The main purpose of this article is to study the calculating problem of one kind fourth power mean of the generalized Kloosterman's sums and provide an accurate calculating formula for it utilizing analytical methods and character sums' properties. Simultaneously, the work also provides a fresh and valuable approach for researching the related power mean problem.
Citation: Junfeng Cui, Li Wang. The generalized Kloosterman's sums and its fourth power mean[J]. AIMS Mathematics, 2023, 8(11): 26590-26599. doi: 10.3934/math.20231359
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The main purpose of this article is to study the calculating problem of one kind fourth power mean of the generalized Kloosterman's sums and provide an accurate calculating formula for it utilizing analytical methods and character sums' properties. Simultaneously, the work also provides a fresh and valuable approach for researching the related power mean problem.
Let q be an integer with q>1. The generalized Kloosterman's sums S(m,n,χ;q) are defined as follows for all integers m, n, and any Dirichlet character χmodq:
S(m,n,χ;q)=q−1∑a=1χ(a)e(ma+n¯aq), |
where ¯a denotes a⋅¯a≡1modq, e(y)=e2πiy, and i2=−1.
Classical Kloosterman's sums (see H. D. Kloosterman [8]) are S(m,n,χ0;q)=S(m,n;q) if χ=χ0 is the main character modulo q. It is common knowledge that for a prime p,
p−1∑a=1e(a+n¯ap)=−2√pcos(θ(n)), |
where the angles θ(n) is equi-distributed in [0,π] with respected to the Sato-Tate measure 2πsin2(θ)dθ. Additional information can be found in the provided reference [7].
In the study of analytical number theory, Kloosterman's sums is crucial. As a result, some scholars have studied the various properties of S(m,n;q), and obtained a series of significant results. For instance, S. Chowla [7] or T. Estermann [3] proved the upper bound estimate:
q∑a=1′e(ma+n¯aq)≪(m,n,q)12⋅d(q)⋅q12, |
where q∑a=1′ denotes the summation over all 1≤a≤q such that (a,q)=1, d(n) denotes the Dirichlet divisor function, (m,n,q) denotes the greatest common factor of m, n, and q.
H. Saliˊe [11] demonstrated that for any odd prime p, one had the identity
p−1∑m=0|p−1∑a=1e(a+m¯ap)|4=2p3−3p2−3p. |
The proofs of this result can also be found in H. Iwaniec [5].
W. P. Zhang [14] used the elementary methods to prove a generalized result. In other words, for any integer n with (n,q)=1, one has the identity
q∑m=1|q∑a=1′e(ma+n¯aq)|4=3ω(q)q2ϕ(q)∏p‖q(23−13p−43p(p−1)), |
where ϕ(q) is Euler function, ω(q) denotes the number of all different prime divisors of q, p‖q denotes the product over all prime divisors of q with p∣q and p2∤q.
There are also some equivalent results for the generalized Kloosterman's sums. For instance, A. V. Malyshev [10] demonstrated that for any odd prime p, then
p−1∑a=1χ(a) e(ma+n¯ap)≪(m,n,p)12⋅p12. |
W. P. Zhang [13] or J. H. Li and Y. N. Liu [9] used the different methods to prove the identity
p−1∑m=1|p−1∑a=1χ(a)e(ma+n¯ap)|4={2p3−3p2−3p−1 if χ is the principal character modulo p;3p3−8p2 if χ is the Legendre symbol modulo p;2p3−7p2 if χ is a complex character modulo p, |
where p is an odd prime.
Some of the other results related to Kloosterman's sums can also be found in references [2,6,12,15,16]. To save space, all the results are no longer listed here.
In this paper, we consider the calculating problem of the fourth power mean
p−1∑m=1|p−1∑a=1χ1(a)e(ma+¯ap)|2⋅|p−1∑b=1χ2(b)e(mb+¯bp)|2, | (1.1) |
where p is an odd prime, χ1 and χ2 are two Dirichlet characters modulo p.
This issue is undoubtedly a generalization of [13]. In reality, if χ1=χ2 be the principal characters modulo p, then (1.1) becomes [11]. When χ1=χ2 or ¯χ2, then (1.1) corresponds to [13]. As a result, the mean value (1.1), which extends the fourth power means of the generalized Kloosterman's sums, is significant.
The primary objective of this research is to investigate the calculating issues in (1.1) and provide a precise formula utilizing the analytic methods and the properties of the classical Gauss sums. The specific content of the argument is as follows.
Theorem 1. Let p>3 be an odd prime, χ1 and χ2 are characters modulo p (not all principal character modulo p). Then we have the identity
p−1∑m=1|p−1∑a=1χ1(a)e(ma+¯ap)|2⋅|p−1∑b=1χ2(b)e(mb+¯bp)|2={p2(3p−8)ifχ1=χ2istheLegendresymbolmodulop;p2(2p−7)if χ1=χ2or¯χ2isnottheLegendresymbolmodulop;p2(p−6)ifallχ1≠χ2andχ1χ2notχ0andχ1(−1)=χ2(−1);p2(p−4)ifallχ1≠χ2andχ1χ2notχ0andχ1(−1)≠χ2(−1);p(p2−4p−2)ifoneofχ1andχ2isχ0andχ1(−1)=χ2(−1);p(p2−2p−2)ifoneofχ1andχ2isχ0andχ1(−1)≠χ2(−1). |
Some notes: It is evident that our result is a broadening and extension of the findings from [13]. This theorem merely took into account the fact that p is an odd prime. The question of whether there is a calculation formula that corresponds to the universal composite number q is still unresolved.
In this section, in order to make the structure of the theorem proof clear and the content complete, we first introduce three simple lemmas. Hereinafter, the properties of the classical Gauss sums and character sums, as well as some understanding of analytic number theory, elementary number theory, and those topics in general may all be found in numerous number theory textbooks, including [1] and [4]. Therefore, we have omitted more detailed content. The first is as follows.
Lemma 1. Let p be a prime, χ1 and χ2 be two fixed Dirichlet characters modulo p (not all principal characters). Then we have the identity
|p−1∑m=1χ0(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2={p2(p−2)2ifχ1χ2=χ0;p2ifallχ1,χ2andχ1χ2nottheprincipalcharactersmodulop;pifoneofχ1andχ2istheprincipalcharactermodulop, |
where χ0 denotes the principal character modulo p.
Proof. For any integer n, note that the trigonometric identity
p−1∑m=0e(mnp)={p if p∣n,0 if p∤n. |
From the definition of the classical Gauss sums and note that the identity
τ(¯χ1)τ(¯χ2)=χ1(−1)p, |
if χ1χ2=χ0, we have
p−1∑m=1(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))=p−1∑a=1p−1∑b=1χ1(a)χ2(b)p−1∑m=1e(m(a+b)+¯a+¯bp)=p−1∑a=1p−1∑b=1χ1(a)χ2(b)p∑m=1e(m(a+b)+¯a+¯bp)−τ(¯χ1)τ(¯χ2)=pp−1∑a=1p−1∑b=1p∣(a+b)χ1(a)χ2(b)−τ(¯χ1)τ(¯χ2)=pp−1∑a=1χ1(a)χ2(p−a)−τ(¯χ1)τ(¯χ2)={χ2(−1)p(p−1)−χ1(−1)p if χ1χ2=χ0,−τ(¯χ1)τ(¯χ2) if χ1χ2≠χ0, | (2.1) |
where τ(χ)=p−1∑a=1χ(a)e(ap) denotes the classical Gauss sums.
Note that for any non-principal character χmodp, we have |τ(χ)|=√p. From (2.1) we may immediately deduce Lemma 1.
Lemma 2. Let p be a prime, χ1 and χ2 be two fixed Dirichlet characters modulo p (not all principal character modulo p). Then we have the identity
∑χmodp|p−1∑a=1χ1(a+1)χ2(¯a+1)¯χ((a+1)(¯a+1))|2={(p−1)(2p−5)ifχ1=χ2;(p−1)(p−4)ifχ1≠χ2andχ1(−1)=χ2(−1);(p−1)(p−2)ifχ1≠χ2andχ1(−1)≠χ2(−1). |
Proof. From the orthogonality of the characters modulo p we have
∑χmodp|p−1∑a=1χ1(a+1)χ2(¯a+1)¯χ((a+1)(¯a+1))|2=(p−1)p−1∑a=1p−1∑b=1(a+1)(¯a+1)≡(b+1)(¯b+1)modpχ1(a+1)χ2(¯a+1)¯χ1(b+1)¯χ2(¯b+1)=(p−1)p−1∑a=1p−1∑b=1(a−b)(¯a−b)≡0modpχ1(a+1)χ2(¯a+1)¯χ1(b+1)¯χ2(¯b+1)=(p−1)(p−2)+(p−1)p−2∑a=1χ1(a)¯χ2(a)−(p−1). | (2.2) |
If χ1=χ2, then χ1¯χ2=χ0, in this time, we have
p−2∑a=1χ1(a)¯χ2(a)=p−2. | (2.3) |
If χ1≠χ2, then χ1¯χ2≠χ0, in this time, we have
p−2∑a=1χ1(a)¯χ2(a)=p−1∑a=1χ1(a)¯χ2(a)−χ1(−1)χ2(−1)={−1 if χ1(−1)=χ2(−1),1 if χ1(−1)≠χ2(−1). | (2.4) |
Combining formula (2.2)–(2.4) we have the identity
∑χmodp|p−1∑a=1χ1(a+1)χ2(¯a+1)¯χ((a+1)(¯a+1))|2={(p−1)(2p−5) if χ1=χ2;(p−1)(p−4) if χ1≠χ2 and χ1(−1)=χ2(−1);(p−1)(p−2) if χ1≠χ2 and χ1(−1)≠χ2(−1). |
This proves Lemma 2.
Lemma 3. Let p be a prime, χ1 and χ2 be two fixed Dirichlet characters modulo p. Then for any non-principal Dirichlet character χmodp, we have the identity
p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))=τ(χ)τ(¯χ1χ2χ)p−1∑a=1χ1(a+1)χ2(¯a+1)¯χ((a+1)(¯a+1)). |
Proof. From the properties of the classical Gauss sums we have
p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))=p−1∑a=1p−1∑b=1χ1(a)χ2(b)p−1∑m=1χ(m)e(m(a+b)+¯a+¯bp)=τ(χ)p−1∑a=1p−1∑b=1χ1(a)χ2(b)¯χ(a+b)e(¯a+¯bp)=τ(χ)p−1∑a=1χ1(a)¯χ(a+1)p−1∑b=1χ1(b)χ2(b)¯χ(b)e(¯b(¯a+1)p)=τ(χ)τ(¯χ1χ2χ)p−1∑a=1χ1(a)¯χ(a+1)χ1(¯a+1)χ2(¯a+1)¯χ(¯a+1)=τ(χ)τ(¯χ1χ2χ)p−1∑a=1χ1(a+1)χ2(¯a+1)¯χ((a+1)(¯a+1)). |
This proves Lemma 3.
This section is the most important part of the article, which is the proof of the main theorem. First from the orthogonality of the characters modulo p we have
∑χmodp|p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2=(p−1)p−1∑m=1|p−1∑a=1χ1(a)e(ma+¯ap)|2⋅|p−1∑b=1χ2(b)e(mb+¯bp)|2. | (3.1) |
On the other hand, note that if χ=χ0, then |τ(χ)|=1. If χ≠χ0, then |τ(χ)|=√p. So if χ1χ2=χ0 and χ1=χ2, then χ1=χ2=(∗p) is the Legendre symbol modulo p. From Lemmas 1–3 we have
∑χmodp|p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2=|p−1∑m=1χ0(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2+∑χmodpχ≠χ0|p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2=p2(p−2)2+∑χmodpχ≠χ0p2⋅|p−1∑a=1χ1(a+1)χ2(¯a+1)¯χ((a+1)(¯a+1))|2=p2(p−2)2+p2⋅∑χmodp|p−1∑a=1χ1(a+1)χ2(¯a+1)¯χ((a+1)(¯a+1))|2−p2|p−1∑a=1χ1(a+1)χ2(¯a+1)|2=p2(p−2)2+p2⋅(p−1)(2p−5)−p2=p2(p−1)(3p−8). | (3.2) |
If χ1χ2=χ0 and χ1 is not the Legendre symbol modulo p, then we have
∑χmodp|p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2=p2(p−2)2+p2⋅(p−1)(p−4)−p2=p2(p−1)(2p−7). | (3.3) |
If χ1=χ2 is a complex character modulo p, then we have
∑χmodp|p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2=|p−1∑m=1χ0(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2+∑χmodpχ≠χ0, χ21|p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2+p⋅|p−1∑a=1χ1(a+1)χ1(¯a+1)¯χ21((a+1)(¯a+1))|2=p2+p2∑χmodp|p−1∑a=1χ1(a+1)χ1(¯a+1)¯χ((a+1)(¯a+1))|2−p2⋅|p−1∑a=1χ1(a+1)χ1(¯a+1)χ0((a+1)(¯a+1))|2−p2⋅|p−1∑a=1χ1(a+1)χ1(¯a+1)¯χ21((a+1)(¯a+1))|2+p⋅|p−1∑a=1χ1(a+1)χ1(¯a+1)¯χ21((a+1)(¯a+1))|2=p2(p−1)(2p−5)+2p2−2p3=p2(p−1)(2p−7). | (3.4) |
If all χ1, χ2, and χ1χ2 are not principal character modulo p, χ1≠χ2 and χ1(−1)=χ2(−1), then we have
∑χmodp|p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2=|p−1∑m=1χ0(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2+∑χmodpχ≠χ0, χ1χ2|p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2+p⋅|p−1∑a=1χ1(a+1)χ2(¯a+1)¯χ1χ2((a+1)(¯a+1))|2=p2+p2∑χmodp|p−1∑a=1χ1(a+1)χ2(¯a+1)¯χ((a+1)(¯a+1))|2−p2⋅|p−1∑a=1χ1(a+1)χ2(¯a+1)χ0((a+1)(¯a+1))|2−p2⋅|p−1∑a=1χ1(a+1)χ2(¯a+1)¯χ1χ2((a+1)(¯a+1))|2+p⋅|p−1∑a=1χ1(a+1)χ2(¯a+1)¯χ1χ2((a+1)(¯a+1))|2=p2+p2(p−1)(p−4)+p2−2p3=p2(p−1)(p−6). | (3.5) |
If all χ1, χ2, and χ1χ2 are not principal character modulo p, χ1≠χ2 and χ1(−1)≠χ2(−1), then we have
∑χmodp|p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2=p2+p2(p−1)(p−2)+p2−2p3=p2(p−1)(p−4). | (3.6) |
Similarly, if one of χ1 and χ2 is principal character modulo p and χ1(−1)χ2(−1)=1, then we also have
∑χmodp|p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2=p+p2⋅(p2−5p+4)+p−p2−p2=p(p−1)(p2−4p−2). | (3.7) |
If one of χ1 and χ2 is principal character modulo p and χ1(−1)χ2(−1)=−1, then we also have
∑χmodp|p−1∑m=1χ(m)(p−1∑a=1χ1(a)e(ma+¯ap))(p−1∑b=1χ2(b)e(mb+¯bp))|2=p+p2⋅(p−1)(p−2)+p−p2−p2=p(p−1)(p2−2p−2). | (3.8) |
Now combining (3.1)–(3.8) we have the identity
p−1∑m=1|p−1∑a=1χ1(a)e(ma+¯ap)|2⋅|p−1∑b=1χ2(b)e(mb+¯bp)|2={p2(3p−8) if χ1=χ2 is the Legendre symbol modulo p;p2(2p−7) if χ1=χ2or¯χ2 is not the Legendre symbol modulo p;p2(p−6) if all χ1≠χ2 and χ1χ2 not χ0 and χ1(−1)=χ2(−1);p2(p−4) if all χ1≠χ2 and χ1χ2 not χ0 and χ1(−1)≠χ2(−1);p(p2−4p−2) if one of χ1 and χ2 is χ0 and χ1(−1)=χ2(−1);p(p2−2p−2) if one of χ1 and χ2 is χ0 and χ1(−1)≠χ2(−1). |
This completes the proof of our theorem.
The primary result of this research is to provide an exact calculating formula for the fourth power mean of one kind generalized Kloosterman's sums
p−1∑m=1|p−1∑a=1χ1(a)e(ma+¯ap)|2⋅|p−1∑b=1χ2(b)e(mb+¯bp)|2, |
where p be an odd prime, χ1 and χ2 is any fixed Dirichlet characters modulo p.
This proof strategies of the paper are not only novel, but they also serve as a good point of reference for problems that may arise in subsequent study.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors would like to thank the referees for their very helpful and detailed comments. This work is supported by the N. S. F. (12126357) of China.
The authors declare that there are no conflicts of interest regarding the publication of this paper.
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