One kind power mean of the hybrid Gauss sums

Qi Lan; Zhang Wenpeng

doi:10.1515/math-2018-0052

Article Open Access

One kind power mean of the hybrid Gauss sums

Qi Lan and Zhang Wenpeng

Published/Copyright: May 30, 2018

Published by

Become an author with De Gruyter Brill

Submit Manuscript Author Information Explore this Subject

$Open Mathematics$

From the journal Open Mathematics Volume 16 Issue 1

Abstract

In this paper, we use the analysis method and the properties of trigonometric sums to study the computational problem of one kind power mean of the hybrid Gauss sums. After establishing some relevant lemmas, we give an exact computational formula for it. As an application of our result, we give an exact formula for the number of solutions of one kind diagonal congruence equation mod p, where p be an odd prime.

Keywords: Gauss sums; Power mean; Analysis method; Computational formula

MSC 2010: 11L03; 11L05

1 Introduction

As usual, let q ≥ 3 be a positive integer. For any positive integer n ≥ 2, the classical n-th Gauss sums G(m, n; q) is defined by

G(m,n;q)=∑a=0q−1e(manq),

where e(y) = e^{2π iy}.

Many mathematical scholars have studied the arithmetical properties concerning G(m, n; q) and have obtained various interesting results, see references [1, 2, 3, 4, 5, 6, 7, 8, 9] and [11]. For example, Shimeng Shen and Wenpeng Zhang [2] studied the computational problem of the number M_n(p) of solutions of the congruence equation

x14+x24+⋯+xn4≡0modp,0≤xi≤p−1,i=1,2,⋯,n,

and proved the following conclusions:

Let p be a prime with p = 8k + 5, U_n(p) = M_n(p) − pⁿ⁻¹. Then for any positive integer n ≥ 5, one has the fourth-order linear recurrence formula

Un(p)=−2pUn−2(p)+4pα(p)Un−3(p)−(9p2−pα2(p))Un−4(p),

where the first four terms are U₁(p) = 0, U₂(p) = −(p − 1), U₃(p) = 3(p − 1)α(p) and U₄(p) = −7p(p − 1) + (p − 1)α²(p).

If p = 8k + 1, then for any positive integer n ≥ 5, one has the fourth-order linear recurrence formula

Un(p)=6pUn−2(p)+4pα(p)Un−3(p)−(p2−pα2(p))Un−4(p),

where the first four terms are U₁(p) = 0, U₂(p) = 3(p − 1), U₃(p) = 3(p − 1)α(p), U₄(p) = 17p(p − 1)+ (p − 1)α²(p), and α(p)=∑a=1p−12(a+a¯p),(∗p) denotes the Legendre’s symbol mod p, and a denotes the multiplicative inverse of a mod p.

Xiaoxue Li and Jiayuan Hu [3] obtained the identity

∑b=1p−1|∑a=0p−1e(ba4p)|2⋅|∑c=1p−1e(bc+c¯p)|2={3p3−3p2−3p+p(τ2(χ¯4)+τ2(χ4)),ifp≡5mod8;,3p3−3p2−3p−pτ2(χ¯4)−pτ2(χ4)+2τ5(χ¯4)+2τ5(χ4),ifp≡1mod8,

where χ₄ denotes any fourth-order character mod p,τ(χ)=∑a=1p−1χ(a)e(ap) denotes the classical Gauss sums.

At the same time, Xiaoxue Li and Jiayuan Hu [3] also pointed out that how to compute the exact value of τ²(χ₄) + τ²(χ₄) and τ⁵(χ₄) + τ⁵(χ₄) are two meaningful problems.

Let A(k, p) = τ^k(χ₄) + τ^k(χ₄). Zhuoyu Chen and Wenpeng Zhang [9] studied the computational problem of A(k, p), and obtained two interesting linear recurrence formulas. That is, let p be an odd prime with p ≡ 1 mod 4. Then for any positive integer k, one has the linear recurrence formulas

A(2k+2,p)=2p⋅α(p)⋅A(2k,p)−p2⋅A(2k−2,p)

and

A(2k+3,p)=2p⋅α(p)⋅A(2k+1,p)−p2⋅A(2k−1,p),

where A(0, p) = 2, A(1, p) = G(1) − p,G(1) = G(1, 4; p), A(2, p) = 2pα(p), A(3, p) = p⋅(2α−(−1)p−14p)⋅(G(1)−p).

In this paper, as a note of [2] and [9], we shall consider the computational problem of one kind hybrid power mean of two different Gauss sums

∑m=1p−1|∑a=0p−1e(ma3p)|2h⋅|∑b=0p−1e(mb4p)|2k,(1)

where p = 12r + 1 is an odd prime, k and h are two non-negative integers.

What we are interested in is whether there exits an exact computational formula for (1). Through researches mentioned above we found that for some special prime p we can give an an efficient method to compute the value of (1). The main purpose of this paper is to illustrate this point. That is, we shall prove the following main results:

Theorem 1.1

Letpbe a prime withp = 24r + 13. Then for any positive integershandk, we have the identity

∑m=1p−1|∑a=0p−1e(ma3p)|2h⋅|∑b=0p−1e(mb4p)|2k=12⋅pk2⋅[(3p+2α(p))k+(3p−2α(p))k]⋅∑m=1p−1|∑a=0p−1e(ma3p)|2h,

whereα(p)=∑a=1p−12(a+a¯p),(∗p)denotes the Legendre’s symbol mod p, adenotes the multiplicative inverse ofa mod p.

From Theorem 1.1 we may immediately deduce the corollaries as follows.

Corollary 1.2

Ifpis a prime withp = 24r + 13, then for any positive integerk, we have

∑m=1p−1|∑b=0p−1e(mb4p)|2k=12(p−1)⋅pk2⋅[(3p+2α(p))k+(3p−2α(p))k].

Corollary 1.3

Ifpis a prime withp = 24r + 13, then for any positive integerk, we have

∑m=1p−1|∑a=0p−1e(ma3p)|2⋅|∑b=0p−1e(mb4p)|2k=(p−1)⋅pk+22⋅[(3p−2α(p))k+(3p+2α(p))k].

Corollary 1.4

Ifp = 24r + 13 is an odd prime, then for any positive integerk, we have

∑m=1p−1|∑a=0p−1e(ma3p)|4⋅|∑b=0p−1e(mb4p)|2k=3(p−1)⋅pk+42⋅[(3p+2α(p))k+(3p−2α(p))k].

Corollary 1.5

Ifp = 24r + 13 is an odd prime, then for any positive integerk, we have

∑m=1p−1|∑a=0p−1e(ma3p)|6⋅|∑b=0p−1e(mb4p)|2k=12(p−1)(18p+d2)⋅pk+42⋅[(3p−2α(p))k+(3p+2α(p))k],

wheredis uniquely determined by 4p = d² + 27b²andd ≡ 1 mod 3.

Let k and h be two positive integers, p is a prime with p = 24r + 13, and M(h, k; p) denotes the number of solutions of the congruence equation

x13+⋯+xh3+y14+⋯+yk4≡z13+⋯+zh3+w14+⋯+wk4modp,

where 0 ≤ x_i, z_iy_j, w_j ≤ p − 1, i = 1, 2, ⋯, h, j = 1, 2, ⋯, k.

Then from Theorem 1.1 we can give an exact computational method for M(h, k; p). In particular, we have the following:

Corollary 1.6

Ifp = 24r + 13 is an odd prime, then for any positive integerk, we have

M(2,k;p)=p2k+3+3(p−1)⋅pk+22⋅[(3p−2α(p))k+(3p+2α(p))k].

If prime p = 24r + 1, then the situation is more complex, we can only give an effective calculation method one by one. Theorem 1.7 indicates some examples of it.

Theorem 1.7

Ifpis an odd prime withp = 24r + 1, then we have the identities

Some notes: If 3 ∤ (p − 1), then for any integer m with (m, p) = 1, we have

|∑a=0p−1e(ma3p)|=|∑a=0p−1e(map)|=0.

If prime p = 4r + 3, then we have

|∑a=0p−1e(ma4p)|=|1+∑a=1p−1e(ma2p)+∑a=1p−1(ap)e(ma2p)|=p.

So in these cases, the problem we are studying is trivial.

2 Some simple lemmas

To prove our main results, we first propose several simple lemmas. During the proof process, we will apply some analytic number theory knowledge and the properties of character and trigonometric sums, all of which can be found in [1].

Lemma 2.1

Ifpis an odd prime with 3|(p − 1), ψis any third-order character mod p, then we have

τ3(ψ)+τ3(ψ¯)=dp,

whereτ(ψ) denotes the classical Gauss sums, dis uniquely determined by 4p = d² + 27b²andd ≡ 1 mod 3.

Proof

See [7] or [11]. □

Lemma 2.2

Ifpis an odd prime withp ≡ 1 mod 4, ψis any fourth-order character mod p, then we have

τ2(ψ)+τ2(ψ¯)=p⋅∑a=1p−1(a+a¯p)=2p⋅α(p),

where(∗p)is the Legendre’s symbol mod p,α(p)=∑a=1p−12(a+a¯p)is an integer and it satisfies the identity (see Theorem 4-11 in [10])

p=α2+β2≡(∑a=1p−12(a+a¯p))2+(∑a=1p−12(a+ra¯p))2,

andris any quadratic non-residue mod p.

Proof

In fact this is Lemma 2 of [9], so its proof is omitted. □

Lemma 2.3

Ifpis a prime withp ≡ 5 mod 8, then for any positive integerk, we have the identity

∑m=1p−1|∑a=0p−1e(ma4p)|2k=12(p−1)⋅pk2⋅[(3p−2α(p))k+(3p+2α(p))k].

Proof

If p = 8h + 5, then for any fourth-order character ψmod p and any integer m with (m, p) = 1, applying the properties of the classic Gauss sums we have ψ(−1) = −1 and

B(m)=∑a=0p−1e(ma4p)=1+∑a=1p−1(1+ψ(a)+ψ2(a)+ψ¯(a))e(map)=∑a=0p−1e(map)+∑a=1p−1χ2(a)e(map)+∑a=1p−1(ψ(m)+ψ¯(m))e(map)=χ2(m)p+ψ¯(m)τ(ψ)+ψ(m)τ(ψ¯),(2)

where χ2=(∗p) denotes the Legendre’s symbol mod p.

Note that ψ(m)τ(ψ) + ψ(m)τ(ψ) = − (ψ(m)τ(ψ) + ψ(m)τ(ψ)) (that is, it is a pure imaginary number) and ψ² = χ₂, from (2) and Lemma 2.2 we have

|B(m)|2=|χ2(m)p+ψ¯(m)τ(ψ)+ψ(m)τ(ψ¯)|2=p+|ψ¯(m)τ(ψ)+ψ(m)τ(ψ¯)|2=3p−χ2(m)(τ2(ψ)+τ2(ψ¯))=3p−2χ2(m)pα(p).(3)

So for any positive integer k, from (3) and binomial theorem we have

|B(m)|2k=(3p−2χ2(m)pα(p))k=∑i=0k(ki)(3p)k−i(−2χ2(m)pα(p))i=∑i=0[k2](k2i)3k−2i⋅pk−i⋅(2α(p))2i−χ2(m)∑i=0[k−12](k2i+1)(3p)k−2i−1⋅(2pα(p))2i+1.(4)

Note that ∑m=1p−1χ2(m)=0, from (4) we may immediately deduce that

∑m=1p−1|B(m)|2k=∑m=1p−1∑i=0[k2](k2i)3k−2i⋅pk−i⋅(2α(p))2i−∑m=1p−1χ2(m)∑i=0[k−12](k2i+1)(3p)k−2i−1⋅(2pα(p))2i+1=(p−1)⋅∑i=0[k2](k2i)3k−2i⋅pk−i⋅(2α(p))2i=12(p−1)pk2⋅[(3p+2α(p))k+(3p−2α(p))k].

This proves Lemma 2.3. □

Lemma 2.4

Ifpis a prime withp ≡ 1 mod 8, then for any positive integerk, we haveS₁(p) = 0, S₂(p) = 3p(p − 1), S₃(p) = 6p(p − 1)α(p), and fork ≥ 4, S_k(p) satisfy the fourth-order linear recurrence formula

Sk(p)=6pSk−2(p)+8pα(p)Sk−3(p)+p(4α2(p)−p)Sk−4(p),

whereSk(p)=∑m=1p−1Bk(m)=∑m=1p−1(∑a=0p−1e(ma4p))k.

Proof

If p = 8r + 1, then ψ(−1) = 1, so ψ(m)τ(ψ) + ψ(m)τ(ψ) is a real number. From (2) and Lemma 2.2 we have

|B(m)|2=B2(m)=3p+2χ2(m)pα(p)+2ψ(m)pτ(ψ)+2ψ¯(m)pτ(ψ¯).(5)

From (2) and (5) we can deduce that

B4(m)−6pB2(m)−8pα(p)B(m)+p2−4pα2(p)=0.(6)

Note that the identities

∑m=1p−1B(m)=0,∑m=1p−1B2(m)=3p(p−1),∑m=1p−1B3(m)=6p(p−1)α(p).(7)

If n ≥ 4, then from (6) we have

∑m=1p−1Bn(m)=∑m=1p−1Bn−4(m)B4(m)=∑m=1p−1Bn−4(m)(6pB2(m)+8pα(p)B(m)−p2+4pα2(p))=6p∑m=1p−1Bn−2(m)+8pα(p)∑m=1p−1Bn−3(m)+(4pα2(p)−p2)∑m=1p−1Bn−4(m).(8)

So for any integer k ≥ 4, from (7), (8) we know that S_k(p) satisfy the fourth-order linear recurrence formula

Sk(p)=6pSk−2(p)+8pα(p)Sk−3(p)+p(4α2(p)−p)Sk−4(p).

This proves Lemma 2.4. □

Lemma 2.5

Ifpis an odd prime withp ≡ 1 mod 3, then for any integermwith (m, p) = 1, we haveM₁(p) = 0, M₂(p) = 2p(p − 1), M₃(p) = dp(p − 1), and for allh ≥ 4, M_h(p) satisfy the linear recurrence formula

Mh(p)=3pMh−2(p)+dpMh−3(p),(9)

whereMh(p)=∑m=1p−1Gh(m)=∑m=1p−1(∑a=0p−1e(ma3p))h,wheredis uniquely determined by 4p = d² + 27b²andd ≡ 1 mod 3

Proof

It can be found in reference [4]. Here we give a simple proof. Let λ be any third-order character mod p. Then for any integer m with p ∤ m, from the definition and properties of the classical Gauss sums we have

G(m)=∑a=0p−1e(ma3p)=∑a=0p−1(1+λ(a)+λ¯(a))e(map)=∑a=0p−1e(map)+∑a=0p−1λ(a)e(map)+∑a=0p−1λ¯(a)e(map)=λ¯(m)τ(λ)+λ(m)τ(λ¯).(10)

Note that τ(λ)τ(λ) = p, λ³ = χ₀, the principal character mod p, from (10) and Lemma 2.1 we may deduce that

G3(m)=τ3(λ)+τ3(λ¯)+3p(λ¯(m)τ(λ)+λ(m)τ(λ¯))=dp+3pG(m).(11)

From (11) and the definition of M_h(p) we can deduce Lemma 2.5. □

3 Proofs of the main results

Now we will use the lemmas in section 2 to prove our main theorems. If p = 12r + 1, let ψ be any fourth-order character mod p, and λ be any third-order character mod p. Then note that

∑m=1p−1λi(m)ψj(m)=0,i=0,1,2,j=0,1,2,3,(i,j)≠(0,0).

So the value of the power mean in (1) only depend on the constant terms in |G(m)|^2h and |B(m)|^2k, those terms are independent of m. So we have

∑m=1p−1|∑a=0p−1e(ma3p)|2h⋅|∑b=0p−1e(mb4p)|2k=1p−1(∑m=1p−1|∑a=0p−1e(ma3p)|2h)⋅(∑m=1p−1|∑b=0p−1e(mb4p)|2k).(12)

If p = 24r + 13, then p ≡ 5 mod 8 and p ≡ 1 mod 3, from Lemma 2.3 we have

This proves Theorem 1.1.

If p = 24r + 1, then we can calculate the value of S_k(p) by Lemma 2.4 for any even number k ≥ 2. We can also calculate the value of M_h(p) by Lemma 2.5 for any even number h ≥ 2. In fact this time note that G(m) and B(m) are both real numbers. So from (11) we have

∑m=1p−1|∑a=0p−1e(ma3p)|2=M2(p)=2p(p−1).(13)

∑m=1p−1|∑a=0p−1e(ma3p)|4=M4(p)=6p2(p−1).(14)

∑m=1p−1|∑a=0p−1e(ma3p)|6=M6(p)=(p−1)p2(18p+d2).(15)

From Lemma 2.4 we have

∑m=1p−1|∑a=0p−1e(ma4p)|2=S2(p)=3p(p−1).(16)

∑m=1p−1|∑a=0p−1e(ma4p)|4=S4(p)=(p−1)p(17p+4α2(p)).(17)

∑m=1p−1|∑a=0p−1e(ma3p)|6=S6(p)=3(p−1)p2(33p+28α2(p)).(18)

Then from (12)-(18) we may immediately deduce the identities

This completes the proof of Theorem 1.7.

Competing interests: The authors declare that there are no conflicts of interest regarding the publication of this paper.
Author’s contributions: All authors read and approved the final manuscript.

Acknowledgement

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 was supported by the N. S. F. (Grant No. 11771351) of P. R. China.

References

[1] Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, 1976.10.1007/978-1-4757-5579-4Search in Google Scholar

[2] Shen S. M., Zhang W. P., On the quartic Gauss sums and their recurrence property, Advances in Difference Equations, 2017, 2017: 43.10.1186/s13662-017-1097-2Search in Google Scholar

[3] Li X. X., Hu J. Y., The hybrid power mean quartic Gauss sums and Kloosterman sums, Open Mathematics, 2017, 15, 151– 156.10.1515/math-2017-0014Search in Google Scholar

[4] S. Chowla, J. Cowles and M. Cowles, On the number of zeros of diagonal cubic forms, Journal of Number Theory, 9 (1977), 502-506.10.1016/0022-314X(77)90010-5Search in Google Scholar

[5] Weil A., On some exponential sums, Proc. Nat. Acad. Sci. U.S.A., 1948, 34, 204–207.10.1073/pnas.34.5.204Search in Google Scholar PubMed PubMed Central

[6] Zhang W. P., Liu H. N., On the general Gauss sums and their fourth power mean, Osaka Journal of Mathematics, 2005, 42, 189–199.Search in Google Scholar

[7] Zhang W. P., On the number of the solutions of one kind congruence equation mod p, Journal of Northwest University (Natural Science Edition), 2016, 46(3), 313–316.Search in Google Scholar

[8] Ren G. L., On the fourth power mean of one kind special quadratic Gauss sums, Journal of Northwest University (Natural Science Edition), 2016, 46(3), 321–324.Search in Google Scholar

[9] Chen Z. Y., Zhang W. P., On the fourth-order linear recurrence formula related to classical Gauss sums, Open Mathematics, 2017, 15, 1251–1255.10.1515/math-2017-0104Search in Google Scholar

[10] Zhang W. P., Li H. L., Elementary Number Theory, Shaanxi Normal University Press, Xi’an, 2013.Search in Google Scholar

[11] Zhang W. P., Hu J. Y., The number of solutions of the diagonal cubic congruence equation modp, Mathematical Reports, 2018, 20(1), 70–76.Search in Google Scholar

Received: 2018-02-22

Accepted: 2018-04-19

Published Online: 2018-05-30

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

Articles in the same Issue

https://doi.org/10.1515/math-2018-0052

Keywords for this article

Gauss sums; Power mean; Analysis method; Computational formula

Creative Commons

BY-NC-ND 4.0