On Certain Sum Involving Quadratic Residue

Abstract

Let p be a prime and ${\mathbb{F}}_p$ be the set of integers modulo p. Let ${\chi }_p$ be a function defined on ${\mathbb{F}}_p$ such that ${\chi }_p\left(0\right)=0$ and for $a\in {\mathbb{F}}_p\backslash \left\{0\right\}$, set ${\chi }_p\left(a\right)=1$ if a is a quadratic residue modulo p and ${\chi }_p\left(a\right)=-1$ if a is a quadratic non-residue modulo p. Note that ${\chi }_p\left(a\right)=\left({\displaystyle \frac{a}{p}}\right)$ is indeed the Legendre symbol. The image of ${\chi }_p$ in the set of real numbers. In this paper, we consider the following sum ${\sum }_{x\in {\mathbb{F}}_p}{\chi }_p((x-a_1)(x-a_2)\dots (x-a_t))$ where $a_1,a_2,\dots ,a_t$ are distinct elements in ${\mathbb{F}}_p$.

1. Introduction

Let p be a prime and ${\mathbb{F}}_p$ be the set of integers modulo p. Let $Q_p\subseteq {\mathbb{F}}_p$ be the set of all quadratic residues modulo p. For $A_1,A_2,\cdots ,A_l\subseteq {\mathbb{F}}_p$, set

$$\begin{array}{c}\hfill A_1+A_2+\cdots +A_l=\left\{\sum _{1\le i\le l}{a}_i:a_i\in A_i\mathrm{for}\hspace{0.17em}\mathrm{all}\hspace{0.17em}i\right\}.\end{array}$$

The set $Q_p$ is said to have an l-decomposition if

$$\begin{array}{c}\hfill Q_p=A_1+A_2+\cdots +A_l.\end{array}$$

It is shown by Sárközy [1] that for sufficiently large p, $Q_p$ has no 3-decomposition $A_1+A_2+A_3=Q_p$ with $|A_i|\hspace{0.17em}\ge 2$ for $i=1,2,3$. In the same paper, he conjectured that for sufficiently large p, $Q_p$ has no 2-decomposition $A_1+A_2=Q_p$ with $|A_i|\hspace{0.17em}\ge 2$ for $i=1,2$. He also showed that if p is a very large prime and $A_1+A_2=Q_p$ with $|A_i|\hspace{0.17em}\ge 2$, then,

$$\begin{array}{c}\hfill \frac{\sqrt{p}}{3lnp}\le \left|A_i\right|\le \sqrt{p}lnp,\mathrm{for}\hspace{0.17em}i=1,2.\end{array}$$

The bounds are improved by Shparlinski [2]. In fact, Shparlinski showed that for any prime p, there exist positive constants $c_1$ and $c_2$ such that

$$\begin{array}{c}\hfill c_1\sqrt{p}\le \left|A_i\right|\le c_2\sqrt{p},\mathrm{for}\hspace{0.17em}i=1,2.\end{array}$$

Later, Shkredov [3] showed that for any prime p,

$$\begin{array}{c}\hfill \left(\frac{1}{6}+o\left(1\right)\right)\sqrt{p}\le \left|A_i\right|\le \left(3+o\left(1\right)\right)\sqrt{p},\mathrm{for}\hspace{0.17em}i=1,2.\end{array}$$

Here, $o\left(1\right)$ can be considered as a function $h\left(p\right)$ with the property ${lim}_{p\to \infty }h\left(p\right)=0$. Recently, Chen and Yan [4] showed that for any prime p,

$$\begin{array}{c}\hfill \left(\frac{7-\sqrt{17}}{16}\right)\sqrt{p}+1\le \left|A_i\right|\le \left(\frac{7+\sqrt{17}}{4}\right)\sqrt{p}-6.63,\mathrm{for}\hspace{0.17em}i=1,2.\end{array}$$

They also showed that for any prime p, $Q_p$ has no 3-decomposition $A_1+A_2+A_3=Q_p$ with $|A_i|\hspace{0.17em}\ge 2$ for $i=1,2,3$. In their proof, they considered the sum

$$\begin{array}{c}\hfill \sum _{x\in {\mathbb{F}}_p}{\chi }_p((x-a_1)(x-a_2)\dots (x-a_t)),\end{array}$$

(1)

where $a_1,a_2,\cdots ,a_t$ are distinct elements in ${\mathbb{F}}_p$. Here, ${\chi }_p$ is a function defined on ${\mathbb{F}}_p$ such that ${\chi }_p\left(0\right)=0$, and for $a\in {\mathbb{F}}_p\backslash \left\{0\right\}$, ${\chi }_p\left(a\right)=1$ if a is a quadratic residue modulo p and ${\chi }_p\left(a\right)=-1$ if a is a quadratic non-residue modulo p. Note that ${\chi }_p\left(a\right)=\left({\displaystyle \frac{a}{p}}\right)$ is indeed the Legendre symbol. Note that $xy$ is a quadratic residue modulo p if both x and y are quadratic residue modulo p or both x and y are quadratic non-residue modulo p. Furthermore, $xy$ is a quadratic non-residue modulo p if x is a quadratic residue modulo p and y is quadratic non-residue modulo p. Therefore, ${\chi }_p\left(xy\right)={\chi }_p\left(x\right){\chi }_p\left(y\right)$ for all $x,y\in {\mathbb{F}}_p\backslash \left\{0\right\}$. Since ${\chi }_p\left(0\right)=0$, it is not hard to see that ${\chi }_p\left(xy\right)={\chi }_p\left(x\right){\chi }_p\left(y\right)$ for all $x,y\in {\mathbb{F}}_p$. Chen and Yan [4] also conjectured that for distinct elements $a_1,a_2,a_3,a_4$,

$$\begin{array}{c}\hfill \sum _{x\in {\mathbb{F}}_p}{\chi }_p\left((x-a_1)(x-a_2)(x-a_3)(x-a_4)\right)<2\sqrt{p}.\end{array}$$

(2)

They claimed that if the above equation is true, then they can show that for any prime p,

$$\begin{array}{c}\hfill \left(\frac{1}{4}\right)\sqrt{p}+O\left(1\right)\le \left|A_i\right|\le 2\sqrt{p}+O\left(1\right),\mathrm{for}\hspace{0.17em}i=1,2.\end{array}$$

Here, $O\left(1\right)$ can be considered as a function $h\left(p\right)$ with the property that there is a positive integer $n_0$ such that $\left|h\right(p\left)\right|\le c$ for all $p\ge n_0$. Chen and Yan gave the exact values of the sum for $1\le t\le 2$.

In this paper, we consider the sum ${\sum }_{x\in {\mathbb{F}}_p}{\chi }_p((x-a_1)(x-a_2)\dots (x-a_t))$ where $a_1,a_2,\dots ,a_t$ are distinct elements in ${\mathbb{F}}_p$. We proved some relations involving the sum (see Theorems 1, 2, 4, 5 and 6). We also give the exact values of the sum for $p-3\le t\le p$ (see Lemmas 2 and 3, and Corollaries 1 and 2).

2. Main Results

Let $A\subseteq {\mathbb{F}}_p$ and $A\ne ⌀$. We set

$$\begin{array}{c}\hfill f_p\left(A\right)=\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{a\in A}(x-a)\right).\end{array}$$

If $A={\mathbb{F}}_p$, then, ${\prod }_{a\in A}(x-a)=0$ for all $x\in {\mathbb{F}}_p$. Therefore, $f_p\left(A\right)=0$.

For $p=2$, we have ${\mathbb{F}}_2=\{0,1\}$. Note that

$$\begin{array}{c}\hfill f_2\left(\left\{0\right\}\right)=\sum _{x\in {\mathbb{F}}_2}{\chi }_2\left(x\right)=1.\end{array}$$

Similarly, $f_2\left(\left\{1\right\}\right)=1$. Hence, the following proposition follows.

Proposition 1.

$f_2\left({\mathbb{F}}_p\right)=0$ and $f_2\left(\left\{0\right\}\right)=f_2\left(\left\{1\right\}\right)=1$.

From now onwards, we shall assume that p is an odd prime. For convenience, we set $f_p(⌀)=p$. We shall need the following lemma which is a well-known result.

Lemma 1.

$$\begin{array}{c}\hfill f_p\left(A\right)=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}\hspace{0.17em}\left|A\right|=1;\hfill \\ -1,\hfill & \mathrm{if}\hspace{0.17em}\left|A\right|=2.\hfill \end{array}\right.\end{array}$$

Lemma 2.

If $\left|A\right|=p$, then, $f_p\left(A\right)=0$.

Proof. 

This follows from ${\prod }_{a\in A}(x-a)=0$ for all $x\in {\mathbb{F}}_p$. □

Lemma 3.

If $\left|A\right|=p-1$, then,

$$\begin{array}{c}\hfill f_p\left(A\right)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}\hspace{0.17em}p\equiv 1mod4;\hfill \\ -1,\hfill & \mathrm{if}\hspace{0.17em}p\equiv 3mod4.\hfill \end{array}\right.\end{array}$$

Proof. 

Let $b\in {\mathbb{F}}_p\backslash A$. Now, ${\prod }_{a\in A}(x-a)=0$ for all $x\in A$. Let r be a primitive element in ${\mathbb{F}}_p$. Note that the set $\{b-a:a\in A\}$ has $p-1$ elements and none of them is zero. So,

$$\begin{array}{c}\hfill \{b-a:a\in A\}=\{r^i:1\le i\le p-1\}.\end{array}$$

Since ${\chi }_p\left(r^i\right)={(-1)}^i$,

$$\begin{array}{cc}\hfill f_p\left(A\right)& =\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{a\in A}(x-a)\right)={\chi }_p\left(\prod _{a\in A}(b-a)\right)\hfill \\ \hfill & ={\chi }_p\left(\prod _{1\le i\le p-1}{r}^i\right)={\chi }_p\left(r^{\frac{p(p-1)}{2}}\right)\hfill \\ \hfill & ={(-1)}^{\frac{p(p-1)}{2}}={(-1)}^{\frac{p-1}{2}},\hfill \end{array}$$

where the last equality follows from the fact that p is odd. Hence, $f_p\left(A\right)=1$ if $p\equiv 1mod4$ and $f_p\left(A\right)=-1$ if $p\equiv 3mod4$ . □

We introduce the following notation: Let $A\subseteq {\mathbb{F}}_p$ and $b\in {\mathbb{F}}_p\backslash A$. We set

$$\begin{array}{c}\hfill f_p(b;A)=\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left({(x-b)}^2\prod _{a\in A}(x-a)\right).\end{array}$$

Clearly, ${\chi }_p\left({(x-b)}^2{\prod }_{a\in A}(x-a)\right)={\chi }_p\left({\prod }_{a\in A}(x-a)\right)$ if and only if $x\ne b$.

Lemma 4.

Let $A\subseteq {\mathbb{F}}_p$ and $b\in {\mathbb{F}}_p\backslash A$. Then,

$$f_p\left(A\right)=f_p(b;A)+{\chi }_p\left(\prod _{a\in A}(b-a)\right).$$

Proof. 

Note that

$$\begin{array}{cc}\hfill f_p(b;A)& =\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left({(x-b)}^2\prod _{a\in A}(x-a)\right)\hfill \\ \hfill & =\sum _{x\in {\mathbb{F}}_p\backslash (A\cup \left\{b\right\})}{\chi }_p\left({(x-b)}^2\prod _{a\in A}(x-a)\right)\hfill \\ \hfill & =\sum _{x\in {\mathbb{F}}_p\backslash (A\cup \left\{b\right\})}{\chi }_p\left(\prod _{a\in A}(x-a)\right)\hfill \\ \hfill & =\sum _{x\in {\mathbb{F}}_p\backslash A}{\chi }_p\left(\prod _{a\in A}(x-a)\right)-{\chi }_p\left(\prod _{a\in A}(b-a)\right).\hfill \end{array}$$

The lemma follows by noting that

$$f_p\left(A\right)=\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{a\in A}(x-a)\right)=\sum _{x\in {\mathbb{F}}_p\backslash A}{\chi }_p\left(\prod _{a\in A}(x-a)\right).$$

□

Lemma 5.

Let $B\subseteq {\mathbb{F}}_p$ and $1\le \left|B\right|\le p-1$. Then,

$$\sum _{a\in {\mathbb{F}}_p\backslash B}{f}_p(B\cup \left\{a\right\})+\sum _{a\in B}{f}_p(a;B\backslash a)=0.$$

Proof. 

Let $y\in {\mathbb{F}}_p$. By Lemma 1,

$$\begin{array}{c}\hfill f_p\left(\left\{y\right\}\right)=\sum _{a\in {\mathbb{F}}_p}{\chi }_p\left(a-y\right)=0.\end{array}$$

Note that ${\chi }_p(-1)={(-1)}^{\frac{p-1}{2}}$ as $-1$ is a quadratic residue modulo p if and only if $p\equiv 1mod4$. Therefore,

$$\begin{array}{c}\hfill 0={\chi }_p(-1)\sum _{a\in {\mathbb{F}}_p}{\chi }_p\left(a-y\right)=\sum _{a\in {\mathbb{F}}_p}{\chi }_p\left(y-a\right).\end{array}$$

This implies that

$$\begin{array}{cc}\hfill 0& ={\chi }_p\left(\prod _{b\in B}(y-b)\right)\left(\sum _{a\in {\mathbb{F}}_p}{\chi }_p\left(y-a\right)\right)\hfill \\ \hfill & =\sum _{a\in {\mathbb{F}}_p}{\chi }_p\left(\left(y-a\right)\prod _{b\in B}(y-b)\right)\hfill \\ \hfill & =\sum _{a\in {\mathbb{F}}_p\backslash B}{\chi }_p\left(\left(y-a\right)\prod _{b\in B}(y-b)\right)+\sum _{a\in B}{\chi }_p\left(\left(y-a\right)\prod _{b\in B}(y-b)\right)\hfill \end{array}$$

(3)

Note that for each $a\in {\mathbb{F}}_p\backslash B$,

$$\begin{array}{c}\hfill f_p(B\cup \left\{a\right\})=\sum _{y\in {\mathbb{F}}_p}{\chi }_p\left(\left(y-a\right)\prod _{b\in B}(y-b)\right).\end{array}$$

Also, for each $a\in B$,

$$\begin{array}{cc}\hfill f_p(a;B\backslash a)& =\sum _{y\in {\mathbb{F}}_p}{\chi }_p\left({(y-a)}^2\prod _{b\in B\backslash \left\{a\right\}}(y-b)\right)\hfill \\ \hfill & =\sum _{y\in {\mathbb{F}}_p}{\chi }_p\left(\left(y-a\right)\prod _{b\in B}(y-b)\right).\hfill \end{array}$$

The lemma follows from Equation (3), by summing up all the $y\in {\mathbb{F}}_p$. □

Lemma 6.

Let $1\le k\le p-1$. Then

$$(k+1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k+1\end{array}}{f}_p\left(B\right)+(p-k)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k-1\end{array}}{f}_p\left(B\right)=0.$$

Proof. 

Consider the sum

$$\begin{array}{c}\hfill \sum _{\begin{array}{c}A\subseteq {\mathbb{F}}_p,\\ \left|A\right|=k\end{array}}\left(\sum _{a\in {\mathbb{F}}_p\backslash A}{f}_p(A\cup \left\{a\right\})+\sum _{a\in A}{f}_p(a;A\backslash a)\right).\end{array}$$

By Lemma 5, the sum in the bracket is zero. Hence, the above sum is zero.

Let $B\subseteq {\mathbb{F}}_p$ with $\left|B\right|=k+1$. For each $b\in B$, let $A\left(b\right)=B\backslash \left\{b\right\}$. Then $\left|A\right(b\left)\right|=k$ and $B=A\left(b\right)\cup \left\{b\right\}$. Thus, $f_p\left(B\right)$ is counted $k+1$ times in

$$\begin{array}{c}\hfill \sum _{\begin{array}{c}A\subseteq {\mathbb{F}}_p,\\ \left|A\right|=k\end{array}}\sum _{a\in {\mathbb{F}}_p\backslash A}{f}_p(A\cup \left\{a\right\}).\end{array}$$

so,

$$\begin{array}{c}\hfill \sum _{\begin{array}{c}A\subseteq {\mathbb{F}}_p,\\ \left|A\right|=k\end{array}}\sum _{a\in {\mathbb{F}}_p\backslash A}{f}_p(A\cup \left\{a\right\})=(k+1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k+1\end{array}}{f}_p\left(B\right).\end{array}$$

On the other hand, let $C\subseteq {\mathbb{F}}_p$ with $\left|C\right|=k-1$. For each $b\in {\mathbb{F}}_p\backslash C$, let $A^{\prime }\left(b\right)=C\cup \left\{b\right\}$. Then $|A^{\prime }\left(b\right)|=k$. Thus,

$$\begin{array}{cc}\hfill \sum _{\begin{array}{c}A\subseteq {\mathbb{F}}_p,\\ \left|A\right|=k\end{array}}\sum _{a\in A}{f}_p(a;A\backslash a)& =\sum _{\begin{array}{c}C\subseteq {\mathbb{F}}_p,\\ \left|C\right|=k-1\end{array}}\sum _{a\in {\mathbb{F}}_p\backslash C}{f}_p(a;C).\hfill \end{array}$$

By Lemma 4,

$$\begin{array}{c}\hfill f_p(a;C)=f_p\left(C\right)-{\chi }_p\left(\prod _{c\in C}(a-c)\right).\end{array}$$

Therefore,

$$\begin{array}{cc}\hfill \sum _{a\in {\mathbb{F}}_p\backslash C}{f}_p(a;C)& =(p-k+1)f_p\left(C\right)-\sum _{a\in {\mathbb{F}}_p\backslash C}{\chi }_p\left(\prod _{c\in C}(a-c)\right)\hfill \\ \hfill & =(p-k+1)f_p\left(C\right)-\sum _{a\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{c\in C}(a-c)\right)\hfill \\ \hfill & =(p-k+1)f_p\left(C\right)-f_p\left(C\right)\hfill \\ \hfill & =(p-k)f_p\left(C\right).\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill 0& =\sum _{\begin{array}{c}A\subseteq {\mathbb{F}}_p,\\ \left|A\right|=k\end{array}}\left(\sum _{a\in {\mathbb{F}}_p\backslash A}{f}_p(A\cup \left\{a\right\})+\sum _{a\in A}{f}_b(a;A\backslash a)\right)\hfill \\ \hfill & =(k+1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k+1\end{array}}{f}_p\left(B\right)+(p-k)\sum _{\begin{array}{c}C\subseteq {\mathbb{F}}_p,\\ \left|C\right|=k-1\end{array}}{f}_p\left(C\right).\hfill \end{array}$$

□

Theorem 1.

$$\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,B\ne ⌀\end{array}}{f}_p\left(B\right)=-p.$$

Proof. 

By Lemma 6,

$$\begin{array}{c}\hfill \sum _{1\le k\le p-1}\left((k+1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k+1\end{array}}{f}_p\left(B\right)+(p-k)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k-1\end{array}}{f}_p\left(B\right)\right)=0.\end{array}$$

By Lemma 2, $f\left(A\right)=0$ if $\left|A\right|=p$. Therefore,

$$\begin{array}{cc}\hfill \sum _{1\le k\le p-1}(k+1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k+1\end{array}}{f}_p\left(B\right)& =\sum _{1\le k\le p-2}(k+1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k+1\end{array}}{f}_p\left(B\right)\hfill \\ \hfill & =(p-1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-1\end{array}}{f}_p\left(B\right)+\sum _{1\le k\le p-3}(k+1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k+1\end{array}}{f}_p\left(B\right).\hfill \end{array}$$

By Lemma 1, $f\left(A\right)=0$ if $\left|A\right|=1$. Recall that $f(⌀)=p$. So,

$$\begin{array}{cc}\hfill \sum _{1\le k\le p-1}(p-k)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k-1\end{array}}{f}_b\left(B\right)& =(p-1)p+\sum _{2\le k\le p-1}(p-k)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k-1\end{array}}{f}_b\left(B\right)\hfill \\ \hfill & =(p-1)p+\sum _{3\le k\le p-1}(p-k)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k-1\end{array}}{f}_b\left(B\right)\hfill \\ \hfill & =(p-1)p+\sum _{1\le k\le p-3}(p-k-2)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k+1\end{array}}{f}_b\left(B\right).\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill 0& =(p-1)p+(p-1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-1\end{array}}{f}_p\left(B\right)+\sum _{1\le k\le p-3}(k+1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k+1\end{array}}{f}_p\left(B\right)+\hfill \\ \hfill & \sum _{1\le k\le p-3}(p-k-2)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k+1\end{array}}{f}_b\left(B\right)\hfill \\ \hfill & =(p-1)p+(p-1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-1\end{array}}{f}_p\left(B\right)+\sum _{1\le k\le p-3}(p-1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k+1\end{array}}{f}_p\left(B\right)\hfill \\ \hfill & =(p-1)p+(p-1)\left(\sum _{2\le k\le p-1}\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k\end{array}}{f}_p\left(B\right)\right).\hfill \end{array}$$

It follows from Lemmas 1 and 2 that

$$\begin{array}{cc}\hfill \sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,B\ne ⌀\end{array}}{f}_p\left(B\right)& =\sum _{2\le k\le p-1}\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=k\end{array}}{f}_p\left(B\right)\hfill \\ \hfill & =-p.\hfill \end{array}$$

□

Theorem 2.

For $0\le l\le \frac{p-1}{2}$,

$$\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-2l\end{array}}{f}_p\left(B\right)=0.$$

Proof. 

By Lemma 2,

$$\begin{array}{c}\hfill \sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p\end{array}}{f}_p\left(B\right)=0.\end{array}$$

So, the theorem holds for $l=0$. Suppose $l>0$. Assume that the theorem holds for $l-1$. By Lemma 6,

$$\begin{array}{c}\hfill (p-2l+2)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-2l+2\end{array}}{f}_p\left(B\right)+(2l-1)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-2l\end{array}}{f}_b\left(B\right)=0.\end{array}$$

By induction,

$$\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-2l+2\end{array}}{f}_p\left(B\right)=0.$$

Thus,

$$\begin{array}{c}\hfill \sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-2l\end{array}}{f}_p\left(B\right)=0.\end{array}$$

This completes the proof of the theorem. □

Theorem 3.

For $1\le l\le \frac{p-1}{2}$,

$$\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-2l+1\end{array}}{f}_p\left(B\right)={(-1)}^{\frac{p-2l+1}{2}}p\left(\genfrac{}{}{0pt}{}{\frac{p-1}{2}}{l-1}\right).$$

Proof. 

By Lemma 3,

$$\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-1\end{array}}{f}_p\left(B\right)={(-1)}^{\frac{p-1}{2}}p.$$

So, the theorem holds for $l=1$. Suppose $l>1$. Assume that the theorem holds for $l-1$. By Lemma 6,

$$\begin{array}{c}\hfill (p-2l+3)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-2l+3\end{array}}{f}_p\left(B\right)+(2l-2)\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-2l+1\end{array}}{f}_p\left(B\right)=0.\end{array}$$

By induction,

$$\sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-2l+3\end{array}}{f}_p\left(B\right)={(-1)}^{\frac{p-2l+3}{2}}p\left(\genfrac{}{}{0pt}{}{\frac{p-1}{2}}{l-2}\right).$$

Thus,

$$\begin{array}{cc}\hfill \sum _{\begin{array}{c}B\subseteq {\mathbb{F}}_p,\\ \left|B\right|=p-2l+1\end{array}}{f}_p\left(B\right)& ={(-1)}^{\frac{p-2l+1}{2}}p\left(\frac{p-2l+3}{2l-2}\right)\left(\genfrac{}{}{0pt}{}{\frac{p-1}{2}}{l-2}\right)\hfill \\ \hfill & ={(-1)}^{\frac{p-2l+1}{2}}p\left(\frac{\left(\frac{p-2l+3}{2}\right)}{l-1}\right)\left(\frac{\left(\frac{p-1}{2}\right)!}{\left(\frac{p-2l+3}{2}\right)!(l-2)!}\right)\hfill \\ \hfill & ={(-1)}^{\frac{p-2l+1}{2}}p\left(\frac{\left(\frac{p-1}{2}\right)!}{\left(\frac{p-2l+1}{2}\right)!(l-1)!}\right)\hfill \\ \hfill & ={(-1)}^{\frac{p-2l+1}{2}}p\left(\genfrac{}{}{0pt}{}{\frac{p-1}{2}}{l-1}\right).\hfill \end{array}$$

This completes the proof of the theorem. □

Theorem 4.

Let $1\le k\le p-2$ and $B\subseteq {\mathbb{F}}_p$ with $\left|B\right|=p-k$. Let $A={\mathbb{F}}_p\backslash B$. Then

$$f_p\left(B\right)={(-1)}^{\frac{p-1}{2}}\sum _{a\in A}{\chi }_p\left(\prod _{b\in (A\backslash \{a\left\}\right)}(a-b)\right).$$

In particular, when $k=1$, $f_p\left(B\right)={(-1)}^{\frac{p-1}{2}}$.

Proof. 

Note that

$$\begin{array}{c}\hfill f_p\left(B\right)=\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{b\in B}(x-b)\right)=\sum _{a\in A}{\chi }_p\left(\prod _{b\in B}(a-b)\right).\end{array}$$

For a fixed $a_0\in A$, by Lemma 3,

$$\begin{array}{cc}\hfill f_p(B\cup (A\backslash \left\{a_0\right\}))& ={(-1)}^{\frac{p-1}{2}}.\hfill \end{array}$$

On the other hand,

$$\begin{array}{cc}\hfill f_p(B\cup (A\backslash \left\{a_0\right\}))& =\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{b\in B\cup (A\backslash \left\{a_0\right\})}(x-b)\right)\hfill \\ \hfill & ={\chi }_p\left(\prod _{b\in B\cup (A\backslash \left\{a_0\right\})}(a_0-b)\right).\hfill \end{array}$$

Therefore,

$$\begin{array}{c}\hfill {\chi }_p\left(\prod _{b\in B\cup (A\backslash \left\{a_0\right\})}(a_0-b)\right)={(-1)}^{\frac{p-1}{2}},\end{array}$$

which is equivalent to

$$\begin{array}{c}\hfill {\chi }_p\left(\prod _{b\in B}(a_0-b)\right){\chi }_p\left(\prod _{b\in (A\backslash \left\{a_0\right\})}(a_0-b)\right)={(-1)}^{\frac{p-1}{2}}.\end{array}$$

Since ${\chi }_p\left(x\right)=\pm 1$ for $x\in {\mathbb{F}}_p\backslash \left\{0\right\}$, we have

$$\begin{array}{c}\hfill {\chi }_p\left(\prod _{b\in B}(a_0-b)\right)={(-1)}^{\frac{p-1}{2}}{\chi }_p\left(\prod _{b\in (A\backslash \left\{a_0\right\})}(a_0-b)\right).\end{array}$$

Hence,

$$\begin{array}{c}\hfill f_p\left(B\right)={(-1)}^{\frac{p-1}{2}}\sum _{a\in A}{\chi }_p\left(\prod _{b\in (A\backslash \{a\left\}\right)}(a-b)\right).\end{array}$$

□

Corollary 1.

If $\left|A\right|=p-2$, then

$$\begin{array}{c}\hfill f_p\left(A\right)=\left\{\begin{array}{cc}\pm 2,\hfill & \mathrm{if}\hspace{0.17em}p\equiv 1mod4;\hfill \\ 0,\hfill & \mathrm{if}\hspace{0.17em}p\equiv 3mod4.\hfill \end{array}\right.\end{array}$$

Furthermore, if $p\equiv 1mod4$ and ${\mathbb{F}}_p\backslash A=\{b_1,b_2\}$, then $f_p\left(A\right)=2$ if $b_1-b_2$ is a quadratic residue modulo p and $f_p\left(A\right)=-2$, otherwise.

Proof. 

Let ${\mathbb{F}}_p\backslash A=\{b_1,b_2\}$. By Theorem 4,

$$\begin{array}{c}\hfill f_p\left(A\right)={(-1)}^{\frac{p-1}{2}}\left({\chi }_p(b_1-b_2)+{\chi }_p(b_2-b_1)\right).\end{array}$$

Suppose $p\equiv 3mod4$. Then ${\chi }_p(-1)=-1$. Therefore, ${\chi }_p(b_1-b_2)+{\chi }_p(b_2-b_1)={\chi }_p(b_1-b_2)+{\chi }_p(-1){\chi }_p(b_1-b_2)={\chi }_p(b_1-b_2)-{\chi }_p(b_1-b_2)=0$. Thus, $f_p\left(A\right)=0$.

Suppose $p\equiv 1mod4$. Then ${\chi }_p(-1)=1$ and

$$\begin{array}{c}\hfill f_p\left(A\right)={\chi }_p(b_1-b_2)+{\chi }_p(b_2-b_1)=2{\chi }_p(b_1-b_2).\end{array}$$

Hence, the lemma follows. □

Corollary 2.

If $\left|A\right|=p-3$, then $f_p\left(A\right)=1$ or $-3$.

Proof. 

Let ${\mathbb{F}}_p\backslash A=\{b_1,b_2,b_3\}$. By Theorem 4,

$$\begin{array}{c}\hfill f_p\left(A\right)={(-1)}^{\frac{p-1}{2}}\left({\chi }_p(b_1-b_2){\chi }_p(b_1-b_3)+{\chi }_p(b_2-b_1){\chi }_p(b_2-b_3)+{\chi }_p(b_3-b_1){\chi }_p(b_3-b_2)\right).\end{array}$$

Case 1. Suppose $p\equiv 1mod4$. Then

$$\begin{array}{c}\hfill f_p\left(A\right)={\chi }_p(b_1-b_2){\chi }_p(b_1-b_3)+{\chi }_p(b_2-b_1){\chi }_p(b_2-b_3)+{\chi }_p(b_3-b_1){\chi }_p(b_3-b_2),\end{array}$$

${\chi }_p(-1)=1$ and ${\chi }_p(-x)={\chi }_p\left(x\right)$ for all $x\in {\mathbb{F}}_p$. We distinguish several cases.

Case 1.1. ${\chi }_p(b_1-b_2)$ and ${\chi }_p(b_1-b_3)$ are of the same sign. This implies that ${\chi }_p(b_2-b_1)$ and ${\chi }_p(b_3-b_1)$ are of the same sign as ${\chi }_p(b_1-b_2)$. Now, if ${\chi }_p(b_1-b_2)=1$, then ${\chi }_p(b_1-b_3)={\chi }_p(b_3-b_1)={\chi }_p(b_2-b_1)=1$. If ${\chi }_p(b_3-b_2)=1$, then $f_p\left(A\right)=3$. If ${\chi }_p(b_3-b_2)=-1$, then $f_p\left(A\right)=-1$.

Next, if ${\chi }_p(b_1-b_2)=-1$, then ${\chi }_p(b_1-b_3)={\chi }_p(b_3-b_1)={\chi }_p(b_2-b_1)=-1$. If ${\chi }_p(b_3-b_2)=1$, then $f_p\left(A\right)=-1$. If ${\chi }_p(b_3-b_2)=-1$, then $f_p\left(A\right)=3$.

Case 1.2. ${\chi }_p(b_1-b_2)$ and ${\chi }_p(b_1-b_3)$ are of different sign. This means ${\chi }_p(b_1-b_3)=-{\chi }_p(b_1-b_2)$. So,

$$\begin{array}{cc}\hfill & {\chi }_p(b_2-b_1){\chi }_p(b_2-b_3)+{\chi }_p(b_3-b_1){\chi }_p(b_3-b_2)\hfill \\ \hfill =& {\chi }_p(b_1-b_2){\chi }_p(b_2-b_3)-{\chi }_p(b_1-b_2){\chi }_p(b_3-b_2)\hfill \\ \hfill =& {\chi }_p(b_1-b_2)\left({\chi }_p(b_2-b_3)-{\chi }_p(b_3-b_2)\right)=0.\hfill \end{array}$$

Hence, $f_p\left(A\right)={\chi }_p(b_1-b_2){\chi }_p(b_1-b_3)=-1$.

Case 2. Suppose $p\equiv 3mod4$. Then

$$\begin{array}{c}\hfill f_p\left(A\right)=-{\chi }_p(b_1-b_2){\chi }_p(b_1-b_3)-{\chi }_p(b_2-b_1){\chi }_p(b_2-b_3)-{\chi }_p(b_3-b_1){\chi }_p(b_3-b_2),\end{array}$$

${\chi }_p(-1)=-1$ and ${\chi }_p(-x)=-{\chi }_p\left(x\right)$ for all $x\in {\mathbb{F}}_p$. We distinguish several cases.

Case 2.1. ${\chi }_p(b_1-b_2)$ and ${\chi }_p(b_1-b_3)$ are of the same sign. This implies that ${\chi }_p(b_2-b_1)$ and ${\chi }_p(b_3-b_1)$ are of opposite sign from ${\chi }_p(b_1-b_2)$. So, ${\chi }_p(b_2-b_1)={\chi }_p(b_3-b_1)$ and

$$\begin{array}{cc}\hfill & {\chi }_p(b_2-b_1){\chi }_p(b_2-b_3)+{\chi }_p(b_3-b_1){\chi }_p(b_3-b_2)\hfill \\ \hfill =& {\chi }_p(b_2-b_1){\chi }_p(b_2-b_3)+{\chi }_p(b_2-b_1){\chi }_p(b_3-b_2)\hfill \\ \hfill =& {\chi }_p(b_2-b_1)\left({\chi }_p(b_2-b_3)+{\chi }_p(b_3-b_2)\right)=0,\hfill \end{array}$$

where the last equation follows from ${\chi }_p(b_2-b_3)=-{\chi }_p(b_3-b_2)$. Hence, $f_p\left(A\right)=-{\chi }_p(b_1-b_2){\chi }_p(b_1-b_3)=-1$.

Case 2.2. ${\chi }_p(b_1-b_2)$ and ${\chi }_p(b_1-b_3)$ are of different sign. Suppose ${\chi }_p(b_1-b_2)=1$. Then ${\chi }_p(b_1-b_3)=-1$, ${\chi }_p(b_3-b_1)=1$ and ${\chi }_p(b_2-b_1)=-1$. If ${\chi }_p(b_2-b_3)=1$, then ${\chi }_p(b_3-b_2)=-1$ and $f_p\left(A\right)=3$. If ${\chi }_p(b_2-b_3)=-1$, then $f_p\left(A\right)=-1$.

Suppose ${\chi }_p(b_1-b_2)=-1$. Then ${\chi }_p(b_1-b_3)=1$, ${\chi }_p(b_3-b_1)=-1$ and ${\chi }_p(b_2-b_1)=1$. If ${\chi }_p(b_2-b_3)=1$, then ${\chi }_p(b_3-b_2)=-1$ and $f_p\left(A\right)=-1$. If ${\chi }_p(b_2-b_3)=-1$, then $f_p\left(A\right)=3$.

This completes the proof of the corollary. □

Theorem 5.

Let $1\le k\le p-2$ and $B\subseteq {\mathbb{F}}_p$ with $\left|B\right|=p-k$. If $A\subseteq {\mathbb{F}}_p$ and $\left|A\right|=p-k$, then

$$f_p\left(A\right)\equiv f_p\left(B\right)mod4.$$

Proof. 

Let $B^c={\mathbb{F}}_p\backslash B=\{b_1,b_2,\cdots ,b_k\}$ and $A^c={\mathbb{F}}_p\backslash A=\{a_1,a_2,\cdots ,a_k\}$. By Theorem 4,

$$\begin{array}{cc}\hfill f_p\left(B\right)& ={(-1)}^{\frac{p-1}{2}}\sum _{a\in B^c}{\chi }_p\left(\prod _{b\in B^c\backslash \left\{a\right\}}(a-b)\right)\hfill \\ \hfill & ={(-1)}^{\frac{p-1}{2}}\sum _{1\le i\le k}{\chi }_p\left(\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i\end{array}}(b_i-b_j)\right)\hfill \\ \hfill & ={(-1)}^{\frac{p-1}{2}}\sum _{1\le i\le k}\left(\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i\end{array}}{\chi }_p(b_i-b_j)\right).\hfill \end{array}$$

Similarly,

$$\begin{array}{cc}\hfill f_p\left(A\right)& ={(-1)}^{\frac{p-1}{2}}\sum _{1\le i\le k}\left(\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i\end{array}}{\chi }_p(a_i-a_j)\right).\hfill \end{array}$$

For $i\ne j$, we set $c_{ij}={\chi }_p(b_i-b_j)$ and $d_{ij}={\chi }_p(a_i-a_j)$. Then

$$\begin{array}{cc}\hfill f_p\left(B\right)& ={(-1)}^{\frac{p-1}{2}}\sum _{1\le i\le k}\left(\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i\end{array}}{c}_{ij}\right);\hfill \\ \hfill f_p\left(A\right)& ={(-1)}^{\frac{p-1}{2}}\sum _{1\le i\le k}\left(\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i\end{array}}{d}_{ij}\right).\hfill \end{array}$$

Note that $c_{ij}={\chi }_p(-1)c_{ji}={(-1)}^{\frac{p-1}{2}}{c}_{ji}$. Similarly, $d_{ij}={(-1)}^{\frac{p-1}{2}}{d}_{ji}$. If $c_{ij}=d_{ij}$ for all $i,j$ with $i\ne j$, then $f_p\left(A\right)=f_p\left(B\right)$ and the theorem holds. So, we may assume that $c_{i_0{j}_0}\ne d_{i_0{j}_0}$ for some $i_0,j_0$ with $i_0\ne j_0$.

Note that

$$\begin{array}{cc}\hfill f_p\left(B\right)& ={(-1)}^{\frac{p-1}{2}}\sum _{\begin{array}{c}1\le i\le k,\\ i\ne i_0,j_0\end{array}}\left(\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i\end{array}}{c}_{ij}\right)+{(-1)}^{\frac{p-1}{2}}{c}_{i_0{j}_0}\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i_0,j_0\end{array}}{c}_{i_{0}j}+{(-1)}^{\frac{p-1}{2}}{c}_{j_0{i}_0}\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i_0,j_0\end{array}}{c}_{j_{0}j}\hfill \end{array}$$

Now, set

$$\begin{array}{cc}\hfill f_p{\left(B\right)}_1& ={(-1)}^{\frac{p-1}{2}}\sum _{1\le i\le k}\left(\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i\end{array}}{c}_{ij}^{\prime }\right),\hfill \end{array}$$

where $c_{ij}^{\prime }=c_{ij}$ for all $i,j$ such that $(i,j)\ne (i_0,j_0),(j_0,i_0)$, $c_{i_0{j}_0}^{\prime }=d_{i_0{j}_0}$ and $c_{j_0{i}_0}^{\prime }=d_{j_0{i}_0}$. Then

$$\begin{array}{cc}\hfill f_p{\left(B\right)}_1& ={(-1)}^{\frac{p-1}{2}}\sum _{\begin{array}{c}1\le i\le k,\\ i\ne i_0,j_0\end{array}}\left(\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i\end{array}}{c}_{ij}\right)+{(-1)}^{\frac{p-1}{2}}{d}_{i_0{j}_0}\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i_0,j_0\end{array}}{c}_{i_{0}j}+{(-1)}^{\frac{p-1}{2}}{d}_{j_0{i}_0}\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i_0,j_0\end{array}}{c}_{j_{0}j},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill f_p\left(B\right)-f_p{\left(B\right)}_1& ={(-1)}^{\frac{p-1}{2}}\left(\left(c_{i_0{j}_0}-d_{i_0{j}_0}\right)\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i_0,j_0\end{array}}{c}_{i_{0}j}+\left(c_{j_0{i}_0}-d_{j_0{i}_0}\right)\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i_0,j_0\end{array}}{c}_{j_{0}j}\right).\hfill \end{array}$$

Now, $c_{i_0{j}_0}-d_{i_0{j}_0}=\pm 2$ and $c_{j_0{i}_0}-d_{j_0{i}_0}={(-1)}^{\frac{p-1}{2}}\left(c_{i_0{j}_0}-d_{i_0{j}_0}\right)=\pm 2$. Since ${\prod }_{\begin{array}{c}1\le j\le k,\\ j\ne i_0,j_0\end{array}}{c}_{i_{0}j}=\pm 1$ and ${\prod }_{\begin{array}{c}1\le j\le k,\\ j\ne i_0,j_0\end{array}}{c}_{j_{0}j}=\pm 1$, we have

$$\begin{array}{cc}\hfill f_p\left(B\right)-f_p{\left(B\right)}_1& =\pm 4or0,\hfill \end{array}$$

i.e., $f_p\left(B\right)-f_p{\left(B\right)}_1\equiv 0mod4$. If $c_{ij}^{\prime }=d_{ij}$ for all $i,j$ with $i\ne j$, then $f_p\left(A\right)=f_p{\left(B\right)}_1$ and the theorem holds. If $c_{i_1{j}_1}^{\prime }\ne d_{i_1{j}_1}$ for some $i_1,j_1$ with $i_1\ne j_1$, then we set

$$\begin{array}{cc}\hfill f_p{\left(B\right)}_2& ={(-1)}^{\frac{p-1}{2}}\sum _{1\le i\le k}\left(\prod _{\begin{array}{c}1\le j\le k,\\ j\ne i\end{array}}{c}_{ij}^{″}\right),\hfill \end{array}$$

where $c_{ij}^{″}=c_{ij}^{\prime }$ for all $i,j$ such that $(i,j)\ne (i_1,j_1),(j_1,i_1)$, $c_{i_1{j}_1}^{″}=d_{i_1{j}_1}$ and $c_{j_1{i}_1}^{″}=d_{j_1{i}_1}$. By using similar argument, it can be shown that $f_p{\left(B\right)}_1-f_p{\left(B\right)}_2\equiv 0mod4$. So, by continuing this process, we conclude that $f_p\left(A\right)\equiv f_p\left(B\right)mod4$. □

Corollary 3.

Let $1\le k\le \frac{p-3}{2}$ and $A_{max},A_{min}\subseteq {\mathbb{F}}_p$ be such that

$$\begin{array}{c}\hfill f_p\left(A_{max}\right)=\underset{\begin{array}{c}A\subseteq {\mathbb{F}}_p,\\ \left|A\right|=2k+1\end{array}}{max}{f}_p\left(A\right);\\ \hfill & \\ \hfill f_p\left(A_{min}\right)=\underset{\begin{array}{c}A\subseteq {\mathbb{F}}_p,\\ \left|A\right|=2k+1\end{array}}{min}{f}_p\left(A\right).\end{array}$$

Then $f_p\left(A_{max}\right)=-f_p\left(A_{min}\right)$ and $f_p\left(A_{min}\right)\equiv 0\equiv f_p\left(A_{max}\right)mod2$.

Proof. 

Let $A_{max}=\{a_1,a_2,\dots ,a_{2k+1}\}$. Then

$$\begin{array}{c}\hfill f_p\left(A_{max}\right)=\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{1\le i\le 2k+1}(x-a_i)\right).\end{array}$$

Let r be a primitive element in ${\mathbb{F}}_p$. Then ${\chi }_p\left(r\right)=-1$ and

$$\begin{array}{cc}\hfill -f_p\left(A_{max}\right)& =\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(r\right){\chi }_p\left(\prod _{1\le i\le 2k+1}(x-a_i)\right)\hfill \\ \hfill & =\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{1\le i\le 2k+1}(rx-r{a}_i)\right)\hfill \\ \hfill & =\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{1\le i\le 2k+1}(x-r{a}_i)\right)\ge f_p\left(A_{min}\right).\hfill \end{array}$$

Thus, $f_p\left(A_{max}\right)\le -f_p\left(A_{min}\right)$. Let $A_{min}=\{b_1,b_2,\cdots ,b_{2k+1}\}$. Then

$$\begin{array}{c}\hfill f_p\left(A_{min}\right)=\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{1\le i\le 2k+1}(x-b_i)\right)\end{array}$$

and

$$\begin{array}{cc}\hfill -f_p\left(A_{min}\right)& =\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(r\right){\chi }_p\left(\prod _{1\le i\le 2k+1}(x-b_i)\right)\hfill \\ \hfill & =\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{1\le i\le 2k+1}(rx-r{b}_i)\right)\hfill \\ \hfill & =\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{1\le i\le 2k+1}(x-r{b}_i)\right)\le f_p\left(A_{max}\right).\hfill \end{array}$$

Hence, $f_p\left(A_{max}\right)\ge -f_p\left(A_{min}\right)$ and $f_p\left(A_{max}\right)=-f_p\left(A_{min}\right)$. It follows from Theorem 5 that $f_p\left(A_{max}\right)-f_p\left(A_{min}\right)\equiv 0mod4$. So, $2f_p\left(A_{max}\right)\equiv 0mod4$ and thus $f_p\left(A_{max}\right)\equiv 0mod2$. Finally, $f_p\left(A_{min}\right)=-f_p\left(A_{max}\right)\equiv 0mod2$. □

Theorem 6.

Let $1\le k\le p-2$ and $B\subseteq {\mathbb{F}}_p$ with $\left|B\right|=p-k$. Let $A={\mathbb{F}}_p\backslash B$. Then,

$$f_p\left(B\right)={(-1)}^{\frac{p-1}{2}}\left(\sum _{a\in A}{f}_p(A\backslash \left\{a\right\})+\sum _{b\in B}{f}_p(A\cup \left\{b\right\})\right).$$

Proof. 

By Theorem 4,

$$f_p\left(B\right)={(-1)}^{\frac{p-1}{2}}\sum _{a\in A}{\chi }_p\left(\prod _{b\in A\backslash \left\{a\right\}}(a-b)\right).$$

On the other hand,

$$\begin{array}{cc}\hfill f_p(A\backslash \left\{a\right\})& =\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left(\prod _{b\in A\backslash \left\{a\right\}}(x-b)\right)\hfill \\ \hfill & ={\chi }_p\left(\prod _{b\in A\backslash \left\{a\right\}}(a-b)\right)+\sum _{x\in {\mathbb{F}}_p\backslash \left\{a\right\}}{\chi }_p\left(\prod _{b\in A\backslash \left\{a\right\}}(x-b)\right)\hfill \\ \hfill & ={\chi }_p\left(\prod _{b\in A\backslash \left\{a\right\}}(a-b)\right)+\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left({(x-a)}^2\prod _{b\in A\backslash \left\{a\right\}}(x-b)\right).\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill f_p\left(B\right)& ={(-1)}^{\frac{p-1}{2}}\sum _{a\in A}\left(f_p(A\backslash \left\{a\right\})-\sum _{x\in {\mathbb{F}}_p}{\chi }_p\left({(x-a)}^2\prod _{b\in (A\backslash \{a\left\}\right)}(x-b)\right)\right)\hfill \\ \hfill & ={(-1)}^{\frac{p-1}{2}}\sum _{a\in A}\left(f_p(A\backslash \left\{a\right\})-f_p(a;A\backslash \left\{a\right\})\right).\hfill \end{array}$$

By Lemma 5,

$$\sum _{a\in {\mathbb{F}}_p\backslash A}{f}_p(A\cup \left\{a\right\})+\sum _{a\in A}{f}_p(a;A\backslash a)=0.$$

Hence,

$$\begin{array}{cc}\hfill f_p\left(B\right)& ={(-1)}^{\frac{p-1}{2}}\left(\sum _{a\in A}{f}_p(A\backslash \left\{a\right\})+\sum _{a\in {\mathbb{F}}_p\backslash A}{f}_p(A\cup \left\{a\right\})\right),\hfill \end{array}$$

and the lemma follows by noting that $B={\mathbb{F}}_p\backslash A$. □

3. Concluding Remark

In this paper, we proved some relations involving the sum in Equation (1) (see Theorems 1, 2, 4, 5 and 6). We hope that these relations may give some insights on how to tackle the conjecture of Chen and Yan [4] (see Equation (2)).