One kind of character sum modulo a prime p and its recurrence formula

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R E S E A R C H

Open Access

One kind of character sum modulo a prime

p

and its recurrence formula

Wenpeng Zhang

1,2

_{and Zhuoyu Chen}

2*

*_{Correspondence:}

[email protected]

2_{School of Mathematics, Northwest} University, Xi’an, P.R. China Full list of author information is available at the end of the article

Abstract

The aim of this paper is to use an analytic method and the properties of the classical Gauss sums to research the computational problem of one kind of character sum of polynomials modulo an odd primepand obtain several meaningful third- and fourth-order linear recurrence formulae for them.

MSC: 11L10; 11L40.

Keywords: The classical Gauss sums; Character sums of polynomials; Linear recurrence formula; Analytic method

1 Introduction

Letq≥3 be an integer, and letχbe a nonprincipal charactermodq. Then for any integral coeﬃcient polynomialf(x), we deﬁne the character sum of the polynomial as

N(χ,f;q) =

q

a=1

χf(a).

This sum plays an extremely signiﬁcant role in analytic number theory, so they have aroused the interest and favor of a great deal of number theorists. A lot of works con-nected withN(χ,f;q) can be found in [1–9] and [10–12]. In fact, the sumsN(χ,f;q) are a particular case of the general character sums of the polynomials

N+M

a=N+1

χf(a),

whereMandNare any positive integers. Ifq=pis an odd prime, then Weil (see [2] and [6]) obtained following signiﬁcant conclusion:

Suppose thatχis aqth-order charactermodpandf(x) is not a perfectqth powermod

p. Then we have the estimate

N+M

x=N+1

χf(x)p12_ln_p_, ₍₁₎

where the constant in “” depends only on the degree off(x). The estimate in (1) is the best possible.

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Now for any odd prime pand any nonprincipal character χmod p, we consider the following problem: for any positive integerskandh, let

Mk(h,χ;p) = p–1

a1=0

p–1

a2=0

· · · p–1

ak=0

χah₁+ah₂+· · ·+ah_k

and

Nk(h,χ;p) =Mk(h,χ;p) +Mk(h,χ;p).

We inquire if there exists an accurate computational formula forNk(h,χ;p)?

About this contents, from our personal perspective, it appears that none had researched it yet; at least so far, we have not seen any related results before. The problem is meaningful, since it can help scholars to ﬁnd out more exact information of the character sums.

In our paper, applying analytic methods and the properties of the classical Gauss sums, we researched the problem of calculatingNk(h,χ;p) and obtained a signiﬁcant linear

re-currence formula for it. We will prove the following:

Theorem 1 Let p be an odd prime,let h be a positive integer,and letχbe any Dirichlet charactermodp such thatχh₌_χ_0,_{the principal character}_mod_p_._{Then for any positive}

integer k,we obtain the identity

Nk(h,χ;p) =Mk(h,χ;p) = p–1

a1=0

p–1

a2=0

· · · p–1

ak=0

χah

1+ah2+· · ·+ahk

= 0.

Theorem 2 Let p be an odd prime with p≡1mod 3,and letχbe any third-order char-actermodp.Then we have the identities

N1(3,χ;p) = 2(p– 1); N2(3,χ;p) = (p– 1)d; N3(3,χ;p) = 6p(p– 1)

and

Nk(3,χ;p) = 3p·Nk–2(3,χ;p) +dp·Nk–3(3,χ;p)

for all integers k≥4,where d is uniquely determined by4p=d2_{+ 27}_b2_{and d}_≡₁_mod _3.

Theorem 3 Let p= 8h+ 5be a prime,and letχ be any fourth-order character modp.

Then we have the identities

N1(4,χ;p) = 2(p– 1), N2(4,χ;p) = –4(p– 1)α,

N3(4,χ;p) = 2(p– 1)p– 2α2

and

Nk+4(4,χ;p) = –2pNk+2(4,χ;p) + 8pαNk+1(4,χ;p) –p

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for all k≥0,where N0(4,χ;p) = 0,α= we have the identities

N1(4,χ;p) = 2(p– 1), N2(4,χ;p) = 4(p– 1)α, N3(4,χ;p) = 2(p– 1)2α2–p

From the methods of proving the theorems, we can also infer the following:

Corollary 1 Let p be a prime with p≡1mod 3,and letχ be a third-order character

modp.Then for any integer k≥0,we have the identity

Then we have the identity

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2 Several lemmas

In this section, we give several lemmas, which are essential in the proofs of our theorems. Hereinafter, we are going to use some properties of the classical Gauss sums, which can be found in some analytic number theory books, such as [13]; so we will not repeat them here. For convenience, ﬁrst, we give the deﬁnition of the classical Gauss sumsτ(χ) as follows: for any integerq> 1, letχbe any Dirichlet charactermodq. Then the famous Gauss sum

τ(χ) is deﬁned as

τ(χ) =

q

a=1

χ(a)e

a q ,

wheree(y) =e2πiy_{. With this mark, we have the following:}

Lemma 1 Given p be an odd prime with p≡1mod 3,and letψbe any third-order char-actermodp.Then we have the identity

τ3(ψ) +τ3(ψ) =dp,

where d is uniquely determined by4p=d2_{+ 27}_b2_{and d}_≡₁_mod _3.

Proof See Lemma 3 of [9] or references [14] and [10].

Lemma 2 Given p be an odd prime with p≡1mod 3,and for any integer b with(b,p) = 1,

let U(b,p) =_ap–1₌₀e(ba_p3).Then we have the identity

U3(b,p) =dp+ 3p·U(b,p),

where d is the same as in Lemma1.

Proof Letχ be any third-order charactermodp. Thenχ2=χ, and from Lemma1and the properties of Gauss sums we have

U(b,p) =

p–1

a=0

e

ba3

p = 1 +

p–1

a=1

e

ba3

p

= 1 +

p–1

a=1

1 +χ(a) +χ(a)e

ba p

=

p–1

a=0

e

ba

p +

p–1

a=1

χ(a)e

ba

p +

p–1

a=1

χ(a)e

ba p

=χ(b)τ(χ) +χ(b)τ(χ). (2)

Note thatχ3₌_χ

0andτ(χ)τ(χ) =p, so from (2) we immediately infer that

U3(b,p) =χ(b)τ(χ) +χ(b)τ(χ)3

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=τ3(χ) +τ3(χ) + 3p·U(b,p) =dp+ 3p·U(b,p).

This proves Lemma2.

Lemma 3 Given p be an odd prime with p≡1mod 4,and letψ be any fourth-order charactermodp.Then we have the identity

τ2(ψ) +τ2(ψ) =√p·

p)denotes the Legendre symbolmodp,and a denotes the

multi-plicative inverse of amod p,that is,aa≡1mod p.

Proof In fact, this is Lemma 2.2 in [15]. Therefore we omit its proof.

Lemma 4 Let p be an odd prime with p≡1mod 4,and for any integer b with(b,p) = 1,

let A(b,p) =_ap–1₌₀e(ba_p4).Then we have the identities

A4(b,p) = 2pC(p) + 2A2(b,p) + 8pαA(b,p) + 4pα2–C2(p)p2,

where C(p) = –3if p= 8k+ 5and C(p) = 1if p= 8k+ 1.

Proof Letψbe a fourth-order character modp. Thenψ2₌_χ

2(the Legendre symbol mod

p), and by the properties of Gauss sums we get

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Now, according to (4), we have

A2(b,p) –C(p)p2=A4(b,p) – 2C(p)pA2(b,p) +C2(p)p2

=2√pA(b,p) + 2√p·α2= 4pA2(b,p) + 2αA(b,p) +α2. (5)

Applying (5), we immediately deduce that

A4(b,p) = 2pC(p) + 2A2(b,p) + 8pαA(b,p) + 4pα2–C2(p)p2.

This proves Lemma4.

Lemma 5 Let p be an odd prime with p≡1mod 3,and letχbe any third-order character

modp.Then we have the identities

M1(3,χ;p) =p– 1; M2(3,χ;p) = p– 1

p ·τ

3₍_χ_); _M_3(3,_χ_;_p_{) = 3}_p₍_p_{– 1)}

and

Mk(3,χ;p) = 3p·Mk–2(3,χ;p) +dp·Mk–3(3,χ;p)

for all integers k≥4,where d is the same as in Lemma1.

Proof It is obvious thatM1(3,χ;p) =p– 1. According to (2) and the properties of Gauss sums, we get

M2(3,χ;p) = 1

τ(χ)

p–1

b=1

χ(b)

p–1

c=0

p–1

d=0

e

b(c3₊_d3₎

p

= 1

τ(χ)

p–1

b=1

χ(b)χ(b)τ(χ) +χ(b)τ(χ)2

= 1

τ(χ)

p–1

b=1

χ(b)χ2(b)τ2(χ) + 2p+χ2(b)τ2(χ)

=τ 2₍_χ₎

τ(χ) ·(p– 1) =

p– 1

p ·τ

3₍_χ_), ₍₆₎

M3(3,χ;p) = 1

τ(χ)

p–1

b=1

χ(b)

p–1

c=0

p–1

d=0

p–1

e=0

e

b(c3₊_d3₊_e3₎

p

= 1

τ(χ)

p–1

b=1

χ(b)χ(b)τ(χ) +χ(b)τ(χ)3

= 1

τ(χ)

p–1

b=1

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For any integerk≥4, according to Lemma1, we have we have the identities

M1(4,χ;p) =p– 1;

For every integer k≥0,we obtain the fourth-order linear recurrence formula

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Suppose thatχis a fourth-order charactermodp. Then we get

In the same way, applying (3) and the orthogonality of the charactersmodp, we also have

M3(4,χ;p) = 1 we have the identities

M1(4,χ;p) =p– 1;

For every integer k≥0,we have the fourth-order linear recurrence formula

Mk+4(4,χ;p) = 6pMk+2(4,χ;p) + 8pαMk+1(4,χ;p) +p

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Proof Ifp= 8h+ 1, then from Lemma4we have

It is not complicated to prove that

M1(4,χ;p) =

Similarly, combined with the method of proving (12), we also get

M3(4,χ;p) = 1

3 Proofs of the theorems

In this section, we complete the proofs of our theorems. First of all, we prove Theorem1 by mathematical induction. Ifk= 1, then note thatχh₌_χ_{0. According to the properties}

of character sumsmodp, we obtain the identity

p–1

Suppose that the conclusion holds for an integerk=m≥1, that is,

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Then fork=m+ 1, combining (17) and (18) with the properties of the complete residue systemmodp, we have

p–1

a1=0

p–1

a2=0

· · · p–1

am=0

p–1

am+1=0

χah₁+ah₂+· · ·+ah_m+ah_m₊₁

=

p–1

a1=0

p–1

a2=0

· · · p–1

am=0

χah₁+ah₂+· · ·+ah_m

+

p–1

a1=0

p–1

a2=0

· · · p–1

am=0

p–1

am+1=1

χah₁+ah₂+· · ·+ah_m+ah_m₊₁

=

p–1

a1=0

p–1

a2=0

· · · p–1

am=0

χah₁+a₂h+· · ·+ah_m+ 1

p–1

am+1=1

χh(am+1)

= 0.

Thus the conclusion is also correct fork=m+ 1. This proves Theorem1.

Now, we are going to prove Theorem2. From Lemmas1and5, noting that by deﬁnition

Nk(h,χ;p) =Mk(h,χ;p) +Mk(h,χ;p), we get

N1(3,χ;p) = 2(p– 1), N3(3,χ;p) = 6p(p– 1),

N2(3,χ;p) =

p– 1

p ·τ

3₍_χ_{) +}p– 1

p ·τ

3₍_χ_{) =}_d₍_p_{– 1)}

and

Nk(3,χ;p) = 3p·Nk–2(3,χ;p) +dp·Nk–3(3,χ;p)

for all integersk≥4. This proves Theorem2.

Suppose thatpis an odd prime withp= 8h+ 5 and thatχis a fourth-order character

modp. Then applying Lemmas3and6, we have

N1(4,χ;p) = 2(p– 1); (19)

N2(4,χ;p) = –2(p√– 1)

p ·τ

2₍_χ_{) –}2(p_√– 1)

p ·τ

2₍_χ_{) = –4(}_p_{– 1)}_·_α_. ₍₂₀₎

By the identitiesτ(χ)2=τ2₍_χ_),_τ2₍_χ₎_τ2₍_χ_{) =}_τ2₍_χ₎_τ₍_χ₎2₌_p2_{, and}

τ4(χ) +τ4(χ) =τ2(χ) +τ2(χ)2– 2τ2(χ)τ2(χ) = 4pα2– 2p2,

we have

N3(4,χ;p) = –

p– 1

p ·τ

4₍_χ_{) –}p– 1

p ·τ

4₍_χ₎

= –p– 1

p

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For every integerk≥0, we get the fourth-order linear recurrence formula

Nk+4(4,χ;p)

= –2pNk+2(4,χ;p) + 8pαNk+1(4,χ;p) –p

9p– 4α2Nk(4,χ;p), (22)

whereN0(4,χ;p) = 0, andαis the same as in Lemma3. Now Theorem3follows from (19)–(22).

In the same way, using Lemmas3and7, we can also deduce Theorem4. To prove Corollary1, applying Lemma5, we get

M4(3,χ;p) = 3(p– 1)τ3(χ) +dp(p– 1).

Then from this formula and Lemma1, by the identities|τ(χ)|=√pandτ3₍_χ_{) =}_τ3₍_χ_{) we} have

p–1

a=0

p–1

b=0

p–1

c=0

p–1

c=0

χa3+b3+c3+d3

=3(p– 1)τ3(χ) +dp(p– 1)

= (p– 1)3τ3(χ) +dp= (p– 1)9p3+ 3dpτ3(χ) +τ3(χ)+d2p2

1 2

= (p– 1)9p3+ 3d2p2+d2p2

1

2 _{= (}_p_{– 1)}_p·₉_p_{+ 4}_d2_.

This proves Corollary1.

Corollaries2and3follow from Lemma6and the identity|τ(χ)|=√p.

Next, we are going to prove Corollary4. Ifp= 8k+ 5, then noting thatτ(χ)2=τ2₍_χ_{) and}

|α+β|2=|α|2+αβ+αβ+|β|2, from Lemmas6and3we have

M4(4,χ;p)=4√p(p– 1)τ2(χ) + 8pα(p– 1)

= 4√p(p– 1)τ2(χ) + 2√pα

= 4√p(p– 1)p2+ 4pα2+ 2√pατ2(χ) +τ2(χ)12

= 4√p(p– 1)p2+ 8pα2

1

2 _{= 4}_p₍_p_{– 1)}·_p_{+ 8}_α2_. ₍₂₃₎

In the same way, ifp= 8k+ 1, then from Lemmas7and3, by the method of proving (23) we have

M4(4,χ;p)=12√p(p– 1)τ2(χ) + 8pα(p– 1)

= 4√p(p– 1)3τ2(χ) + 2√pα

= 4√p(p– 1)9p2+ 4pα2+ 6√pατ2(χ) +τ2(χ)12

= 4√p(p– 1)9p2+ 16pα2

1

2 _{= 4}_p₍_p_{– 1)}·₉_p_{+ 16}_α2_. ₍₂₄₎

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Acknowledgements

The authors would like to thank the editors and referees for their very helpful and detailed comments, which have signiﬁcantly improved the presentation of this paper.

Funding

This work is supported by the N. S. F. (11771351) and (11826205) of P. R. China.

Competing interests

The authors declare that there are no conﬂicts of interest regarding the publication of this paper.

Authors’ contributions

Both authors have equally contributed to this work. Both authors read and approved the ﬁnal manuscript.

Author details

1_{School of Science, Xi’an Technological University, Xi’an, P.R. China.}2_{School of Mathematics, Northwest University, Xi’an,}

P.R. China.

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Received: 18 February 2019 Accepted: 29 March 2019

References

1. Burgess, D.A.: On character sums and primitive roots. Proc. Lond. Math. Soc.12, 179–192 (1962) 2. Burgess, D.A.: On Dirichlet characters of polynomials. Proc. Lond. Math. Soc.13, 537–548 (1963)

3. Granville, A., Soundararajan, K.: Large character sums: pretentious characters and the Pólya–Vinogradov theorem. J. Am. Math. Soc.20, 357–384 (2007)

4. Zhang, W.P., Yi, Y.: On Dirichlet characters of polynomials. Bull. Lond. Math. Soc.34, 469–473 (2002) 5. Zhang, W.P., Yao, W.L.: A note on the Dirichlet characters of polynomials. Acta Arith.115, 225–229 (2004) 6. Weil, A.: On some exponential sums. Proc. Natl. Acad. Sci. USA34, 204–207 (1948)

7. Bourgain, J., Garaev, M.Z., Konyagin, S.V., Shparlinski, I.E.: On the hidden shifted power problem. SIAM J. Comput.41, 1524–1557 (2012)

8. Zhang, W.P., Liu, H.N.: On the general Gauss sums and their fourth power mean. Osaka J. Math.42, 189–199 (2005) 9. Zhang, W.P.: On the number of the solutions of one kind congruence equation modp. J. Northwest Univ. Nat. Sci.46,

313–316 (2016)

10. Zhang, W.P., Hu, J.Y.: The number of solutions of the diagonal cubic congruence equation modp. Math. Rep.20, 73–80 (2018)

11. Zhang, H., Zhang, W.P.: The fourth power mean of two-term exponential sums and its application. Math. Rep.19, 75–83 (2017)

12. Chen, L., Hu, J.Y.: A linear recurrence formula involving cubic Gauss sums and Kloosterman sums. Acta Math. Sinica (Chin. Ser.)61, 67–72 (2018)

13. Apostol, T.M.: Introduction to Analytic Number Theory. Springer, New York (1976)

14. Berndt, B.C., Evans, R.J.: The determination of Gauss sums. Bull. Am. Math. Soc.5, 107–128 (1981)