R E S E A R C H
Open Access
On the quartic Gauss sums and their
recurrence property
Shimeng Shen and Wenpeng Zhang
**Correspondence:
School of Mathematics, Northwest University, Xi’an, Shaanxi, P.R. China
Abstract
The main purpose of this paper is, using the method of trigonometric sums and the properties of Gauss sums, to study the computational problem of one kind of congruence equation modulo an odd prime and give some interesting fourth-order linear recurrence formulas.
MSC: 11L05
Keywords: quartic Gauss sums; trigonometric sums; recurrence formulas
1 Introduction
Letq≥ be a positive integer. For any positive integer kand integerm, thekth Gauss sumsG(m,k;q) are defined as
G(m,k;q) =
q
a= e
mak
q
,
wheree(y) =eπiy.
Concerning these sums, many authors had studied their properties and obtained a series of interesting results, see [, ]. In fact, from Weil’s important work [], one can get the upper bound estimate
p–
a= χ(a)e
mak
p
k√p,
wherepis an odd prime,χdenotes a Dirichlet charactermodpandkdenotes the Big O
notation dependent onk.
We also mention that the fourth power mean value ofG(m,k;q) was well explored by Zhang and Liu [], in which some sharp asymptotic formulas can be found.
On the other hand, Yang and Tang [] studied a number of solutions of the congruence equation
x+y≡cmodn with (xy,n) =
and gave an exact computational formula for it.
Letsbe a positive integer andpbe an odd prime withp≡mod. LetMs denote a
number of solutions of the equation
X+X+X+· · ·+Xs=
in the finite fieldGF(p), and letUs=Ms–ps–. Chowla et al. [] proved thatUssatisfies
the linear recurrence
Us– pUs––pdUs–= ,
whereU= ,U= p– andU= (p– )dwithdbeing uniquely determined by p= d+ bandd≡mod.
Some related results can also be found in [–].
Now we consider a similar problem: Letnbe a positive integer andpbe an odd prime withp≡mod. LetMn(p) denote a number of solutions of the congruence equation
x +x+· · ·+xn–+xn≡modp, () where ≤xi≤p– ,i= , , . . . ,n.
It is natural to ask whether there exists an exact computational formula forMn(p) when
nis a positive integer andpis an odd prime?
As far as we know, it seems that no one has studied this problem yet, at least we have not seen any related result before. The problem is interesting because it can help us to understand more accurate information of the quartic Gauss sums.
In this paper, we shall use the method of trigonometric sums and the properties of Gauss sums to study this problem and give some interesting computational formulas. For the sake of convenience, first we letB(p) =pa–=(a+a
p ),adenotes the solution of the equation
ax≡modp, and (∗p) denotes the Legendre symbolmodp. Then we have the following theorem.
Theorem Let p= k+ be a prime,Un(p) =Mn(p) –pn–.Then,for any positive integer
n≥,we have the fourth-order linear recurrence formula
Un(p) = –pUn–(p) + pB(p)Un–(p) –
p–pB(p)Un–(p),
where the first four terms are U(p) = ,U(p) = –(p– ),U(p) = (p– )B(p)and U(p) =
–p(p– ) + (p– )B(p).
Theorem Let p= k+ be a prime,Un(p) =Mn(p) –pn–.Then,for any positive integer
n≥,we have the fourth-order linear recurrence formula
Un(p) = pUn–(p) + pB(p)Un–(p) –
p–pB(p)Un–(p),
where the first four terms are U(p) = ,U(p) = (p– ),U(p) = (p– )B(p)and U(p) =
p(p– ) + (p– )B(p).
Corollary Let p= k+ be an odd prime.Then we have
M(p) =p+ (p– )B(p) – p(p– ), M(p) =p– p(p– )B(p).
Corollary Let p= k+ be an odd prime.Then we have
M(p) =p+ (p– )B(p) + p(p– ), M(p) =p+ p(p– )B(p).
Note the estimate for character sums
B(p)=
p–
a=
a+a p
≤
√ p,
from Corollary and Corollary we can also deduce the following corollary.
Corollary Let p be an odd prime with p≡mod.Then we have the asymptotic formu-las
M(p) =p+O(p) and M(p) =p+O(p
).
Remark Letpbe an odd prime withp≡mod, it is clear that we have the identity (see Theorems - in [])
p=
p–
a=
a+a p
+
p–
a=
a+ra p
=α+β,
whereris a quadratic non-residuemodp, that is to say, (rp) = –. The above identity implies that|B(p)|is a constant depending only onpand|B(p)| ≤√p.
For primep= k+ and positivebwith ≤b≤p– , we have the identity
A(b) =
p–
a= e
ba
p
=i
b p
√
p, i= –.
It is very easy to prove thatMn–(p) =pn–and Mn(p) =pn–+ (–)npn–(p– ).
2 Several lemmas
In this section, we give some lemmas which are necessary in the proofs of our theorems. Hereinafter, we shall use some properties of the classical Gauss sums, all of them can be found in reference [], so they will not be repeated here. First we have the following lemma.
Lemma Let p be an odd prime with p≡mod, (∗p) =χdenotes the Legendre symbol
modp.Then,for any integer b with(b,p) = ,we have the identities
(I) A(b) =
p–
a= e
ba
p
=
b p
√ p+
p–
a=
a p
e
ba p
(II) A(b) =R(p)·p+
Proof From the definition of the quadratic residuemodpand the properties of the Legen-dre symbol, we have
A(b) =
From () and () we may immediately deduce formula (I) of Lemma . Now we prove formula (II) of Lemma . It is clear that from (I) we have
A(b) =p+
From the properties of the reduced residue systemmodpand (), we have
Combining () and (), we have the identity
This proves formula (II) of Lemma .
Similarly, from (I) and () we can also prove that ifp= k+ , then we have
and let
Nk(p) = p–
b= Ak(b).
Then we have the identities
N(p) = , N(p) =R(p)p(p– ) and N(p) = p(p– )
Proof Note the trigonometric identity
p–
From Lemma and the properties of Gauss sums we have
p–
Similarly, from () and () we have
p–
where we have used the identity
+ p
It is clear that our Lemma obtained a fourth-order linear recurrence formula for the quartic Gauss sumsA(b).
3 Proofs of the theorems
Now we shall complete the proofs of our main results. First we prove Theorem . Ifp= k+ is an odd prime, then, for any positive integern, from () and the definition ofMn(p)
we have
From Lemma we have
U(p) =M(p) –p=
p
p–
b= A(b)
= –pU(p) + pU(p)B(p) – (p– )
p–B(p)
= –p(p– ) + (p– )B(p). ()
Ifk≥, then from () and Lemma we have
Uk(p) =Mk(p) –pk–=
p
p–
b=
Ak(b) =
p
p–
b=
Ak–(b)A(b)
=
p
p–
b=
Ak–(b) –pA(b) + pA(b)
p–
a=
a+a p
– p–
p–
a=
a+a p
p–
b= Ak–(b)
= –pUk–(p) + p
p–
a=
a+a p
Uk–(p)
– p–p
p–
a=
a+a p
Uk–(p). ()
Now Theorem follows from formulas ()-(). Ifp= k+ , then from () and Lemma we have
U(p) =M(p) – =
p
p–
b=
A(b) = , ()
U(p) =M(p) –p=
p
p–
b=
A(b) = (p– ), ()
U(p) =M(p) –p=
p
p–
b=
A(b) = (p– )
p–
a=
a+a p
. ()
From Lemma we also have
U(p) =M(p) –p=
p
p–
b= A(b)
= pU(p) + pU(p)
p–
a=
a+a p
– (p– ) p–
p–
a=
a+a p
Ifk≥, then from () and Lemma we also have
Uk(p) =Mk(p) –pk–=
p
p–
b=
Ak(b) =
p
p–
b=
Ak–(b)A(b) = pUk–(p)
+ p
p–
a=
a+a p
Uk–(p) – p–p
p–
a=
a+a p
Uk–(p). ()
Now Theorem follows from formulas ()-().
Competing interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Authors’ contributions
WPZ put forward the main ideas, and SMS carried on the actual operation and finishing process. All authors read and approved the final manuscript.
Acknowledgements
The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the N.S.F. (11371291) of P.R. China.
Received: 16 November 2016 Accepted: 1 February 2017
References
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