R E S E A R C H
Open Access
An identity involving the mean value of
two-term character sums
Xiaoyan Guo
*and Jingzhe Wang
*Correspondence:
[email protected] Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R. China
Abstract
The main purpose of this paper is, using the properties of Gauss sums and the estimate for character sums, to study the mean value problem of the two-term character sums and give an interesting identity for it.
MSC: 11M20
Keywords: two-term character sums; mean value; identity; analytic method; Gauss sums
1 Introduction
Letq≥ be an integer,χ be a non-principal charactermodq. For any integernwith (n,q) = , the two-term character sumsN(n,r,s,χ;q) are defined as
N(n,r,s,χ;q) =
q
a=
χar+nas,
wherer>sare two fixed positive integers. These sums play a very important role in the study of analytic number theory, so they caused many number theorists’ interest and favor. Some works related toN(n,r,s,χ;q) can be found in [–]. If fact, the sumsN(n,r,s,χ;q) is a special case of the general character sums of the polynomials
N+M
a=N+
χf(a), ()
whereMandNare any positive integers, andf(x) is a polynomial. Ifq=pis an odd prime, then Weil (see []) obtained the following important conclusion: Letχ be a qth-order character modp. Iff(x) is not a perfectqth powermodp, then we have the estimate
N+M
x=N+
χf(x)plnp, ()
where ‘’ constant depends only on the degree off(x).
Now we are concerned about whether there exists a computational formula for the mean value
∗
χmodq q
a=
χar+nas
k
, ()
wheres>r≥ are two integers. In this paper, we shall use the analytic method and the properties of Gauss sums to study this problem, and give an exact computational formula for (). That is, we shall prove the following theorem.
Theorem Let q> be an odd integer.Then,for any real number k and integer n with
(n,q) = ,we have the identity
∗
χmodq
q
a=
χa+na
k
=J(q)·qk· pq
–
p– +
pk(p– )
,
where ∗χmodqdenotes the summation over all primitive charactersmodq,pαq denotes
that pα|q and pα+q,and J(q) =
d|qμ(d)φ( q
d)denotes the number of all primitive char-actersmodq.
Theorem Let q> be an odd integer.Then,for any real number k and integer n with
(n,q) = ,we have the identity
∗
χmodq
q
a=
χa+na
k
=J(q)·qk· pq,p–
–
p– +
pk(p– )
×
pq,|p–
–
p– +
pk(p– )
.
From these two theorems we may immediately deduce the following corollaries.
Corollary Let q be an odd square-full number(that is,for any prime p, if p|q,then
p|q). Then, for any real number k and integer n with (n,q) = , we have the
iden-tity
∗
χmodq
q
a=
χa+na
k
= ∗
χmodq q
a=
χa+na
k
=J(q)·qk.
Corollary Let q be an odd square-free number(that is,for any prime p,p†q).Then,
for any real number k and integer n with(n,q) = ,we have the identity
∗
χmodq
q
a=
χa+na
k
=qk· p|q
p– +
pk
.
Corollary Let q be an odd square-full number.Then,for any integer n with(n,q) = ,we
have the identity
∗
χmodq
q
a=
χa+na
–
= ∗
χmodq q
a=
χa+na
–
=J(q)
For general charactersχmodq,whether there exists an identity for
χmodq
q
a=
χar+nas
k
is an interesting open problem,where r>s are two positive integers.
2 Several lemmas
In this section, we shall give several lemmas, which are necessary in the proof of our the-orems. Hereinafter, we shall use many properties of character sums and Gauss sums, all of which can be found in references [] and [], so they will not be repeated here. First we have the following lemmas.
Lemma Let q> be an integer,J(q)denotes the number of all primitive characters modq.
Then we have the identity
J(q) =
d|a
μ(d)·φ
q d
,
whereμ(n)is the Möbius function,φ(n)is the Euler function.
Proof This is a well-known result, here we give a simple proof. It is clear that for any non-principal characterχmodq, there exists one and only oned|qand a primitive character χ∗ moddsuch thatχ=χ∗·χ, whereχdenotes the principal charactermodq. So, from these properties, we have
φ(q) =
χmodq
=
d|q J(d).
From this identity and the Möbius inversion formula, we may immediately deduce
J(q) =
d|a
μ(d)·φ
q d
.
This proves Lemma .
Lemma Let p be an odd prime,i≥be an integer and q=pi.Then,for any real number k and integer n with(n,q) = ,we have the identities
∗
χmodq
q
a=
χa+na
k
= ∗
χmodq
qk=φ(q)·qk–
and
∗
χmodp p
a=
χa+na
k
Proof For any primitive characterχmodq, from the properties of Gauss sums, we know that
q
a=
χa+na= τ(χ)
q
a=
q
b= χ(b)e
b(a+na)
q
=
τ(χ)
q
a=
q
b= χ(ba)e
ba(a+na)
q
=
τ(χ)
q
b= χ(b)e
nb
q q
a= χ(a)e
ba
q
=τ(χ) τ(χ)
q
b= χ(b)e
nb
q
=χ(n)τ(χ)·τ(χ )
τ(χ) . ()
Note thatq=piandi≥, so ifχis a primitive charactermodq, thenχis also a primitive character modq, so that|τ(χ)|=|τ(χ)|=|τ(χ)|=√q. From these identities, Lemma and formula (), we may immediately deduce that
∗
χmodq
q
a=
χa+na
k
= ∗
χmodq
qk=φ(q)·qk–.
This proves the first formula of Lemma .
Now we prove the second formula. Ifχ is the Legendre symbolmodp, then we have |τ(χ)|= . So, from Lemma and the method of proving (), we have
∗
χmodp p
a=
χa+na
k
= + (p– )pk.
This completes the proof of Lemma .
Lemma Let p> be an odd prime,α≥be an integer and q=pα.Then,for any real
number k and n with(n,q) = ,we have the identities
∗
χmodq
q
a=
χa+na
k
=φ(q)·qk–
and
∗
χmodp p
a=
χa+na
k
=
+ (p– )·pk if p≡mod;
Proof From the properties of primitive characters modq and the method of proving Lemma , we have
q
a=
χa+na= τ(χ) q a= q b= χ(b)e
b(a+na)
q = τ(χ) q b= χ(b)e
nb q q a= χ(a)e
ba
q
=χ(n)τ(χ
)·τ(χ)
τ(χ) . ()
Sinceχis a primitive charactermodqand (,q) = , soχandχare also two primitive characters modq. So, from Lemma , () and the properties of Gauss sums, we can deduce the first identity of Lemma .
Note that if (,p– ) = ,χis the Legendre symbolmodp, then for any non-principal character χmodp, we have |τ(χ)|=√p and|τ(χ
)|= . So, from Lemma and the method of proving (), we have
∗
χmodp p a=
χa+na
k
= + (p– )pk. ()
If |p– , then there exist two -order charactersmodp, so from the method of proving (), we have the identity
∗
χmodp p a=
χa+na
k
= + (p– )pk. ()
Combining () and (), we can deduce the second identity of Lemma .
Lemma Let q> and q> be two integers with(q,q) = ,χmodqandχmodq.
Then,for any integer n with(n,qq) = ,we have
qq
a= χχ
a+na
= q a= χ a+na
· q b= χ b+nb
.
Proof From the properties of complete residue systemqq, we have
qq
a= χχ
a+na
= q a= q b=
χχ(aq+bq)+n(aq+bq)
= q a= χ
(aq+bq)+n(aq+bq)
q
b= χ
(aq+bq)+n(aq+bq)
= q a= χ
(aq)+n(aq)
q
b= χ
(bq)+n(bq)
= q a= χ a+na
q
b= χ
b+nb
= q a= χ a+na
· q b= χ b+nb
.
This proves Lemma .
3 Proof of the theorems
In this section, we shall complete the proof of our theorems. First we prove Theorem . For any odd numberq> , it is clear that there exist two integersMandNsuch thatq=M·N, whereMis a square-free number, andNis a square-full number. Now, for any primitive character χmodq, there exist two primitive charactersχmodMandχmodN such thatχ=χχ. Note thatJ(N) = φ
(N)
N , so from these properties, Lemma , Lemma and
Lemma , we have
∗
χmodq q a=
χa+na k = ∗
χmodM M a=
χa+na k · ∗
χmodN N a=
χa+na
k
=
p|M
+ (p– )·pk pαN
φpα·pα(k–)
=J(q)·qk· pq
–
p– +
pk(p– )
,
wherepαqdenotes thatpα|qandpα+q. This proves Theorem .
Now we prove Theorem . From Lemma , Lemma and the method of proving The-orem , we have
∗
χmodq q a=
χa+na k = ∗
χmodM M a=
χa+na k · ∗
χmodN N a=
χa+na
k
=
p|M
+ (p– )·pk pαN
φpα·pα(k–)
=J(q)·qk· pq,p–
–
p– +
pk(p– )
×
pq,|p–
–
p– +
pk(p– )
.
Competing interests
The authors did not provide this information.
Authors’ contributions
The authors did not provide this information.
Acknowledgements
The authors would like to thank the referee for their very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the National Natural Science Foundation of P.R. China (No. 11071194).
Received: 6 July 2013 Accepted: 17 October 2013 Published:12 Nov 2013
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