On Classical Gauss Sums and Some of Their Properties

Abstract

The goal of this paper is to solve the computational problem of one kind rational polynomials of classical Gauss sums, applying the analytic means and the properties of the character sums. Finally, we will calculate a meaningful recursive formula for it.

1. Introduction

Let $q\ge 3$ be an integer. For any Dirichlet character $\chi modq$, according to the definition of classical Gauss sums $\tau \left(\chi \right)$, we can write

$$\begin{array}{c}\hfill \tau \left(\chi \right)=\sum _{a=1}^q\chi \left(a\right)e\left(\frac{a}{q}\right),\end{array}$$

where $e\left(y\right)=e^{2\pi iy}$.

Since this sum appears in numerous classical number theory problems, and it has a close connection with the trigonometric sums, we believe that classical Gauss sums play a crucial part in analytic number theory. Because of this phenomenon, plenty of experts have researched Gauss sums. Meanwhile, more conclusions have been obtained as regards their arithmetic properties. Such as the following results provided by Chen and Zhang [1]:

Let p be an odd prime with $p\equiv 1mod4$, $\lambda$ be any fourth-order character $modp$. Then one has the identity

$${\tau }^2\left(\lambda \right)+{\tau }^2\left(\overline{\lambda }\right)=\sqrt{p}·\sum _{a=1}^{p-1}\left(\frac{a+\overline{a}}{p}\right)=2\sqrt{p}·\alpha ,$$

where $\left(\frac{\ast }{p}\right)={\chi }_2$ denotes the the Legendre’s symbol $modp$ (please see Reference [1,2] for its definition and related properties), and ${\displaystyle \alpha =\sum _{a=1}^{{\displaystyle \frac{p-1}{2}}}\left({\displaystyle \frac{a+\overline{a}}{p}}\right)}$.

If p is a prime with $p\equiv 1mod3$, $\psi$ is any third-order character $modp$, then Zhang and Hu [3] had already obtained an analogous result (see Lemma 1). However, perhaps the most beautiful and important property of Gauss sums $\tau \left(\chi \right)$ is that $\left|\tau \left(\chi \right)\right|=\sqrt{q}$, for any primitive character $\chi modq$.

Reference [2] and References [4,5,6,7,8,9,10,11,12,13] have a good deal of various elementary properties of Gauss sums. In this paper, the following rational polynomials of Gauss sums attract our attention.

$$\begin{array}{c}\hfill U_k(p,\chi )=\frac{{\tau }^{3k}\left(\chi \right)}{{\tau }^{3k}\left(\overline{\chi }\right)}+\frac{{\tau }^{3k}\left(\overline{\chi }\right)}{{\tau }^{3k}\left(\chi \right)},\end{array}$$

(1)

where p is an odd prime, k is a non-negative integer, $\chi$ is any non-principal character $modp$.

Observing the basic properties of Equation (1), we noticed that hardly anyone had published research in any academic papers to date. We consider that the question is significant. In addition, the regularity of the value distribution of classical Gauss sums could be better revealed. Presently, we will explain certain properties discovered in our investigation. See that $U_k(p,\chi )$ has some good properties. In fact, for some special character $\chi modp$, the second-order linear recurrence formula for $U_k(p,\chi )$ for all integers $k\ge 0$ may be found similarly.

The goal of this paper is to use the analytic method and the properties of the character sums to solve the computational problem of $U_k(p,\chi )$, and to calculate two recursive formulae, which are listed hereafter:

Theorem 1.

Let p be a prime with $p\equiv 1mod12$, ψ be any third-order character $modp$. Then, for any positive integer k, we can deduce the following second-order linear recursive formulae

$$U_{k+1}(p,\psi )=\frac{d^2-2p}{p}·U_k(p,\psi )-U_{k-1}(p,\psi ),$$

where the initial values $U_0(p,\psi )=2$ and $U_1(p,\psi )=\frac{d^2-2p}{p}$, d is uniquely determined by $4p=d^2+27b^2$ and $d\equiv 1mod3$.

So we can deduce the general term

$$U_k(p,\psi )={\left(\frac{d^2-2p+3dbi\sqrt{3}}{2p}\right)}^k+{\left(\frac{d^2-2p-3dbi\sqrt{3}}{2p}\right)}^k,i^2=-1.$$

Theorem 2.

Let p be a prime with $p\equiv 7mod12$, ψ be any third-order character $modp$. Then, for any positive integer k, we will obtain the second-order linear recursive formulae

$$U_{k+1}(p,\psi )=\frac{i\left(2p-d^2\right)}{p}·U_k(p,\psi )-U_{k-1}(p,\psi ),$$

where the initial values $U_0(p,\psi )=2$, $U_1(p,\psi )=\frac{i\left(2p-d^2\right)}{p}$ and $i^2=-1$.

Similarly, we can also deduce the general term

$$U_k(p,\psi )=i^k{\left(\frac{2p-d^2+\sqrt{8p^2-4p{d}^2+d^4}}{2p}\right)}^k+i^k{\left(\frac{2p-d^2-\sqrt{8p^2-4p{d}^2+d^4}}{2p}\right)}^k.$$

2. Several Lemmas

We have used five simple and necessary lemmas to prove our theorems. Hereafter, we will apply relevant properties of classical Gauss sums and the third-order character $modp$, all of which can be found in books concerning elementary and analytic number theory, such as in References [2,10], so we will not duplicate the related contents.

Lemma 1.

If p is any prime with $p\equiv 1mod3$, ψ is any third-order character $modp$, then, we have the equation

$${\tau }^3\left(\psi \right)+{\tau }^3\left(\overline{\psi }\right)=dp,$$

where $\tau \left(\psi \right)$ denotes the classical Gauss sums, d is uniquely determined by $4p=d^2+27b^2$ and $d\equiv 1mod3$.

Proof. 

See References [3] or [8]. □

Lemma 2.

Let p be a prime with $p\equiv 1mod3$, ψ be any third-order character $modp$, ${\chi }_2=\left(\frac{\ast }{p}\right)$ denotes the Legendre’s symbol $modp$. The following identity holds

$$\begin{array}{c}\hfill {\tau }^2\left(\overline{\psi }\right)=\left(\frac{-1}{p}\right)\psi \left(4\right)\tau \left({\chi }_2\right)\tau \left(\psi {\chi }_2\right).\end{array}$$

Proof. 

Firstly, using the properties of Gauss sums, we get

$$\begin{array}{cc}& \sum _{a=1}^{p-1}\overline{\psi }\left(a(a+1)\right)=\frac{1}{\tau \left(\psi \right)}\sum _{b=1}^{p-1}\psi \left(b\right)\sum _{a=1}^{p-1}\overline{\psi }\left(a\right)e\left(\frac{b(a+1)}{p}\right)\hfill \\ =\hfill & \frac{{\tau }^2\left(\overline{\psi }\right)}{\tau \left(\psi \right)}=\frac{{\tau }^3\left(\overline{\psi }\right)}{p}.\hfill \end{array}$$

(2)

On the other side, we get the sums

$$\begin{array}{cc}& \sum _{a=1}^{p-1}\overline{\psi }\left(a(a+1)\right)=\psi \left(4\right)\sum _{a=0}^{p-1}\overline{\psi }\left(4a^2+4a\right)\hfill \\ =\hfill & \psi \left(4\right)\sum _{a=0}^{p-1}\overline{\psi }\left({(2a+1)}^2-1\right)=\psi \left(4\right)\sum _{a=0}^{p-1}\overline{\psi }\left(a^2-1\right)\hfill \\ =\hfill & \frac{\psi \left(4\right)}{\tau \left(\psi \right)}\sum _{b=1}^{p-1}\psi \left(b\right)\sum _{a=0}^{p-1}e\left(\frac{b(a^2-1)}{p}\right)=\frac{\psi \left(4\right)}{\tau \left(\psi \right)}\sum _{b=1}^{p-1}\psi \left(b\right)e\left(\frac{-b}{p}\right)\sum _{a=0}^{p-1}e\left(\frac{b{a}^2}{p}\right)\hfill \\ =\hfill & \frac{\psi \left(4\right)\tau \left({\chi }_2\right)}{\tau \left(\psi \right)}\sum _{b=1}^{p-1}\psi \left(b\right){\chi }_2\left(b\right)e\left(\frac{-b}{p}\right)=\frac{\psi \left(4\right){\chi }_2(-1)\tau \left({\chi }_2\right)\tau \left(\psi {\chi }_2\right)}{\tau \left(\psi \right)}.\hfill \end{array}$$

(3)

Combining Equations (2) and (3), we obtain

$$\begin{array}{c}\hfill {\tau }^2\left(\overline{\psi }\right)=\left(\frac{-1}{p}\right)\psi \left(4\right)\tau \left({\chi }_2\right)\tau \left(\psi {\chi }_2\right).\end{array}$$

Now, Lemma 2 has been proved. □

Lemma 3.

Let p be a prime with $p\equiv 1mod6$, χ be any sixth-order character $modp$. Then, about classical Gauss sums $\tau \left(\chi \right)$, the following holds:

$${\tau }^3\left(\chi \right)+{\tau }^3\left(\overline{\chi }\right)=\left\{\begin{array}{cc}{\displaystyle p^{\frac{1}{2}}\left(d^2-2p\right)}\hfill & \mathit{if}p=12h+1,\hfill \\ {\displaystyle -i·p^{\frac{1}{2}}\left(d^2-2p\right)}\hfill & \mathit{if}p=12h+7,\hfill \end{array}\right.$$

where $i^2=-1$, d is uniquely determined by $4p=d^2+27b^2$ and $d\equiv 1mod3$.

Proof. 

Since $p\equiv 1mod6$, $\psi$ is a third-order character $modp$. Any sixth-order character $\chi modp$ can be denoted as $\chi =\psi {\chi }_2$ or $\chi =\overline{\psi }{\chi }_2$. Note that ${\psi }^3\left(4\right)=1$, ${\overline{\psi }}^3\left(4\right)=1$ and ${\chi }_2^3={\chi }_2$, from Lemma 2 we deduce

$$\begin{array}{c}\hfill {\tau }^6\left(\overline{\psi }\right)=\left(\frac{-1}{p}\right){\tau }^3\left({\chi }_2\right){\tau }^3\left(\psi {\chi }_2\right)\end{array}$$

(4)

and

$$\begin{array}{c}\hfill {\tau }^6\left(\psi \right)=\left(\frac{-1}{p}\right){\tau }^3\left({\chi }_2\right){\tau }^3\left(\overline{\psi }{\chi }_2\right).\end{array}$$

(5)

Adding Equations (4) and (5), and then applying Lemma 1 we have

$$\begin{array}{cc}& \left(\frac{-1}{p}\right){\tau }^3\left({\chi }_2\right)\left({\tau }^3\left(\psi {\chi }_2\right)+{\tau }^3\left(\overline{\psi }{\chi }_2\right)\right)={\tau }^6\left(\overline{\psi }\right)+{\tau }^6\left(\psi \right)\hfill \\ =\hfill & {\left({\tau }^3\left(\overline{\psi }\right)+{\tau }^3\left(\psi \right)\right)}^2-2p^3=d^2{p}^2-2p^3.\hfill \end{array}$$

(6)

Note that ${\chi }_2$ is a real character $modp$, $\overline{\psi }{\chi }_2=\overline{\psi {\chi }_2}$, and $\tau \left({\chi }_2\right)=\sqrt{p}$. If $p\equiv 1mod4$; $\tau \left({\chi }_2\right)=i·\sqrt{p}$, $i^2=-1$, if $p\equiv 3mod4$. From Equation (6) we may immediately prove the sum

$$\begin{array}{c}\hfill {\tau }^3\left(\psi {\chi }_2\right)+{\tau }^3\left(\overline{\psi }{\chi }_2\right)=\left\{\begin{array}{cc}{\displaystyle p^{\frac{1}{2}}\left(d^2-2p\right)}\hfill & \mathrm{if}p=12h+1,\hfill \\ {\displaystyle -i·p^{\frac{1}{2}}\left(d^2-2p\right)}\hfill & \mathrm{if}p=12h+7.\hfill \end{array}\right.\end{array}$$

(7)

Let $\chi =\psi {\chi }_2$, then $\chi$ is a sixth-order character $modp$ and $\overline{\psi }{\chi }_2=\overline{\chi }$. From Equation (7) we can deduce the sum term

$$\begin{array}{c}\hfill {\tau }^3\left(\chi \right)+{\tau }^3\left(\overline{\chi }\right)=\left\{\begin{array}{cc}{\displaystyle p^{\frac{1}{2}}\left(d^2-2p\right)}\hfill & \mathrm{if}p=12h+1,\hfill \\ {\displaystyle -i·p^{\frac{1}{2}}\left(d^2-2p\right)}\hfill & \mathrm{if}p=12h+7.\hfill \end{array}\right.\end{array}$$

The proof of Lemma 3 has been completed. □

Lemma 4.

Let p be a prime with $p\equiv 7mod12$, ψ be any three-order character $modp$. Then, we compute the sum term

$$\frac{{\tau }^3\left(\overline{\psi }\right)}{{\tau }^3\left(\psi \right)}+\frac{{\tau }^3\left(\psi \right)}{{\tau }^3\left(\overline{\psi }\right)}=\frac{i·(2p-d^2)}{p}.$$

Proof. 

Let $\psi$ be a three-order character $modp$. Then, for any six-order character $\chi modp$, we must have $\chi =\psi {\chi }_2$ or $\chi =\overline{\chi }{\chi }_2$. Without loss of generality we suppose that $\chi =\psi {\chi }_2$, then note that $\psi (-1)=1$, ${\chi }_2(-1)=-1$ and Theorem 7.5.4 in Reference [10], we acquire

$$\sum _{a=0}^{p-1}e\left(\frac{b{a}^2}{p}\right)={\chi }_2\left(b\right)·\sqrt{p},(p,b)=1.$$

Using the properties of Gauss sums we can write

$$\begin{array}{cc}& \sum _{a=0}^{p-1}\chi \left(a^2-1\right)=\frac{1}{\tau \left(\overline{\chi }\right)}\sum _{b=1}^{p-1}\overline{\chi }\left(b\right)\sum _{a=0}^{p-1}e\left(\frac{b(a^2-1)}{p}\right)\hfill \\ =\hfill & \frac{1}{\tau \left(\overline{\chi }\right)}\sum _{b=1}^{p-1}\overline{\chi }\left(b\right)e\left(\frac{-b}{p}\right)\sum _{a=0}^{p-1}e\left(\frac{b{a}^2}{p}\right)=\frac{\sqrt{p}}{\tau \left(\overline{\chi }\right)}\sum _{b=1}^{p-1}\overline{\chi }\left(b\right){\chi }_2\left(b\right)e\left(\frac{-b}{p}\right)\hfill \\ =\hfill & \frac{\overline{\chi }(-1){\chi }_2(-1)\sqrt{p}\tau \left(\overline{\chi }{\chi }_2\right)}{\tau \left(\overline{\chi }\right)}=\frac{\sqrt{p}\tau \left(\overline{\chi }{\chi }_2\right)}{\tau \left(\overline{\chi }\right)}.\hfill \end{array}$$

(8)

Noting that ${\overline{\chi }}^2={\overline{\psi }}^2=\psi$, we can deduce

$$\begin{array}{cc}& \sum _{a=0}^{p-1}\chi \left(a^2-1\right)=\sum _{a=0}^{p-1}\chi \left({(a+1)}^2-1\right)=\sum _{a=1}^{p-1}\chi \left(a\right)\chi (a+2)\hfill \\ =\hfill & \frac{1}{\tau \left(\overline{\chi }\right)}\sum _{b=1}^{p-1}\overline{\chi }\left(b\right)\sum _{a=1}^{p-1}\chi \left(a\right)e\left(\frac{b(a+2)}{p}\right)=\frac{\tau \left(\chi \right)}{\tau \left(\overline{\chi }\right)}\sum _{b=1}^{p-1}{\overline{\chi }}^2\left(b\right)e\left(\frac{2b}{p}\right)\hfill \\ =\hfill & \frac{\overline{\psi }\left(2\right)\tau \left(\chi \right)\tau \left(\psi \right)}{\tau \left(\overline{\chi }\right)}.\hfill \end{array}$$

(9)

Obviously, $\overline{\chi }{\chi }_2=\overline{\psi }$ and ${\psi }^3\left(2\right)=1$, applying Equations (8) and (9) we have

$$\begin{array}{c}\hfill {\tau }^3\left(\chi \right)=p^{\frac{3}{2}}·\frac{{\tau }^3\left(\overline{\psi }\right)}{{\tau }^3\left(\psi \right)}.\end{array}$$

(10)

Similarly, we can see

$$\begin{array}{c}\hfill {\tau }^3\left(\overline{\chi }\right)=p^{\frac{3}{2}}·\frac{{\tau }^3\left(\psi \right)}{{\tau }^3\left(\overline{\psi }\right)}.\end{array}$$

(11)

Combining Equation (10), Equation (11) and Lemma 3 we compute

$$\begin{array}{c}\hfill \frac{{\tau }^3\left(\psi \right)}{{\tau }^3\left(\overline{\psi }\right)}+\frac{{\tau }^3\left(\overline{\psi }\right)}{{\tau }^3\left(\psi \right)}=\frac{1}{p^{\frac{3}{2}}}\left({\tau }^3\left(\chi \right)+{\tau }^3\left(\overline{\chi }\right)\right)=\frac{i·(2p-d^2)}{p}.\end{array}$$

This completes the proof of Lemma 4. □

Lemma 5.

Let p be a prime with $p\equiv 1mod12$, ψ be any three-order character $modp$. Then, we obtain the sum term

$$\frac{{\tau }^3\left(\overline{\psi }\right)}{{\tau }^3\left(\psi \right)}+\frac{{\tau }^3\left(\psi \right)}{{\tau }^3\left(\overline{\psi }\right)}=\frac{d^2-2p}{p}.$$

Proof. 

From Lemma 3 and the method of proving Lemma 4 we can easily deduce Lemma 5. □

3. Proofs of the Theorems

In this section, we prove our two theorems. For Theorem 1, since $p\equiv 1mod12$, $\psi$ is a third-order character $modp$, then for any positive integer k, let

$$U_k\left(p\right)=\frac{{\tau }^{3k}\left(\psi \right)}{{\tau }^{3k}\left(\overline{\psi }\right)}+\frac{{\tau }^{3k}\left(\overline{\psi }\right)}{{\tau }^{3k}\left(\psi \right)}.$$

From Lemma 5 we have

$$\begin{array}{c}\hfill U_1\left(p\right)=\frac{{\tau }^3\left(\overline{\psi }\right)}{{\tau }^3\left(\psi \right)}+\frac{{\tau }^3\left(\psi \right)}{{\tau }^3\left(\overline{\psi }\right)}=\frac{d^2-2p}{p}\end{array}$$

(12)

and

$$\begin{array}{cc}& \frac{d^2-2p}{p}·U_k\left(p\right)=U_k\left(p\right)U_1\left(p\right)=\left(\frac{{\tau }^{3k}\left(\psi \right)}{{\tau }^{3k}\left(\overline{\psi }\right)}+\frac{{\tau }^{3k}\left(\overline{\psi }\right)}{{\tau }^{3k}\left(\psi \right)}\right)·\left(\frac{{\tau }^3\left(\overline{\psi }\right)}{{\tau }^3\left(\psi \right)}+\frac{{\tau }^3\left(\psi \right)}{{\tau }^3\left(\overline{\psi }\right)}\right)\hfill \\ =\hfill & \frac{{\tau }^{3k+3}\left(\overline{\psi }\right)}{{\tau }^{3k+3}\left(\psi \right)}+\frac{{\tau }^{3k+3}\left(\psi \right)}{{\tau }^{3k+3}\left(\overline{\psi }\right)}+\frac{{\tau }^{3k-3}\left(\overline{\psi }\right)}{{\tau }^{3k-3}\left(\psi \right)}+\frac{{\tau }^{3k-3}\left(\psi \right)}{{\tau }^{3k-3}\left(\overline{\psi }\right)}=U_{k+1}\left(p\right)+U_{k-1}\left(p\right).\hfill \end{array}$$

(13)

Combining Equations (12) and (13) we may immediately compute the second-order linear recursive formula

$$\begin{array}{c}\hfill U_{k+1}\left(p\right)=\frac{d^2-2p}{p}·U_k\left(p\right)-U_{k-1}\left(p\right)\end{array}$$

(14)

with initial values $U_0\left(p\right)=2$ and $U_1\left(p\right)=\frac{d^2-2p}{p}$.

Note that the two roots of the equation ${\lambda }^2-\frac{d^2-2p}{p}\lambda +1=0$ are

$${\lambda }_1=\frac{d^2-2p+3dbi\sqrt{3}}{2p}\mathrm{and}{\lambda }_2=\frac{d^2-2p-3dbi\sqrt{3}}{2p}.$$

So from Equation (14) and its initial values we may immediately deduce the general term

$$U_k(p,\psi )={\left(\frac{d^2-2p+3dbi\sqrt{3}}{2p}\right)}^k+{\left(\frac{d^2-2p-3dbi\sqrt{3}}{2p}\right)}^k,$$

where $i^2=-1$. Now Theorem 1 has been finished.

Similarly, from Lemma 4 and the method of proving Theorem 1 we can easily obtain Theorem 2. Now, we have completed all the proofs of our Theorems.

4. Conclusions

The main results of this paper are Theorem 1 and 2. They give a new second-order linear recurrence formula for Equation (1) with the third-order character $\psi modp$. Therefore, we can calculate the exact value of Equation (1). Note that $|\tau \left(\overline{\psi }\right)/\tau \left(\psi \right)|=1$, so $\tau \left(\overline{\psi }\right)/\tau \left(\psi \right)$ is a unit root, thus, the results in this paper profoundly reveal the distributional properties of two different Gauss sums quotients on the unit circle.

For the other $modp$ characters, for example, the fifth-order character $\chi modp$ with $p\equiv 1mod5$, we naturally ask whether there exists a similar formula as presented in our theorems. This is still an open problem. It will be the content of our future investigations.