Some estimate of character sums and its applications

Abstract

The main purpose of this paper is using the elementary methods and the properties of Gauss sums to give a sharp estimate for some character sums. Then using this estimate to prove the existence of some special primitive roots $modp$, an odd prime, and to prove that for any integer n with $(n,p)=1$, if p is large enough, then there exist two primitive roots α and β of p such that both $\alpha +n\beta$ and $n\overline{\alpha }+\overline{\beta }$ are also primitive roots of p, where $\overline{a}$ satisfies $\overline{a}a\equiv 1modp$. Let $N(n,p)$ denote the number of all pairs $(\alpha ,\beta )$ of primitive roots of p such that both $\alpha +n\beta$ and $n\overline{\alpha }+\overline{\beta }$ are also primitive roots of p. Then we can give an interesting asymptotic formula for $N(n,p)$.

MSC:11M20.

1 Introduction

Let $q>1$ be an integer. For any integer n with $(n,q)=1$, from the well-known Euler-Fermat theorem, we have $n^{\varphi (q)}\equiv 1modq$, where $\varphi (q)$ is Euler ϕ-function. That is, $\varphi (q)$ denotes the number of all integers $1\le a\le q$ with $(a,q)=1$. Let k be the smallest positive integer such that $n^k\equiv 1modq$. If $k=\varphi (q)$, then n is called a primitive root of q. If q has a primitive root, then each reduced residue system $modq$ can be expressed as a geometric progression. This gives a powerful tool that can be used in problems involving reduced residue systems. Unfortunately, not all modulo q have primitive roots. In fact primitive roots exist only for the following several cases:

$$q=1,2,4,p^{\alpha },2p^{\alpha },$$

where p is an odd prime and $\alpha \ge 1$.

Many people have studied the properties of primitive roots and related problems, and obtained many interesting results; see [1–6] and [7]. For example, Juping Wang [4] proved that Golomb’s conjecture is true for almost all $q=p^n$. That is, there exist two primitive elements α and β in finite fields ${\mathbf{F}}_q$ such that $\alpha +\beta =1$. Cohen and Mullen [2] established a generalization of Golomb’s conjecture by proving the existence of $q_0>0$ such that, whenever $q>q_0$, there exist $\alpha ,\beta \in {\mathbf{F}}_q$ with $\gamma \alpha +\delta \beta =\epsilon$, where γ, δ and ε are arbitrary non-zero elements of ${\mathbf{F}}_q$.

In this paper, we consider the existence of some special primitive roots of p, such as all α, β, $\alpha +\beta$ and $\overline{\alpha }+\overline{\beta }$ are primitive roots of p, where $\overline{a}$ satisfies $\overline{a}a\equiv 1modp$. Furthermore, for any integer n with $(n,p)=1$, are there primitive roots α and β of p such that both $\alpha +n\beta$ and $n\overline{\alpha }+\overline{\beta }$ are also two primitive roots of p? Let $N(n,p)$ denote the number of all pairs $(\alpha ,\beta )$ of primitive roots of p such that both $\alpha +n\beta$ and $n\overline{\alpha }+\overline{\beta }$ are also primitive roots of p. How about the asymptotic properties of $N(n,p)$?

In this paper, we shall use the elementary methods and estimate for character sums to study this problem, and prove the following conclusion.

Theorem Let p be an odd prime, then for any integer n with $(n,p)=1$, we have the asymptotic formula

$$N(n,p)=\frac{{\varphi }^3(p-1)}{p-1}\prod _{q|p-1}(1-\frac{1}{q-1}+\frac{1}{{(q-1)}^2})+\theta \cdot \frac{{\varphi }^4(p-1)}{{(p-1)}^4}\cdot p^{\frac{3}{2}}\cdot {16}^{\omega (p-1)},$$

where $|\theta |\le 5$, $\omega (m)$ denotes the number of all distinct prime divisors of m, ${\prod }_{q|p-1}$ denotes the product over all distinct prime divisors of $p-1$.

Taking $n=\pm 1$, from our theorem we may immediately deduce the following two corollaries.

Corollary 1 Let p be a prime large enough, then there exist two primitive roots α and β of p such that both $\alpha +\beta$ and $\overline{\alpha }+\overline{\beta }$ are also primitive roots of p.

Corollary 2 Let p be a prime large enough, then there exist two primitive roots α and β of p such that both $\alpha -\beta$ and $\overline{\beta }-\overline{\alpha }$ are also primitive roots of p.

2 Several lemmas

In this section, we shall give several lemmas, which are necessary in the proof of our theorem. Dirichlet characters and Gauss sums are used in this paper, please refer to [8] for more details. First we have the following lemma.

Lemma 1 Let p be an odd prime, and let ${\chi }_1$ and ${\chi }_2$ be $modp$ (not all principal characters). Then, for any m with $(m,p)=1$, we have the estimate

$$|\sum _{a=1}^{p-1}{\chi }_1(a){\chi }_2(ma+1)|\le \sqrt{p}.$$

Proof If ${\chi }_1={\chi }_0$ is the principal character $modp$, then we have

$$\begin{array}{rcl}\sum _{a=1}^{p-1}{\chi }_1(a){\chi }_2(ma+1)& =& \sum _{a=1}^{p-1}{\chi }_2(ma+1)\\ =& \sum _{a=0}^{p-1}{\chi }_2(ma+1)-1=\sum _{a=0}^{p-1}{\chi }_2(a)-1=-1.\end{array}$$

(2.1)

If ${\chi }_2={\chi }_0$, then we have

$$\sum _{a=1}^{p-1}{\chi }_1(a){\chi }_2(ma+1)=\sum _{a=1}^{p-1}{\chi }_1(a){\chi }_0(ma+1)=\sum _{a=1}^{p-1}{\chi }_1(a)-{\chi }_1(-\overline{m})=-{\chi }_1(-\overline{m}).$$

(2.2)

If neither ${\chi }_1$ nor ${\chi }_2$ is the principal character $modp$, then from the properties of Gauss sums we have

$$\begin{array}{rcl}\sum _{a=1}^{p-1}{\chi }_1(a){\chi }_2(ma+1)& =& \frac{1}{\tau ({\overline{\chi }}_2)}\sum _{a=1}^{p-1}{\chi }_1(a)\sum _{b=1}^{p-1}{\overline{\chi }}_2(b)e\left(\frac{b(ma+1)}{p}\right)\\ =& \frac{1}{\tau ({\overline{\chi }}_2)}\sum _{b=1}^{p-1}{\overline{\chi }}_2(b)\sum _{a=1}^{p-1}{\chi }_1(a)e\left(\frac{mba+b}{p}\right)\\ =& \frac{1}{\tau ({\overline{\chi }}_2)}\sum _{b=1}^{p-1}{\overline{\chi }}_2(b)e\left(\frac{b}{p}\right){\overline{\chi }}_1(mb)\tau ({\chi }_1)\\ =& {\overline{\chi }}_1(m)\frac{\tau ({\chi }_1)}{\tau (\overline{{\chi }_2})}\tau ({\overline{\chi }}_1{\overline{\chi }}_2).\end{array}$$

(2.3)

Note that both ${\chi }_1$ and ${\chi }_2$ are non-principal characters $modp$, for any character $\chi modp$, $|\tau (\chi )|\le \sqrt{p}$ and $|\tau ({\chi }_1)|=|\tau ({\chi }_2)|=|\tau ({\overline{\chi }}_2)|=\sqrt{p}$. So from (2.3) we may immediately deduce the inequality

$$|\sum _{a=1}^{p-1}{\chi }_1(a){\chi }_2(ma+1)|\le \sqrt{p}.$$

(2.4)

Now Lemma 1 follows from identities (2.1), (2.2) and estimate (2.4). □

Lemma 2 Let p be an odd prime, and let ${\chi }_1$, ${\chi }_2$, ${\chi }_3$ and ${\chi }_4$ be four non-principal characters $modp$ with ${\chi }_3{\chi }_4\ne {\chi }_0$ (or ${\chi }_1\ne {\chi }_4$). Then, for any integer n with $(n,p)=1$, we have the estimate

$$|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(a){\chi }_2(b){\chi }_3(a+nb){\chi }_4(n\overline{a}+\overline{b})|\le p^{\frac{3}{2}}.$$

Proof From the properties of reduced residue system $modp$, we have

$$\begin{array}{c}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(a){\chi }_2(b){\chi }_3(a+nb){\chi }_4(n\overline{a}+\overline{b})\hfill \\ =\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(ab){\chi }_2(b){\chi }_3(ab+nb){\chi }_4(n\overline{a}\overline{b}+\overline{b})\hfill \\ =\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1{\chi }_2{\chi }_3{\overline{\chi }}_4(b){\chi }_1{\overline{\chi }}_4(a){\chi }_3{\chi }_4(a+n).\hfill \end{array}$$

(2.5)

It is clear that for any character $\chi modp$, we have

$$\sum _{a=1}^{p-1}\chi (a)=\{\begin{array}{cc}p-1\hfill & \text{if }\chi \text{ is the principal character }modp\text{,}\hfill \\ 0\hfill & \text{otherwise.}\hfill \end{array}$$

(2.6)

So, if ${\chi }_1{\chi }_2{\chi }_3{\overline{\chi }}_4\ne {\chi }_0$, the principal character $modp$, then from (2.5) and (2.6) we have the identity

$$|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(a){\chi }_2(b){\chi }_3(a+nb){\chi }_4(n\overline{a}+\overline{b})|=0.$$

(2.7)

If ${\chi }_1{\chi }_2{\chi }_3{\overline{\chi }}_4={\chi }_0$, note that ${\chi }_1{\overline{\chi }}_4$ and ${\chi }_3{\chi }_4$ not all principal characters $modp$, so from (2.5), (2.6) and Lemma 1 we have the estimate

$$\begin{array}{c}|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(a){\chi }_2(b){\chi }_3(a+nb){\chi }_4(n\overline{a}+\overline{b})|\hfill \\ =|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1{\chi }_2{\chi }_3{\overline{\chi }}_4(b){\chi }_1{\overline{\chi }}_4(a){\chi }_3{\chi }_4(a+n)|\hfill \\ =(p-1)\cdot |\sum _{a=1}^{p-1}{\chi }_1{\overline{\chi }}_4(a){\chi }_3{\chi }_4(a+n)|\le p^{\frac{3}{2}}.\hfill \end{array}$$

(2.8)

Combining (2.7) and (2.8), we may immediately deduce Lemma 2. □

Lemma 3 Let p be an odd prime. Then, for any integer c with $(c,p)=1$, we have the identity

$$\frac{\varphi (p-1)}{p-1}\sum _{h|p-1}\frac{\mu (h)}{\varphi (h)}\underset{(h,k)=1}{\sum _{k=1}^h}e\left(\frac{kindc}{h}\right)=\{\begin{array}{cc}1\hfill & \mathit{\text{if}}c\mathit{\text{is a primitive root of}}p,\hfill \\ 0\hfill & \mathit{\text{otherwise}},\hfill \end{array}$$

where indc denotes the index of c relative to some fixed primitive root of p, $\mu (n)$ is the Möbius function.

Proof See Proposition 2.2 of reference [9]. □

3 Proof of the theorem

In this section, we shall complete the proof of our theorem. First we write ${\chi }_{s,h}(c)=e(\frac{sindc}{h})$. It is clear that ${\chi }_{s,h}(c)$ is a Dirichlet character $modp$. For any integer n with $(n,p)=1$, from Lemma 3 we have

$$\begin{array}{c}N(n,p)\hfill \\ =\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\sum _{h|p-1}\sum _{i|p-1}\sum _{j|p-1}\sum _{k|p-1}\frac{\mu (h)}{\varphi (h)}\frac{\mu (i)}{\varphi (i)}\frac{\mu (j)}{\varphi (j)}\frac{\mu (k)}{\varphi (k)}\hfill \\ \times \underset{(h,h_1)=1}{\sum _{h_1=1}^h}\underset{(i,i_1)=1}{\sum _{i_1=1}^i}\underset{(j,j_1)=1}{\sum _{j_1=1}^j}\underset{(k,k_1)=1}{\sum _{k_1=1}^k}{\chi }_{h_1,h}(a){\chi }_{i_1,i}(b){\chi }_{j_1,j}(a+nb){\chi }_{k_1,k}(n\overline{a}+\overline{b})\hfill \\ =\frac{{\varphi }^4(p-1)}{{(p-1)}^2}+2\frac{{\varphi }^4(p-1)}{{(p-1)}^3}\underset{h>1}{\sum _{h|p-1}}\frac{\mu (h)}{\varphi (h)}\underset{(h,s)=1}{\sum _{s=1}^h}\sum _{a=1}^{p-1}{\chi }_{s,h}(a)\hfill \\ +2\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{u>1}{\sum _{u|p-1}}\frac{\mu (u)}{\varphi (u)}\underset{(v,u)=1}{\sum _{v=1}^u}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{v,u}(a+nb)\hfill \\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{h>1}{\sum _{h|p-1}}\underset{u>1}{\sum _{u|p-1}}\frac{\mu (h)}{\varphi (h)}\frac{\mu (u)}{\varphi (u)}\underset{(h,s)=1}{\sum _{s=1}^h}\underset{(v,u)=1}{\sum _{v=1}^u}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{v,u}(a){\chi }_{s,h}(b)\hfill \\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{h>1}{\sum _{h|p-1}}\underset{u>1}{\sum _{u|p-1}}\frac{\mu (h)}{\varphi (h)}\frac{\mu (u)}{\varphi (u)}\underset{(h,s)=1}{\sum _{s=1}^h}\underset{(v,u)=1}{\sum _{v=1}^u}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{v,u}(a+nb){\chi }_{s,h}(n\overline{a}+\overline{b})\hfill \\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{h>1}{\sum _{h|p-1}}\underset{u>1}{\sum _{u|p-1}}\frac{\mu (h)}{\varphi (h)}\frac{\mu (u)}{\varphi (u)}\underset{(h,s)=1}{\sum _{s=1}^h}\underset{(v,u)=1}{\sum _{v=1}^u}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{v,u}(a){\chi }_{s,h}(n\overline{a}+\overline{b})\hfill \\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{h>1}{\sum _{h|p-1}}\underset{u>1}{\sum _{u|p-1}}\frac{\mu (h)}{\varphi (h)}\frac{\mu (u)}{\varphi (u)}\underset{(h,s)=1}{\sum _{s=1}^h}\underset{(v,u)=1}{\sum _{v=1}^u}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{v,u}(b){\chi }_{s,h}(n\overline{a}+\overline{b})\hfill \\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{h>1}{\sum _{h|p-1}}\underset{u>1}{\sum _{u|p-1}}\frac{\mu (h)}{\varphi (h)}\frac{\mu (u)}{\varphi (u)}\underset{(h,s)=1}{\sum _{s=1}^h}\underset{(v,u)=1}{\sum _{v=1}^u}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{s,h}(a){\chi }_{v,u}(a+nb)\hfill \\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{h>1}{\sum _{h|p-1}}\underset{u>1}{\sum _{u|p-1}}\frac{\mu (h)}{\varphi (h)}\frac{\mu (u)}{\varphi (u)}\underset{(h,s)=1}{\sum _{s=1}^h}\underset{(v,u)=1}{\sum _{v=1}^u}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{s,h}(b){\chi }_{v,u}(a+nb)\hfill \\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{i>1}{\sum _{i|p-1}}\underset{j>1}{\sum _{j|p-1}}\underset{k>1}{\sum _{k|p-1}}\frac{\mu (i)}{\varphi (i)}\frac{\mu (j)}{\varphi (j)}\frac{\mu (k)}{\varphi (k)}\hfill \\ \times \underset{(i,i_1)=1}{\sum _{i_1=1}^i}\underset{(j,j_1)=1}{\sum _{j_1=1}^j}\underset{(k,k_1)=1}{\sum _{k_1=1}^k}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{i_1,i}(b){\chi }_{j_1,j}(a+nb){\chi }_{k_1,k}(n\overline{a}+\overline{b})\hfill \\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{h>1}{\sum _{h|p-1}}\underset{j>1}{\sum _{j|p-1}}\underset{k>1}{\sum _{k|p-1}}\frac{\mu (h)}{\varphi (h)}\frac{\mu (j)}{\varphi (j)}\frac{\mu (k)}{\varphi (k)}\hfill \\ \times \underset{(h,h_1)=1}{\sum _{h_1=1}^h}\underset{(j,j_1)=1}{\sum _{j_1=1}^j}\underset{(k,k_1)=1}{\sum _{k_1=1}^k}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{h_1,h}(a){\chi }_{j_1,j}(a+nb){\chi }_{k_1,k}(n\overline{a}+\overline{b})\hfill \\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{h>1}{\sum _{h|p-1}}\underset{i>1}{\sum _{i|p-1}}\underset{k>1}{\sum _{k|p-1}}\frac{\mu (h)}{\varphi (h)}\frac{\mu (i)}{\varphi (i)}\frac{\mu (k)}{\varphi (k)}\hfill \\ \times \underset{(h,h_1)=1}{\sum _{h_1=1}^h}\underset{(i,i_1)=1}{\sum _{i_1=1}^i}\underset{(k,k_1)=1}{\sum _{k_1=1}^k}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{h_1,h}(a){\chi }_{i_1,i}(b){\chi }_{k_1,k}(n\overline{a}+\overline{b})\hfill \\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{h>1}{\sum _{h|p-1}}\underset{i>1}{\sum _{i|p-1}}\underset{j>1}{\sum _{j|p-1}}\frac{\mu (h)}{\varphi (h)}\frac{\mu (i)}{\varphi (i)}\frac{\mu (j)}{\varphi (j)}\hfill \\ \times \underset{(h,h_1)=1}{\sum _{h_1=1}^h}\underset{(i,i_1)=1}{\sum _{i_1=1}^i}\underset{(j,j_1)=1}{\sum _{j_1=1}^j}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{h_1,h}(a){\chi }_{i_1,i}(b){\chi }_{j_1,j}(a+nb)\hfill \\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{h>1}{\sum _{h|p-1}}\underset{i>1}{\sum _{i|p-1}}\underset{j>1}{\sum _{j|p-1}}\underset{k>1}{\sum _{k|p-1}}\frac{\mu (h)}{\varphi (h)}\frac{\mu (i)}{\varphi (i)}\frac{\mu (j)}{\varphi (j)}\frac{\mu (k)}{\varphi (k)}\hfill \\ \times \underset{(h,h_1)=1}{\sum _{h_1=1}^h}\underset{(i,i_1)=1}{\sum _{i_1=1}^i}\underset{(j,j_1)=1}{\sum _{j_1=1}^j}\underset{(k,k_1)=1}{\sum _{k_1=1}^k}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{h_1,h}(a){\chi }_{i_1,i}(b){\chi }_{j_1,j}(a+nb){\chi }_{k_1,k}(n\overline{a}+\overline{b}).\hfill \end{array}$$

(3.1)

Now we estimate each term in (3.1) respectively. It is clear that for any integer $h>1$ and $(s,h)=1$, we have the identity

$$\sum _{a=1}^{p-1}{\chi }_{s,h}(a)=0.$$

(3.2)

For any three non-principal characters ${\chi }_1$, ${\chi }_2$ and ${\chi }_{3}modp$, from Lemma 2 we have

$$|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(a){\chi }_2(b){\chi }_3(a+nb)|=|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1{\chi }_2{\chi }_3(b){\chi }_1(a){\chi }_3(a+n)|\le p^{\frac{3}{2}},$$

(3.3)

$$|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(a){\chi }_2(b){\chi }_3(n\overline{a}+\overline{b})|\le p^{\frac{3}{2}},$$

(3.4)

$$|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(a){\chi }_2(a+nb){\chi }_3(n\overline{a}+\overline{b})|\le p^{\frac{3}{2}},$$

(3.5)

$$|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(b){\chi }_2(a+nb){\chi }_3(n\overline{a}+\overline{b})|\le p^{\frac{3}{2}}.$$

(3.6)

From Lemma 1 we know that for any two non-principal characters ${\chi }_1$ and ${\chi }_{2}modp$, we have the estimates

$$|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(a){\chi }_2(a+nb)|\le p^{\frac{3}{2}},$$

(3.7)

$$|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(b){\chi }_2(a+nb)|\le p^{\frac{3}{2}},$$

(3.8)

$$|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(a){\chi }_2(n\overline{a}+\overline{b})|\le p^{\frac{3}{2}},$$

(3.9)

$$|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(b){\chi }_2(n\overline{a}+\overline{b})|\le p^{\frac{3}{2}},$$

(3.10)

$$|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(a+nb){\chi }_2(n\overline{a}+\overline{b})|=|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_1(b){\overline{\chi }}_2(b){\overline{\chi }}_1(a){\chi }_1{\chi }_2(a+n)|\le p^{\frac{3}{2}}.$$

(3.11)

If four characters ${\chi }_{h_1,h}$, ${\chi }_{i_1,i}$, ${\chi }_{j_1,j}$ and ${\chi }_{k_1,k}$ in the last term of (3.1) do not satisfy the condition of Lemma 2, this time from Lemma 2 and the orthogonality of characters $modp$ we have

$$\begin{array}{c}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{h_1,h}(a){\chi }_{i_1,i}(b){\chi }_{j_1,j}(a+nb){\chi }_{k_1,k}(n\overline{a}+\overline{b})\hfill \\ =\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_{h_1,h}{\chi }_{i_1,i}{\chi }_{j_1,j}{\overline{\chi }}_{k_1,k}(b){\chi }_{h_1,h}{\overline{\chi }}_{k_1,k}(a){\chi }_{j_1,j}{\chi }_{k_1,k}(a+n)\hfill \\ =\{\begin{array}{cc}(p-1){\sum }_{a=1}^{p-1}{\chi }_{h_1,h}{\overline{\chi }}_{k_1,k}(a){\chi }_{j_1,j}{\chi }_{k_1,k}(a+n)\hfill & \text{if }{\chi }_{h_1,h}{\chi }_{i_1,i}{\chi }_{j_1,j}{\overline{\chi }}_{k_1,k}={\chi }_0;\hfill \\ 0\hfill & \text{if }{\chi }_{h_1,h}{\chi }_{i_1,i}{\chi }_{j_1,j}{\overline{\chi }}_{k_1,k}\ne {\chi }_0,\hfill \end{array}\hfill \\ =\{\begin{array}{cc}\le p^{\frac{3}{2}}\hfill & \text{if }{\chi }_{i_1,i}{\chi }_{j_1,j}\ne {\chi }_0\text{ or }{\chi }_{h_1,h}\ne {\chi }_{k_1,k};\hfill \\ (p-1)\cdot {\sum }_{a=1}^{p-1}{\chi }_0(a+n)\hfill & \text{if }{\chi }_{h_1,h}={\chi }_{i_1,i}={\chi }_{k_1,k}\text{ and }{\chi }_{j_1,j}={\overline{\chi }}_{k_1,k};\hfill \\ 0\hfill & \text{if }{\chi }_{h_1,h}{\chi }_{i_1,i}{\chi }_{j_1,j}{\overline{\chi }}_{k_1,k}\ne {\chi }_0.\hfill \end{array}\hfill \end{array}$$

(3.12)

Note that the identity ${\sum }_{h|p-1}|\mu (h)|=2^{\omega (p-1)}$, ${\chi }_{h_1,h}={\chi }_{i_1,i}={\chi }_{k_1,k}$ and ${\chi }_{i_1,i}={\overline{\chi }}_{j_1,j}$ implies $h=i=j=k$, $h_1=i_1=k_1$ and $j_1=k-h_1$. So from (3.1)-(3.12) we have

$$\begin{array}{rcl}N(n,p)& =& \frac{{\varphi }^4(p-1)}{{(p-1)}^2}+R_1\cdot \frac{{\varphi }^4(p-1)}{{(p-1)}^4}\cdot p^{\frac{3}{2}}\cdot {16}^{\omega (p-1)}\\ +\frac{{\varphi }^4(p-1)}{{(p-1)}^4}\underset{h>1}{\sum _{h|p-1}}\frac{{|\mu (h)|}^4}{{\varphi }^4(h)}\underset{(s,h)=1)}{\sum _{s=1}^h}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_0(a+n)\\ =& \frac{{\varphi }^3(p-1)}{p-1}\prod _{q|p-1}(1-\frac{1}{q-1}+\frac{1}{{(q-1)}^2})+R\cdot \frac{{\varphi }^4(p-1)}{{(p-1)}^4}\cdot p^{\frac{3}{2}}\cdot {16}^{\omega (p-1)},\end{array}$$

where we have used the identity ${\sum }_{h|p-1}\frac{{|\mu (h)|}^4}{{\varphi }^3(h)}=\frac{p-1}{\varphi (p-1)}{\prod }_{q|p-1}(1-\frac{1}{q-1}+\frac{1}{{(q-1)}^2})$, and $|R|\le 5$. This completes the proof of our theorem.