Citation: Yan Zhao, Wenpeng Zhang, Xingxing Lv. A certain new Gauss sum and its fourth power mean[J]. AIMS Mathematics, 2020, 5(5): 5004-5011. doi: 10.3934/math.2020321
[1] | Xingxing Lv, Wenpeng Zhang . The generalized quadratic Gauss sums and its sixth power mean. AIMS Mathematics, 2021, 6(10): 11275-11285. doi: 10.3934/math.2021654 |
[2] | Jin Zhang, Wenpeng Zhang . A certain two-term exponential sum and its fourth power means. AIMS Mathematics, 2020, 5(6): 7500-7509. doi: 10.3934/math.2020480 |
[3] | Xuan Wang, Li Wang, Guohui Chen . The fourth power mean of the generalized quadratic Gauss sums associated with some Dirichlet characters. AIMS Mathematics, 2024, 9(7): 17774-17783. doi: 10.3934/math.2024864 |
[4] | Xue Han, Tingting Wang . The hybrid power mean of the generalized Gauss sums and the generalized two-term exponential sums. AIMS Mathematics, 2024, 9(2): 3722-3739. doi: 10.3934/math.2024183 |
[5] | Jinmin Yu, Renjie Yuan, Tingting Wang . The fourth power mean value of one kind two-term exponential sums. AIMS Mathematics, 2022, 7(9): 17045-17060. doi: 10.3934/math.2022937 |
[6] | Junfeng Cui, Li Wang . The generalized Kloosterman's sums and its fourth power mean. AIMS Mathematics, 2023, 8(11): 26590-26599. doi: 10.3934/math.20231359 |
[7] | Wenpeng Zhang, Yuanyuan Meng . On the fourth power mean of one special two-term exponential sums. AIMS Mathematics, 2023, 8(4): 8650-8660. doi: 10.3934/math.2023434 |
[8] | Xiaoxue Li, Wenpeng Zhang . A note on the hybrid power mean involving the cubic Gauss sums and Kloosterman sums. AIMS Mathematics, 2022, 7(9): 16102-16111. doi: 10.3934/math.2022881 |
[9] | He Yanqin, Zhu Chaoxi, Chen Zhuoyu . A sum analogous to Kloosterman sum and its fourth power mean. AIMS Mathematics, 2020, 5(3): 2569-2576. doi: 10.3934/math.2020168 |
[10] | Xiaoge Liu, Yuanyuan Meng . On the k-th power mean of one kind generalized cubic Gauss sums. AIMS Mathematics, 2023, 8(9): 21463-21471. doi: 10.3934/math.20231093 |
Let q≥3 is a positive integer. For any integral coefficient polynomial f(x) and any Dirichlet character χmodq, we define a generalized Gauss sums G(f,χ;q) as
G(f,χ;q)=q∑a=1χ(a)e(f(a)q), |
where as usual, e(y)=e2πiy.
If f(x)=x, then G(f,χ;q)=τ(χ) is the classical Gauss sums. That is,
τ(χ)=q∑a=1χ(a)e(aq). |
A very important property of τ(χ) is that if χ is a primitive character mod q, then we have the identity |τ(χ)|=√q (see Theorem 8.15 in [1]).
If f(x)=ax2+bx, then G(f,χ;q) becomes the generalized quadratic Gauss sums. Many authors have studied its properties and obtained interesting results. For example, if q=p is an odd prime, then from A. Weil's important work [2] we can get the estimate
|G(f,χ;p)|≤2⋅√p |
for all integers a and b with (a,b,p)=1.
Let p be an odd prime, α be a positive integer α≥2, and λ be a primitive character modulo pα. Zhang Wenpeng and Lin Xin [3] proved that for any integer n with (n,p)=1, we have
pα∑m=1′|pα∑a=1λ(a)e(ma2+napα)|4=p2αϕ(pα)(α+1−5p−1); |
If λ is any non-primitive character modulo pα, then we have the identity
pα∑m=1′|pα∑a=1λ(a)e(ma2+napα)|4=p2αϕ(pα), |
where ϕ(n) denotes the Euler function.
Li Xiaoxue and Xu Zhefeng [4] also studied the special case q=p, an odd prime, and obtained several interesting identities.
On the other hand, Lv Xingxing and Zhang Wenpeng [5] introduced a new sum analogous to Kloosterman sum as follows:
K(m,n,r,χ;q)=q∑a=1χ(ma+n¯a)e(raq), | (1.1) |
where m, n and r are integers, ¯a denotes a⋅¯a≡1modq.
Using properties of the character modulo p and analytic methods, they proved the following results: For any odd prime p with p≡3mod4 and integers m and n with (mn,p)=1, one has the identity
∑χmodp|p−1∑a=1χ(a)e(ma2p)|2⋅|p−1∑b=1χ(b+¯b)|2=(p−1)(3p2−6p−1) |
and
∑χmodp|p−1∑a=1χ(a)e(ma2p)|2⋅|p−1∑b=1χ(b+¯b)e(nbp)|2=(p−1)(p2−2p−1)+p(p−1)(p−2∑b=2e(n(b+¯b)p)+p−2∑b=2e(n(b−¯b)p)). |
From the second formula and the estimate for Kloosterman sums, Lv Xingxing and Zhang Wenpeng [5] deduced the asymptotic formula
∑χmodp|p−1∑a=1χ(a)e(ma2p)|2⋅|p−1∑b=1χ(b+¯b)e(nbp)|2=p3+O(p52). |
Shane Chern [6] also studied the properties of K(m,n,r,χ;q), and obtained the identity
∑χmodp|p−1∑m=1|p−1∑a=1χ(ma+n¯a)e(kap)|2|2=(p−1)(p4−7p3+17p2−5p−25), |
where p is an odd prime, n and k are integers with (nk,p)=1.
Some other papers related to Gauss sums and two-term exponential sums can also be found in [5,6,7,8,9,10,11,12,13], but we will not go into further detail here.
Inspired by [5] and [6], it is natural to study the 2k-th power means
p−1∑m=0∑χmodp|p−1∑a=0χ(a2+1)e(map)|2k, k≥2. | (1.2) |
To the best knowledge of the authors, sums of the type (1.2) have not been previously investigated but the authors think that this problem is meaningful, and can be viewed as a new combination of additive and multiplicative functions.
In this paper, we will use the number of the solutions of some congruence equations mod p and the properties of the Legendre symbol to study the problem of calculating (1.2) with k=2, and give two exact calculating formulae according to p≡1mod4 or p≡3mod4. That is, we will prove the following results:
Theorem 1. Let p be an odd prime with p≡3mod4, then we have
∑χmodpp−1∑m=0|p−1∑a=0χ(a2+1)e(map)|4=p(p−1)(3p2−4p+4). |
Theorem 2. Let p be an odd prime with p≡1mod4, then we have
∑χmodpp−1∑m=0|p−1∑a=0χ(a2+1)e(map)|4=3p(p−1)(p2−6p+10). |
Some notes: For k≥3, whether there exists an exact calculating formula for (1.2) is an open problem. It is also interesting to ask whether a general formula analogous to Theorems 1 and 2 can be obtained for all integers q≥3 and k=2.
In this section, we first give four basic lemmas, which are necessary in the proof of our theorems. Of course, in order to prove these lemmas, we need some knowledge of elementary and analytic number theory. They can be found in reference [1], and we do not repeat them all here. First we have the following:
Lemma 1. For any odd prime p, we have the identity
p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab≡cdmodpa+b≡c+dmodp1=2p2−p. |
Proof. Note that if a, b, c pass through a reduced residue system mod p, then da, db and cd also pass through a reduced residue system mod p, providing (d,p)=1. So from these properties we have
p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab≡cdmodpa+b≡c+dmodp1=4(p−1)+1+p−1∑a=1p−1∑b=1p−1∑c=1p−1∑d=1ab≡cdmodpa+b≡c+dmodp1=4p−3+p−1∑a=1p−1∑b=1p−1∑c=1p−1∑d=1ab≡cmodpa+b≡c+1modp1=4p−3+(p−1)p−1∑a=1p−1∑b=1a+b≡ab+1modp1=4p−3+(p−1)p−1∑a=1p−1∑b=1(a−1)(b−1)≡0modp1=4p−3+(p−1)(2p−1∑a=11−1)=4p−3+(p−1)(2p−3)=2p2−p. |
This proves Lemma 1.
Lemma 2. If p is an odd prime, then we have the identity
p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab+cd≡2modpa+b≡c+dmodp1=p2+(−1p)⋅p. |
Proof. It is clear that if d passes through a reduced residue system mod p, then ¯d also passes through a reduced residue system mod p. From the properties of the reduced residue system and quadratic residue mod p we have
p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab+cd≡2modpa+b≡c+dmodp1=p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=1ab+cd≡2modpa+b≡c+dmodp1+p−1∑a=0p−1∑b=0p−1∑c=0ab≡2modpa+b≡cmodp1=p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=1ab+c≡2¯d2modpa+b≡c+1modp1+p−1∑a=0p−1∑b=0ab≡2modp1=p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=1ab+c≡2d2modpa+b≡c+1modp1+(p−1)=p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab+c≡2dmodpa+b≡c+1modp(1+(dp))−p−1∑a=0p−1∑b=0p−1∑c=0ab+c≡0modpa+b≡c+1modp1+(p−1)=p−1∑a=0p−1∑b=0p−1∑c=0a+b≡c+1modp1+p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab+c≡2dmodpa+b≡c+1modp(dp)−p−1∑a=0p−1∑b=0ab+a+b≡1modp1+(p−1)=p2+p−1∑a=0p−1∑b=0p−1∑d=0ab+a+b≡1+2dmodp(dp)−p−1∑a=1p−1∑b=1ab≡2modp1+(p−1). | (2.1) |
It is clear that if a passes through a complete residue system mod p, then a+1 also passes through a complete residue system mod p. So we make the change of variables a to a−1 and b to b−1 modulo p to change the congruence from ab+a+b≡cmodp to ab≡c+1modp. Then from (2.1) and the orthogonality property for the Dirichlet character modulo p we have
p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab+cd≡2modpa+b≡c+dmodp1=p2+p∑a=1p∑b=1p−1∑d=0ab≡2+2dmodp(dp)=p2+(2p)p∑a=1p∑b=1(ab−2p)=p2+(2p)p−1∑a=1p∑b=1(ab−2p)+(2p)p∑b=1(−2p)=p2+(2p)p−1∑a=1p∑b=1(b−2p)+(−1p)⋅p=p2+(−1p)⋅p. |
This proves Lemma 2.
Lemma 3. If p is an odd prime, then we have the identity
p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab+cd≡2modpab≡cdmodpa+b≡c+dmodp1=2p−4. |
Proof. It is clear that the conditions ab≡cdmodp and ab+cd≡2modp are equivalent to ab≡cd≡1modp. So we have
p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab+cd≡2modpab≡cdmodpa+b≡c+dmodp1=p−1∑a=1p−1∑b=1p−1∑c=1p−1∑d=1ab≡cd≡1modpa+b≡c+dmodp1=p−1∑a=1p−1∑c=1a+¯a≡c+¯cmodp1=p−1∑a=1p−1∑c=1(a−c)(ac−1)≡0modp1=p−1∑a=1p−1∑c=1a≡cmodp1+p−1∑a=1p−1∑c=1ac≡1modp1−p−1∑a=1p−1∑c=1a≡cmodpac≡1modp1=p−1+p−1−2=2p−4. |
This proves Lemma 3.
Lemma 4. If p is an odd prime with p≡1mod4, then we have the identity
p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0(a2+1)(b2+1)≡(c2+1)(d2+1)≡0modpa+b≡c+dmodp1=16p−26. |
Proof. Since p≡1mod4, there exists an integer k satisfying the congruence (±k)2≡−1modp. In fact, these are all such solutions modulo p. Choosing 0≤h≤p−1 for which a+b≡c+d≡hmodp, we have
p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0(a2+1)(b2+1)≡(c2+1)(d2+1)≡0modpa+b≡c+dmodp1=p−1∑h=0(p−1∑a=0p−1∑b=0(a2+1)(b2+1)≡0modpa+b≡hmodp1)2=p−1∑h=0(p−1∑a=0p−1∑b=0a2+1≡0modpa+b≡hmodp1+p−1∑a=0p−1∑b=0b2+1≡0modpa+b≡hmodp1−p−1∑a=0p−1∑b=0a2+1≡b2+1≡0modpa+b≡hmodp1)2=p−1∑h=0(2p−1∑b=0k+b≡hmodp1+2p−1∑b=0−k+b≡hmodp1−p−1∑a=0p−1∑b=0a2+1≡b2+1≡0modpa+b≡hmodp1)2=p−1∑h=0h≠−2k,2k,0(2+2−0)2+(4−p−1∑a=0p−1∑b=0a2+1≡b2+1≡0modpa+b≡0modp1)2+(4−p−1∑a=0p−1∑b=0a2+1≡b2+1≡0modpa+b≡2kmodp1)2+(4−p−1∑a=0p−1∑b=0a2+1≡b2+1≡0modpa+b≡−2kmodp1)2=16(p−3)+(4−2)2+(4−1)2+(4−1)2=16p−26. |
In this section, we use the four basic lemmas of the previous section to prove our theorems. First we prove Theorem 1. If p≡3mod4, then (−1p)=−1 and p∤(a2+1) for all 0≤a≤p−1. Then from Lemma 1, Lemma 2, Lemma 3, the trigonometric identity
p−1∑a=0e(map)={p,if p∣m;0,if p∤m |
and the orthogonality of characters modp
∑χmodpχ(a)={p−1,if a≡1modp;0,otherwise |
we have
1p(p−1)∑χmodpp−1∑m=0|p−1∑a=0χ(a2+1)e(map)|4=1pp−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0(a2+1)(b2+1)≡(c2+1)(d2+1)modpp−1∑m=0e(m(a+b−c−d)p)=p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0a2b2+a2+b2+1≡c2d2+c2+d2+1modpa+b≡c+dmodp1=p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0(ab−1)2+(a+b)2≡(cd−1)2+(c+d)2modpa+b≡c+dmodp1=p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0(ab−1)2≡(cd−1)2modpa+b≡c+dmodp1=p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0(ab−cd)(ab+cd−2)≡0modpa+b≡c+dmodp1=p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab≡cdmodpa+b≡c+dmodp1+p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab+cd≡2modpa+b≡c+dmodp1−p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0ab−cd≡ab+cd−2≡0modpa+b≡c+dmodp1=2p2−p+p2−p−(2p−4)=3p2−4p+4. |
This proves Theorem 1.
Now we prove Theorem 2. Note that if p≡1mod4, then (−1p)=1. If a2+1≡0modp, then for any character χmodp, we have χ(a2+1)=0. So from Lemma 4 and the methods of proving Theorem 1 we have
1p(p−1)∑χmodpp−1∑m=0|p−1∑a=0χ(a2+1)e(map)|4=p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0(a2+1)(b2+1)≡(c2+1)(d2+1)modpa+b≡c+dmodp1−p−1∑a=0p−1∑b=0p−1∑c=0p−1∑d=0(a2+1)(b2+1)≡(c2+1)(d2+1)≡0modpa+b≡c+dmodp1=2p2−p+p2+p−(2p−4)−(16p−26)=3(p2−6p+10). |
This completes the proof of all of our results.
The authors express their sincere thanks to the reviewer for pointing out the errors in our paper.
This work is supported by the Xian Science and Technology Planning Project (2019KJWL28) and N. S. F. (11771351) of P. R. China.
The authors declare no conflict of interest.
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