On the quotient set of the distance set

TL;DR

This paper investigates the quotient set of distances in finite field vector spaces, establishing size conditions under which the quotient set covers the entire field or includes all quadratic residues, with results optimal in general.

Contribution

It provides new size thresholds for subsets of finite field vector spaces to ensure their distance quotient sets cover the entire field or include quadratic residues, extending previous understanding.

Findings

For even dimensions, large enough sets produce a quotient set equal to the entire field.

For odd dimensions, the quotient set contains zero and all nonzero quadratic residues.

Results are shown to be optimal in general.

Abstract

Let ${\Bbb F}_q$ be a finite field of order $q.$ We prove that if $d\ge 2$ is even and $E \subset {\Bbb F}_q^d$ with $|E| \ge 9q^{\frac{d}{2}}$ then $$ {\Bbb F}_q=\frac{\Delta(E)}{\Delta(E)}=\left\{ \frac{a}{b}: a \in \Delta(E), b \in \Delta(E) \backslash \{0\} \right\},$$ where $$ \Delta(E)=\{||x-y||: x,y \in E\}, \ ||x||=x_1^2+x_2^2+\cdots+x_d^2.$$ If the dimension $d$ is odd and $E\subset \mathbb F_q^d$ with $|E|\ge 6q^{\frac{d}{2}},$ then $$ \{0\}\cup\mathbb F_q^+ \subset \frac{\Delta(E)}{\Delta(E)},$$ where $\mathbb F_q^+$ denotes the set of nonzero quadratic residues in $\mathbb F_q.$ Both results are, in general, best possible, including the conclusion about the nonzero quadratic residues in odd dimensions.

Equations

$${\mathbb{F}}_q=\frac{\Delta (E)}{\Delta (E)}=\left\{\frac{a}{b}:a\in \Delta (E),b\in \Delta (E)\backslash \{0\}\right\},$$

$$\Delta (E)=\{||x-y||:x,y\in E\},||x||=x_1^2+x_2^2+\cdots +x_d^2.$$

$$\{0\}\cup {\mathbb{F}}_q^{+}\subset \frac{\Delta (E)}{\Delta (E)},$$

$$\Delta (E)=\{||x-y||:x,y\in E\},||x||=x_1^2+\cdots +x_d^2,$$

$$|\Delta (E)|>\frac{q}{2}.$$

$$\frac{\Delta (E)}{\Delta (E)}:=\left\{\frac{a}{b}:a\in \Delta (E),b\in \Delta (E)\backslash \{0\}\right\}.$$

$${\mathbb{F}}_q=\frac{\Delta (E)}{\Delta (E)}.$$

$$\{0\}\cup {\mathbb{F}}_q^{+}\subset \frac{\Delta (E)}{\Delta (E)}.$$

$$\left\{\frac{a}{b}:a\in \Delta (E),b\in \Delta (E)\backslash \{0\}\right\}={\mathbb{F}}_p.$$

$$\nu (t)=\sum _{x,y\in {\mathbb{F}}_q^d}E(x)E(y)S_t(x-y),$$

$$S_t=\{x\in {\mathbb{F}}_q^d:||x||=t\}.$$

$${\nu }^2(0)<\sum _{t\in {\mathbb{F}}_q}\nu (t)\nu (rt).$$

$$\hat{f}(m)=q^{-d}\sum _{x\in {\mathbb{F}}_q^d}\chi (-x\cdot m)f(x),$$

$$\text{(FOURIER INVERSION)}f(x)=\sum _{m\in {\mathbb{F}}_q^d}\chi (x\cdot m)\hat{f}(m)$$

$$\text{(PLANCHEREL)}\sum _{m\in {\mathbb{F}}_q^d}{|\hat{f}(m)|}^2=q^{-d}\sum _{x\in {\mathbb{F}}_q^d}{|f(x)|}^2.$$

$$\nu (t)$$

$$=q^{-d}{|E|}^2|S_t|+q^{2d}\sum _{m\ne \vec{0}}\widehat{S_t}(m)|\hat{E}(m){|}^2.$$

$$\sum _{t\in {\mathbb{F}}_q}\nu (t)\nu (rt)$$

$$=\sum _{t\in {\mathbb{F}}_q}\left(q^{-d}{|E|}^2|S_t|+q^{2d}\sum _{m\ne \vec{0}}{\hat{S}}_t(m)|\hat{E}(m){|}^2\right)\left(q^{-d}{|E|}^2|S_{rt}|+q^{2d}\sum _{m^{\prime }\ne \vec{0}}{\hat{S}}_{rt}(m^{\prime })|\hat{E}(m^{\prime }){|}^2\right)$$

$$=q^{-2d}|E{|}^4\sum _{t\in {\mathbb{F}}_q}|S_t||S_{rt}|+q^d|E{|}^2\sum _{m^{\prime }\ne \vec{0}}|\hat{E}(m^{\prime }){|}^2\sum _{t\in {\mathbb{F}}_q}|S_t|{\hat{S}}_{rt}(m^{\prime })$$

$$+q^d|E{|}^2\sum _{m\ne \vec{0}}|\hat{E}(m){|}^2\sum _{t\in {\mathbb{F}}_q}|S_{rt}|{\hat{S}}_t(m)+q^{4d}\sum _{m,m^{\prime }\ne \vec{0}}|\hat{E}(m){|}^2|\hat{E}(m^{\prime }){|}^2\sum _{t\in {\mathbb{F}}_q}\widehat{S_t}(m){\hat{S}}_{rt}(m^{\prime })$$

$$:=I+II+III+IV.$$

$$|S_t|=q^{d-1}+\lambda (t)q^{\frac{d-2}{2}}\eta \left((-1{)}^{d/2}\right),$$

$$\widehat{S_{j}}(m)=q^{-1}\delta_{0}(m)+q^{-d-1}\eta^{d}(-1)G^{d}\sum_{s\in{\mathbb{F}}_{q}^{*}}\eta^{d}(s)\chi\Big{(}js+\frac{\|m\|}{4s}\Big{)},$$

$$I:=q^{-2d}|E{|}^4\sum _{t\in {\mathbb{F}}_q}|S_t||S_{rt}|=q^{-2d}|E{|}^4\left(q^{2d-1}+q^{d-2}\sum _{t\in {\mathbb{F}}_q}\lambda (t)\lambda (rt)\right)$$

$$=q^{-2d}|E{|}^4\left(q^{2d-1}+q^{d-2}{\lambda }^2(0)+q^{d-2}\sum _{t\ne 0}\lambda (t)\lambda (rt)\right)$$

$$=q^{-2d}|E{|}^4\left(q^{2d-1}+q^{d-2}(q-1{)}^2+q^{d-2}(q-1)\right).$$

$$I=q^{-1}|E{|}^4+q^{-d}|E{|}^4-q^{-d-1}|E{|}^4.$$

$$\sum _{t\in {\mathbb{F}}_q}{\hat{S}}_{rt}(m)=0$$

$$\sum _{t\in {\mathbb{F}}_q}\lambda (t)\widehat{S_{rt}}(m)=q\widehat{S_0}(m),$$

$$q^{-d}\sum _{t\in {\mathbb{F}}_q}\sum _{x\in {\mathbb{F}}_q^d}\chi (-x\cdot m)S_{rt}(x)=q^{-d}\sum _{x\in {\mathbb{F}}_q^d}\chi (-x\cdot m)\sum _{t\in {\mathbb{F}}_q}{S}_{rt}(x)$$

Full text

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where

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If the dimension $d$ is odd and $E\subset\mathbb{F}_{q}^{d}$ with $|E|\geq 6q^{\frac{d}{2}},$ then

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where $\mathbb{F}_{q}^{+}$ denotes the set of nonzero quadratic residues in $\mathbb{F}_{q}.$ Both results are, in general, best possible, including the conclusion about the nonzero quadratic residues in odd dimensions.

Key words and phrases: Quotient set, distance set, finite field

1. Introduction

The Erdős-Falconer distance problem in vector spaces over finite fields asks for the smallest possible size of

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given $E\subset{\mathbb{F}}_{q}^{d}$, $d\geq 2$. This problem was introduced by Bourgain, Katz and Tao in [2]. Here ${\mathbb{F}}_{q}$ denotes the finite field with $q$ elements and ${\mathbb{F}}_{q}^{d}$ is the $d$-dimensional vector space over this field.

In [6], the first listed author and Misha Rudnev proved that if $E\subset{\mathbb{F}}_{q}^{d}$, $d\geq 2$, with $|E|>2q^{\frac{d+1}{2}}$, then $\Delta(E)={\mathbb{F}}_{q}$. Hart, the first two listed authors, and Rudnev [4] showed that, in a sense, this result is best possible when $d$ is odd. More precisely, for any $c\in(0,1)$ and any $q$ sufficiently large with respect to $c$, they construct subsets $E\subseteq\mathbf{F}_{q}^{d}$ with $|E|>\frac{c}{2}q^{\frac{d+1}{2}}$ but $|\Delta(E)|<cq$. This construction does not appear to generalize to the even dimensional case. In [3], Chapman, the first listed author, Erdoğan, Hart and the second listed author proved that if $q$ is prime, $q\equiv 3\mod 4$ and if $E\subset\mathbb{F}_{q}^{2}$ with $|E|\geq Cq^{\frac{4}{3}}$ for a sufficiently large constant $C>0$, then

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This result was extended to two dimensional vector spaces over arbitrary finite fields in [1]. In even dimensions $d\geq 2$, it is reasonable to conjecture that if $|E|\geq Cq^{\frac{d}{2}}$ with a sufficiently large $C$, then $|\Delta(E)|>\frac{1}{2}q$, but this conjecture currently remains open. The exponent $\frac{d}{2}$ cannot be improved. To see this, let $q=p^{2}$, $p$ prime, and let $E={\mathbb{F}}_{p}^{d}\subset{\mathbb{F}}_{q}^{d}$. Then $|E|=q^{\frac{d}{2}}$, yet $\Delta(E)={\mathbb{F}}_{p}$. When $q$ is a prime and $d\geq 4$, the sharpness of $\frac{d}{2}$ can be demonstrated using Lagrangian subspaces ([4]). In two dimensions, the sharpness of $\frac{d}{2}=1$ is easily demonstrated by taking a suitable subset of a straight line.

The purpose of this paper is to show that under the assumption $|E|\geq Cq^{\frac{d}{2}}$, taking the quotient of the elements of $\Delta(E)$ recovers all of ${\mathbb{F}}_{q}$ for $d$ even, and at least all the square elements of ${\mathbb{F}}_{q}$ when $d$ is odd. More precisely, for $E\subseteq\mathbf{F}_{q}^{d}$ we define

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Our main results are the following.

Theorem 1.1**.**

Let $E\subset{\mathbb{F}}_{q}^{d}$ $d$ even. Then if $|E|\geq 9q^{\frac{d}{2}}$ , we have

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Theorem 1.2**.**

Let $d\geq 3$ be an odd integer and $E\subset\mathbb{F}_{q}^{d}.$ Then if $|E|\geq 6q^{\frac{d}{2}}$, we have

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1.1. Sharpness of results

The results are in general sharp up to constants. To see this, we once again take $q=p^{2}$ and $E={\mathbb{F}}_{p}^{2}$. Then $|E|=q^{\frac{d}{2}}$, yet

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The statement about the squares in Theorem 1.2 is also sharp. The example in [4] (page 15) that illustrates the sharpness of the exponent $\frac{d+1}{2}$ yields a set of size $cq^{\frac{d+1}{2}}$, with $c$ sufficiently small, such that $\Delta(E)\subset\{{(a-a^{\prime})}^{2}:a,a^{\prime}\in A\}$, where $A$ is a suitable arithmetic progression in ${\mathbb{F}}_{q}$. In particular, $\Delta(E)$ is a subset of the squares, so the ratios of the elements of $\Delta(E)$ are also squares.

2. Proof of Theorem 1.1

For $t\in\mathbb{F}_{q}$, let

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where

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It is clear that $\displaystyle 0\in\frac{\Delta(E)}{\Delta(E)}$ unless $\Delta(E)=\{0\}.$ Thus it suffices to prove that for each $r\neq 0$ there exists $t\in\Delta(E)\setminus\{0\}$ such that $tr\in\Delta(E).$ Since $t\in\Delta(E)$ if and only if $\nu(t)>0$, we must show that for any $r\in{\mathbb{F}}_{q}^{*}$,

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We shall need the following standard Fourier analytic preliminaries. Given $f:{\mathbb{F}}_{q}^{d}\to{\mathbb{C}}$, define the Fourier transform $\widehat{f}$ by the formula

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where $\chi$ is a non-trivial principal character on ${\mathbb{F}}_{q}$. We shall use the following calculation repeatedly.

Lemma 2.1**.**

With the notation above,

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and

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By Fourier inversion,

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It follows that for $r\in\mathbb{F}_{q}^{*}$,

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We shall invoke the explicit value of $|S_{t}|$ which can be deduced by Theorem 6.26 in [7].

Lemma 2.2**.**

Let $S_{t}\subset\mathbb{F}_{q}^{d}$ denote the sphere with radius $t\in\mathbb{F}_{q}.$ Then if $d\geq 2$ is even,

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*where $\eta$ is the quadratic character of $\mathbb{F}_{q}^{*},$ $\lambda(t)=-1$ for $t\in\mathbb{F}_{q}^{*},$ and $\lambda(0)=q-1.$

We also use the following result which was given as Lemma 4 in [5].

Lemma 2.3**.**

Let $S_{j}$ be a sphere in ${\mathbb{F}}_{q}^{d}$, $d\geq 2$. Then for any $m\in{\mathbb{F}}_{q}^{d},$ we have

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where $G$ denotes the Gauss sum, $\eta$ is the quadratic character of $\mathbb{F}_{q}^{*},$ and $\delta_{0}(m)=1$ if $m=(0,\ldots,0)$ and $\delta_{0}(m)=0$ otherwise.

2.1. A lower bound of $\displaystyle\sum_{t\in\mathbb{F}_{q}}\nu(t)\nu(rt)$ for even dimensions $d\geq 2$

Since $\displaystyle\sum_{t\in\mathbb{F}_{q}}\lambda(rt)=0$ for $r\neq 0$, it follows from Lemma 2.2 that

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Hence, we obtain

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In order to estimate the remaining terms, we need the following calculations.

Lemma 2.4**.**

Suppose that $m\not=\vec{0}$ in $\mathbb{F}_{q}^{d},d\geq 2.$ Then for any $r\neq 0,$ we have

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and

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where $\lambda(t)$ is defined as in Lemma 2.2.

To see this, observe that the left hand side of (2.4) equals

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since $m\not=(0,\ldots,0)$. Hence the equation (2.4) follows. By the definition of $\lambda(t)$,

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Then the equation (2.5) follows by (2.4). This completes the proof of Lemma 2.4.

We shall also need the following orthogonality lemma.

Lemma 2.5**.**

Suppose that $r\in\mathbb{F}_{q}^{*}$ and $m,m^{\prime}\in\mathbb{F}_{q}^{d}.$ If $d\geq 2$ is even, then we have

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The proof shall be given at the end of the paper (see Lemma 4.2). With the lemmas in tow, we are ready to handle terms $II,III$ and $IV$. In view of Lemmas 2.2 and 2.4, if $m^{\prime}\neq\vec{0},$ then

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Using this equation, it follows that

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By the same argument, it is not difficult to see that $II=III$. Namely, we have

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We now move on to term $\displaystyle IV:=q^{4d}\sum_{m,m^{\prime}\neq\vec{0}}|\widehat{E}(m)|^{2}|\widehat{E}(m^{\prime})|^{2}\sum_{t\in\mathbb{F}_{q}}\widehat{S_{t}}(m)\widehat{S_{rt}}(m^{\prime})$. Using Lemma 2.5, we can write $IV=A+B$, where

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and

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It follows that

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Combining this with equations (2.3), (2.6), we obtain that if $d\geq 2$ is even and $r\neq 0,$ then

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Notice that each term above is a real number. It follows that

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where we used the Plancherel theorem which states $\displaystyle\sum_{m\in\mathbb{F}_{q}^{d}}|\widehat{E}(m)|^{2}=q^{-d}|E|.$ By definitions of $A^{\prime}$ and $B^{\prime}$, we see that

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We also see from Lemma 2.3 that if $d\geq 2$ is even, then

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Thus we conclude that if $d\geq 2$ is even and $r\neq 0$, then

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2.2. An upper bound of $\nu^{2}(0)$ for even dimensions $d\geq 2$

It follows that

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By Lemma 2.3, notice that if $d\geq 2$ is even, then

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Then we see that

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By the Plancherel theorem and the orthogonality of $\chi$,

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Since $\widehat{E}(\vec{0})=q^{-d}|E|,$ we can write

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We shall use the following explicit form of the Gauss sum $G.$

Lemma 2.6** ([7], Theorem 5.15).**

Let $\mathbb{F}_{q}$ be a finite field with $q=p^{\ell}$ for an odd prime $p$ and $\ell\in\mathbb{N}.$ Then the Gauss sum $G$ satisfies that

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Observe from Lemma 2.6 that if the dimension $d$ is even, then $G^{d}=\pm q^{d/2}$ where the sign depends on $d$and $q$. Combining this observation with (2.8), and considering the sign of each term, we see that if $d$ is even, then

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Assuming that $|E|\geq q^{d/2},$ we see that

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Since $\nu(0)$ is a non-negative real number, it follows that if $|E|\geq q^{d/2},$ then

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Since $\sum\limits_{\|m\|=0,m\neq\vec{0}}|\widehat{E}(m)|^{2}\leq\sum\limits_{m\in\mathbb{F}_{q}^{d}}|\widehat{E}(m)|^{2}=q^{-d}|E|,$ we conclude that if $|E|\geq q^{d/2}$, then

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Now we are ready to complete the proof of Theorem 1.1.

2.3. Complete proof of Theorem 1.1

We must show that (2.1) holds. By (2.7) and (2.9), it is enough to show that if $|E|\geq 9q^{d/2},$ then

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It suffices to show that

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If $|E|\geq 9q^{d/2}$, then we see that

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so it is sufficient to show that

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Observe that if $|E|\geq 9q^{d/2}(\geq\sqrt{12}q^{d/2})$, then

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Consequently, it suffices to show that

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which holds if $q>16.$ Therefore, when $q\leq 16$, it suffices to prove the statement of Theorem 1.1. More precisely, it remains to show that if $|E|\geq 9q^{d/2}$ and $q\leq 16$, then $\mathbb{F}_{q}=\frac{\Delta(E)}{\Delta(E)}.$ Since $9q^{d/2}>2q^{(d+1)/2}$ for $q\leq 16,$ it will be enough to prove that if $|E|>2q^{(d+1)/2}$, then $\Delta(E)=\mathbb{F}_{q}.$ In [6], this was proved by the first listed author and Misha Rudnev. Thus the proof of Theorem 1.1 is complete.

3. Proof of Theorem 1.2

We proceed as in the proof of Theorem 1.1. As seen in (2.2), for $r\in\mathbb{F}_{q}^{+}$, we can write that

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The following explicit value of $|S_{t}|$ is given as Theorem 6.27 in [7].

Lemma 3.1**.**

Let $S_{t}\subset\mathbb{F}_{q}^{d}$ denote the sphere with radius $t\in\mathbb{F}_{q}.$ If $d\geq 3$ is odd, then

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where $\eta$ denotes the quadratic character of $\mathbb{F}_{q}^{*}$ and $\eta(0)=0.$

We record from Lemma 2.3 that if $d\geq 3$ is odd, then for any $m\in{\mathbb{F}}_{q}^{d},$

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3.1. Estimate of $\displaystyle\sum_{t\in\mathbb{F}_{q}}\nu(t)\nu(rt)$ for odd dimensions $d\geq 3$

Since $\displaystyle\sum_{t\in\mathbb{F}_{q}^{*}}\eta(t)=0$ (by the orthogonality of $\eta$) and $\eta(0)=0,$ it follows from Lemma 3.1 that

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[TABLE]

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Since $\eta(r)=1$ (by the assumption that $r\in\mathbb{F}_{q}^{+}$), we have

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In order to estimate the second term $\mbox{II},$ we begin by proving the following result.

Lemma 3.2**.**

Let $S_{j}$ be the sphere in $\mathbb{F}_{q}^{d}$ for odd $d\geq 3.$ Then for $r\neq 0$ and $m\neq\vec{0}$, we have

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To prove this lemma, recall from (2.4) of Lemma 2.4 that $\displaystyle\sum_{t\in\mathbb{F}_{q}}\widehat{S}_{rt}(m)=0$ for $r\neq 0$ and $m\neq\vec{0}.$ By Lemma 3.1,

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By using the value of $\widehat{S_{rt}}(m)$ in (3.1), we can write

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Since $\eta(0)=0$ and $\eta(a)=\eta(a^{-1})$ for $a\neq 0,$ a simple change of variables yields

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and thus we have

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which completes the proof of Lemma 3.2.

By Lemma 3.2 and the orthogonality of $\chi,$ we see that

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Now observe from Lemma 2.6 that $G^{d+1}\in\mathbb{R}$ for odd $d$ and so both $\mbox{II}_{1}$ and $\mbox{II}_{2}$ are real numbers. Furthermore, both values ​​are real numbers with the same sign. Hence, $\mbox{II}=\mbox{II}_{1}-\mbox{II}_{2}\geq\min\{-|\mbox{II}_{1}|,-|\mbox{II}_{2}|\}.$ Since

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which is same as $-|E|^{3},$ we obtain that

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By the same argument, it is not hard to see that $II=III$ and we also have

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In order to estimate the forth term $\mbox{IV},$ we shall need the following orthogonality lemma.

Lemma 3.3**.**

Suppose that $r\in\mathbb{F}_{q}^{*}$ and $m,m^{\prime}\in\mathbb{F}_{q}^{d}.$ If $d\geq 3$ is odd, then we have

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where $\eta$ denotes the quadratic character of $\mathbb{F}_{q}^{*}.$

The proof shall be given at the end of the paper (see Lemma 4.2). By the definition of the term IV and Lemma 3.3, it follows that

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[TABLE]

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Since $\eta(r)=1$ (by our assumption that $r$ is a square number in $\mathbb{F}_{q}^{*}$), the second term above is positive. Thus we have

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By the Plancherel theorem,

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Putting this together with all other estimates, we obtain that if $d\geq 3$ is odd and $r$ is a square number, then

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3.2. Estimate of $\nu^{2}(0)$ for odd dimensions $d\geq 3$

Recall that we can write

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Since $|S_{0}|=q^{d-1}$ for odd $d\geq 3$ (see Lemma 3.1),

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To estimate $R$, observe that

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By the equation (3.1), we see that if $d\geq 3$ is odd and $m\neq\vec{0}$, then

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Since $\displaystyle\left|\sum_{s\neq 0}\eta(s)\chi\left(\frac{\|m\|}{4s}\right)\right|=\sqrt{q}$ for $\|m\|\neq 0$ and [math] otherwise, we see

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Hence we obtain

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We have seen that $\nu(0):=M+R\leq q^{-1}|E|^{2}+q^{\frac{d-1}{2}}|E|$ which in turn implies

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since $\nu(0)$ is a non-negative integer.

3.3. Complete proof of Theorem 1.2

Let $d\geq 3$ be odd. Suppose that $r$ is a square number in $\mathbb{F}_{q}^{*}.$ We must show that if $E\subset\mathbb{F}_{q}^{d}$ with $|E|\geq 6q^{d/2},$ then

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By (3.2) and (3.3), it will be enough to show that if $|E|\geq 6q^{d/2}$, then

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Note that to prove this it suffices to show that

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If $|E|\geq 6q^{d/2}(\geq 6q^{(d-1)/2})$, then we see that

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Hence it is sufficient to show that if $|E|\geq 6q^{d/2},$ then

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Observe that if $|E|\geq 6q^{d/2}(\geq\sqrt{24}q^{d/2}),$ then

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In conclusion, it is enough to prove that if $|E|\geq 6q^{d/2},$ then

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which is clearly true provided that $q>4.$ For this reason, it suffices to prove the statement of Theorem 1.2 in the case when $q\leq 4$ and $|E|\geq 6q^{d/2}.$ In other words, our task is to prove that if $|E|\geq 6q^{d/2}$ for $q\leq 4$, then $\mathbb{F}_{q}=\frac{\Delta(E)}{\Delta(E)}.$ Since $6q^{d/2}>2q^{(d+1)/2}$ for $q\leq 4$, it will be enough to show that if $|E|>2q^{(d+1)/2}$, then $\Delta(E)=\mathbb{F}_{q}.$ This is a well-known result on the Erdős-Falconer distance problem due to the first listed author and Misha Rudnev [6]. Thus we finish the proof of Theorem 1.2.

4. Proof of Lemmas 2.5 and 3.3

We begin by proving the following lemma.

Lemma 4.1**.**

Let $r\in\mathbb{F}_{q}^{*}$ and $m,m^{\prime}\in\mathbb{F}_{q}^{d}.$ Then we have

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Proof.

By Lemma 2.3, we have

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and

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Since $\displaystyle\sum_{t\in\mathbb{F}_{q}}A(t)D(t)=0=\sum_{t\in\mathbb{F}_{q}}B(t)C(t)$ by the orthogonality of $\chi,$ we have

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[TABLE]

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Thus the statement follows by the orthogonality of $\chi.$ ∎

As a corollary of Lemma 4.1, one can deduce Lemmas 2.5 and 3.3 which can be restated as follows.

Lemma 4.2**.**

Suppose that $r\in\mathbb{F}_{q}^{*}$ and $m,m^{\prime}\in\mathbb{F}_{q}^{d}.$ If $d\geq 2$ is even, then we have

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On the other hand, if $d\geq 3$ is odd, then we have

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Proof.

Suppose that $d\geq 2$ is even. Then $\eta^{d}=1.$ By Lemma 2.6, we see that $G^{2d}=q^{d}$ for even $d\geq 2.$ Thus the statement follows by Theorem 4.1.

Next, assume that $d\geq 3$ is odd. Then $\eta^{d}=\eta.$ Hence, by Theorem 4.1 it suffices to show that $G^{2d}\eta(-1)=q^{d}$ for odd $d\geq 3.$ This equality follows by combining Lemma 2.6 with the facts that $\eta(-1)=1$ for $q\equiv 1~{}(\mbox{mod}~{}4),$ and $\eta(-1)=-1$ for $q\equiv 3~{}(\mbox{mod}~{}4).$ ∎