A new fourth power mean of two-term exponential sums

Chen Li; Wang Xiao

doi:10.1515/math-2019-0034

Artikel Open Access

A new fourth power mean of two-term exponential sums

Chen Li und Wang Xiao

Veröffentlicht/Copyright: 16. Mai 2019

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Abstract

The main purpose of this paper is to use analytic methods and properties of quartic Gauss sums to study a special fourth power mean of a two-term exponential sums modp, with p an odd prime, and prove interesting new identities. As an application of our results, we also obtain a sharp asymptotic formula for the fourth power mean.

Keywords: Two-term exponential sums; fourth power mean; elementary method; identity; asymptotic formula

MSC 2010: 11L05; 11L07

1 Introduction

Let q ≥ 3 be an integer. For any integer m and n, the two-term exponential sum G(k, h, m, n; q) is defined as

G(k,h,m,n;q)=∑a=0q−1emak+nahq,

where as usual, e(y) = e^2πiy, k and h are positive integers with k ≠ h.

Many scholars have studied various elementary properties of G(k, h, m, n; q) and obtained a series of results. For example, from the A. Weil’s important work [2], one can get the general upper bound estimate

∑a=1p−1χ(a)emak+nap≪p,

where p is an odd prime, χ is any Dirichlet character mod p and (m, n, p) = 1.

Zhang Han and Zhang Wenpeng [3] proved the identity

∑m=1p−1∑a=0p−1ema3+nap4=2p3−p2 if 3∤p−1,2p3−7p2 if 3∣p−1,

where p be an odd prime.

Zhang Han and Zhang Wenpeng [4] also obtained

∑m=0p−1∑a=1p−1ema5+nap4=3p3−p28+2−1p+4−3p−3p if 5∤p−1,3p3+Op2 if 5∣p−1,

where ∗p = χ₂ denotes the Legendre symbol mod p.

Some other related mean value papers can also be found in [5] - [13]. If someone is interested in this field, please refer to these references. However, regarding the fourth power mean

∑m=1p−1∑a=0p−1ema4+ap4, (1)

it seems that it hasn’t been studied yet, at least so far we haven’t seen any related papers. We think one of the reasons for this may be that the methods used in the past are not suitable for studying this situation, or perhaps 4 is not a prime number, so it is difficult to study (1).

In this paper we will use analytic methods and properties of quartic Gauss sums to study this problem and solve it completely. That is, we will prove the following two results.

Theorem 1

Let p > 3 be a prime with p ≡ 3mod 4, then we have

∑m=1p−1∑a=0p−1ema4+ap4=2p2p−2 if p=12h+7,2p3 if p=12h+11.

Theorem 2

Let p be a prime with p ≡ 1mod 4, then we have

∑m=1p−1∑a=0p−1ema4+ap4=2pp2−10p−2α2 if p=24h+1,2pp2−4p−2α2 if p=24h+5,2pp2−6p−2α2 if p=24h+13,2pp2−8p−2α2 if p=24h+17,

where α=α(p)=∑a=1p−12a+a¯p is an integer satisfying the identity (state displayed identity), where r is any quadratic non-residue modp : see Theorem 4-11 in [16].

p=α2+β2≡∑a=1p−12a+a¯p2+∑a=1p−12a+ra¯p2,

and r is any quadratic non-residue mod p.

From these two theorems we may immediately deduce the following:

Corollary

For any odd prime p, we have the asymptotic formula

∑m=1p−1∑a=0p−1ema4+ap4=2p3+Op2.

2 Several Lemmas

To prove our theorems, we first need to give several necessary lemmas. Hereafter, we will use many properties of the classical Gauss sums, the fourth-order character mod p and the quartic Gauss sums. All of these contents can be found in any Elementary Number Theory or Analytic Number Theory book, such as references [1], [14] or [16]. These contents will not be repeated here. First we have the followings:

Lemma 1

If p is a prime with p ≡ 1mod 4, and λ is any fourth-order character mod p, then we have

τ2(λ)+τ2λ¯=p⋅∑a=1p−1a+a¯p=2p⋅α.

Proof

In fact this is Lemma 2 of [15], so its proof is omitted.

Lemma 2

If p is a prime with p ≡ 1mod 4, then we have the identity

∑m=1p−1∑a=0p−1ema4+ap2∑c=0p−1e−mc4−cp=2+χ2(7)p2+2pα if p≡5mod8,2+χ2(7)p2−6pα if p≡1mod8,

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

Proof

First applying trigonometric identity

∑m=1qenmq=q if q∣n,0 if q∤n, (2)

we have

∑m=1p−1∑a=0p−1ema4+ap2∑c=0p−1e−mc4−cp=∑m=0p−1∑a=0p−1ema4+ap2∑c=0p−1e−mc4−cp=∑a=0p−1∑b=0p−1∑c=0p−1∑m=0p−1ema4+b4−c4+a+b−cp=p∑a=0p−1∑b=0p−1∑c=0p−1a4+b4≡c4modp⁡ea+b−cp=p∑a=0p−1∑b=0p−1a4+b4≡0modp⁡ea+bp+p∑a=0p−1∑b=0p−1a4+b4≡1modp⁡∑c=1p−1ec(a+b−1)p=p−p∑a=0p−1a4+1≡0modp⁡1+p2∑a=0p−1∑b=0p−1a4+b4≡1modpa+b≡1modp⁡1−p∑a=0p−1∑b=0p−1a4+b4≡1modp⁡1. (3)

Let λ be a fourth-order character mod p, if p ≡ 5mod 8, then note that λ(− 1) = −1 we have

−p∑a=0p−1a4+1≡0modp⁡1=0. (4)

Noting the identity λ χ₂ = λ and

B(m)=∑a=0p−1ema4p=χ2(m)p+λ¯(m)τ(λ)+λ(m)τλ¯. (5)

Applying (5) and Lemma 1 we have

p∑a=0p−1∑b=0p−1a4+b4≡1modp⁡1=∑a=0p−1∑b=0p−1∑m=0p−1ema4+b4−1p=p2+∑m=1p−1∑a=0p−1ema4p2e−mp=p2+∑m=1p−12χ2(m)pα−p+2λ(m)pτ(λ)+2λ¯(m)pτλ¯e−mp=p2+2pα+p−2pτ2(λ)−2pτ2λ¯=p2+p−2pα. (6)

It is clear that the congruences a⁴ + b⁴ ≡ 1mod p and a + b ≡ 1mod p imply that ab(2a² + 3ab + 2b²) ≡ 0mod p and a + b ≡ 1mod p. So we have

p2∑a=0p−1∑b=0p−1a4+b4≡1modpa+b≡1modp⁡1=p2∑a=0p−1∑b=0p−1ab2a2+3ab+2b2≡0modpa+b≡1modp⁡1=2p2+p2∑a=1p−1∑b=1p−12a2+3ab+2b2≡0modpa+b≡1modp⁡1=2p2+p2∑a=1p−1∑b=1p−12a2+3a+2≡0modpb(a+1)≡1modp⁡1=2p2+p2∑a=0p−14a+32≡−7modp⁡1=2p2+p2∑a=0p−1a2≡−7modp⁡1=3+7p⋅p2. (7)

If p ≡ 1mod 8, then noting that λ(− 1) = 1 we have

−p∑a=0p−1a4+1≡0modp⁡1=−4p. (8)

Applying (5), Lemma 1 and note that τ(λ)τ(λ) = p we have

p∑a=0p−1∑b=0p−1a4+b4≡1modp⁡1=∑a=0p−1∑b=0p−1∑m=0p−1ema4+b4−1p=p2+∑m=1p−1∑a=0p−1ema4p2e−mp=p2+∑m=1p−13p+2χ2(m)pα+2λ(m)pτ(λ)+2λ¯(m)pτλ¯e−mp=p2+2pα−3p+2pτ2(λ)+2pτ2λ¯=p2−3p+6pα. (9)

Combining (3), (4), (6)-(9) we have the identity

∑m=1p−1∑a=0p−1ema4+ap2∑c=0p−1e−mc4−cp=2+χ2(7)p2+2pα if p≡5mod8,2+χ2(7)p2−6pα if p≡1mod8.

This proves Lemma 2.

Lemma 3

If p is a prime with p ≡ 3mod 4, then we have the identity

∑m=1p−1∑a=0p−1ema4+ap2∑c=0p−1e−mc4−cp=2−χ2(7)p2.

Proof

If p = 4h + 3, then χ₂(− 1) = −1 and τ(χ₂) = i p , i² = −1. For any integer m with (m, p) = 1, we have

∑a=0p−1ema4p=1+∑a=1p−11+χ2(a)ema2p=∑a=0p−1ema2p=χ2(m)τ(χ2) (10)

and

−p∑a=0p−1a4+1≡0modp⁡1=0. (11)

From (10), (11) and the method of proving Lemma 2 we have

∑m=1p−1∑a=0p−1ema4+ap2∑c=0p−1e−mc4−cp=2−χ2(7)p2.

This proves Lemma 3.

Lemma 4

If p is a prime with p ≡ 1mod 4, then we have the identity

∑m=1p−1∑a=0p−1ema4+ap2∑c=0p−1e−mc4−cp∑d=1p−1e−md4−dp=p2p2−22p−χ2(7)p−4α2+6α if p=24h+1,p2p2−10p−χ2(7)p−4α2−2α if p=24h+5,p2p2−14p−χ2(7)p−4α2−2α if p=24h+13,p2p2−18p−χ2(7)p−4α2+6α if p=24h+17.

Proof

From identity (2) we have

∑m=1p−1∑a=0p−1ema4+ap2∑c=0p−1e−mc4−cp∑d=1p−1e−md4−dp=∑a=0p−1∑b=0p−1∑c=0p−1∑d=1p−1∑m=0p−1ema4+b4−c4−d4+a+b−c−dp=p∑a=0p−1∑b=0p−1∑c=0p−1a4+b4≡c4+1modp⁡∑d=1p−1ed(a+b−c−1)p=p2∑a=0p−1∑b=0p−1∑c=0p−1a4+b4≡c4+1modpa+b≡c+1modp⁡1−p∑a=0p−1∑b=0p−1∑c=0p−1a4+b4≡c4+1modp⁡1. (12)

It is clear that the congruences a⁴ + b⁴ ≡ c⁴ + 1mod p and a + b ≡ c + 1mod p imply that (a − 1)(b − 1)(2a² + 3ab + 2b² − a − b + 1) ≡ 0mod p. So we have

∑a=0p−1∑b=0p−1∑c=0p−1a4+b4≡c4+1modpa+b≡c+1modp⁡1=∑a=0p−1∑b=0p−1a4+b4≡(a+b−1)4+1modp⁡1=∑a=0p−1∑b=0p−1(a−1)(b−1)2a2+2b2+3ab−a−b+1≡0modp⁡1=2p−1+∑a=0p−1∑b=0p−12a2+2b2+3ab−a−b+1≡0modpa≠1,b≠1⁡1=2p−1+∑a=0p−1∑b=0p−12a2+2b2+3ab−a−b+1≡0modp⁡1−2∑a=0p−1a2+a+1≡0modp⁡1=2p−1+∑a=0p−1∑b=0p−1(4a+3b−1)2≡−7b2+2b−7modp⁡1−2∑a=2p−1a3≡1modp⁡1=2p+1+∑a=0p−1∑b=0p−1a2≡−7b2+2b−7modp⁡1−2∑a=1p−1a3≡1modp⁡1=2p+1+∑b=0p−11+−7b2+2b−7p−2∑a=1p−1a3≡1modp⁡1=3p+1+7p∑b=0p−1(7b−1)2+48p−2∑a=1p−1a3≡1modp⁡1=3p+1+7p∑b=0p−1b2+48p−2∑a=1p−1a3≡1modp⁡1=3p+1−7p−2∑a=1p−1a3≡1modp⁡1, (13)

where we have used the identity ∑b=0p−1b2+48p=−1 .

If p = 24h + 13 or p = 24h + 1, then a³ ≡ 1mod p has three solutions. So from (13) we have

∑a=0p−1∑b=0p−1∑c=0p−1a4+b4≡c4+1modpa+b≡c+1modp⁡1=3p−5−7p. (14)

If p = 24h + 5 or p = 24h + 17, then a³ ≡ 1mod p has one solution. So from (13) we have

∑a=0p−1∑b=0p−1∑c=0p−1a4+b4≡c4+1modpa+b≡c+1modp⁡1=3p−1−7p. (15)

If p = 8h + 5, then applying (5), Lemma 1 and noting that τ(λ)τ(λ) = − p we have

p∑a=0p−1∑b=0p−1∑c=0p−1a4+b4≡c4+1modp⁡1=∑a=0p−1∑b=0p−1∑c=0p−1∑m=0p−1ema4+b4−c4−1p=∑m=0p−1∑a=0p−1ema4p2∑b=0p−1e−mb4pe−mp=p3+∑m=1p−13p−χ2(m)τ2(λ)−χ2(m)τ2λ¯×χ2(m)p+λ¯(m)τ(λ)+λ(m)τλ¯e−mp=p3+9p2+4pα2+2pα. (16)

If p = 8h + 1, then applying (5), Lemma 1 and noting that τ(λ)τ(λ) = p we have

p∑a=0p−1∑b=0p−1∑c=0p−1a4+b4≡c4+1modp⁡1=∑a=0p−1∑b=0p−1∑c=0p−1∑m=0p−1ema4+b4−c4−1p=∑m=0p−1∑a=0p−1ema4p2∑b=0p−1e−mb4pe−mp=p3+∑m=1p−13p+2χ2(m)pα+2λ(m)pτ(λ)+2λ¯(m)pτλ¯×χ2(m)p+λ¯(m)τ(λ)+λ(m)τλ¯e−mp=p3+17p2+4pα2−6pα. (17)

Combining (12), (14) - (17) we have the identity

∑m=1p−1∑a=0p−1ema4+ap2∑c=0p−1e−mc4−cp∑d=1p−1e−md4−dp=p2p2−14p−χ2(7)p−4α2−2α if p=24h+13,p2p2−22p−χ2(7)p−4α2+6α if p=24h+1,p2p2−10p−χ2(7)p−4α2−2α if p=24h+5,p2p2−18p−χ2(7)p−4α2+6α if p=24h+17.

This proves Lemma 4.

Lemma 5

If p is a prime with p ≡ 3mod 4, then we have the identity

∑m=1p−1∑a=0p−1ema4+ap2∑c=0p−1e−mc4−cp∑d=1p−1e−md4−dp=p22p−6+χ2(7) if p=12h+7,p22p−2+χ2(7) if p=12h+11.

Proof

Noting (11) and χ₂(− 1) = −1, from the methods of proving Lemma 4 we can easily deduce Lemma 5.

3 Proofs of the theorems

Now we prove our main results. First we prove Theorem 2. If p = 24h + 1, then from Lemma 2 and Lemma 4 we have

∑m=1p−1∑a=0p−1ema4+ap4−=∑m=1p−1∑a=0p−1ema4+ap2∑c=0p−1e−ma4−cp∑d=1p−1e−md4−dp +∑m=1p−1∑a=0p−1ema4+ap2∑c=0p−1e−ma4−cp =p2p2−22p−χ2(7)p−4α2+6α+2p2+χ2(7)p2−6pα =2pp2−10p−2α2. (18)

Similarly, if p = 24h + 5, then from Lemma 2 and Lemma 4 we have

∑m=1p−1∑a=0p−1ema4+ap4=2pp2−4p−2α2. (19)

If p = 24h + 13, then we have

∑m=1p−1∑a=0p−1ema4+ap4=2pp2−6p−2α2. (20)

If p = 24h + 17, then we have

∑m=1p−1∑a=0p−1ema4+ap4=2pp2−8p−2α2. (21)

Now Theorem 2 follows from (18) - (21).

Similarly, from Lemma 3, Lemma 5 and the methods of proving Theorem 2 we can also deduce Theorem 1. This completes the proofs of all of our results.

Dedicated to our supervisor Professor Zhang Wenpeng for his 60th birthday.

Acknowledgement

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

References

[1] Apostol T.M., Introduction to Analytic Number Theory, 1976, Springer-Verlag, New York.10.1007/978-1-4757-5579-4Suche in Google Scholar

[2] Weil A., On some exponential sums, Proc. Nat. Acad. Sci. U.S.A., 1948, 34, 204–20710.1007/978-1-4757-1705-1_48Suche in Google Scholar

[3] Zhang H., Zhang W.P., The fourth power mean of two-term exponential sums and its application, Mathematical Reports, 2017, 19, 75–83Suche in Google Scholar

[4] Zhang H., Zhang W.P., On the fourth power mean of the two-term exponential sums, The Scientific World Journal, 2014, 2014, Article ID: 72484010.1155/2014/724840Suche in Google Scholar PubMed PubMed Central

[5] Zhang W.P., Han D., On the sixth power mean of the two-term exponential sums, Journal of Number Theory, 2014, 136, 403–41310.1016/j.jnt.2013.10.022Suche in Google Scholar

[6] Cochrane T., Pinner C., Using Stepanov’s method for exponential sums involving rational functions, Journal of Number Theory, 2006, 116, 270–29210.1016/j.jnt.2005.04.001Suche in Google Scholar

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

[8] Cochrane T., Zheng Z.Y., Bounds for certain exponential sums, Asian J. Math., 2000, 4, 757–77410.4310/AJM.2000.v4.n4.a3Suche in Google Scholar

[9] Zhang W.P., On the fourth power mean of the general Kloosterman sums, Journal of Number Theory, 2016, 169, 315–32610.1016/j.jnt.2016.05.018Suche in Google Scholar

[10] Zhang W.P., Shen S.M., A note on the fourth power mean of the generalized Kloosterman sums, Journal of Number Theory, 2017, 174, 419–42610.1016/j.jnt.2016.11.020Suche in Google Scholar

[11] Zhang W.P., Hu J.Y., The number of solutions of the diagonal cubic congruence equation mod p, Mathematical Reports, 2018, 20, 73–80Suche in Google Scholar

[12] Han D., A Hybrid mean value involving two-term exponential sums and polynomial character sums, Czechoslovak Mathematical Journal, 2014, 64, 53–6210.1007/s10587-014-0082-0Suche in Google Scholar

[13] Shen S.M., Zhang W.P., On the quartic Gauss sums and their recurrence property, Advances in Difference Equations, 2017, 2017: 4310.1186/s13662-017-1097-2Suche in Google Scholar

[14] Hua L.K., Introduction to Number Theory, 1979, Science Press, Beijing.Suche in Google Scholar

[15] Chen Z.Y., Zhang W.P., On the fourth-order linear recurrence formula related to classical Gauss sums, Open Mathematics, 2017, 15, 1251–125510.1515/math-2017-0104Suche in Google Scholar

[16] Zhang W.P., Li H.L., Elementary Number Theory, 2008, Shaanxi Normal University Press, Xi’an, Shaanxi.Suche in Google Scholar

Received: 2018-09-18

Accepted: 2019-01-09

Published Online: 2019-05-16

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Two-term exponential sums; fourth power mean; elementary method; identity; asymptotic formula

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