Research article

A certain new Gauss sum and its fourth power mean

  • Received: 13 April 2020 Accepted: 01 June 2020 Published: 09 June 2020
  • MSC : 11L03, 11L07

  • The main purpose of this paper is using the elementary methods and the properties of the Legendre symbol to study the computational problem of the fourth power mean of a certain generalized quadratic Gauss sum, and give two exact calculating formulae for it.

    Citation: Yan Zhao, Wenpeng Zhang, Xingxing Lv. A certain new Gauss sum and its fourth power mean[J]. AIMS Mathematics, 2020, 5(5): 5004-5011. doi: 10.3934/math.2020321

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  • The main purpose of this paper is using the elementary methods and the properties of the Legendre symbol to study the computational problem of the fourth power mean of a certain generalized quadratic Gauss sum, and give two exact calculating formulae for it.


    Let q3 is a positive integer. For any integral coefficient polynomial f(x) and any Dirichlet character χmodq, we define a generalized Gauss sums G(f,χ;q) as

    G(f,χ;q)=qa=1χ(a)e(f(a)q),

    where as usual, e(y)=e2πiy.

    If f(x)=x, then G(f,χ;q)=τ(χ) is the classical Gauss sums. That is,

    τ(χ)=qa=1χ(a)e(aq).

    A very important property of τ(χ) is that if χ is a primitive character mod q, then we have the identity |τ(χ)|=q (see Theorem 8.15 in [1]).

    If f(x)=ax2+bx, then G(f,χ;q) becomes the generalized quadratic Gauss sums. Many authors have studied its properties and obtained interesting results. For example, if q=p is an odd prime, then from A. Weil's important work [2] we can get the estimate

    |G(f,χ;p)|2p

    for all integers a and b with (a,b,p)=1.

    Let p be an odd prime, α be a positive integer α2, and λ be a primitive character modulo  pα. Zhang Wenpeng and Lin Xin [3] proved that for any integer n with (n,p)=1, we have

    pαm=1|pαa=1λ(a)e(ma2+napα)|4=p2αϕ(pα)(α+15p1);

    If λ is any non-primitive character modulo pα, then we have the identity

    pαm=1|pαa=1λ(a)e(ma2+napα)|4=p2αϕ(pα),

    where ϕ(n) denotes the Euler function.

    Li Xiaoxue and Xu Zhefeng [4] also studied the special case q=p, an odd prime, and obtained several interesting identities.

    On the other hand, Lv Xingxing and Zhang Wenpeng [5] introduced a new sum analogous to Kloosterman sum as follows:

    K(m,n,r,χ;q)=qa=1χ(ma+n¯a)e(raq), (1.1)

    where m, n and r are integers, ¯a denotes a¯a1modq.

    Using properties of the character modulo p and analytic methods, they proved the following results: For any odd prime p with p3mod4 and integers m and n with (mn,p)=1, one has the identity

    χmodp|p1a=1χ(a)e(ma2p)|2|p1b=1χ(b+¯b)|2=(p1)(3p26p1)

    and

    χmodp|p1a=1χ(a)e(ma2p)|2|p1b=1χ(b+¯b)e(nbp)|2=(p1)(p22p1)+p(p1)(p2b=2e(n(b+¯b)p)+p2b=2e(n(b¯b)p)).

    From the second formula and the estimate for Kloosterman sums, Lv Xingxing and Zhang Wenpeng [5] deduced the asymptotic formula

    χmodp|p1a=1χ(a)e(ma2p)|2|p1b=1χ(b+¯b)e(nbp)|2=p3+O(p52).

    Shane Chern [6] also studied the properties of K(m,n,r,χ;q), and obtained the identity

    χmodp|p1m=1|p1a=1χ(ma+n¯a)e(kap)|2|2=(p1)(p47p3+17p25p25),

    where p is an odd prime, n and k are integers with (nk,p)=1.

    Some other papers related to Gauss sums and two-term exponential sums can also be found in [5,6,7,8,9,10,11,12,13], but we will not go into further detail here.

    Inspired by [5] and [6], it is natural to study the 2k-th power means

    p1m=0χmodp|p1a=0χ(a2+1)e(map)|2k,  k2. (1.2)

    To the best knowledge of the authors, sums of the type (1.2) have not been previously investigated but the authors think that this problem is meaningful, and can be viewed as a new combination of additive and multiplicative functions.

    In this paper, we will use the number of the solutions of some congruence equations mod p and the properties of the Legendre symbol to study the problem of calculating (1.2) with k=2, and give two exact calculating formulae according to p1mod4 or p3mod4. That is, we will prove the following results:

    Theorem 1. Let p be an odd prime with p3mod4, then we have

    χmodpp1m=0|p1a=0χ(a2+1)e(map)|4=p(p1)(3p24p+4).

    Theorem 2. Let p be an odd prime with p1mod4, then we have

    χmodpp1m=0|p1a=0χ(a2+1)e(map)|4=3p(p1)(p26p+10).

    Some notes: For k3, whether there exists an exact calculating formula for (1.2) is an open problem. It is also interesting to ask whether a general formula analogous to Theorems 1 and 2 can be obtained for all integers q3 and k=2.

    In this section, we first give four basic lemmas, which are necessary in the proof of our theorems. Of course, in order to prove these lemmas, we need some knowledge of elementary and analytic number theory. They can be found in reference [1], and we do not repeat them all here. First we have the following:

    Lemma 1. For any odd prime p, we have the identity

    p1a=0p1b=0p1c=0p1d=0abcdmodpa+bc+dmodp1=2p2p.

    Proof. Note that if a, b, c pass through a reduced residue system mod p, then da, db and cd also pass through a reduced residue system mod p, providing (d,p)=1. So from these properties we have

    p1a=0p1b=0p1c=0p1d=0abcdmodpa+bc+dmodp1=4(p1)+1+p1a=1p1b=1p1c=1p1d=1abcdmodpa+bc+dmodp1=4p3+p1a=1p1b=1p1c=1p1d=1abcmodpa+bc+1modp1=4p3+(p1)p1a=1p1b=1a+bab+1modp1=4p3+(p1)p1a=1p1b=1(a1)(b1)0modp1=4p3+(p1)(2p1a=111)=4p3+(p1)(2p3)=2p2p.

    This proves Lemma 1.

    Lemma 2. If p is an odd prime, then we have the identity

    p1a=0p1b=0p1c=0p1d=0ab+cd2modpa+bc+dmodp1=p2+(1p)p.

    Proof. It is clear that if d passes through a reduced residue system mod p, then ¯d also passes through a reduced residue system mod p. From the properties of the reduced residue system and quadratic residue mod p we have

    p1a=0p1b=0p1c=0p1d=0ab+cd2modpa+bc+dmodp1=p1a=0p1b=0p1c=0p1d=1ab+cd2modpa+bc+dmodp1+p1a=0p1b=0p1c=0ab2modpa+bcmodp1=p1a=0p1b=0p1c=0p1d=1ab+c2¯d2modpa+bc+1modp1+p1a=0p1b=0ab2modp1=p1a=0p1b=0p1c=0p1d=1ab+c2d2modpa+bc+1modp1+(p1)=p1a=0p1b=0p1c=0p1d=0ab+c2dmodpa+bc+1modp(1+(dp))p1a=0p1b=0p1c=0ab+c0modpa+bc+1modp1+(p1)=p1a=0p1b=0p1c=0a+bc+1modp1+p1a=0p1b=0p1c=0p1d=0ab+c2dmodpa+bc+1modp(dp)p1a=0p1b=0ab+a+b1modp1+(p1)=p2+p1a=0p1b=0p1d=0ab+a+b1+2dmodp(dp)p1a=1p1b=1ab2modp1+(p1). (2.1)

    It is clear that if a passes through a complete residue system mod p, then a+1 also passes through a complete residue system mod p. So we make the change of variables a to a1 and b to b1 modulo p to change the congruence from ab+a+bcmodp to abc+1modp. Then from (2.1) and the orthogonality property for the Dirichlet character modulo p we have

    p1a=0p1b=0p1c=0p1d=0ab+cd2modpa+bc+dmodp1=p2+pa=1pb=1p1d=0ab2+2dmodp(dp)=p2+(2p)pa=1pb=1(ab2p)=p2+(2p)p1a=1pb=1(ab2p)+(2p)pb=1(2p)=p2+(2p)p1a=1pb=1(b2p)+(1p)p=p2+(1p)p.

    This proves Lemma 2.

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

    p1a=0p1b=0p1c=0p1d=0ab+cd2modpabcdmodpa+bc+dmodp1=2p4.

    Proof. It is clear that the conditions abcdmodp and ab+cd2modp are equivalent to abcd1modp. So we have

    p1a=0p1b=0p1c=0p1d=0ab+cd2modpabcdmodpa+bc+dmodp1=p1a=1p1b=1p1c=1p1d=1abcd1modpa+bc+dmodp1=p1a=1p1c=1a+¯ac+¯cmodp1=p1a=1p1c=1(ac)(ac1)0modp1=p1a=1p1c=1acmodp1+p1a=1p1c=1ac1modp1p1a=1p1c=1acmodpac1modp1=p1+p12=2p4.

    This proves Lemma 3.

    Lemma 4. If p is an odd prime with p1mod4, then we have the identity

    p1a=0p1b=0p1c=0p1d=0(a2+1)(b2+1)(c2+1)(d2+1)0modpa+bc+dmodp1=16p26.

    Proof. Since p1mod4, there exists an integer k satisfying the congruence (±k)21modp. In fact, these are all such solutions modulo p. Choosing 0hp1 for which a+bc+dhmodp, we have

    p1a=0p1b=0p1c=0p1d=0(a2+1)(b2+1)(c2+1)(d2+1)0modpa+bc+dmodp1=p1h=0(p1a=0p1b=0(a2+1)(b2+1)0modpa+bhmodp1)2=p1h=0(p1a=0p1b=0a2+10modpa+bhmodp1+p1a=0p1b=0b2+10modpa+bhmodp1p1a=0p1b=0a2+1b2+10modpa+bhmodp1)2=p1h=0(2p1b=0k+bhmodp1+2p1b=0k+bhmodp1p1a=0p1b=0a2+1b2+10modpa+bhmodp1)2=p1h=0h2k,2k,0(2+20)2+(4p1a=0p1b=0a2+1b2+10modpa+b0modp1)2+(4p1a=0p1b=0a2+1b2+10modpa+b2kmodp1)2+(4p1a=0p1b=0a2+1b2+10modpa+b2kmodp1)2=16(p3)+(42)2+(41)2+(41)2=16p26.

    In this section, we use the four basic lemmas of the previous section to prove our theorems. First we prove Theorem 1. If p3mod4, then (1p)=1 and p(a2+1) for all 0ap1. Then from Lemma 1, Lemma 2, Lemma 3, the trigonometric identity

    p1a=0e(map)={p,if pm;0,if pm

    and the orthogonality of characters modp

    χmodpχ(a)={p1,if a1modp;0,otherwise

    we have

    1p(p1)χmodpp1m=0|p1a=0χ(a2+1)e(map)|4=1pp1a=0p1b=0p1c=0p1d=0(a2+1)(b2+1)(c2+1)(d2+1)modpp1m=0e(m(a+bcd)p)=p1a=0p1b=0p1c=0p1d=0a2b2+a2+b2+1c2d2+c2+d2+1modpa+bc+dmodp1=p1a=0p1b=0p1c=0p1d=0(ab1)2+(a+b)2(cd1)2+(c+d)2modpa+bc+dmodp1=p1a=0p1b=0p1c=0p1d=0(ab1)2(cd1)2modpa+bc+dmodp1=p1a=0p1b=0p1c=0p1d=0(abcd)(ab+cd2)0modpa+bc+dmodp1=p1a=0p1b=0p1c=0p1d=0abcdmodpa+bc+dmodp1+p1a=0p1b=0p1c=0p1d=0ab+cd2modpa+bc+dmodp1p1a=0p1b=0p1c=0p1d=0abcdab+cd20modpa+bc+dmodp1=2p2p+p2p(2p4)=3p24p+4.

    This proves Theorem 1.

    Now we prove Theorem 2. Note that if p1mod4, then (1p)=1. If a2+10modp, then for any character χmodp, we have χ(a2+1)=0. So from Lemma 4 and the methods of proving Theorem 1 we have

    1p(p1)χmodpp1m=0|p1a=0χ(a2+1)e(map)|4=p1a=0p1b=0p1c=0p1d=0(a2+1)(b2+1)(c2+1)(d2+1)modpa+bc+dmodp1p1a=0p1b=0p1c=0p1d=0(a2+1)(b2+1)(c2+1)(d2+1)0modpa+bc+dmodp1=2p2p+p2+p(2p4)(16p26)=3(p26p+10).

    This completes the proof of all of our results.

    The authors express their sincere thanks to the reviewer for pointing out the errors in our paper.

    This work is supported by the Xian Science and Technology Planning Project (2019KJWL28) and N. S. F. (11771351) of P. R. China.

    The authors declare no conflict of interest.



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