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On the fourth-order linear recurrence formula related to classical Gauss sums

  • Chen Zhuoyu and Zhang Wenpeng EMAIL logo
Published/Copyright: October 5, 2017
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Open Mathematics
From the journal Open Mathematics Volume 15 Issue 1

Abstract

Let p be an odd prime with p ≡ 1 mod 4, k be any positive integer, ψ be any fourth-order character mod p. In this paper, we use the analytic method and the properties of character sums mod p to study the computational problem of G(k, p) = τk(ψ)+τk(ψ), and give an interesting fourth-order linear recurrence formula for it, where τ(ψ) denotes the classical Gauss sums.

Keywords: The classical Gauss sums; Fourth-order linear recurrence formula; Analytic method; Character sums
MSC 2010: 11L05; 11L07; 11T24

1 Introduction

Let q≥3 be a positive integer. For any positive integer k≥2, the k-th Gauss sums G(m, k; q) is defined as

A(m,k;q)=∑a=1qemakq,

where e(y) = e2πiy.

Recently, some scholars have studied the properties of A(m, k; q), and obtained many interesting results. For example, Shen Shimeng and Zhang Wenpeng [1] proved a recurrence formula related to A(m, 4; p). Li Xiaoxue and Hu Jiayuan [2] studied the computational problem of the hybrid power mean

∑b=1p−1∑a=0p−1eba4p2⋅∑c=1p−1ebc+c¯p2, (1)

where p is an odd prime with p ≡ 1 mod 4. They proved the identity

∑b=1p−1∑a=0p−1eba4p2⋅∑c=1p−1ebc+c¯p2=3p3−3p2−3p+pτ2χ¯4+τ2χ4,if p≡5 mod 8;3p3−3p2−3p−pτ2χ¯4−pτ2χ4+2τ5χ¯4+2τ5χ4,if p≡1 mod 8,

where χ4 denotes any fourth-order character mod p,τ(χ)=∑a=1p−1χ(a)e(ap) denotes the classical Gauss sums, and c denotes the multiplicative inverse of c mod p.

Li Xiaoxue and Hu Jiayuan [2] also computed the exact value of τ2(χ4)+τ2(χ4) and τ5(χ4)+τ5(χ4).

Other works related to k-th Gauss sums and the generalized k-th Gauss sums can be found in [3-10].

Now let p be an odd prime with p ≡ 1 mod 4, ψ be any fourth-order character mod p, and we also define G(k, p) as

G(k,p)=τk(ψ)+τk(ψ¯).

In this paper, we employ a remark of [2] by using the analytic method and the properties of the classical Gauss sums to study the computational problem of G(k, p) for any positive integer k, and give an interesting fourth-order linear recurrence formula for G(k, p). That is, we will prove the following result.

Theorem

Let p be an odd prime with p ≡ 1 mod 4. Then for any positive integer k, we have the linear recurrence formulae

G2k+2,p=2p⋅α⋅G2k,p−p2⋅G2k−2,p

and

G2k+3,p=2p⋅α⋅G2k+1,p−p2⋅G2k−1,p,

where G(0, p) = 2, G(1, p) = A(1) − p , A(1) = A(1, 4; p), G(2, p) = 2 p α, G(3, p) = p . (2α−−1p−14p)⋅(A1−p),α=∑a=1p−12(a+a¯p)and(∗p) is the Legendre’s symbol mod p.

From these recurrence formulae, we can obtain the exact value of G(k, p) for all positive integer k. Hence, we can also deduce the following corollaries:

Corollary 1.1

Let p be an odd prime with p ≡ 5 mod 8. Then we have:

∑b=1p−1|∑a=0p−1eba4p|2⋅|∑c=1p−1ebc+c¯p|2=3p3−3p2+2p32α−3p.

Corollary 1.2

For odd prime p with p ≡ 1 mod 8, we have the identity

∑b=1p−1|∑a=0p−1eba4p|2⋅|∑c=1p−1e(bc+c¯p)|2=3p3−3p2−2p32α−3p+Iψ22p+2pα4pα2−p2+2p32α,

where I(ψ) = 1 if G(1, p) is positive, and I(ψ) = −1 if G(1, p) is negative.

Corollary 1.3

For any prime p with p ≡ 5 mod 8, we have the identities

|G1,p|=2p−2pαand|G5,p|=2p−2pα⋅|4pα2−p2+2p32α|.

Corollary 1.4

Let p be an odd prime with p ≡ 5 mod 8, then we have

|∑a=0p−1ea4p|=3p−2pα.

It is clear that Corollary 1.1 and Corollary 1.2 improved the results of [2].

2 Several Lemmas

In this section, we need several simple lemmas, which are necessary in the proof of our theorem. Hereinafter, we will use many properties of the classical Gauss sums, all of which can be found in [11]; so they will not be repeated here. First we have the following:

Lemma 2.1

Let p be an odd prime with p ≡ 1 mod 4. Then for any quadratic non-residue r mod p, we have the identity

p=α2+β2≡∑a=1p−12(a+a¯p)2+∑a=1p−12(a+ra¯p)2,

where (∗p) denotes the Legendre’s symbol mod p, and a denotes the multiplicative inverse of a mod p.

Proof

This is a well known result. See Theorem 4-11 in [12].□

Lemma 2.2

Let p be an odd prime with p ≡ 1 mod 4, ψ be any fourth-order character mod p. Then we have the identity

G2,p=τ2(ψ)+τ2(ψ¯)=p⋅∑a=1p−1(a+a¯p)=2p⋅α.

Proof

Let ψ be any fourth-order character mod p, it is clear that ψ2 = (∗p) = χ2, the Legendre’s symbol mod p. Then from the definition and properties of the classical Gauss sums we have

τ2ψ=∑a=1p−1∑b=1p−1ψabea+bp=∑a=1p−1ψa∑b=1p−1χ2be(ba+1p)=τχ2⋅∑a=1p−1ψaχ2a+1. (2)

Similarly, we also have

τ2(ψ¯)=τ(χ2)⋅∑a=1p−1ψ¯(a)χ2(a+1). (3)

As ψ is a fourth-order character mod p, for any integer m with (m, p) = 1, we have

1+ψm+χ2m+ψ¯m=4, if m satisfies m≡c4 mod p for some integer c with (c,p)=1;0, if otherwise.

Note that for prime p with p ≡ 1 mod 4, one has identity τ(χ2) = p . From (2) and (3) we have

τ2(ψ)+τ2(ψ¯)=p⋅∑a=1p−1ψ(a)χ2(a+1)+p⋅∑a=1p−1ψ¯(a)χ2(a+1)=p⋅∑a=1p−1(1+ψ(a)+χ2(a)+ψ¯(a))χ2(a)−p∑a=1p−1χ2(a+1)−p∑a=1p−1χ2(a)χ2(a+1)=p⋅∑a=1p−1χ2(a4+1)−p⋅∑a=1p−1χ2(a)−p⋅∑a=1p−1χ2(1+a¯)+p=p⋅∑a=1p−11+χ2(a)χ2(a2+1)+2p=p⋅∑a=1p−1(1+χ2(a))χ2(a+1)+p⋅∑a=1p−1χ2(a+a¯)+2p=p⋅∑a=1p−1(a+a¯p)=2p⋅α.

This proves Lemma 2.2.□

Lemma 2.3

Let p be an odd prime with p ≡ 1 mod 4, ψ be any fourth-order character mod p. Then we have the identities

τ(ψ)+τ(ψ¯)=A(1)−pandτ(ψ)τ(ψ¯)=−1p−14⋅p,

where A(m)=∑a=0p−1e(ma4p).

Proof

As ψ is a fourth-order character mod p, from the definition of the classical Gauss sums, we have:

A(1)=∑a=0p−1ea4p=1+∑a=1p−1(1+ψ(a)+χ2a+ψ¯a)eap=∑a=0p−1eap+∑a=1p−1ψ(a)eap+∑a=1p−1χ2(a)eap+∑a=1p−1ψ¯(a)eap=τ(ψ)+p+τ(ψ¯). (4)

Note the identities:

τ(ψ)¯=∑a=1p−1ψ¯(a)e−ap=ψ¯−1⋅∑a=1p−1ψ¯(a)eap=ψ¯−1⋅τ(ψ¯),

τ(ψ)⋅τ(ψ) = p and ψ(−1) = (−1)p−14. Hence,

τ(ψ)⋅τ(ψ¯)=(−1)p−14⋅τ(ψ)⋅τ(ψ)¯=(−1)p−14⋅p. (5)

Now Lemma 2.3 follows from equations (4) and (5).□

3 Proof of the main results

In this section, we shall complete the proof of our theorem. Let G(k, p) = τk(ψ) + τk(ψ), then for any integer k ≥ 1, note that τ2(ψ)τ2(ψ) = p2. From Lemmas 2.1 and 2.2 we have

2p⋅α⋅G2k,p=G2,p⋅G2k,p=τ2(ψ)+τ2(ψ¯)τ2k(ψ)+τ2k(ψ¯)=τ2k+2(ψ)+τ2k+2(ψ¯)+p2τ2k−2(ψ)+p2τ2k−2(ψ¯)=G2k+2,p+p2⋅G2k−2,p.

So we have

G(2k+2,p)=2p⋅α⋅G2k,p−p2⋅G(2k−2,p). (6)

Similarly, from Lemma 2.2 and Lemma 2.3 we also have

2p⋅α⋅G2k+1,p=G2,p⋅G2k+1,p=τ2(ψ)+τ2(ψ¯)τ2k+1(ψ)+τ2k+1(ψ¯)=τ2k+3(ψ)+τ2k+3(ψ¯)+p2τ2k−1(ψ)+p2τ2k−1(ψ¯)=G2k+3,p+p2⋅G(2k−1,p),

The last equation implies:

G(2k+3,p)=2p⋅α⋅G(2k+1,p)−p2⋅G(2k−1,p) (7)

for all integer k ≥ 1.

Now our Theorem follows from equations (6) and (7).

Proof of Corollary 1.3

For any prime p with p ≡ 5 mod p, note that ψ(−1) = −1 and G(1,p) = −G(1, p); so G(1, p) is pure imaginary. Thus, from Lemma 2.2 we have

|G(1,p)|=|τ2(ψ)+τ2(ψ¯)−2p|=2p−2pα. (8)

Applying the main Theorem proved and Lemma 2.1 together with τ(ψ)τ(7ψ = −p, we have G(4, p) = G2(2, p) − 2p2 = 2p(α2 − β2) and

G1,pG4,p=G5,p−pG3,p=G5−pG1,pG2,p+p.

Hence

G5,p=G1,p2pα2−β2−2p32α+p2=G1,p4pα2−p2−2p32α. (9)

Now Corollary 1.3 follows from equations (8) and (9).□

Proof of Corollary 1.4

we note that A(1) = G(1, p) + p and G(1, p) is pure imaginary; so from (8) we have

|A(1)|=∑a=0p−1ea4p=|G1,p|2+p=3p−2pα.

This completes the proof.□

  1. Competing interests

    The authors declare that they have no competing interests.

Acknowledgement

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.

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Received: 2017-3-14
Accepted: 2017-8-1
Published Online: 2017-10-5

© 2017 Zhuoyu and Wenpeng

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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https://doi.org/10.1515/math-2017-0104
Keywords for this article
The classical Gauss sums; Fourth-order linear recurrence formula; Analytic method; Character sums
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