1. IntroductionFor any integer q≥2 and any Dirichlet character χ mod q, the definition of the classical Gauss sums Gm,χ;q is(1)Gm,χ;q=∑a=1qχaemaq,where m is an integer and ey=e2πiy.
This sum and its properties are of great significance to the analytic number theory, and many number theory problems are closely related to them. Therefore, it is necessary to study the various properties of Gm,χ;q and related sums. In this paper, we consider the generalized k-th Gauss sums Gm,k,χ;q, which is defined as follows:(2)Gm,k,χ;q=∑a=1qχaemakq,where k is any positive integer and m is an integer with m,q=1.
It is clear that this sum is a generalization of the classical Gauss sums Gm,χ;q. In fact, Gm,1,χ;q=Gm,χ;q. Of course, the value of Gm,k,χ;q is irregular as χ varies. However, some scholars have found that Gm,k,χ;q has good value distribution properties in some problems of weighted mean value, even if we can get their exact calculation formulae for some 2kth power mean. In addition, there are some good upper bound estimates for Gm,k,χ;q. For example, for any integer n with n,q=1, from the general result of Cochrane and Zheng [1], we can deduce(3)Gm,k,χ;q≤2ωqq1/2,where ωq denotes the number of distinct prime divisors of q. The case that q is a prime is due to Weil [2].
For k=2, by the results of Zhang [3], let n be any integer with n,p=1, and there are the following two identities:(4)1p−1∑χmodpGn,2,χ;p4=3p2−6p−1+4npp,if p≡1mod4,3p2−6p−1,if p≡3mod4,1p−1∑χmodpGn,2,χ;p6=10p3−25p2−4p−1, if p≡3mod4,where ∗/p denotes the Legendre symbol modulo p.
Zhang and Liu [4] have studied the sum(5)∑χmodpGn,3,χ;q4and obtained the following calculation formula:(6)∑χmodpG1,3,χ;p4=5p3−18p2+20p+1+U5p+5pU−5U3−4U2+4U,where p is a prime with 3|p−1 and U=∑a=1pea3/p is a real constant.
However, the value of U was not given in [4], and the form of U was not concise. Now, for any integer 1≤m≤p−1, we let(7)Em,p=1p−1∑χmodpGm,3,χ;p4−5p2+13p+1.
In this paper, we use the analytic methods and the properties of the classical Gauss sums to study the calculating problem of the nth power mean of Em,p and give two recurrence formulae for it. That is, we shall prove the following main results.
Theorem 1.Let p be an odd prime with 3|p−1. Then, for any positive integer n and integer m with m,p=1, we have the third-order linear recurrence formula:(8)Enm,p=Up,d⋅En−3m,p+Vp,d⋅En−2m,p,where Up,d and Vp,d are defined as(9)Up,d=64 dp+d5−12d4+60d3−160d2+240 d−384p2−5d3−48d2+180 d−320p3+5 d−24p4,Vp,d=3p⋅p2+8p−4 dp+4d2−16 d+16,and 4p=d2+27b2, where d is uniquely determined by d≡1 mod 3 (see [5]).
Theorem 2.Let p be an odd prime with 3|p−1. Then, for any positive integer n≥3, we have the recurrence formula(10)∑m=1p−1Enm,p=Up,d∑m=1p−1En−3m,p+Vp,d∑m=1p−1En−2m,p,where the first three terms of ∑m=1p−1Enm,p are(11)∑m=1p−1Em,p=0,∑m=1p−1E2m,p=2pp−1p−2 d+42,∑m=1p−1E3m,p=p−1Up,d.
Theorem 3.Let p be an odd prime with 3|p−1. Then, for any positive integer n≥1, we have the third-order linear recurrence formula(12)∑m=1p−1E−nm,p=1Up,d⋅∑m=1p−1E−n−3m,p−Vp,dUp,d⋅∑m=1p−1E−n−1m,p,where the initial values of ∑m=1p−1E−nm,p are(13)∑m=1p−1E−1m,p=2pp−1p−2 d+42−p−1Vp,dUp,d,∑m=1p−1E−2m,p=p−1V2p,d−2pp−1p−2 d+42Vp,dU2p,d,∑m=1p−1E−3m,p=p−1Up,d+2pp−1p−2 d+42V2p,d−p−1V3p,dU3p,d.
Taking n=4 in Theorem 3, we may immediately deduce the following corollary.
Corollary 1.Let p be an odd prime with 3|p−1; then, we have the identity(14)1p−1∑m=1p−11Em,p4=2pp−2 d+42−2Vp,dU2p,d+V4p,d−2pp−2 d+42V3p,dU4p,d.
2. Several LemmasIn this section, we give three lemmas which are necessary in the proofs of our theorems. In the process of proving our lemmas, we need some knowledge of the analytic number theory; all of which can be found in [6–8], so it is not necessary to repeat them here.
Lemma 1.Let p be an odd prime with p≡1mod3. Then, for any third-order character λ mod p, we have(15)τ3λ+τ3λ¯=dp ,where d is uniquely determined by 4p=d2+27b2 and d≡1mod3.
Proof.This result can be found in [9] or [10].
Lemma 2.Let p be an odd prime with 3|p−1. Then, for any cubic character λmodp, we have the identities(16)Em,p=λ¯m∑a=1p−1λ¯a3−12⋅τλ+λm∑b=1p−1λb3−12⋅τλ¯=λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−22.
Proof.For any integer 1≤a≤p−1, it is easy to show that(17)1+λa+λ2a=3,if a is a cubic residue mod p,0,otherwise.
From the properties of the cubic character modulo p, we have(18)λ2=λ¯,λ−1=1,τλ¯=τλ¯.
So, we have the identity(19)∑a=1p−1λa3−1=1τλ¯∑b=1p−1∑a=1p−1λ¯beba3−1p=1τλ¯∑b=1p−1λ¯be−bp∑a=1p−11+λa+λ¯aebap=1τλ¯∑b=1p−1λ¯be−bp−1+λ¯bτλ+λbτλ¯=1τλ¯−λ−1τλ¯+τλ∑b=1p−1λbe−bp+τλ¯∑b=1p−1e−bp=1τλ¯−τλ¯+τ2λ−τλ¯=τ3λp−2.
Therefore,(20)Em,p=λ¯mτλ∑a=1p−1λ¯a3−12+λmτλ¯∑a=1p−1λa3−12=λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−22.
This proves Lemma 2.
Lemma 3.Let p be an odd prime with 3|p−1. Then, for any cubic character λ modulo p, we have the identity(21)E3m,p=Up,d+Vp,d⋅Em,p,where(22)Up,d=64 dp+d5−12d4+60d3−160d2+240 d−384p2−5d3−48d2+180 d−320p3+5 d−24p4,Vp,d=3p⋅p2+8p−4 dp+4d2−16 d+16,and Em,p is defined as the same as in Lemma 2.
Proof.From Lemma 2, we have(23)E3m,p=λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−223=τλτ3λ¯p−223+τλ¯τ3λp−223+3p⋅τ3λ¯p−22τ3λp−22⋅Em,p.
Note that τλτλ¯=p, and from (17), we have(24)τ6λ+τ6λ¯=τ3λ+τ3λ¯2−2τ3λτ3λ¯=d2p2−2p3.
Therefore,(25)τ3λ¯p−22τ3λp−22=4−4τ3λ¯p+τ6λ¯p2⋅4−4τ3λp+τ6λp2=16−16pτ3λ+τ3λ¯+4p2τ6λ+τ6λ¯+16p2τ3λτ3λ¯−4p3τ3λτ6λ¯+τ3λ¯τ6λ+1p4τ6λτ6λ¯=16−16p⋅dp+4p2⋅d2p2−2p3+16p2⋅p3−4p3⋅p3⋅dp+1p4⋅p6=p2+8p−4 dp+4d2−16 d+16.
In addition,(26)τλτ3λ¯p−223=τ3λτ3λ¯p−26=τ3λτ18λ¯p6+36τ12λ¯p4+144τ6λ¯p2+64−12τ15λ¯p5+24τ12λ¯p4−16τ9λ¯p3−144τ9λ¯p3+96τ6λ¯p2−192τ3λ¯p=64τ3λ+240pτ3λ¯−160τ6λ¯+60pτ9λ¯−12p2τ12λ¯+1p3τ15λ¯−192p2.
Using the method similar to (24), we obtain(27)τ9λ+τ9λ¯=τ6λ+τ6λ¯τ3λ+τ3λ¯−p3τ3λ+τ3λ¯=d3p3−3 dp4,τ12λ+τ12λ¯=d4p4−4d2p5+2p6,τ15λ+τ15λ¯=d5p5−5d3p6+5 dp7.
Combining formulae (23)–(27), we have(28)E3m,p=64τ3λ+τ3λ¯+240pτ3λ¯+τ3λ−160τ6λ¯+τ6λ+60pτ9λ¯+τ9λ−12p2τ12λ¯+τ12λ+1p3τ15λ¯+τ15λ−384p2+3p⋅p2+8p−4 dp+4d2−16 d+16⋅Em,p=5 d−24p4−5d3−48d2+180 d−320p3+d5−12d4+60d3−160d2+240 d−384p2+64 dp+3p⋅p2+8p−4 dp+4d2−16 d+16⋅Em,p=Up,d+Vp,d⋅Em,p.
This proves Lemma 3.
3. Proofs of the TheoremsNow, we shall complete the proofs of our main results. Firstly, we prove Theorem 1. Let p be an odd prime with 3|p−1, χ be any Dirichlet character modulo p, and λ be a cubic character modulo p. Then, from the properties of the classical Gauss sums and (17), we have(29)∑χmodpGm,3,χ;p4=∑χmodp∑a=1p−1χaema3p4=∑χmodp∑a=1p−1∑b=1p−1∑c=1p−1∑d=1p−1χabc d¯ema3+b3−c3−d3p=p−1∑a=1p−1∑b=1p−1∑c=1p−1∑d=1p−1ab≡c dmodpema3+b3−c3−d3p=p−1∑a=1p−1∑b=1p−1∑c=1p−1∑d=1p−1ab≡cmodpemd3a3+b3−c3−1p=p−1∑a=1p−1∑b=1p−1∑d=1p−1emd3a3−1b3−1p=p−1∑a=1p−1∑b=1p−1∑d=1p−11+λd+λ¯dem da3−1b3−1p=p−1∑a=1p−1∑b=1p−1∑d=1p−1em da3−1b3−1p+p−1λ¯mτλ∑a=1p−1∑b=1p−1λ¯a3−1b3−1+p−1λmτλ¯∑a=1p−1∑b=1p−1λa3−1b3−1=p−1∑a=1p−1∑b=1p−1∑d=1p−1em da3−1b3−1p+p−1Em,p,where Em,p is the same as in Lemma 2.
For any integer n, we have the trigonometric identity(30)∑a=1p−1enap=p−1,if n,p=p,−1,if n,p=1.
From (30), we have the identity(31)∑a=1p−1∑b=1p−1∑d=1p−1em da3−1b3−1p=2p−1∑a=1p−1∑b=1p−11−p−1a3≡1modp∑a=1p−1∑b=1p−11a3≡1modp−b3≡1modp∑a=1p−1∑b=1p−11a3≡1modpb3≡1modp=6p−12−9p−1−p−42=5p2−13p−1.
Combining (29)–(31), we have(32)∑χmodpGm,3,χ;p4=∑χmodp∑a=1p−1χaema3p4=p−15p2−13p−1+p−1Em,p.
For any positive integer n, from Lemma 3, we have(33)Enm,p=E3m,p⋅En−3m,p=Up,d+Vp,d⋅Em,p⋅En−3m,p=Up,d⋅En−3m,p+Vp,d⋅En−2m,p,where(34)Up,d=64 dp+d5−12d4+60d3−160d2+240 d−384p2−5d3−48d2+180 d−320p3+5 d−24p4,Vp,d=3p⋅p2+8p−4 dp+4d2−16 d+16.
This proves Theorem 1.
Now, we prove Theorem 2. From Theorem 1, we have(35)∑m=1p−1Enm,p=Up,d∑m=1p−1En−3m,p+Vp,d∑m=1p−1En−2m,p.
From Lemma 2, we have(36)∑m=1p−1Em,p=∑m=1p−1λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−22=0,(37)∑m=1p−1E2m,p=∑m=1p−1λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−222=∑m=1p−1λmτ2λτ3λ¯p−24+∑m=1p−1λ¯mτ2λ¯τ3λp−24+∑m=1p−12pτ3λ¯p−22τ3λp−22=2pp−1p−2 d+42.
From (36) and Lemma 3, we have(38)∑m=1p−1E3m,p=∑m=1p−1λ¯mτλτ3λ¯p−22+λmτλ¯τ3λp−223=∑m=1p−1Up,d+Vp,dEm,p=p−1Up,d.
Now, Theorem 2 follows from (35)–(38).
Finally, we prove Theorem 3. For any integer n≥1, from Lemma 3, we have(39)∑m=1p−1E−nm,p=1Up,d⋅∑m=1p−1E−n−3m,p−Vp,dUp,d⋅∑m=1p−1E−n−1m,p,(40)E2m,p=Up,d⋅E−1m,p+Vp,d,Em,p=Up,d⋅E−2m,p+Vp,d⋅E−1m,p,E0m,p=Up,d⋅E−3m,p+Vp,d⋅E−2m,p,E−1m,p=Up,d⋅E−4m,p+Vp,d⋅E−3m,p.
Therefore, we have(41)∑m=1p−1E−1m,p=∑m=1p−1E2m,p−Vp,dUp,d=2pp−1p−2 d+42−p−1Vp,dUp,d,(42)∑m=1p−1E−2m,p=∑m=1p−1Em,p−Vp,d⋅E−1m,pUp,d=p−1V2p,d−2pp−1p−2 d+42Vp,dU2p,d,(43)∑m=1p−1E−3m,p=∑m=1p−11−Vp,d⋅E−2m,pUp,d=p−1Up,d+2pp−1p−2 d+42V2p,d−p−1V3p,dU3p,d.
Combining (39)–(43), we may immediately complete the proof of Theorem 3.