On the two-term exponential sums and character sums of polynomials

Yuankui Ma; Wenpeng Zhang

doi:10.1515/math-2019-0107

Article Open Access

On the two-term exponential sums and character sums of polynomials

Yuankui Ma and Wenpeng Zhang

Published/Copyright: November 2, 2019

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$Open Mathematics$

From the journal Open Mathematics Volume 17 Issue 1

Abstract

The main aim of this paper is to use the analytic methods and the properties of the classical Gauss sums to research the computational problem of one kind hybrid power mean containing the character sums of polynomials and two-term exponential sums modulo p, an odd prime, and acquire several accurate asymptotic formulas for them.

Keywords: two-term exponential sums; character sums of polynomials; hybrid power mean; analytic method; asymptotic formula

MSC 2010: 11L03; 11L40

1 Introduction

Let q ≥ 3 be an integer and χ be a Dirichlet character modulo q. For any positive integers N and M with M > N, and rational coefficient polynomial f(x) of x with degree n, the character sums of polynomials mod q is defined by

S(χ,f;q)=∑a=N+1N+Mχ(f(a)).

It is well known that the upper bound estimate of S(χ, f; q) is a particularly vital classical problem in analytic number theory. Any substantial progress in this area will certainly play a valuable role in promoting the development of analytic number theory. For this reason, a great number of scholars have researched the estimate problem of S(χ, f; q), and obtained a series of meaningful results. For instance, Pólya and Vinogradov’s ground breaking work (see [1]: Theorem 8.21 and Theorem 13.15) proved that for any non-principal character χ mod q, one has the estimate

∑a=N+1N+Mχa≪q12ln⁡q,

where the symbol A ≪ B denotes |A| < cB for some constant c.

Suppose that q = p is an odd prime. A. Weil’s [2] proved a particularly significant conclusion: Let χ be a q-th character mod p, and polynomial f(x) is not a perfect q-th power mod p, then the estimate

∑x=N+1N+Mχf(x)≪p12ln⁡p (1.1)

can be acquired, where the estimate p12 in (1.1) is the best one. Actually, Zhang Wenpeng and Yi Yuan [3] gave a series of polynomials f(x) = (x − r)^m(x − s)ⁿ such that

∑a=1qχ(a−r)m(a−s)n=q,

where (r − s, q) = 1, m, n and χ satisfy several special conditions.

The minor term ln p in (1.1) is difficult to improve, and it cannot even be improved to ln^λ p, where 0 < λ < 1 is any fixed real number.

A lot of results associated with character sums of polynomials can be found in various analytic number theory books, such as [4, 5], and several papers about character sums of polynomials can be found in [6, 7, 8, 9, 10]. We are not going to list them one by one.

On the other hand, for any integers m and n, the two-term exponential sums G(m, n, r, s; q) is defined as

G(m,n,r,s;q)=∑a=0qemar+nasq,

where r > s ≥ 1 are integers, and e(y) = e^2πiy.

It is necessary to research two-term exponential sums. In fact, if r = p is an odd prime, they are closely related to Fourier analysis on finite fields. Because of this, a lot of researchers have discussed the various properties of G(m, n, r, s; q), and obtained a great number of meaningful results, see [11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]. For instance, Zhang Wenpeng and Han Di [24] researched the sixth power mean of the two-term exponential sums, and obtained an exact computational formula.

Zhang Han and Zhang Wenpeng [25] proved a significant and precise formula

∑m=1p−1∑a=0p−1ema3+nap4=2p3−p2 if 3∤p−1,2p3−7p2 if 3|p−1,

where p indicates an odd prime with (n, p) = 1.

In this paper, we are going to consider the computational problem of the hybrid power mean containing character sums of polynomials and two-term exponential sums

H(r,s,t,χ;p)=∑m=1p−1∑a=1p−1χar+mas2⋅∑b=0p−1embt+bp2. (1.2)

Han Di [26] studied the asymptotic properties of the hybrid mean value involving the two-term exponential sums and polynomial character sums, and proved the following asymptotic formula

∑m=1p−1∑a=1p−1emak+nap2⋅∑a=1p−1χma+a¯2=2p3+O|k|p2 if 2∣k,2p3+O|k|p52 if 2∤k,

where p is an odd prime, χ denotes any non-principal even Dirichlet character mod p, and a represents the multiplicative inverse of a mod p. That is, aa ≡ 1 mod p.

If we take k = −1 in this theorem, one can deduce the asymptotic formula

∑m=1p−1∑a=1p−1ema+a¯p2⋅∑b=1p−1χmb+b¯2=2p3+Op2.

Du Xiaoying [27] researched a similar problem, and proved the following conclusion:

Let p > 3 be a prime with (3, p − 1) = 1. Then for any non-principal even character χmod p, one has the identity

∑m=0p−1∑a=1p−1χma3+a2⋅∑b=1p−1emb3+bp2=2pp2−p−1−p2+3p∑u=1p−1χ¯(u)∑a=1p−1(a−1)(a3−u2)p,

where ∗p denotes the Legendre symbol mod p.

According to this formula, Du Xiaoying [27] deduced the following asymptotic formula:

∑m=0p−1∑a=1p−1χma3+a2⋅∑b=1p−1emb3+bp2=2p3+Op2.

The main aim of this paper is to use the analytic methods and the properties of the classical Gauss sums to research the computational problem of (1.2) for special integers r = 4, s = 1 and t = 3 or 4. We will give several sharp asymptotic formulas for (1.2). That is, we will prove the following two main results:

Theorem 1

Let p be an odd prime with p ≡ 1 mod 3 and χ be any Dirichlet character mod p. If χ³ ≠ χ₀ and χ⁴ ≠ χ₀, then we acquire the asymptotic formula

∑m=1p−1∑a=1p−1χ3a4+ma2⋅∑b=0p−1emb3+bp2=3p3+E(p),

where the error term E(p) satisfies the estimate |E(p)| ≤ 18 ⋅ p².

If χ³ ≠ χ₀ and χ⁴ = χ₀, then we acquire the asymptotic formula

∑m=1p−1∑a=1p−1χ3a4+ma2⋅∑b=0p−1emb3+bp2=2p3+E1(p),

where the error term E₁(p) satisfies the estimate |E(p)| ≤ 15 ⋅ p².

Theorem 2

Let p be an odd prime with p ≡ 1 mod 3, χ be any Dirichlet character mod p. If χ³ ≠ χ₀ and χ⁴ ≠ χ₀, then we obtain the asymptotic formula

∑m=1p−1∑a=1p−1χ3a4+ma2⋅∑b=0p−1emb4+bp2=3p3+W(p),

where the error term W(p) satisfies the estimate |W(p)| ≤ 27 ⋅ p52 .

If χ³ ≠ χ₀ and χ⁴ = χ₀, then we obtain the asymptotic formula

∑m=1p−1∑a=1p−1χ3a4+ma2⋅∑b=0p−1emb4+bp2=2p3+W1(p),

where the error term W₁(p) satisfies the estimate |W₁(p)| ≤ 23 ⋅ p52 .

Some notes: Above all, if 3 ∤ (p − 1) in Theorem 1, then for any non-principal character χ mod p, if χ⁴ = χ₀, then we acquire the identity

∑a=1p−1χa4+ma=1.

Therefore, in this case, the result is trivial. That is,

∑m=1p−1∑a=1p−1χa4+ma2⋅∑b=0p−1emb3+bp2=p2.

If χ⁴ ≠ χ₀, then we acquire the identity

∑a=1p−1χa4+ma=p.

In this case, we acquire the identity

∑m=1p−1∑a=1p−1χa4+ma2⋅∑b=0p−1emb3+bp2=p3.

Secondly, if 3 | (p − 1) and χ is not a third character mod p (that is, there is not any character χ₁ mod p such that χ = χ13 ), then we obtain the identity

∑a=1p−1χa4+ma=0.

Therefore, we did not discuss this special case.

Third, our methods can also be applied to the general hybrid power mean H(3, 1, k, χ; p) for all integers k ≥ 3. Actually, if 3 | k, then the asymptotic formula for H(3, 1, k, χ; p) is the same as in Theorem 1. If 3 ∤ k, then the asymptotic formula for H(3, 1, k, χ; p) is the same as in Theorem 2.

Finally, it is worthwhile to improve the error term in Theorem 2.

2 Some lemmas

In this part, firstly, we introduce several simple properties related to classical Gauss sums mod q, which is defined as

τ(χ)=∑a=1qχ(a)eaq,where e(y)=e2πiy.

If χ is a primitive character mod q, then one has the identities

∑a=1qχ(a)enaq=χ¯(n)τ(χ)and |τ(χ)|=q.

The other properties of τ(χ) can also be found in a great number of analytic number theory text books, such as [1] or [4] and [5], here we will not repeat them.

Lemma 1

Let p be an odd prime with p ≡ 1 mod 3 and m be any integer with (m, p) = 1. If χ is not a third character mod p, then we acquire

∑a=1p−1χa4+ma2=0;

If χ is a third character mod p and χ⁴ ≠ χ₀, then we acquire the identity

∑a=1p−1χa4+ma2=3p+λ¯(m)∑a=1p−1χ(a)∑b=1p−1λb4a3−1λ¯(b−1)+λ(m)∑a=1p−1χ(a)∑b=1p−1λ¯b4a3−1λ(b−1).

If χ is a third character mod p and χ⁴ = χ₀, then we acquire the identity

∑a=1p−1χa4+ma2=2p+1+λ¯(m)∑a=1p−1χ(a)∑b=1p−1λb4a3−1λ¯(b−1)+λ(m)∑a=1p−1χ(a)∑b=1p−1λ¯b4a3−1λ(b−1),

where λ is a third-order character mod p. That is, λ ≠ χ₀ and λ³ = χ₀.

Proof

If χ is not a third character mod p, then there exists an integer r such that r³ ≡ 1 mod p and χ(r) ≠ 1. According to the properties of the reduced residue system mod p, we obtain

∑a=1p−1χa4+ma=∑a=1p−1χ(ra)4+mra=χ(r)∑a=1p−1χr3a4+ma=χ(r)∑a=1p−1χa4+ma

1−χ(r)⋅∑a=1p−1χa4+ma=0. (2.1)

Since χ(r) ≠ 1, applying with (2.1) we obtain the identity

∑a=1p−1χa4+ma2=0. (2.2)

If χ is a third character mod p, for any integer m with (m, p) = 1, we obtain the identity

∑a=0p−1ema3p=1+∑a=1p−11+λ(a)+λ¯(a)emap=λ(m)τλ¯+λ¯(m)τ(λ),

where λ is any third-order character mod p.

According to the properties of Gauss sums and reduced residue system, we obtain

∑a=1p−1χa4+ma2=1p∑a=1p−1∑b=1p−1χab¯∑c=0p−1∑d=1p−1edc3ba3−1+mdb−1p=∑a=1p−1∑b=1p−1ba3≡1modp⁡χab¯∑d=0p−1emdb−1p−∑a=1p−1∑b=1p−1ba3≡1modp⁡χab¯+1p∑a=1p−1∑b=1p−1χab¯∑d=1p−1λd(ba3−1))τλ¯+λ¯d(ba3−1τ(λ)emdb−1p=p∑a=1p−1a3≡1modp⁡χ(a)−∑a=1p−1χ4(a)+τ(λ)τλ¯p∑a=1p−1∑b=1p−1χab¯λba3−1λ¯m(b−1)+τ(λ)τλ¯p∑a=1p−1∑b=1p−1χab¯λ¯ba3−1λm(b−1)=3p+λ¯(m)∑a=1p−1∑b=1p−1χaλb4a3−1λ¯(b−1)+λ(m)∑a=1p−1∑b=1p−1χaλ¯b4a3−1λ(b−1). (2.3)

If χ⁴ ≠ χ₀, we can use the identity τ(λ)τ(λ) = p.

If χ⁴ = χ₀, then we get

∑a=1p−1χa4+ma2=2p+1+λ¯(m)∑a=1p−1∑b=1p−1χaλb4a3−1λ¯(b−1)+λ(m)∑a=1p−1∑b=1p−1χaλ¯b4a3−1λ(b−1). (2.4)

Now, Lemma 1 follows from (2.2), (2.3) and (2.4).

Lemma 2

Let p be an odd prime, then we obtain the identity

∑m=0p−1∑a=0p−1ema3+ap2=p(p−2) if 3∣(p−1),p2 if 3∤(p−1).

Proof

For any positive integer q > 1, applying with the trigonometric identity

∑m=1qenmq=q if q∣n,0 if q∤n

and the properties of the reduced residue system mod p, we get

∑m=0p−1∑a=0p−1ema3+ap2=∑m=0p−11+∑a=1p−1ema3+ap2=p+∑a=1p−1∑m=0p−1ema3+ap+∑a=1p−1∑b=1p−1∑m=0p−1emb3a3−1+b(a−1)p=p+p∑a=1p−1a3≡1modp⁡∑b=1p−1eb(a−1)p=p(p−2) if 3∣(p−1),p2 if 3∤(p−1).

This proves Lemma 2.

Lemma 3

Let p be an odd prime, then we obtain the identity

∑m=0p−1∑a=0p−1ema4+ap2=p(p−3) if 4∣(p−1),p(p−1) if 4∤(p−1).

Proof

It is not difficult to prove above formula by the same method of proving Lemma 2, so we omit details of the proof.

Lemma 4

Let p be an odd prime with p ≡ 1 mod 3, then for any third-order character λ mod p, we acquire

∑m=1p−1λ(m)∑a=0p−1ema3+ap2=−τ2λ¯.

Proof

Applying with the properties of Gauss sums, we acquire

∑m=1p−1λ(m)∑a=0p−1ema3+ap2=∑m=1p−1λ(m)+2∑a=1p−1∑m=1p−1λ(m)ema3+ap+∑a=1p−1∑b=1p−1∑m=1p−1λ(m)emb3a3−1+b(a−1)p=2τ(λ)∑a=1p−1eap+τ(λ)∑a=1p−1λ¯a3−1∑b=1p−1eb(a−1)p=−2τ(λ)−τ(λ)∑a=1p−1λ¯a3−1=−τ(λ)∑a=0pλ¯a3−1. (2.5)

It is obvious that

∑a=0pλa3−1=2+∑a=1p−11+λ(a)+λ¯(a)λ(a−1)=∑a=0p−1λ(a−1)+∑a=0p−1λ¯(a−1)+∑a=1p−1λ(a)λ(a−1)=1τλ¯∑b=1p−1λ¯(b)∑a=1p−1λ(a)eb(a−1)p=τ(λ)τλ¯∑b=1p−1λ¯2(b)e−bp=τ2(λ)τλ¯=τ3(λ)τ(λ)τλ¯=τ3(λ)p. (2.6)

Combining (2.5) and (2.6), we have

∑m=1p−1λ(m)∑a=0p−1ema3+ap2=−τ2λ¯.

This proves Lemma 4.

Lemma 5

Let p be an odd prime with p ≡ 1 mod 3, then for any third-order character λmod p, we can acquire the estimate

∑m=1p−1λ(m)∑a=0p−1ema4+ap2≤p32+p.

Proof

Note that τ(λ)τ(λ) = p, and λ⁴ = λ, from the method of proving Lemma 4, we obain

∑m=1p−1λ(m)∑a=0p−1ema4+ap2=∑a=1p−1∑b=1p−1∑m=1p−1λ(m)emb4a4−1+b(a−1)p+∑m=1p−1λ(m)+∑a=1p−1∑m=1p−1λ(m)ema4+ap+∑a=1p−1∑m=1p−1λ(m)e−ma4−ap=2τ(λ)τλ¯+τ(λ)∑a=1p−1λ¯a4−1∑b=1p−1λ¯(b)eb(a−1)p=2p+p∑a=1p−1λ¯a4−1λ(a−1)=2p+p∑a=2p−1λ¯a3+a2+a+1=p1−λ(2)+p∑a=0p−1λ¯a3+a2+a+1. (2.7)

On the other hand, for any integer b with (b, p) = 1, note that the identity

∑a=0p−1eba2p=1+∑a=1p−11+χ2(a)ebap=χ2(b)τ(χ2),

where χ₂ = ∗p denotes the Legendre symbol mod p. We also deduce the identity

∑a=0p−1λ¯a3+a2+a+1=∑a=0p−1λ¯(a+1)3−2(a+1)2+2(a+1)=∑a=1p−1λ¯2a¯2−2a¯+1=∑a=0p−1λ¯2a2−2a+1−1=λ(2)∑a=0p−1λ¯(2a−1)2+1−1=λ(2)∑a=0p−1λ¯a2+1−1=λ(2)τ(λ)∑b=1p−1λ(b)∑a=0p−1eba2+bp−1=λ(2)τ(χ2)τ(λ)∑b=1p−1λ(b)χ2(b)ebp−1=λ(2)τ(χ2)τλχ2τ(λ)−1. (2.8)

Now Lemma 5 follows from (2.7), (2.8) and the identity |τ(χ)| = p .

Lemma 6

Let p be an odd prime with p ≡ 1 mod 3, then for any non-principal third character χmod p, we can obtain the estimate

∑a=1p−1χ(a)∑b=1p−1λb4a3−1λ¯(b−1)≤6p,

where λ is any third-order character mod p.

Proof

Since 3 | (p − 1), there exists an integer 1 < r < p − 1 such that r³ ≡ 1 mod p. For any integer m with (m, p) = 1, note that λ(m) + λ(mr) + λ(r² m) = λ(m)(1 + λ(r) + λ(r²)) = 0, and combining with Lemma 1, we know that for any third character χmod p with χ ≠ χ₀, we obtain the identity

∑a=1p−1χa4+ma2+∑a=1p−1χa4+mra2+∑a=1p−1χa4+mr2a2=9p+λ¯(m)+λ¯(rm)+λ¯r2m∑a=1p−1χ(a)∑b=1p−1λb4a3−1λ¯(b−1)+λ(m)+λ(rm)+λr2m∑a=1p−1χ(a)∑b=1p−1λ¯b4a3−1λ(b−1)=9p. (2.9)

In summary, for i = 0, 1, 2, according to (2.9) we acquire the estimate

∑a=1p−1χa4+mria2≤9p. (2.10)

On the other hand, from Lemma 1 we also obtain

∑a=1p−1χa4+ma4+∑a=1p−1χa4+mra4+∑a=1p−1χa4+mr2a4=27p2+6pλ¯(m)+λ¯(rm)+λ¯r2m∑a=1p−1χ(a)∑b=1p−1λb4a3−1λ¯(b−1)+6pλ(m)+λ(rm)+λr2m∑a=1p−1χ(a)∑b=1p−1λ¯b4a3−1λ(b−1)+λ(m)+λ(rm)+λr2m∑a=1p−1χ(a)∑b=1p−1λb4a3−1λ¯(b−1)2+λ¯(m)+λ¯(rm)+λ¯r2m∑a=1p−1χ(a)∑b=1p−1λ¯b4a3−1λ(b−1)2+6⋅∑a=1p−1χ(a)∑b=1p−1λb4a3−1λ¯(b−1)2=27p2+6⋅∑a=1p−1χ(a)∑b=1p−1λb4a3−1λ¯(b−1)2. (2.11)

Combining (2.10) and (2.11) we may immediately deduce the estimate

∑a=1p−1χ(a)∑b=1p−1λb4a3−1λ¯(b−1)≤6p.

This proves Lemma 6.

3 Proofs of the theorems

In this section, we are going to prove our main theorem. Let p be an odd prime with p ≡ 1 mod 3. For any non-principal third character χmod p with χ⁴ ≠ χ₀, from Lemma 1, Lemma 2 and Lemma 4 we acquire

∑m=1p−1∑a=1p−1χa4+ma2⋅∑b=0p−1emb3+bp2=∑m=1p−1λ¯(m)∑b=0p−1emb3+bp2∑a=1p−1χ(a)∑b=1p−1λb4a3−1λ¯(b−1)+∑m=1p−1λ(m)∑b=0p−1emb3+bp2∑a=1p−1χ(a)∑b=1p−1λ¯b4a3−1λ(b−1)+3p∑m=1p−1∑b=0p−1emb3+bp2=3p2(p−2)−τ2λ¯∑a=1p−1χ(a)∑b=1p−1λb4a3−1λ¯(b−1)−τ2λ∑a=1p−1χ(a)∑b=1p−1λ¯b4a3−1λ(b−1). (3.1)

Note that |τ(λ)|² = p. From (3.1) and Lemma 6 we may instantly deduce the asymptotic formula

∑m=1p−1∑a=1p−1χa4+ma2⋅∑b=0p−1emb3+bp2=3p3+E(p), (3.2)

where the error term E(p) satisfies the estimate |E(p)| ≤ 18 p².

If χ ≠ χ₀ is a third character mod p with χ⁴ = χ₀, then from Lemma 1 and the method of proving (3.2), we can also deduce the asymptotic formula

∑m=1p−1∑a=1p−1χa4+ma2⋅∑b=0p−1emb3+bp2=2p3+E1p, (3.3)

where the error term E₁(p) satisfies the estimate |E₁(p)| ≤ 15 p².

It is obviously that Theorem 1 follows from (3.2) and (3.3).

Applying Lemma 1, Lemma 3, Lemma 5, Lemma 6 and the method of proving Theorem 1 we can easily deduce the conclusion of Theorem 2. That is,

∑m=1p−1∑a=1p−1χ3a4+ma2⋅∑b=0p−1emb4+bp2=3p3+W(p) if χ4≠χ0,2p3+W1p if χ4=χ0,

where the error terms W(p) and W₁(p) satisfy the estimates |W(p)| ≤ 27 ⋅ p32 and |W₁(p)| ≤ 23 ⋅ p32 .

This completes the proofs of our all results.

4 Conclusion

The main results of this paper are two theorems. We obtained two sharp asymptotic formulas for the hybrid power mean involving character sums of polynomials and two-term exponential sums. In addition to this, we also acquired the upper bound estimation of the error terms. These results profoundly reveal the law of the value distribution of the character sums of polynomials and two-term exponential sums, and it can also be used for reference in the research of similar problems.

Acknowledgment

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. (11771351) of P. R. China.

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Received: 2019-06-24

Accepted: 2019-09-27

Published Online: 2019-11-02

This work is licensed under the Creative Commons Attribution 4.0 Public License.

Articles in the same Issue

https://doi.org/10.1515/math-2019-0107

Keywords for this article

two-term exponential sums; character sums of polynomials; hybrid power mean; analytic method; asymptotic formula

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