# On the Hybrid Fourth Power Mean Involving Legendre’s Symbol and One Kind Two-Term Exponential Sums

Authors: Jingzhe Wang
Journal: Journal of Mathematics
 (2021)

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## Abstract

The main purpose of this article is using the elementary methods and the properties of the quadratic residue modulo an odd primep to study the calculating problem of the fourth power mean of one kind two-term exponential sums and give an interesting calculating formula for it.


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## Body

## 1. Introduction

Letq≥3 be a fixed integer. For any integer k≥2 and integer m with m,q=1, we define the two-term exponential sums Gm,k;q as follows:(1)Gm,k;q=∑a=0q−1emak+aq,where, as usual, ey=e2πiy and i denotes the imaginary unit, that is i2=−1.Since this kind of sums play a very important role in the study of analytic number theory, so many number theorists and scholars had studied the various properties ofGm,k;q and obtained a series of meaningful research results, we do not want to enumerate here, and interested readers can refer to [1–16]. For example, Zhang and Zhang [1] proved that for any odd prime p, one has(2)∑m=1p−1∑a=0p−1ema3+nap4=2p3−p2,if3∤p−1,2p3−7p2,if3|p−1,where n represents any integer with n,p=1.Shen and Zhang [2] obtained an interesting recurrence formula for(3)Am,4;p=∑a=0p−1ema4p,where p is an odd prime with p≡1mod4.Chen and Zhang [3] proved that for any prime p with p≡5mod8, one has the identity(4)∑m=1p−1∑a=0p−1ema4p2⋅∑a=1p−1ema+a¯p2=3p3−3p2+2p3/2α−3p,where α=αp=∑a=1p−1/2a+a¯/p, ∗/p denotes Legendre’s symbol modulo p, and a⋅a¯≡1modp.Zhang and Han [4] used the elementary method to obtain the identity(5)∑a=1p−1∑n=0p−1en3+anp6=5p4−8p3−p2,where p denotes an odd prime with 3∤p−1.Chen and Wang [5] studied the calculating problem of the fourth power mean of Gm,4;p and proved the following conclusion.Letp>3 be an odd prime, then one has the identity(6)∑m=1p−1∑a=0p−1ema4+ap4=2p3,ifp=12k+11;2p2p−2,ifp=12k+7;2pp2−4p−2α2,ifp=24k+5;2pp2−6p−2α2,ifp=24k+13;2pp2−10p−2α2,ifp=24k+1;2pp2−8p−2α2,ifp=24k+17,where α=∑a=1p−1/2a+a¯/p and ∗/p denotes Legendre’s symbol modulo p.Zhang and Zhang [6] proved that for any prime p, one has the identity(7)∑m=1p−1mp⋅∑a=0p−1ema3+ap4=p2δ−3,ifp≡1mod6;p2δ+3,ifp≡−1mod6,where δ=∑d=1p−1d−1+d¯/p is an integer which satisfies the estimate δ≤2p.Liu and Zhang [7] proved that for any prime p with 3∤p−1, one has the identity(8)∑χmodp∑m=0p−1∑a=1p−1χaema3+ap6=pp−16p3−28p2+39p+5,where ∑χmodp denotes the summation over all Dirichlet characters, modulo p.Inspired by the works in [5, 6], in this paper, we consider the following calculating problem of the 2h-th power mean of the two-term exponential sums:(9)∑m=1p−1mp⋅∑a=0p−1ema4+nap2h,where p is an odd prime and h≥2 is an integer.About this problem, it seems that none had studied it before; at least we have not seen such a result at present. In this paper, we will use the properties of the solutions of the congruence equations and the quadratic residue to study this problem and give an interesting calculating formula for (9) with h=2. That is, we will prove the following result.Theorem 1.
Letp>3 be an odd prime, then we have the identity(10)∑m=1p−1mp⋅∑a=0p−1ema4+ap4=2p3/26α−3p⋅p,ifp=8k+5;2p3/23p⋅p−10α,ifp=8k+1;0,ifp=4k+3,where α=∑a=1p−1/2a+a¯/p and ∗/p denotes Legendre’s symbol modulo p.Note that the estimateα≤p (see [17] for general results), and from this theorem and [5], we may immediately deduce the following several corollaries.Corollary 1.
Letp>3 be an odd prime with p≡1mod4, then we have the asymptotic formula(11)∑m=1p−1mp⋅∑a=0p−1ema4+ap4=2⋅6p⋅p5/2+Op2.Corollary 2.
Letp>3 be an odd prime with p≡1mod4, then we have the asymptotic formula(12)∑m=1p−1∑a=0p−1em2a4+ap4=2p3+2⋅6p⋅p5/2+Op2.Some notes: the constantα=αp in our theorem has a special meaning. In fact, for any prime p with p≡1mod4, one has the identity (see [18])(13)p=12∑a=1p−1a+a¯p2+12∑a=1p−1a+ra¯p2=α2+β2,where r is any quadratic nonresidue modulo p. That is, r/p=−1.For any primep=4k+1 and integer h≥3, we naturally ask whether there is an exact calculating formula for (9)?This is an open problem. We believe this to be true. We even have the following.Conjecture 1.
Letp be an odd prime with p≡1mod4. Then, for any integer h≥2, there are two integers C=Ch,p and D=Dh,p depending only on h and p, such that the identity(14)∑m=1p−1mp⋅∑a=0p−1ema4+ap2h=C⋅p2h+1/2+D⋅p2h−1/2⋅α.


## 2. Several Lemmas

In this section, we will give several necessary lemmas. Of course, the proofs of some lemmas need the knowledge of elementary and analytic number theory. In particular, the properties of the quadratic residues and the Legendres symbol modulop are going to be used. All these can be found in [15, 18–20], and we do not repeat them. First, we have the following lemma.Lemma 1.
Letp be an odd prime with p≡1mod4. Then, for any fourth-order character λmodp, we have the identity(15)τ2λ+τ2λ¯=2p⋅α,where τχ=∑a=1p−1χaea/p denotes the classical Gauss sums, α=αp=∑a=1p−1/2a+a¯/p, and ∗/p=χ2 denotes Legendre’s symbol modulo p.Proof.
See Lemma2 in Chen and Zhang [3].Lemma 2.
Letp be an odd prime with p≡1mod4, then we have the identity(16)∑a=0p−1∑b=0p−1∑c=0p−1χ2a4+b4−c4−1=−p⋅12α+3,ifp=8k+5;p⋅20α−7,ifp=8k+1,where α=αp is defined as in Lemma 1.Proof.
For any odd primep with p≡1mod4, let λ be any fourth-order character modulo p, and(17)Am=∑a=0p−1ema4p.
Then, from the properties of the classical Gauss sums, we have(18)Am=1+∑a=1p−11+λa+χ2a+λ¯a⋅emap=λ¯m⋅τλ+χ2m⋅τχ2+λm⋅τλ¯,where m,p=1. Note that τχ2=p, χ2⋅λ=λ¯, from (18) and Lemma 1, we have(19)A2m=λ¯m⋅τλ+χ2m⋅τχ2+λm⋅τλ¯2=p+2p⋅λm⋅τλ+λ¯m⋅τλ¯+2τλ⋅τλ¯+χ2m⋅2⋅p⋅α,where p≡1mod4. If p=8k+5, then λ−1=−1 and τλ⋅τλ¯=−p, so we have(20)A2m⋅A−m=−p+2p⋅λm⋅τλ+λ¯m⋅τλ¯+χ2m⋅2⋅p⋅α×−λ¯m⋅τλ+χ2m⋅p−λm⋅τλ¯=3χ2mp3/2+3pλ¯mτλ+λmτλ¯−2pαλmτλ+λ¯mτλ¯−2pα.
Ifp=8k+1, then λ−1=1 and τλ⋅τλ¯=p, so we have(21)A2m⋅A−m=3p+2p⋅λm⋅τλ+λ¯m⋅τλ¯+χ2m⋅2⋅p⋅α×λ¯m⋅τλ+χ2m⋅p+λm⋅τλ¯=7χ2mp3/2+5pλ¯mτλ+λmτλ¯+6pα+2pαλmτλ+λ¯mτλ¯.
Now, ifp=8k+5, then from (20) and Lemma 1, we have(22)τχ2∑a=0p−1∑b=0p−1∑c=0p−1χ2a4+b4−c4−1=∑m=1p−1χ2m⋅A2m⋅A−m⋅e−mp=−3p3/2+3p⋅∑m=1p−1λmτλ+λ¯mτλ¯⋅e−mp−2pα⋅∑m=1p−1λ¯mτλ+λmτλ¯⋅e−mp−2p3/2⋅α=−3p3/2−3p⋅τ2λ+τ2λ¯−4p3/2⋅α−2p3/2⋅α=−p3/2⋅12α+3.
Ifp=8k+1, then from (21) and Lemma 1, we have(23)τχ2∑a=0p−1∑b=0p−1∑c=0p−1χ2a4+b4−c4−1=∑m=1p−1χ2m⋅A2m⋅A−m⋅e−mp=−7p3/2+5p⋅∑m=1p−1λmτλ+λ¯mτλ¯⋅e−mp+2pα⋅∑m=1p−1λ¯mτλ+λmτλ¯⋅e−mp+6p3/2⋅α=−7p3/2+5p⋅τ2λ+τ2λ¯+4p3/2⋅α+6p3/2⋅α=p3/2⋅20⋅α−7.
Combining (22) and (23) and τχ2=p, we may immediately deduce Lemma 2.Lemma 3.
Letp be an odd prime with p≡1mod4. Then, we have the identity(24)∑a=0p−1∑b=0p−1χ2a4+1−b4=3p+2α,ifp=8k+5;7p−2α,ifp=8k+1.Proof.
From the methods of proving Lemma2, we have(25)∑a=0p−1∑b=0p−1χ2a4+1−b4=1p⋅∑m=1p−1χ2mAm2⋅emp.
Ifp=8k+5, then note that τλ¯=−τλ¯, and from (18), we have(26)Am2=3p−χ2m⋅τ2λ+τ2λ¯=3p−2χ2m⋅p⋅α,
Then, applying (25), we have(27)∑a=0p−1∑b=0p−1χ2a4+1−b4=1p⋅∑m=1p−1χ2m3p−2χ2m⋅p⋅α⋅emp=3p+2α.
Ifp=8k+1, then Am2=A2m, and from (19) and (25), we have(28)∑a=0p−1∑b=0p−1χ2a4+1−b4=1p⋅∑m=1p−1χ2mAm2⋅emp=∑m=1p−1χ2m3p+2⋅λm⋅τλ+λ¯m⋅τλ¯+χ2m⋅2⋅α⋅emp=7p−2α.
Now, Lemma3 follows from (27) and (28).Lemma 4.
Letp>3 be a prime with p≡1mod4. Then, we have the identity(29)∑a=0p−1∑b=0p−1∑c=0p−1χ2a+b≡c+1modpa4+b4−c4−1=2⋅χ26⋅p−χ22⋅∑b=0p−1χ2b3+b+2.Proof.
From the properties of the complete residue system modulop, we have(30)∑a=0p−1∑b=0p−1∑c=0p−1χ2a+b≡c+1modpa4+b4−c4−1=∑a=0p−1∑b=0p−1∑c=0p−1χ2a+b≡0modpa+c4+b+14−c4−1=∑a=0p−1∑c=0p−1χ2a4+4a3c+6a2c2+4ac3+a4−4a3+6a2−4a=∑a=1p−1∑c=0p−1χ2a4+4a4c+6a4c2+4a4c3+a4−4a3+6a2−4a=∑a=1p−1∑c=0p−1χ22+4c+6c2+4c3−4a+6a2−4a3=χ22⋅∑a=1p−1∑c=0p−1χ24+4c+3c2+c3−4a+3a2−a3=χ22⋅∑a=1p−1∑c=0p−1χ2c+13+c+1−a−13−a−1=χ22⋅∑a=0p−1∑c=0p−1χ2c3+c−a3−a−χ22⋅∑b=0p−1χ2b3+b+2=χ22⋅∑a=0p−1∑c=0p−1χ2c3+3c2a+3ca2+c−χ22⋅∑b=0p−1χ2b3+b+2=χ22⋅∑a=0p−1∑c=0p−1χ2c3+3c2a+c2+4c−χ22⋅∑b=0p−1χ2b3+b+2=χ22⋅∑a=0p−1∑c=1p−1χ2c3+3ca2+4c−χ22⋅∑b=0p−1χ2b3+b+2.
Note that the identity(31)∑a=0p−1a2+np=p−1,ifn,p=p,−1,ifn,p=1,
andχ22=−1, if p=8k+5; χ22=1, if p=8k+1. We have(32)∑a=0p−1∑c=1p−1χ2c3+3ca2+4c=p⋅∑c=1p−1χ23cc3+4c≡0modp−∑c=1p−1χ23c=p⋅∑c=1p−1χ26cc2+1≡0modp=2⋅χ23⋅p.
From (30) and (32), we have(33)∑a=0p−1∑b=0p−1∑c=0p−1χ2a+b≡c+1modpa4+b4−c4−1=2⋅χ26⋅p−χ22⋅∑b=0p−1χ2b3+b+2.
This proves Lemma4.Lemma 5.
Letp>3 be a prime with p≡1mod4. Then, we have the identity(34)∑a=0p−1∑b=0p−1∑c=0p−1χ2a4+b4−c4ea+b−cp=χ22⋅p⋅∑b=0p−1χ2b3+b+2−3p,ifp=8k+5;χ22⋅p⋅∑b=0p−1χ2b3+b+2−7p,ifp=8k+1.Proof.
It is clear that(35)∑a=0p−1∑b=0p−1∑c=0p−1χ2a4+b4−c4ea+b−cp=∑a=1p−1∑b=0p−1∑c=0p−1χ2a4+b4−c4ea+b−cp+∑b=0p−1∑c=0p−1χ2b4−c4eb−cp=p⋅∑b=0p−1∑c=0p−1χ2b+1≡cmodp1+b4−c4−∑b=0p−1∑c=0p−1χ21+b4−c4−1+∑b=0p−1χ2b4−1∑c=0p−1ecb−1p−∑b=0p−1χ2b4−1=p⋅∑b=0p−1χ21+b4−b+14−∑b=0p−1∑c=0p−1χ21+b4−c4−1−∑b=0p−1χ2b4−1=p⋅∑b=0p−1χ24b3+6b2+4b−∑b=0p−1∑c=0p−1χ21+b4−c4−1−∑b=0p−1χ2b4−1.
From (31), we have(36)∑b=0p−1χ2b4−1=∑b=0p−11+χ2b⋅χ2b2−1=∑b=0p−1χ2b2−1+∑b=1p−1χ2b⋅χ2b2−1=−1+∑b=1p−1χ2b−b¯=−1+2⋅−1p−1/4⋅α.
Note thatχ24=1 and 2,p=1, and from the properties of complete residue system modulo p and χ2−1=1, we have(37)∑b=0p−1χ24b3+6b2+4b=χ24∑b=0p−1χ24b3+6b2+4b=χ22⋅∑b=0p−1χ22b3+32b2+42b=χ22⋅∑b=0p−1χ2b3+3b2+4b=χ22⋅∑b=0p−1χ2b+13+b+1−2=χ22⋅∑b=0p−1χ2b3+b−2=χ22⋅∑b=0p−1χ2b3+b+2.
Combining (35)–(37) and Lemma 3, we have(38)∑a=0p−1∑b=0p−1∑c=0p−1χ2a4+b4−c4ea+b−cp=χ22⋅p⋅∑b=0p−1χ2b3+b+2−3p,ifp=8k+5;χ22⋅p⋅∑b=0p−1χ2b3+b+2−7p,ifp=8k+1.
This proves Lemma5.


## 3. Proof of the Theorem

Applying several basic lemmas in Section2, we can easily complete the proof of our theorem. In fact, for any odd prime p>3, if p=4k+3, then from the properties of the reduced residue system modulo p, we have(39)∑m=1p−1mp⋅∑a=0p−1ema4+ap4=∑m=1p−1−mp⋅∑a=0p−1e−ma4+ap4=−∑m=1p−1mp⋅∑a=0p−1e−ma4−ap4=−∑m=1p−1mp⋅∑a=0p−1ema4+ap4.So, in this case, we have the identity(40)∑m=1p−1mp⋅∑a=0p−1ema4+ap4=0.Ifp=4k+1, then we have(41)∑m=1p−1mp⋅∑a=0p−1ema4+ap4=τχ2⋅∑a=0p−1∑b=0p−1∑c=0p−1∑d=0p−1χ2a4+b4−c4−d4ea+b−c−dp=τχ2⋅∑a=0p−1∑b=0p−1∑c=0p−1∑d=1p−1χ2a4+b4−c4−1eda+b−c−1p+τχ2⋅∑a=0p−1∑b=0p−1∑c=0p−1χ2a4+b4−c4ea+b−cp=τχ2⋅p⋅∑a=0p−1∑b=0p−1∑c=0p−1χ2a+b≡c+1modpa4+b4−c4−1−τχ2⋅∑a=0p−1∑b=0p−1∑c=0p−1χ2a4+b4−c4−1+τχ2⋅∑a=0p−1∑b=0p−1∑c=0p−1χ2a4+b4−c4ea+b−cp.Ifp=8k+5, then note that χ22=−1, and from (41), Lemma 2, Lemma 4, and Lemma 5, we may immediately deduce(42)∑m=1p−1mp⋅∑a=0p−1ema4+ap4=p3/2⋅2⋅χ26⋅p−χ22⋅∑b=0p−1χ2b3+b+2+p3/2⋅12α+3+p1/2⋅χ22⋅p⋅∑b=0p−1χ2b3+b+2−3p=2⋅p3/2⋅6α−3p⋅p.Similarly, ifp=8k+1, then note that χ22=1, and from (41), Lemma 2, Lemma 4, and Lemma 5, we may immediately deduce(43)∑m=1p−1mp⋅∑a=0p−1ema4+ap4=p3/2⋅2⋅χ26⋅p−χ22⋅∑b=0p−1χ2b3+b+2−p3/2⋅20α−7+p1/2⋅χ22⋅p⋅∑b=0p−1χ2b3+b+2−7p=2⋅p3/2⋅3p⋅p−10α.Now, the theorem follows from (40), (42), and (43).This completes the proof of our main result.



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