Dirichlet characters of the rational polynomials

Wenjia Guo; Xiaoge Liu; Tianping Zhang; Wenjia Guo; Xiaoge Liu; Tianping Zhang

doi:10.3934/math.2022194

AIMS Mathematics

2022, Volume 7, Issue 3: 3494-3508. doi: 10.3934/math.2022194

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Research article Special Issues

Dirichlet characters of the rational polynomials

School of Mathematics and Statistics, Shaanxi Normal University, Xi'an, Shaanxi, China

Received: 27 July 2021 Accepted: 22 November 2021 Published: 02 December 2021
MSC : 11L05, 11L10

Denote by $\chi$ a Dirichlet character modulo $q\geq 3$ , and $\overline{a}$ means $a\cdot\overline{a} \equiv 1 \bmod q$ . In this paper, we study Dirichlet characters of the rational polynomials in the form

$\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a}),$

where $\sum\limits_{a = 1}^{q}'$ denotes the summation over all $1\le a\le q$ with $(a, q) = 1$ . Relying on the properties of character sums and Gauss sums, we obtain W. P. Zhang and T. T. Wang's identity ^[6] under a more relaxed situation. We also derive some new identities for the fourth power mean of it by adding some new ingredients.

Keywords:

Citation: Wenjia Guo, Xiaoge Liu, Tianping Zhang. Dirichlet characters of the rational polynomials[J]. AIMS Mathematics, 2022, 7(3): 3494-3508. doi: 10.3934/math.2022194

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Abstract

$\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a}),$

1. Introduction and main results

Let $q\geq 3$ be an integer, and $\chi$ be a Dirichlet character modulo $q$ . The characters of the rational polynomial are defined as follows:

$\begin{eqnarray*} && \sum\limits_{x = N+1}^{N+M}\chi\left(f(x)\right), \end{eqnarray*}$

where $M$ and $N$ are any given positive integers, and $f(x)$ is a rational polynomial. For example, when $f(x) = x$ , for any non-principal Dirichlet character $\chi$ mod $q$ , Pólya ^[1] and Vinogradov ^[2] independently proved that

$\begin{eqnarray*} && \left|\sum\limits_{x = N+1}^{N+M}\chi(x)\right| < \sqrt{q}\ln q, \end{eqnarray*}$

and we call it Pólya-Vinogradov inequality.

When $q = p$ is an odd prime, $\chi$ is a $p$ -th order character modulo $p$ , Weil ^[3] proved

$\begin{eqnarray*} && \sum\limits_{x = N+1}^{N+M}\chi\left(f(x)\right)\ll p^{\frac{1}{2}}\ln p, \end{eqnarray*}$

where $f(x)$ is not a perfect $p$ -th power modulo $p$ , $A\ll B$ denotes $|A| < kB$ for some constant $k$ , which in this case depends on the degree of $f$ .

Many authors have obtained numerous results for various forms of $f(x)$ . For example, W. P. Zhang and Y. Yi ^[4] constructed a special polynomial as $f(x) = (x-r)^{m}(x-s)^{n}$ and deduced

$\begin{eqnarray*} && \left|\sum\limits_{a = 1}^{q}\chi((a-r)^{m}(a-s)^{n}) \right| = \sqrt{q}, \end{eqnarray*}$

where $(r-s, q) = 1$ , and $\chi$ is a primitive character modulo $q$ . This shows the power of $q$ in Weil's result is the best possible!

Also, when $\chi$ is a primitive character mod $q$ , W. P. Zhang and W. L. Yao ^[5] obtained

$\begin{eqnarray*} && \sum\limits_{a = 1}^{q}\chi(a^{m}(1-a)^{m}) = \sqrt{q}\overline{\chi}(4^{m}), \end{eqnarray*}$

where $q$ is an odd perfect square and $m$ is any positive integer with $(m, q) = 1$ .

When $q = p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots p^{\alpha_{s}}_{s}$ is a square full number with $p_{i}\equiv 3 \bmod 4$ , $\chi = \chi_{1}\chi_{2}\ldots\chi_{s}$ with $\chi_{i}$ being any primitive even character mod $p^{\alpha_{i}}_{i}$ $(i = 1, 2, \ldots, s)$ , W. P. Zhang and T. T. Wang ^[6] obtained the identity

$\begin{eqnarray} && \left|\sum\limits_{a = 1}^{q}'\chi(ma^{2^{k}-1}+n\overline{a})\right| = \sqrt{q}\prod\limits_{p|q}\left(1+\left(\frac{mn(2^{k}-1)}{p} \right) \right), \end{eqnarray}$

(1.1)

where $a\cdot\overline{a} \equiv 1 \bmod q$ , and $\left(\frac{\ast}{p}\right)$ denotes the Legendre symbol. Besides, $k$ , $m$ and $n$ also satisfying some special conditions. Other related work about Dirichlet characters of the rational polynomials can be found in references ^{[7,8,9,10,11,12,13,14]}. Inspired by these, we will study the sum

$\begin{eqnarray*} && \sum\limits^{q}_{a = 1}'\chi(ma+\overline{a}). \end{eqnarray*}$

Following the way in ^[6], we obtain W. P. Zhang and T. T. Wang's identity (1.1) under a more relaxed situation. Then by adding some new ingredients, we derive some new identities for the fourth power mean of it.

Noting that if $\chi$ is an odd character modulo $q$ , $m$ is a positive integer with $(m, q) = 1$ , we can get

$\begin{eqnarray*} && \sum\limits^{q}_{a = 1}'\chi(ma+\overline{a}) = \sum\limits^{q}_{a = 1}'\chi(-ma+\overline{(-a)}) = -\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a}). \end{eqnarray*}$

That is to say, under this condition,

$\begin{eqnarray*} && \sum\limits^{q}_{a = 1}'\chi(ma+\overline{a}) = 0. \end{eqnarray*}$

So, we will only discuss the case of $\chi$ an even character. To the best of our knowledge, the following identities dealing with arbitrary odd square-full number cases are new and have not appeared before.

Theorem 1.1. Let $q = p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots p^{\alpha_{s}}_{s}$ be an odd square-full number, $\chi_{i}$ be any primitive even character mod $p^{\alpha_{i}}_{i}$ $(i = 1, 2, \ldots, s)$ and $\chi = \chi_{1}\chi_{2}\cdots\chi_{s}$ . Then for any integer $m$ with $(m, q) = 1$ , we have the identity

$\begin{eqnarray*} && \left|\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a})\right| = \sqrt{q}\prod\limits_{p\mid q}\left(1+ \left( \frac{m}{p}\right)\right), \end{eqnarray*}$

where $\prod_{p\mid q}$ denotes the product over all distinct prime divisors $p$ of $q$ .

Remark 1.1. It is obvious that Theorem 1.1 is W. P. Zhang and T. T. Wang's identity (1.1) with $k = n = 1$ by removing the condition $p_{i} \equiv 3 \bmod 4$ $(i = 1, 2, \ldots, s)$ . Besides, using our results, we can directly obtain the absolute values of the sums of Dirichlet characters satisfying some conditions, which avoids complex calculations. What's more, the result of Theorem 1.1 also shows that the order of $q$ in Weil's result can not be improved.

To understand the result better, we give the following examples:

Example 1.1. Let $q = 3^{2}$ , $\chi$ be a Dirichlet character modulo $9$ defined as follows:

$\chi(n) = \left\{ \begin{array}{ll} e^{\frac{2 \pi i \cdot \text{ind}_{2} n}{3}}, & \hbox{if $(n, 9) = 1$}; \\ 0, & \hbox{if $(n, 9) > 1$}. \end{array} \right.$

Obviously, $\chi$ is a primitive even character modulo $9$ . Taking $m = 1, 2$ , then we have

$\begin{align} &\left|\sum\limits^{9}_{a = 1}'\chi(ma+\overline{a})\right| = \left|\sum\limits^{9}_{a = 1}'\chi(a+\overline{a})\right| = |3\chi(2)+3\chi(7)| = |3e^{\frac{2 \pi i }{3}}+3e^{\frac{2\pi i\cdot 4 }{3}}| = 6, \\ &\left|\sum\limits^{9}_{a = 1}'\chi(ma+\overline{a})\right| = \left|\sum\limits^{9}_{a = 1}'\chi(2a+\overline{a})\right| = |2\chi(3)+2\chi(6)+2\chi(9)| = 0. \end{align}$

Example 1.2. Let $q = 5^{2}$ , $\chi$ be a primitive even character modulo $25$ defined as follows:

$\chi(n) = \left\{ \begin{array}{ll} e^{\frac{2 \pi i \cdot \text{ind}_{2} n}{5}}, & \hbox{if $(n, 25) = 1$}; \\ 0, & \hbox{if $(n, 25) > 1$}. \end{array} \right.$

Taking $m = 1, 2$ , then we have

$\begin{align} &\left|\sum\limits^{25}_{a = 1}'\chi(ma+\overline{a})\right| = \left|\sum\limits^{25}_{a = 1}'\chi(a+\overline{a})\right| = |5\chi(2)+5\chi(23)| = |5e^{\frac{2 \pi i }{5}}+5e^{\frac{2\pi i\cdot 11 }{5}}| = 10, \\ &\left|\sum\limits^{25}_{a = 1}'\chi(ma+\overline{a})\right| = \left|\sum\limits^{25}_{a = 1}'\chi(2a+\overline{a})\right| = \left|4\chi(2)+4\chi(3)+4\chi(7)+4\chi(8)+4\chi(12)\right|\\ = &\left|4e^{\frac{2 \pi i}{5}}+4e^{\frac{2 \pi i \cdot 7}{5}}+4e^{\frac{2 \pi i \cdot 5}{5}}+4e^{\frac{2 \pi i \cdot 3}{5}}+4e^{\frac{2 \pi i \cdot 9}{5}}\right| = 0. \end{align}$

Example 1.3. Let $q = 13^{2}$ , $\chi$ be a primitive even character modulo $169$ defined as follows:

$\chi(n) = \left\{ \begin{array}{ll} e^{\frac{2 \pi i \cdot \text{ind}_{2} n}{13}}, & \hbox{if $(n, 169) = 1$}; \\ 0, & \hbox{if $(n, 169) > 1$}. \end{array} \right.$

Taking $m = 1, 2$ , then we have

$\begin{align} & \left|\sum\limits^{169}_{a = 1}'\chi(ma+\overline{a})\right| = \left|\sum\limits^{169}_{a = 1}'\chi(a+\overline{a})\right| = |4\chi(1)+26\chi(2)+4\chi(4)+4\chi(9)+4\chi(12)+4\chi(14)+4\chi(17)\\ &+4\chi(22)+4\chi(25)+4\chi(27)+4\chi(30)+4\chi(35)+4\chi(38)+4\chi(40)+4\chi(43)+4\chi(48)+4\chi(51)\\ &+4\chi(53)+4\chi(56)+4\chi(61)+4\chi(64)+4\chi(66)+4\chi(69)+4\chi(74)+4\chi(77)+4\chi(79)+4\chi(82)|\\ & = \left|8+8e^{\frac{\pi i }{13}}+34e^{\frac{2\pi i }{13}}+8e^{\frac{3\pi i }{13}}+8e^{\frac{4\pi i }{13}}+8e^{\frac{5\pi i }{13}}+8e^{\frac{6\pi i }{13}}+8e^{\frac{7\pi i }{13}}+8e^{\frac{8\pi i }{13}}+8e^{\frac{9\pi i }{13}}+8e^{\frac{10\pi i }{13}}+8e^{\frac{11\pi i }{13}}+8e^{\frac{12\pi i }{13}}\right|\\ & = 26, \end{align}$

$\begin{align} & \left|\sum\limits^{169}_{a = 1}'\chi(ma+\overline{a})\right| = \left|\sum\limits^{169}_{a = 1}'\chi(2a+\overline{a})\right| = |4\chi(2)+4\chi(3)+4\chi(5)+4\chi(8)+4\chi(10)+4\chi(11)+4\chi(15)\\ &+4\chi(16)+4\chi(18)+4\chi(21)+4\chi(23)+4\chi(24)+4\chi(28)+4\chi(29)+4\chi(31)+4\chi(34)+4\chi(36)\\ &+4\chi(37)+4\chi(41)+4\chi(42)+4\chi(44)+4\chi(47)+4\chi(49)+4\chi(50)+4\chi(54)+4\chi(55)+4\chi(57)\\ &+4\chi(60)+4\chi(62)+4\chi(63)+4\chi(67)+4\chi(68)+4\chi(70)+4\chi(73)+4\chi(75)+4\chi(76)+4\chi(80)\\ &+4\chi(81)+4\chi(83)|\\ & = |12+12e^{\frac{\pi i }{13}}+12e^{\frac{2\pi i }{13}}+12e^{\frac{3\pi i }{13}}+12e^{\frac{4\pi i }{13}}+12e^{\frac{5\pi i }{13}}+12e^{\frac{6\pi i }{13}}+12e^{\frac{7\pi i }{13}}+12e^{\frac{8\pi i }{13}}+12e^{\frac{9\pi i }{13}}+12e^{\frac{10\pi i }{13}}\\ &\quad+12e^{\frac{11\pi i }{13}}+12e^{\frac{12\pi i }{13}}| = 0. \end{align}$

The above examples can be easily achieved by our Theorem 1.1. From Theorem 1.1, we may immediately obtain the following two corollaries:

Corollary 1.1. Let $q = p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots p^{\alpha_{s}}_{s}$ be an odd square-full number, $\chi_{i}$ be any primitive even character mod $p^{\alpha}_{i}$ $(i = 1, 2, \ldots, s)$ and $\chi = \chi_{1}\chi_{2}\cdots\chi_{s}$ . Then for any integer $m$ with $(m, q) = 1$ , we have the identity

$\begin{eqnarray*} && \left|\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a})\right| = \left\{ \begin{array}{ll} 2^{\omega(q)}\sqrt{q}, & \hbox{ if $m$ is a quadratic residue modulo $q$}; \\ 0, & \hbox{ otherwise}, \end{array} \right. \end{eqnarray*}$

where $\omega(q)$ denotes the number of all distinct prime divisors of $q$ .

Corollary 1.2. Let $q = p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots p^{\alpha_{s}}_{s}$ be an odd number with $\alpha_{i}\geq 1$ $(i = 1, 2, \ldots, s)$ , $\chi_{i}$ be any primitive even character mod $p^{\alpha_{i}}_{i}$ and $\chi = \chi_{1}\chi_{2}\cdots\chi_{s}$ . Then for any integer $m$ with $(m, q) = 1$ , we have the inequality

$\begin{eqnarray*} && \left|\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a})\right|\leq 2^{\omega(q)}\sqrt{q}. \end{eqnarray*}$

Theorem 1.2. Let $q = p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots p^{\alpha_{s}}_{s}$ be an odd square-full number, $\chi_{i}$ be any primitive even character mod $p^{\alpha_{i}}_{i}$ $(i = 1, 2, \ldots, s)$ and $\chi = \chi_{1}\chi_{2}\cdots\chi_{s}$ . Then for any integers $k$ and $m$ with $k \geq 1$ and $(m, q) = 1$ , we have the identity

$\begin{eqnarray*} && \mathop{{\sum}^{\ast}}_{\chi \bmod q \atop \chi(-1) = 1}\left|\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a})\right|^{2k} = \frac{q^{k}}{2^{\omega(q)}}J(q)\prod\limits_{p\mid q}\left(1+\left(\frac{m}{p} \right) \right)^{2k}, \end{eqnarray*}$

where $J(q)$ denotes the number of primitive characters modulo $q$ , and ${{\sum\limits_{\chi \,\bmod \,q}}^{*}}$ denotes the summation over all primitive characters modulo $q$ .

Example 1.4. Taking $q = 5^{2}$ , $m = 1, 2$ , then we have

$\begin{align} &\mathop{{\sum}^{\ast}}_{\chi \bmod 25 \atop \chi(-1) = 1}\left|\sum\limits^{25}_{a = 1}'\chi(ma+\overline{a})\right|^{2k} = \mathop{{\sum}^{\ast}}_{\chi \bmod 25 \atop \chi(-1) = 1}\left|\sum\limits^{25}_{a = 1}'\chi(a+\overline{a})\right|^{2k} = 8\cdot10^{2k}, \\ &\mathop{{\sum}^{\ast}}_{\chi \bmod 25 \atop \chi(-1) = 1}\left|\sum\limits^{25}_{a = 1}'\chi(ma+\overline{a})\right|^{2k} = \mathop{{\sum}^{\ast}}_{\chi \bmod 25 \atop \chi(-1) = 1}\left|\sum\limits^{25}_{a = 1}'\chi(2a+\overline{a})\right|^{2k} = 0, \end{align}$

which can be easily achieved by our Theorem 1.2.

Taking $k = 2$ in Theorem 1.2, we may immediately obtain the followings:

Corollary 1.3. Let $q = p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots p^{\alpha_{s}}_{s}$ be an odd square-full number, $\chi_{i}$ be any primitive even character mod $p^{\alpha_{i}}_{i}$ $(i = 1, 2, \ldots, s)$ and $\chi = \chi_{1}\chi_{2}\cdots\chi_{s}$ . Then for any integer $m$ with $(m, q) = 1$ , we have the identity

$\begin{eqnarray*} && \mathop{{\sum}^{\ast}}_{\chi \bmod q \atop \chi(-1) = 1}\left|\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a})\right|^{4} = \frac{q^{2}}{2^{\omega(q)}}J(q)\prod\limits_{p\mid q}\left(1+\left(\frac{m}{p} \right) \right)^{4}. \end{eqnarray*}$

Corollary 1.4. Let $q = p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots p^{\alpha_{s}}_{s}$ be an odd square-full number, $\chi_{i}$ be any primitive even character mod $p^{\alpha_{i}}_{i}$ $(i = 1, 2, \ldots, s)$ and $\chi = \chi_{1}\chi_{2}\cdots\chi_{s}$ . Then we have the identity

$\begin{eqnarray*} && \mathop{{\sum}^{\ast}}_{\chi \bmod q \atop \chi(-1) = 1}\left|\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a})\right|^{4} = \left\{ \begin{array}{ll} 8^{\omega(q)}q^{2}J(q), & \hbox{ if $m$ is a quadratic residue modulo $q$}; \\ 0, & \hbox{ otherwise}. \end{array} \right. \end{eqnarray*}$

Theorem 1.3. Let $p$ be an odd prime, $\chi$ be any non-principal character mod $p$ . Then for any integer $m$ with $(m, p) = 1$ , we have the identity

$\begin{eqnarray*} && \sum\limits_{\chi \bmod p \atop \chi(-1) = 1}\left|\sum\limits_{a = 1}^{p-1}\chi(ma+\overline{a}) \right|^{4} = \left\{ \begin{array}{ll} 2p^{3}-6p^{2}+4-4(p^{2}-3p+2)\left(\frac{m}{p} \right)+(p-1)E, & \hbox{if $p\equiv 3 \bmod 4$}; \\ 2p^{3}-6p^{2}+4-4(p^{2}+p-2)\left(\frac{m}{p} \right)+(p-1)E, & \hbox{if $p\equiv 1 \bmod 4$}, \end{array} \right. \end{eqnarray*}$

where

$\begin{eqnarray*} && E = \sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\left(\frac{(a^{2}b-1)(b-1)b}{p} \right)\sum\limits_{d = 1}^{p-1} \left(\frac{(\overline{a}^{2}d-1)(d-1)d}{p} \right). \end{eqnarray*}$

Remark 1.2. From ^[8], we know that when $f(x)$ is a polynomial of odd degree $n\geq 3$ , Weil's estimate (^[15,16])

$\begin{eqnarray*} && \left|\sum\limits_{x = 0}^{p-1}\left(\frac{f(x)}{p} \right)\right|\leq(n-1)\sqrt{p}, \end{eqnarray*}$

implies that $E < 4p^{2}-8p$ . Noting that $\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a})$ can be regarded as a dual form of Kloosterman sums, which defined as $\sum\limits^{q}_{a = 1}'e^{ 2\pi i\frac{ma+\bar{a}}{q} }$ , we can obtain some distributive properties of $\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a})$ from Theorem 1.2 and 1.3.

From Theorem 1.3, we also have the following corollaries:

Corollary 1.5. Let $p$ be an odd prime, $\chi$ be any non-principal character mod $p$ . Then for any quadratic residue $m$ mod $p$ , we have the identity

$\begin{eqnarray*} && \sum\limits_{\chi \bmod p \atop \chi(-1) = 1}\left|\sum\limits_{a = 1}^{p-1}\chi(ma+\overline{a}) \right|^{4} = \left\{ \begin{array}{ll} 2p^{3}-10p^{2}+12p-4+(p-1)E, & \hbox{if $p\equiv 3 \bmod 4$}; \\ 2p^{3}-10p^{2}-4p+12+(p-1)E, & \hbox{if $p\equiv 1 \bmod 4$}. \end{array} \right. \end{eqnarray*}$

Corollary 1.6. Let $p$ be an odd prime, $\chi$ be any non-principal character mod $p$ . Then for any quadratic non-residue $m$ mod $p$ , we have the identity

$\begin{eqnarray*} && \sum\limits_{\chi \bmod p \atop \chi(-1) = 1}\left|\sum\limits_{a = 1}^{p-1}\chi(ma+\overline{a}) \right|^{4} = \left\{ \begin{array}{ll} 2p^{3}-2p^{2}-12p+4+(p-1)E, & \hbox{if $p\equiv 3 \bmod 4$}; \\ 2p^{3}-2p^{2}+4p-4+(p-1)E, & \hbox{if $p\equiv 1 \bmod 4$}. \end{array} \right. \end{eqnarray*}$

2. Some Lemmas

To prove our Theorems, we need some Lemmas as the following:

Lemma 2.1. Let $q$ , $q_{1}$ , $q_{2}$ be integers with $q = q_{1}q_{2}$ and $(q_{1}, q_{2}) = 1$ , $\chi_{i}$ be any non-principal character mod $q_{i}$ $(i = 1, 2)$ . Then for any integer $m$ with $(m, q) = 1$ and $\chi = \chi_{1}\chi_{2}$ , we have the identity

$\begin{eqnarray*} && \sum\limits^{q}_{a = 1}'\chi(ma+\overline{a}) = \sum\limits^{q_{1}}_{b = 1}'\chi_{1}\left(mb+\overline{b}\right) \sum\limits^{q_{2}}_{c = 1}'\chi_{2}\left(mc+\overline{c}\right). \end{eqnarray*}$

Proof. From the properties of Dirichlet characters, we have

$\begin{align} & \sum\limits^{q}_{a = 1}'\chi(ma+\overline{a})\\ = & \sum\limits ^{q_{1}q_{2}}_{a = 1}'\chi_{1}\chi_{2}(ma+\overline{a})\\ = &\sum\limits^{q_{1}}_{b = 1}'\sum\limits^{q_{2}}_{c = 1}'\chi_{1}\chi_{2}\left(m(bq_{2}+cq_{1})+\overline{bq_{2}+cq_{1}}\right)\\ = &\sum\limits^{q_{1}}_{b = 1}'\sum\limits^{q_{2}}_{c = 1}'\chi_{1}\left(m(bq_{2}+cq_{1})+\overline{bq_{2}+cq_{1}}\right) \chi_{2}\left(m(bq_{2}+cq_{1})+\overline{bq_{2}+cq_{1}}\right)\\ = &\sum\limits^{q_{1}}_{b = 1}'\chi_{1}\left(mbq_{2}+\overline{bq_{2}}\right) \sum\limits^{q_{2}}_{c = 1}'\chi_{2}\left(mcq_{1}+\overline{cq_{1}}\right)\\ = &\sum\limits^{q_{1}}_{b = 1}'\chi_{1}\left(mb+\overline{b}\right) \sum\limits^{q_{2}}_{c = 1}'\chi_{2}\left(mc+\overline{c}\right) . \end{align}$

This completes the proof of Lemma $2.1$ .

Lemma 2.2. Let $p$ be an odd prime, $\alpha$ and $m$ be integers with $\alpha\geq 1$ and $(m, p) = 1$ . Then for any primitive even character $\chi$ mod $p^{\alpha}$ , we have the identity

$\begin{eqnarray*} && \sum\limits^{p^{\alpha}}_{a = 1}'\chi(ma+\overline{a}) = \frac{\chi_{1}(m)\tau^{2}(\overline{\chi}_{1})}{\tau(\overline{\chi})} \left(1+\chi^{0}_{2}(m)\frac{\tau^{2}(\chi^{0}_{2}\overline{\chi}_{1})}{\tau^{2}(\overline{\chi}_{1})} \right), \end{eqnarray*}$

where $\chi^{0}_{2} = \left(\frac{\ast}{p}\right)$ , $\tau(\chi) = \sum\limits_{a = 1}^{p^{\alpha}}\chi(a)e\left(\frac{a}{p^{\alpha}}\right)$ , $\chi_{1}$ is a primitive character mod $p^{\alpha}$ and $\chi = \chi^{2}_{1}$ .

Proof. For any primitive even character $\chi$ mod $p^{\alpha}$ , there exists one primitive character $\chi_{1}$ mod $p^{\alpha}$ such that $\chi = \chi^{2}_{1}$ . From the properties of Gauss sum, we can obtain

$\begin{align} & \sum\limits^{p^{\alpha}}_{a = 1}'\chi(ma+\overline{a}) \\ = &\frac{1}{\tau(\overline{\chi})}\sum\limits^{p^{\alpha}}_{a = 1}'\sum\limits_{b = 1}^{p^{\alpha}}\overline{\chi}(b) e\left(\frac{b(ma+\overline{a})}{p^{\alpha}} \right) \\ = &\frac{1}{\tau(\overline{\chi})}\sum\limits_{a = 1}^{p^{\alpha}}\overline{\chi}(a)\sum\limits_{b = 1}^{p^{\alpha}}\overline{\chi}(b) e\left(\frac{b(ma^{2}+1)}{p^{\alpha}} \right) \\ = &\frac{1}{\tau(\overline{\chi})}\sum\limits_{b = 1}^{p^{\alpha}}\overline{\chi}(b)e\left(\frac{b}{p^{\alpha}} \right)\sum\limits_{a = 1}^{p^{\alpha}}\overline{\chi}(a)e\left(\frac{bma^{2}}{p^{\alpha}} \right)\\ = &\frac{1}{\tau(\overline{\chi})}\sum\limits_{b = 1}^{p^{\alpha}}\overline{\chi}(b)e\left(\frac{b}{p^{\alpha}} \right)\sum\limits_{a = 1}^{p^{\alpha}}\overline{\chi}_{1}(a^{2})e\left(\frac{bma^{2}}{p^{\alpha}} \right)\\ = &\frac{1}{\tau(\overline{\chi})}\sum\limits_{b = 1}^{p^{\alpha}}\overline{\chi}(b)e\left(\frac{b}{p^{\alpha}} \right)\sum\limits_{a = 1}^{p^{\alpha}}\left(1+\chi^{0}_{2}(a)\right)\overline{\chi}_{1}(a)e\left(\frac{bma}{p^{\alpha}} \right)\\ = &\frac{1}{\tau(\overline{\chi})}\sum\limits_{b = 1}^{p^{\alpha}}\overline{\chi}(b)e\left(\frac{b}{p^{\alpha}} \right)\sum\limits_{a = 1}^{p^{\alpha}}\overline{\chi}_{1}(a)e\left(\frac{bma}{p^{\alpha}} \right)\\ &+\frac{1}{\tau(\overline{\chi})}\sum\limits_{b = 1}^{p^{\alpha}}\overline{\chi}(b)e\left(\frac{b}{p^{\alpha}} \right)\sum\limits_{a = 1}^{p^{\alpha}}\chi^{0}_{2}(a)\overline{\chi}_{1}(a)e\left(\frac{bma}{p^{\alpha}} \right)\\ : = &B_{1}+B_{2} . \end{align}$

Now we compute $B_{1}$ and $B_{2}$ respectively.

$\begin{align} B_{1} = & \frac{1}{\tau(\overline{\chi})}\sum\limits_{b = 1}^{p^{\alpha}}\overline{\chi}(b)e\left(\frac{b}{p^{\alpha}} \right)\sum\limits_{a = 1}^{p^{\alpha}}\overline{\chi}_{1}(a)e\left(\frac{bma}{p^{\alpha}} \right) \\ = & \frac{1}{\tau(\overline{\chi})}\sum\limits_{b = 1}^{p^{\alpha}}\overline{\chi}(b)\chi_{1}(bm)e\left(\frac{b}{p^{\alpha}} \right)\sum\limits_{a = 1}^{p^{\alpha}}\overline{\chi}_{1}(bma)e\left(\frac{bma}{p^{\alpha}} \right) \\ = &\frac{\chi_{1}(m)\tau(\overline{\chi}_{1})}{\tau(\overline{\chi})}\sum\limits_{b = 1}^{p^{\alpha}}\overline{\chi}(b)\chi_{1}(b) e\left(\frac{b}{p^{\alpha}} \right) \\ = &\frac{\chi_{1}(m)\tau(\overline{\chi}_{1})}{\tau(\overline{\chi})}\sum\limits_{b = 1}^{p^{\alpha}}\overline{\chi}_{1}(b) e\left(\frac{b}{p^{\alpha}} \right)\\ = &\frac{\chi_{1}(m)\tau^{2}(\overline{\chi}_{1})}{\tau(\overline{\chi})} . \end{align}$

Similarly, we have

$\begin{eqnarray*} && B_{2} = \frac{\chi_{1}(m)\chi^{0}_{2}(m)\tau^{2}(\chi^{0}_{2}\overline{\chi}_{1})}{\tau(\overline{\chi})}. \end{eqnarray*}$

Therefore, we can obtain

Lemma 2.3. Let $p$ be an odd prime. Then for any integer $n$ , we have the identity

$\begin{eqnarray*} && \sum\limits_{a = 1}^{p}\left(\frac{a^{2}+n}{p} \right) = \left\{ \begin{array}{ll} -1, & \hbox{if $(n, p) = 1$}; \\ p-1, & \hbox{if $(n, p) = p$}. \end{array} \right. \end{eqnarray*}$

Proof. See Theorem 8.2 of ^[17].

Lemma 2.4. Let $p$ be an odd prime. Then we have the identity

$\begin{eqnarray*} && \sum\limits_{a = 2}^{p-2} \sum\limits_{b = 1}^{p-1} \left(\frac{(a^{2}b-1)(b-1)b}{p} \right) = 2\times(-1)^{\frac{p-1}{2}}+2. \end{eqnarray*}$

Proof. From the properties of character sum, we have

$\begin{align} & \sum\limits_{a = 2}^{p-2} \sum\limits_{b = 1}^{p-1} \left(\frac{(a^{2}b-1)(b-1)b}{p} \right) \\ = &\sum\limits_{b = 1}^{p-1}\left(\frac{b-1}{p} \right)\sum\limits_{a = 2}^{p-2} \left(\frac{(a^{2}b-1)b}{p} \right)\\ = &\sum\limits_{b = 1}^{p-1}\left(\frac{b-1}{p} \right)\sum\limits_{a = 2}^{p-2} \left(\frac{b^{2}(a^{2}-\overline{b})}{p} \right)\\ = &\sum\limits_{b = 1}^{p-1}\left(\frac{b-1}{p} \right)\sum\limits_{a = 2}^{p-2} \left(\frac{a^{2}-\overline{b}}{p} \right)\\ = &\sum\limits_{b = 1}^{p-1}\left(\frac{b-1}{p} \right)\left(\sum\limits_{a = 1}^{p} \left(\frac{a^{2}-\overline{b}}{p}\right)-\left(\frac{1-\overline{b}}{p} \right)-\left(\frac{(p-1)^{2}-\overline{b}}{p} \right) -\left(\frac{p^{2}-\overline{b}}{p} \right) \right)\\ = &\sum\limits_{b = 1}^{p-1}\left(\frac{b-1}{p} \right)\left(-1-2\left(\frac{1-\overline{b}}{p} \right)-\left(\frac{-\overline{b}}{p} \right) \right)\\ = &-\sum\limits_{b = 1}^{p-1}\left(\frac{b-1}{p} \right)-2\sum\limits_{b = 1}^{p-1}\left(\frac{b-1}{p} \right)\left(\frac{1-\overline{b}}{p} \right)-\sum\limits_{b = 1}^{p-1}\left(\frac{b-1}{p} \right)\left(\frac{-\overline{b}}{p} \right) \\ = &-\sum\limits_{b = 0}^{p-2}\left(\frac{b}{p} \right)-2\sum\limits_{b = 1}^{p-1}\left(\frac{b-1}{p} \right)\left(\frac{(1-\overline{b})b^{2}}{p} \right)-\sum\limits_{b = 1}^{p-1}\left(\frac{\overline{b}-1}{p} \right)\\ = &-2\sum\limits_{b = 0}^{p-2}\left(\frac{b}{p} \right)-2\sum\limits_{b = 1}^{p-1}\left(\frac{(b-1)^{2}b}{p} \right)\\ = &-2\left(\sum\limits_{b = 0}^{p-1}\left(\frac{b}{p} \right)-\left(\frac{p-1}{p} \right) \right)-2\times(-1)\\ = &2\times(-1)^{\frac{p-1}{2}}+2 . \end{align}$

This completes the proof of Lemma $2.4$ .

3. Proof of Theorems

Now we come to prove our Theorems.

Firstly, we prove Theorem $1.1$ . With the help of Lemma $2$ in ^[6], when $\alpha\geq 2$ , we have

$\begin{eqnarray*} && \frac{\tau^{2}(\chi^{0}_{2}\overline{\chi}_{1})}{\tau^{2}(\overline{\chi}_{1})} = \left(\frac{1}{p}\right)^{2} = 1, \end{eqnarray*}$

which implies from Lemma $2.2$ , we can obtain

$\begin{eqnarray*} && \left|\sum\limits^{p^{\alpha}}_{a = 1}'\chi(ma+\overline{a})\right| = \left|\frac{\chi_{1}(m)\tau^{2}(\overline{\chi}_{1})}{\tau(\overline{\chi})} \left(1+\left(\frac{m}{p}\right) \right)\right| = \sqrt{p^{\alpha}}\left(1+\left(\frac{m}{p}\right) \right). \end{eqnarray*}$

Then, applying Lemma $2.1$ , we can obtain

$\begin{align} & \left|\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a})\right| \\ = &\left|\sum\limits^{p_{1}^{\alpha_{1}}}_{a_{1} = 1}'\chi_{1}(ma_{1}+\overline{a_{1}})\right|\cdots \left|\sum\limits^{p_{s}^{\alpha_{s}}}_{a_{s} = 1}'\chi_{s}(ma_{s}+\overline{a_{s}})\right| \\ = &\sqrt{q}\prod\limits_{p\mid q}\left(1+ \left( \frac{m}{p}\right)\right). \end{align}$

This completes the proof of Theorem $1.1$ .

Then, from Lemma $2.1$ and Lemma $2.2$ , we can prove Theorem $1.2$ as following:

$\begin{align} & \mathop{{\sum}^{\ast}}_{\chi \bmod q \atop \chi(-1) = 1}\left|\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a})\right|^{2k}\\ = &\mathop{{\sum}^{\ast}}_{\chi_{1} \bmod p_{1}^{\alpha_{1}}\atop \chi_{1}(-1) = 1}\left|\sum\limits^{p_{1}^{\alpha_{1}}}_{a_{1} = 1}'\chi_{1}(ma_{1}+\overline{a_{1}})\right|^{2k}\cdots \mathop{{\sum}^{\ast}}_{\chi_{s} \bmod p_{s}^{\alpha_{s}}\atop \chi_{s}(-1) = 1}\left|\sum\limits^{p_{s}^{\alpha_{s}}}_{a_{s} = 1}'\chi_{s}(ma_{s}+\overline{a_{s}})\right|^{2k}\\ = &\prod\limits_{i = 1}^{s}\left[\frac{1}{2}J(p_{i}^{\alpha_{i}})p^{k\alpha_{i}}_{i}\left|1+\left(\frac{m}{p_{i}}\right)\right|^{2k}\right]\\ = &\frac{q^{k}}{2^{\omega(q)}}J(q)\prod\limits_{p\mid q}\left(1+\left(\frac{m}{p} \right) \right)^{2k}. \end{align}$

Finally, we prove Theorem $1.3$ . For any integer $m$ with $(m, p) = 1$ , we have

$\begin{align} & \sum\limits_{a = 1}^{p-1}\chi(ma+\overline{a})\\ = &\sum\limits_{u = 1}^{p-1}\chi(u)\sum\limits_{a = 1 \atop am+\overline{a}\equiv u \bmod p}^{p-1}1\\ = &\sum\limits_{u = 1}^{p-1}\chi(u)\sum\limits_{a = 1 \atop a^{2}m^{2}-amu+m \equiv 0 \bmod p}^{p-1}1\\ = &\sum\limits_{u = 1}^{p-1}\chi(u)\sum\limits_{a = 0 \atop (2am-u)^{2} \equiv u^{2}-4m \bmod p}^{p-1}1\\ = &\sum\limits_{u = 1}^{p-1}\chi(u)\sum\limits_{a = 0 \atop a^{2} \equiv u^{2}-4m \bmod p}^{p-1}1\\ = &\sum\limits_{u = 1}^{p-1}\chi(u)\left(1+\left(\frac{u^{2}-4m}{p} \right) \right)\\ = &\sum\limits_{u = 1}^{p-1}\chi(u)\left(\frac{u^{2}-4m}{p} \right)\\ = &\chi(2)\sum\limits_{u = 1}^{p-1}\chi(u)\left(\frac{u^{2}-m}{p} \right). \end{align}$

So from the orthogonality of Dirichlet characters and the properties of reduced residue system modulo $p$ , we have

$\begin{align} & \sum\limits_{\chi \bmod p \atop \chi(-1) = 1}\left|\sum\limits_{a = 1}^{p-1}\chi(ma+\overline{a}) \right|^{4}\\ = &\sum\limits_{\chi \bmod p \atop \chi(-1) = 1}\left|\chi(2)\sum\limits_{u = 1}^{p-1}\chi(u)\left(\frac{u^{2}-m}{p} \right) \right|^{2}\left|\chi(2)\sum\limits_{u = 1}^{p-1}\chi(u)\left(\frac{u^{2}-m}{p} \right) \right|^{2}\\ = &\sum\limits_{\chi \bmod p \atop \chi(-1) = 1}\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\sum\limits_{c = 1}^{p-1}\sum\limits_{d = 1}^{p-1}\chi(ac\overline{bd})\left(\frac{a^{2}-m}{p} \right)\left(\frac{b^{2}-m}{p} \right)\left(\frac{c^{2}-m}{p} \right)\left(\frac{d^{2}-m}{p} \right)\\ = &\sum\limits_{\chi \bmod p \atop \chi(-1) = 1}\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\sum\limits_{c = 1}^{p-1}\sum\limits_{d = 1}^{p-1}\chi(ac)\left(\frac{a^{2}b^{2}-m}{p} \right)\left(\frac{b^{2}-m}{p} \right)\left(\frac{c^{2}d^{2}-m}{p} \right)\left(\frac{d^{2}-m}{p} \right)\\ = &\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\sum\limits_{c = 1}^{p-1}\sum\limits_{d = 1}^{p-1}\left(\frac{a^{2}b^{2}-m}{p} \right)\left(\frac{b^{2}-m}{p} \right)\left(\frac{c^{2}d^{2}-m}{p} \right)\left(\frac{d^{2}-m}{p} \right)\sum\limits_{\chi \bmod p \atop \chi(-1) = 1}\chi(ac)\\ = &\frac{p-1}{2}\underset{a\equiv \overline{c} \bmod p}{\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\sum\limits_{c = 1}^{p-1}\sum\limits_{d = 1}^{p-1}}\left(\frac{a^{2}b^{2}-m}{p} \right)\left(\frac{b^{2}-m}{p} \right)\left(\frac{c^{2}d^{2}-m}{p} \right)\left(\frac{d^{2}-m}{p} \right)\\ &+\frac{p-1}{2}\underset{a\equiv -\overline{c} \bmod p}{\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\sum\limits_{c = 1}^{p-1}\sum\limits_{d = 1}^{p-1}}\left(\frac{a^{2}b^{2}-m}{p} \right)\left(\frac{b^{2}-m}{p} \right)\left(\frac{c^{2}d^{2}-m}{p} \right)\left(\frac{d^{2}-m}{p} \right)\\ = &(p-1)\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\sum\limits_{d = 1}^{p-1}\left(\frac{a^{2}b^{2}-m}{p} \right)\left(\frac{b^{2}-m}{p} \right)\left(\frac{\overline{a}^{2}d^{2}-m}{p} \right)\left(\frac{d^{2}-m}{p} \right)\\ = &(p-1)\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\left(1+\left(\frac{b}{p} \right) \right) \left(\frac{a^{2}b-m}{p} \right)\left(\frac{b-m}{p} \right)\sum\limits_{d = 1}^{p-1}\left(1+\left(\frac{d}{p} \right) \right)\left(\frac{\overline{a}^{2}d-m}{p} \right)\left(\frac{d-m}{p} \right)\\ = &(p-1)\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\left(\frac{a^{2}b-1}{p} \right)\left(\frac{b-1}{p} \right)\sum\limits_{d = 1}^{p-1}\left(\frac{\overline{a}^{2}d-1}{p} \right)\left(\frac{d-1}{p} \right)\\ &+(p-1)\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\left(\frac{a^{2}b-1}{p} \right)\left(\frac{b-1}{p} \right)\sum\limits_{d = 1}^{p-1}\left(\frac{m}{p} \right) \left(\frac{(\overline{a}^{2}d-1)(d-1)d}{p} \right)\\ &+(p-1)\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\left(\frac{m}{p} \right)\left(\frac{(a^{2}b-1)(b-1)b}{p} \right)\sum\limits_{d = 1}^{p-1}\left(\frac{\overline{a}^{2}d-1}{p} \right)\left(\frac{d-1}{p} \right)\\ &+(p-1)\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\left(\frac{m}{p} \right)\left(\frac{(a^{2}b-1)(b-1)b}{p} \right)\sum\limits_{d = 1}^{p-1}\left(\frac{m}{p} \right) \left(\frac{(\overline{a}^{2}d-1)(d-1)d}{p} \right)\\ : = & A_{1}+A_{2}+A_{3}+A_{4} . \end{align}$

Now we compute $A_{1}$ , $A_{2}$ , $A_{3}$ , $A_{4}$ respectively. Noticing that $\chi(-1) = 1$ , from the properties of the complete residue system modulo $p$ , we have

$\begin{align} & \sum\limits_{b = 1}^{p-1}\left(\frac{a^{2}b-1}{p} \right)\left(\frac{b-1}{p} \right)\\ = &\sum\limits_{b = 0}^{p-1}\left(\frac{a^{2}b-1}{p} \right)\left(\frac{b-1}{p} \right)-1\\ = &\sum\limits_{b = 0}^{p-1}\left( \frac{4a^{2}}{p} \right)\left(\frac{(a^{2}b-1)(b-1)}{p} \right)-1\\ = &\sum\limits_{b = 0}^{p-1}\left( \frac{(2a^{2}b-a^{2}-1)^{2}-(a^{2}-1)^{2}}{p} \right)-1\\ = &\sum\limits_{b = 0}^{p-1}\left( \frac{b^{2}-(a^{2}-1)^{2}}{p} \right)-1 . \end{align}$

Applying Lemma $2.3$ , we can get

$\begin{align} A_{1} = &(p-1)\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\left(\frac{a^{2}b-1}{p} \right)\left(\frac{b-1}{p} \right)\sum\limits_{d = 1}^{p-1}\left(\frac{\overline{a}^{2}d-1}{p} \right)\left(\frac{d-1}{p} \right)\\ = &(p-1)\sum\limits_{a = 1}^{p-1}\left(\sum\limits_{b = 0}^{p-1}\left( \frac{b^{2}-(a^{2}-1)^{2}}{p} \right)-1\right)\left(\sum\limits_{d = 0}^{p-1}\left( \frac{d^{2}-(\overline{a}^{2}-1)^{2}}{p} \right)-1 \right)\\ = &(p-1)\left[2\sum\limits_{b = 0}^{p-1}\left(\frac{b^{2}}{p} \right)\sum\limits_{d = 0}^{p-1}\left( \frac{d^{2}}{p}\right)+\sum\limits_{a = 2}^{p-2}\sum\limits_{b = 0}^{p-1}\left(\frac{b^{2}-(a^{2}-1)^{2}}{p} \right)\sum\limits_{d = 0}^{p-1}\left(\frac{d^{2}-(\overline{a}^{2}-1)^{2}}{p} \right) \right] \\ &-2(p-1)\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 0}^{p-1}\left(\frac{b^{2}-(a^{2}-1)^{2}}{p} \right)+(p-1)^{2} \\ = &2p^{3}-6p^{2}+4 . \end{align}$

Then, we compute $A_{2}$ . With the aid of Lemma $2.4$ , we have

$\begin{align} A_{2} = &(p-1)\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\left(\frac{a^{2}b-1}{p} \right)\left(\frac{b-1}{p} \right)\sum\limits_{d = 1}^{p-1}\left(\frac{m}{p} \right) \left(\frac{(\overline{a}^{2}d-1)(d-1)d}{p} \right)\\ = &(p-1)\sum\limits_{a = 1}^{p-1}\left[\sum\limits_{b = 0}^{p-1}\left(\frac{b^{2}-(a^{2}-1)^{2}}{p} \right)-1\right]\sum\limits_{d = 1}^{p-1}\left(\frac{m}{p} \right) \left(\frac{(\overline{a}^{2}d-1)(d-1)d}{p} \right)\\ = &(p-1)^{2}\sum\limits_{d = 1}^{p-1}\left(\frac{m}{p} \right) \left(\frac{(d-1)^{2}d}{p} \right)-(p-1)\sum\limits_{a = 2}^{p-2} \sum\limits_{d = 1}^{p-1}\left(\frac{m}{p} \right) \left(\frac{(\overline{a}^{2}d-1)(d-1)d}{p} \right)\\ &+(p-1)^{2}\sum\limits_{d = 1}^{p-1}\left(\frac{m}{p} \right) \left(\frac{\left((p-1)^{2}d-1\right)(d-1)d}{p} \right) -(p-1)\sum\limits_{a = 1}^{p-1}\sum\limits_{d = 1}^{p-1}\left(\frac{m}{p} \right) \left(\frac{(\overline{a}^{2}d-1)(d-1)d}{p} \right)\\ = &(p^{2}-3p+2)\left[ \sum\limits_{d = 1}^{p-1}\left(\frac{m}{p} \right) \left(\frac{(d-1)^{2}d}{p} \right)+ \sum\limits_{d = 1}^{p-1}\left(\frac{m}{p} \right) \left(\frac{\left((p-1)^{2}d-1\right)(d-1)d}{p} \right) \right]\\ &-2(p-1)\sum\limits_{a = 2}^{p-2} \sum\limits_{d = 1}^{p-1}\left(\frac{m}{p} \right) \left(\frac{(a^{2}d-1)(d-1)d}{p} \right)\\ = &2(p^{2}-3p+2)\left(\frac{m}{p} \right)\sum\limits_{d = 1}^{p-1}\left(\frac{(d-1)^{2}d}{p} \right)-4(p-1)\left[(-1)^{\frac{p-1}{2}}+1\right]\left(\frac{m}{p} \right)\\ = &2(p^{2}-3p+2)\left(\frac{m}{p} \right)\sum\limits_{b = 2}^{p-1}\left(\frac{b}{p} \right)-4(p-1)\left[(-1)^{\frac{p-1}{2}}+1\right]\left(\frac{m}{p} \right)\\ = &-2(p^{2}-3p+2)\left(\frac{m}{p} \right)-4(p-1)\left[(-1)^{\frac{p-1}{2}}+1\right]\left(\frac{m}{p} \right). \end{align}$

Similarly, we have

$\begin{eqnarray*} A_{3} = -2(p^{2}-3p+2)\left(\frac{m}{p} \right)-4(p-1)\left[(-1)^{\frac{p-1}{2}}+1\right]\left(\frac{m}{p} \right). \end{eqnarray*}$

Note that

$\begin{eqnarray*} && A_{4} = (p-1)\sum\limits_{a = 1}^{p-1}\sum\limits_{b = 1}^{p-1}\left(\frac{(a^{2}b-1)(b-1)b}{p} \right)\sum\limits_{d = 1}^{p-1} \left(\frac{(\overline{a}^{2}d-1)(d-1)d}{p} \right), \end{eqnarray*}$

which completes the proof of Theorem $1.3$ .

4. Conclusions

Three Theorems are stated in the main results. The Theorem 1.1 obtains an exact computational formula for $\sum\limits^{q}_{a = 1}'\chi(ma+\overline{a}),$ which broadens the scope of $q$ by removing the condition $p \equiv 3 \bmod 4$ in the previous article, where $p$ is the prime divisor of $q$ . The Theorem 1.2 derives a new identity for the mean value of it by adding some different ingredients. What's more, the Theorem 1.3 bridges the fourth power of Dirichlet characters with Legendre symbols of certain polynomials, which may be useful in the related future research. However, due to some technical reasons, we can only deal with the odd square-full number $q$ case.

Acknowledgments

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 National Natural Science Foundation of China (No. 11871317), and the Natural Science Basic Research Plan for Distinguished Young Scholars in Shaanxi Province of China (No. 2021JC-29).

Conflict of interest

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

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沈阳化工大学材料科学与工程学院沈阳 110142

AIMS Mathematics

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AIMS Mathematics

Dirichlet characters of the rational polynomials

Related Papers:

Abstract

1. Introduction and main results

2. Some Lemmas

3. Proof of Theorems

4. Conclusions

Acknowledgments

Conflict of interest

References

Reader Comments

通讯作者: 陈斌, bchen63@163.com

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AIMS Mathematics

Dirichlet characters of the rational polynomials

Related Papers:

Abstract

1. Introduction and main results

2. Some Lemmas

3. Proof of Theorems

4. Conclusions

Acknowledgments

Conflict of interest

References

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

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