(d+2p)=p2p1d=1(d1+¯dp)3(3p)p2. (3.2)

    Note that (3p)=1, if p1mod6; And (3p)=1, if p1mod6. From (3.2) we may immediately deduce our theorem.

    In this article, we obtained an identity for the fourth power mean of a certain two-term exponential sum with the weight of the Legendre's symbol modulo p. This result is the further promotion and extension of [5,6], and it is new contribution in the related fields.

    The authors would like to express their heartfelt thanks to the reviewers for their valuable comments and suggestions. In particular, the calculation errors are pointed out, which makes this paper further improved and perfected.

    This work is supported by the N. S.F. (11771351) of P. R. China and Sci-Tech Program of Xi'an (Grant No. 2020KJWL08) of P. R. China.

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



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