The number of rational points of certain quartic diagonal hypersurfaces over finite fields

1.   Introduction

Let $\mathbb{Z}$ and $\mathbb{Z}^+$ stand for the set of all integers and the set of all positive integers. In this paper, we always let $p$ be an odd prime and $\mathbb{F}_q$ be a finite field of $q = p^s$ elements with $s\in \mathbb Z^+$. Let $f(x_1, \cdots, x_n)$ be a polynomial with $n$ variables in $\mathbb{F}_q$. We set $N(f = 0)$ to be the number of $\mathbb{F}_q$-rational points of the affine hypersurface $f(x_1, \cdots, x_n) = 0$. That is, one has

Calculating the exact value of $N(f = 0)$ is a main topic in finite fields. Generally speaking, it is difficult to give an exact formula for $N(f = 0)$. Studying the explicit formula for $N(f = 0)$ under certain conditions has attracted a lot of authors for many years. See, for instance, ^{[1,2,4,6,7,8,9,10,11,12,16,17,18,19,20,21,22,23,24]}.

In 1977, Chowla, Cowles and Cowles ^[8] got a formula for the rational points of the hypersurface

in $\mathbb{F}_p$. Recently, Zhang and Hu ^[23] presented an explicit formula for the rational points of the hypersurface

In fact, let $\tilde N(c)$ be the rational points of $x_1^3+x_2^3+x_3^3+x_4^3 = c, \ c\in \mathbb{F}_p^*$ with $p\equiv 1\pmod 3$, and $\mathbb{F}_p^{\ast} = \langle g\rangle$ with $g$ being a generator of $\mathbb{F}_p^{\ast}$. Then they showed that

On the other hand, Myerson ^[17] extended the result in ^[8] to the field $\mathbb{F}_q$ and first studied the rational points of the hypersurface

over $\mathbb{F}_q$. Now let $q\equiv 3\pmod 4$. Then $p\equiv 3\pmod 4$ and $\gcd(4, q-1) = 2$. Noticing the fact

it follows from Theorems 6.26 and 6.27 in ^[15] that the following result is true.

Theorem 1.1. Let $q\equiv 3\pmod 4$ and $f(x_1, \cdots, x_n) = x_1^4+\cdots+x_n^4-c\in\mathbb{F}_q[x]$ with $c\in\mathbb{F}_q^*$. If $n$ is even, then

and if $n$ is odd, then

where $\eta$ is the quadratic character of $\mathbb{F}_q$.

By Theorem 1.1, it remains to consider the case $q\equiv 1\pmod 4$. When $q\equiv 1\pmod 4$, it is difficult to give an explicit formula for $N(x_1^4+\cdots+x_n^4 = c)$ with $c\in\mathbb{F}_q^*$ in general. Recently, Zhao and Zhao ^[24] presented an explicit formula for the number of rational points of the hypersurface $x_1^4+x_2^4 = c$ with $c\in\mathbb{F}_q^*.$

In this paper, we investigate the number of rational points of the following quartic diagonal hypersurface: $f_1(x_1, x_2, x_3) = c$ and $f_2(x_1, x_2, x_3) = c$, where $c\in\mathbb{F}_q^*$ and

and

For any primitive element $g$ of $\mathbb{F}_q^*$, we define the index of $c$ with respect to $g$, denoted by ${\rm ind}_g(c)$, to be the unique integer $r\in [1, q-1]$ such that $c = g^r$ (see, for instance, ^[3]). The first main result of this paper can be stated as follows.

Theorem 1.2. Let $F = \mathbb{F}_q$ be a finite field with $q = p^s$ and $c\in \mathbb{F}_q^*.$ Let $g$ be a primitive element of $\mathbb{F}_q^*$.

$\rm(i).$ If either $p\equiv1\pmod8$ or $p\equiv5\pmod8$ and $s$ is even, then

If $p\equiv5\pmod8$ and $s$ is odd, then

where $a+b{\rm i} = (a'+b'{\rm i})^s$ with ${\rm i}^2 = -1$ and $a'$ and $b'$ being integers such that

$\rm(ii).$ Let $p\equiv 3\pmod 4$ and $q\equiv 1\pmod 4.$ If $q\equiv 1\pmod 8,$ then

If $q\equiv 5\pmod 8,$ then

where $r$ is uniquely determined by $q = r^2+4t^2$ with $r\equiv1\pmod4.$

From Theorem 1.1, the following result follows immediately.

Corollary 1.1. Let $F = \mathbb{F}_q$ be a finite field with $q = p^s$ such that $q\equiv 1 \pmod 4$ and $c\in \mathbb{F}_q^*.$ Then $N(f_1 = c) = q^2+O(q).$

Consequently, the second main result of this paper can be stated as follows.

Theorem 1.3. Let $F = \mathbb{F}_q$ be a finite field with $q = p^s$ and $c\in \mathbb{F}_q^*.$ Let $g$ be a primitive element of $\mathbb{F}_q^*$.

$\rm(i).$ Let $p\equiv 1\pmod 4.$ If $s$ is even, then

If $p\equiv 1 \pmod 8$ and $s$ is odd, then

If $p\equiv 5 \pmod 8$ and $s$ is odd, then

with $a+b{\rm i} = (a'+b'{\rm i})^s,$ where $a'$ and $b'$ are integers such that

$\rm(ii).$ Let $p\equiv 3\pmod 4$ and $q\equiv 1\pmod 4.$ If $q\equiv 1\pmod 8,$ then

If $q\equiv 5\pmod 8,$ then

where $r$ is uniquely determined by $q = r^2+4t^2$ with $r\equiv1\pmod4.$

From Theorem 1.2, one can easily deduce the following result.

Corollary 1.2. Let $F = \mathbb{F}_q$ be a finite field with $q = p^s$ such that $q\equiv 1 \pmod 4$ and $c\in \mathbb{F}_q^*.$ Then $N(f_2 = c) = q^3+O(q^{\frac{3}{2}}).$

This paper is organized as follows. First of all, in Section 2, we present several basic concepts including the Gauss sums and Jacobi sums, and give some preliminary lemmas. Then in Section 3, we give the proofs of Theorems 1.1 and 1.2. Finally, in Section 4, we supply some examples to illustrate the validity of our main results.

2.   Preliminary lemmas

In this section, we present several auxiliary lemmas that are needed in the proofs of Theorems 1.1 and 1.2.

Let $\alpha$ be an element of $\mathbb{F}_{q}$. Then the trace and norm of $\alpha$ relative to $\mathbb{F}_p$ are defined by (see, for example, ^[13,14,15])

and

respectively. For the simplicity, we write $\text{Tr}(\alpha)$ and $\mathbb{N}(\alpha)$ for $\text{Tr}_{\mathbb{F}_{q}/{\mathbb{F}_p}}(\alpha)$ and $\mathbb{N}_{\mathbb{F}_{q}/{\mathbb{F}_p}}(\alpha)$, respectively. Let $\chi$ be a multiplicative character of $\mathbb{F}_q$ and $\psi$ an additive character of $\mathbb{F}_q$. Then we define the Gauss sum $G(\chi, \psi)$ by

Let $\chi _1$ and $\chi _2$ be multiplicative characters of $\mathbb{F}_q$. Then the sum

is called a Jacobi sum in $\mathbb{F}_q$. The readers are referred to ^[5] for the basic facts on Jacobi sum.

The character $\psi_0$ represents the trivial additive character such that $\psi_0(x) = 1$ for all $x\in\mathbb{F}_q$ and $\chi _0$ represents the trivial multiplicative character such that $\chi _0(x) = 1$ for all $x\in\mathbb{F}_q$. For any $x\in \mathbb{F}_q$, let

Then we call $\psi_1$ the canonical additive character of $\mathbb{F}_q$. Let $a\in \mathbb{F}_q$. Then we define

for all $x\in \mathbb{F}_q$. For each character $\psi$ of $\mathbb{F}_q$ there is associated the conjugate character $\overline{\psi}$ defined by $\overline{\psi}(x) = \overline{\psi(x)}$ for all $x\in\mathbb{F}_q$.

We give several basic identities about Gauss sums as follows.

Lemma 2.1. ^[13,15] Each of the following is true:

$\rm(i).$ $G(\chi, \psi_{ab}) = \overline{\chi (a)}G(\chi, \psi_b)$ for $a\in \mathbb{F} _q^*$, $b\in \mathbb{F} _q$.

$\rm(ii).$ $G(\overline\chi, \psi) = \chi (-1)\overline {G(\chi, \psi)}.$

$\rm(iii).$ $\left|G(\chi, \psi)\right| = \sqrt{q}$ for $\chi \ne\chi _0$ and $\psi\ne\psi_0.$

Lemma 2.2. ^[13,15] Let $\mathbb{F}_q$ be a finite field with $q = p^s$ and $\eta$ be the quadratic character of $\mathbb{F}_q$. Then

If $\chi _1$ and $\chi _2$ are nontrivial, there exists an important connection between Jacobi sums and Gauss sums that will allow us to determine the value of Jacobi sums.

Lemma 2.3. ^[13,15] If $\chi _1$ and $\chi _2$ are multiplicative characters of $\mathbb{F}_q$ and $\psi$ is a nontrivial additive character of $\mathbb{F}_q$, then

if $\chi _1 \chi _2$ is nontrivial.

Lemma 2.4. Let $\chi$ be a multiplicative character of $\mathbb{F}_q$ of order $n$. Then $\chi (-1) = -1$ if and only if $n$ is even and $\frac{q-1}{n}$ is odd.

Proof. Let $g$ be a generator of $\mathbb{F} _q^*$. Then $g^{q-1} = 1$ and $g^{\frac{q-1}{2}} = -1$.

First, we prove the necessity. Let $\chi (-1) = -1$. For any $x\in \mathbb{F}_q^*$, we have ${\chi }^{q-1}(x) = \chi (x^{q-1}) = \chi (1) = 1$. Thus ${\chi }^{q-1} = \chi _0$. Since ${\rm ord}(\chi) = n$ and ${\chi }^{q-1} = \chi _0$, we obtain that $n|(q-1)$. From $(\chi (-1))^n = \chi ^n(-1) = \chi_0(-1) = 1$ and $\chi (-1) = -1$, one can deduce that $(-1)^n = 1$. Hence $n$ must be even.

On the other hand, since $g^{\frac{q-1}{2}} = -1$, we have $\chi^{\frac{n}{2}}(g) = \pm 1$. But ${\rm ord}(\chi ^{\frac{n}{2}}) = 2$ together with the assumption that $g$ is a generator of $\mathbb{F}_q^*$ implies that $\chi^{\frac{n}{2}}(g)\ne 1$. Thus

Therefore $\frac{q-1}{n}$ is odd as required. The necessity is proved.

Now we prove the sufficiency. Let $n$ be even and $\frac{q-1}{n}$ be odd. Then $\xi: = \chi (g)$ is an $n$th primitive root of unity. Since $n$ is even and ${\rm ord}(\xi) = n$, we have $\xi^{\frac{n}{2}} = -1$. Since $\frac{q-1}{n}$ is odd, i.e., $\frac{q-1}{n}\equiv1 \pmod 2$, one derives that $\frac{q-1}{2}\equiv \frac{n}{2}\pmod n.$ Thus

as desired. Hence the sufficiency part is proved.

This finishes the proof of Lemma 2.4.

If $p\equiv 3\pmod 4,$ then we easily have the following facts.

If $p\equiv 3\pmod 4$ and $q = p^s\equiv 3\pmod 4,$ then there does not exist a multiplicative character $\varphi$ of $\mathbb{F}_q^*$ with ${\rm ord}(\varphi) = 4.$ Otherwise, if $\varphi\in\widehat{\mathbb{F}_q^*}$ with ${\rm ord}(\varphi) = 4,$ where $\widehat{\mathbb{F}_q^*}$ is the dual group consisting of all multiplicative characters of $\mathbb{F} _q^*.$ Since $\widehat{\mathbb{F}_q^*}\cong\mathbb{F}_q^*,$ one has $4\mid(q-1)$ which is a contradiction.

From the above statement, we deduce the following consequence.

Corollary 2.1. Let $\varphi$ be a multiplicative character of $\mathbb{F}_q$ of order $4$ with $q = p^s$. Then the following is true.

$\rm(i).$ If $p\equiv 1\pmod 4,$ then

$\rm(ii).$ If $p\equiv 3\pmod 4,$ then $s$ is even and $\varphi(-1) = 1.$

Proof. (ⅰ). Since $p\equiv1\pmod4$, we have either $p\equiv1\pmod8$ or $p\equiv5\pmod8$.

If $p\equiv1\pmod8$, then $p^s\equiv1\pmod 8$. Hence $\frac{q-1}{4}\equiv0\pmod2$. By Lemma 2.4, we have $\varphi(-1) = 1.$

If $p\equiv5\pmod8$ and $s$ is even, then $p^s\equiv1\pmod8.$ Therefore

Then Lemma 2.4 tells us that $\varphi(-1) = 1.$

If $p\equiv5\pmod8$ and $s$ is odd, then $p^s\equiv5\pmod8.$ It follows that

So by Lemma 2.4, one can deduce that $\varphi(-1) = -1.$ This finishes the proof of (ⅰ).

(ⅱ). From the above discussion, we can easily deduce that $s$ is even.

Let $p\equiv 3\pmod 4.$ It implies $p\equiv 3\pmod 8$ or $p\equiv 5\pmod 8.$ Since $s$ is even, one has $q = p^s\equiv 1\pmod 8.$ It follows that

By Lemma 2.4, one has $\varphi(-1) = 1.$ The proof of (ⅱ) is finished.

Let $\varphi$ be a generator of $\widehat{\mathbb{F} _q^*}.$ Then ${\rm ord}(\varphi) = q-1$ and the multiplicative character $\lambda$ with order $d$ with $d|(q-1)$ has the expression $\lambda = \varphi^{\frac{q-1}{d}t^{\prime}}$, where $0\le t^{\prime} < d$ and $\gcd(t^{\prime}, d) = 1$. Actually, we have $\lambda = \varphi^t$ with $0\le t < q-1$. Hence ${\rm ord} (\lambda) = \frac{q-1}{\gcd(t, q-1)}$. Since ${\rm ord} (\lambda) = d$, one has $d = \frac{q-1}{\gcd(t, q-1)}$. This infers that $\gcd(t, q-1) = \frac{q-1}{d}$, and so $\frac{q-1}{d}\mid t$. Thus one can write $t = \frac{q-1}{d}t^{\prime}$ with $1\le t'\le d$ and $\gcd(t^{\prime}, d) = 1$. That is, one has $\lambda = \varphi^{\frac{q-1}{d}t^{\prime}}$ with $1\le t'\le d$ and $\gcd(t^{\prime}, d) = 1$. Moreover, the number of multiplicative character $\lambda$ with order $d$ is $\phi(d)$, where $\phi$ is Euler's totient function. In the following, we express the number of solutions of $f(x) = b$ as certain character sums.

Lemma 2.5. We have

where $\lambda$ is a multiplicative character of $\mathbb{F}_q^*$ with order $\gcd(n, q-1)$.

Proof. We divide this into the following three cases. Let $\lambda$ be any multiplicative character of $\mathbb{F}_q$ with order $d: = \gcd(n, q-1)$.

CASE 1. $b = 0$. Then $x^n = 0$ has only zero solution $x = 0$ in $\mathbb{F}_q$. That is, one has $N(x^n = 0) = 1$. Since ${\lambda}^0(0) = 1$ and ${\lambda}^j(0) = 0$ for $1\le j \le {d-1}$, it follows that

as desired. So part (ⅰ) is proved in this case.

CASE 2. $b\ne 0$ and $x^n = b$ has a solution in $\mathbb{F}_q$. Let $b = g^k$ and $x = g^y$. Then $x^n = b$ is equivalent to the congruence

Then the congruence (2.1) has exactly $d = \gcd(n, q-1)$ solutions $y$. Hence $x^n = b$ has exactly $d$ solutions in $\mathbb{F}_q$. Namely, $N(x^n = b) = d$.

Let $x_0$ be an element of $\mathbb{F}_q$ with $x_0^n = b$. For any integer $j$ with $0\le j \le {d-1}$, since $d|n$ implying that $\lambda^n = \chi_0$, the trivial multiplicative character, we have

Therefore one derives that

as desired. Hence part (ⅰ) holds in this case.

CASE 3. $b\ne 0$ and $x^n = b$ has no solution in $\mathbb{F}_q$. Then $N(x^n = b) = 0$ and (2.1) has no solution in $\mathbb{F}_q$. Let $b = g^k$. Then $d\nmid k$ and $\lambda(b) = {\lambda}^k(g)\ne 1$ since $\lambda(g)$ is a $d$-th primitive root of unity. Then

which implies that

Since $\lambda(b)\ne 1$, we have

as required.

The proof of Lemma 2.5 is complete.

Let $\chi'$ be a multiplicative character of $\mathbb{F}_p$ and $\psi'$ be an additive character of $\mathbb{F}_p$. From the multiplicativity of the norm and the additivity of the trace, it follows that $\chi'\circ{\mathbb{N}}$ is a multiplicative and $\psi'\circ{\text{Tr}}$ an additive character of $\mathbb{F}_q$. We say that the multiplicative character $\chi$ of $\mathbb{F}_q$ is lifted from the multiplicative character $\chi'$ of $\mathbb{F}_p$ if $\chi = \chi'\circ{\mathbb{N}}.$ We also say that the multiplicative character $\chi'$ of $\mathbb{F}_p$ is lifting to the multiplicative character $\chi$ of $\mathbb{F}_q$. Similarly, we say that the additive character $\psi$ of $\mathbb{F}_q$ is lifted from the additive character $\psi'$ of $\mathbb{F}_p$ if $\psi = \psi'\circ{\text{Tr}}.$ We also say that the additive character $\psi'$ of $\mathbb{F}_p$ is lifting to the additive character $\psi$ of $\mathbb{F}_q$. The following lemma is due to Hasse and Davenport establishes an important relationship between the Gauss sum $G(\chi ^\prime, \psi^\prime)$ of $\mathbb{F}_p$ and the Gauss sum $G(\chi, \psi)$ of $\mathbb{F}_q.$

Lemma 2.6. ^[13,15] Let $\psi^\prime$ be an additive and $\chi ^\prime$ a multiplicative character of $\mathbb{F}_p$, not both of them trivial. Suppose that $\psi^\prime$ and $\chi ^\prime$ are lifting to characters $\psi$ and $\chi$, respectively, of the finite extension field $\mathbb{F}_q$ of $\mathbb{F}_p$ with $[\mathbb{F}_q:\mathbb{F}_p] = s$. Then

The characters of $\mathbb{F}_p$ can be lifted to the characters of $\mathbb{F}_q$, but not all the characters of $\mathbb{F}_q$ can be obtained by lifting a character of $\mathbb{F}_p$. The following result characterizes all the characters of $\mathbb{F}_q$ that can be obtained by lifting a character of $\mathbb{F}_p$.

Lemma 2.7. Let $\chi$ be a multiplicative character of $\mathbb{F}_q$ with $q = p^s$. Then $\chi$ can be lifted from a multiplicative character $\chi'$ of $\mathbb{F}_p$ if and only if $\chi ^{p-1}$ is trivial.

Proof. Let $g$ be a generator of $\mathbb{F}_q^*$. Since

and

one knows that $\mathbb{N}(g)$ is a generator of $\mathbb{F}_p^*$.

First of all, we prove the necessity. Suppose that $\chi$ can be obtained by lifting a multiplicative character $\chi'$ of $\mathbb{F}_p$. It then follows that $\chi (g) = \chi'(\mathbb{N}(g)).$ Since $\chi'$ is a character of $\mathbb{F}_p^*$, we get that the values of $\chi'$ are $(p-1)$-th roots of unity. Hence

Thus $\chi ^{p-1}$ is trivial. The necessity is proved.

Now we prove the sufficiency. Let $\chi^{p-1}$ be trivial. Since $\chi$ is a multiplicative character of $\mathbb{F}_q^*$, $\chi(g)$ must be a $(q-1)$-th roots of unity, say

for some integer $j$ with $0\le j\le q-2$, where throughout this paper, i stands for the 4-th primitive root of unity, i.e. ${\rm i}^4 = 1$ and ${\rm i}^2\ne 1$. Since $\chi ^{p-1}$ is trivial, we have

This implies that $(q-1)\mid j(p-1)$. It follows that ${\frac{q-1}{p-1}}\mid j.$ Therefore we derive that

for some $k\in \mathbb{Z}.$

By substituting $j = k{\frac{q-1}{p-1}}$ in (2.2), we obtain that

Since $\mathbb{N}(g)$ is a generator of $\mathbb{F}_p^*$, one can define a multiplicative character $\chi'$ of $\mathbb{F}_p^*$ by

From (2.3) and (2.4), we get $\chi (g) = \chi'(\mathbb{N}(g)).$ This completes the sufficiency.

This finishes the proof of Lemma 2.7.

For a certain special multiplicative character of $\mathbb{F}_p,$ the following result gives an explicit formula about the associated Jacobi sums.

Lemma 2.8. ^[5] Let $p\equiv1\pmod4$ be an odd prime number and let $\varphi'$ be a multiplicative character of $\mathbb{F}_p$ with $ord(\varphi') = 4.$ If $\theta$ is a generator of $\mathbb{F}_p^*$ with $\varphi'(\theta) = {\rm i}$, then

where $a'$ and $b'$ are integers such that $a'^2+b'^2 = p$, $a'\equiv-1\pmod4$ and $b'\equiv a'{\theta}^{\frac{p-1}{4}} \pmod p$.

For a certain special multiplicative character of $\mathbb{F}_q,$ we establish an explicit formula of the associated Gauss sums.

Lemma 2.9. Let $\mathbb{F}_q$ be a finite field with $q = p^s$, where $p\equiv1\pmod4$ is an odd prime and $s\in \mathbb{Z}^+$. Let $\varphi$ be a multiplicative character of $\mathbb{F}_q$ with ${\rm ord}(\varphi) = 4$ and $g$ be a primitive element of $\mathbb{F}_q^*$ with $\varphi(g) = {\rm i}$. Then

where $a'$ and $b'$ are integers such that

Proof. First we claim that if $\chi = \chi' \circ\mathbb{N}$, then ${\rm ord}(\chi) = {\rm ord}(\chi').$ To prove it, we let ${\rm ord}(\chi') = m$ and ${\rm ord}(\chi) = l$. Since $\mathbb{N}(x) \in \mathbb{F}_p^*$ for all $x\in \mathbb{F}_{q}^*$, we have

It follows that $l|m$. Conversely, let $g'$ be a generator of $\mathbb{F}_p^{*}$. Then $\chi'(g')$ is a primitive $m$-th root of unity. Since the norm map $\mathbb{N}$ is surjective, there exist an element $g\in \mathbb{F}_q^{*}$ such that $\mathbb{N}(g) = g'$. Then we have

Hence one can deduce that $m|l$. So the desired result $l = m$ follows immediately. Namely, we have ${\rm ord}(\chi) = {\rm ord}(\chi')$. The claim is proved.

Since ${\rm ord}(\varphi) = 4$ and $p\equiv 1\pmod 4,$ we derive that $\varphi^{p-1}$ is trivial. By Lemma 2.7, $\varphi$ can be obtained by lifting a character $\varphi'$ of $\mathbb{F}_p^*.$ Morever, the additive character $\psi'_1(x) = \exp\Big(\frac{2\pi{\rm i}x}{p}\Big)$ of $\mathbb{F}_p$ can be lifted to the additive character $\psi_1(x) = \exp\Big(\frac{2\pi{\rm i}\ \text{Tr}(x)}{p}\Big)$ of $\mathbb{F}_q.$ By the famous Hasse-Davenport relation (see Lemma 2.6), we get that

Hence

By Lemma 2.3, the following identity is true:

Since ${\rm ord}(\varphi') = 4,$ one has ${\rm ord}((\varphi')^2) = 2.$ Then by Lemma 2.2, we have

Now we evaluate $J(\varphi', \varphi').$ Since the order of lifting character is invariant, we have ${\rm ord}(\varphi') = {\rm ord}(\varphi) = 4.$ Note that ${\rm ord}(\mathbb{N}(g)) = p-1.$ It means that $\mathbb{N}(g)$ is a primitive element of $\mathbb{F}_p^*$. Noticing that ${\rm i} = \varphi(g) = \varphi'\big(\mathbb{N}(g)\big))$ and ${\rm ord}(\varphi') = 4$, Lemma 2.8 tells that

where $a'$ and $b'$ are integers such that

But $\mathbb{N}(g) = g^{\frac{q-1}{p-1}}$. It then follows from (2.8) that

From (2.6) to (2.9), we deduce that

It follows from (2.7), (2.10) and $q = p^s$ that

where $a'$ and $b'$ are integers satisfying (2.8).

This concludes the proof of Lemma 2.9.

Lemma 2.10. Let $\mathbb{F}_q$ be a finite field with $q = p^s$ elements and $q\equiv1\pmod4$. Let $\varphi$ be a multiplicative character of $\mathbb{F}_q^*$ with ${\rm ord}(\varphi) = 4$ and $c\in {\mathbb F}_q^*$. Then we have

Proof. For $x\in\mathbb{F}_q,$ from the trigonometric identity

we can deduce that

Denote

By Lemma 2.5, we get that

where $\varphi$ is a multiplicative character of $\mathbb{F}_q$ with ${\rm ord}(\varphi) = 4.$ Then $\varphi(g) = \pm{\rm i}$. WLOG, in what follows, we set $\varphi(g) = {\rm i}$.

Note that $\varphi^2 = \eta$ and $\varphi^3 = \overline{\varphi},$ we know that

Since

it follows from the definition of Gauss sum and Lemma 2.1 that

Note that the value of $\eta$ is real and $\eta(x) = \Big(\frac{\mathbb{N}(x)}{p}\Big)$, from the Lemmas 2.1 and 2.2 we deduce that

So the desired result follows immediately.

Lemma 2.11. Let $\mathbb{F}_q$ be a finite field of $q = p^s$ with $p\equiv 3\pmod 4$ and $q\equiv 1\pmod 4.$ Let $\varphi$ be a multiplicative character of $\mathbb{F}_q^*$ with $ord(\varphi) = 4.$ Then

where $r$ is uniquely determined by

Proof. By $p\equiv 3\pmod 4$ and $q\equiv 1\pmod 4$, we can deduce $s$ is even immediately.

First of all, we calculate $N(x_1^4+x_2^4 = -1)$ using Lemma 2.10. Setting $n = 2$ in Lemma 2.10 gives us that

Since $\varphi^3 = \overline{\varphi}$, $\overline{\varphi^2} = \varphi^2 = \eta$, $\varphi^2(x) = \Big(\frac{\mathbb{N}(x)}{p}\Big)$ and Lemma 2.1 implying that

it follows that

By using the following simple facts

we deduce that

Since $-xc$ runs over $\mathbb{F}_q^*$ as $x$ runs through $\mathbb{F}_q^*$, it follows from (2.13), (2.14) and the fact of $\sum\limits_{x\in\mathbb{F}_q^*}\psi_1(-xc) = -1$ that

Since $s$ is even. From Lemma 2.1, 2.2, Corollary 2.1 and (2.15), one has

Consequently, we calculate $N(x_1^4+x_2^4 = -1)$ by using another method. Let $N_n$ be the number of solutions of

over $\mathbb{F}_q.$ From Lemma 6 to Lemma 8 in ^[17], we can easily derive that

and

where $r$ is uniquely determined by

Since

one has

From (2.16) and (2.17), one deduces that

as desired. This completes the proof of Lemma 2.11.

Remark 2.1. From Lemma 2.11, we can write

But it is known that $G^2(\varphi, \psi_1)\overline{G^2(\varphi, \psi_1)} = q^2$. So one has $A = \pm\sqrt{q^2-9r^2q}$ and $B = \pm\sqrt{q^2-(3r-2)^2q}.$

3.   Proofs of Theorems 1.2 and 1.3

In this section, we present the proofs of the main results of this paper. We begin with the proof of Theorem 1.2.

Proof of Theorem 1.2. (ⅰ). Let $p\equiv 1\pmod 4.$ Then Lemma 2.10 applied to $n = 3$ gives us that

Since $\varphi^3 = \overline{\varphi}$, $\varphi^2(x) = \Big(\frac{\mathbb{N}(x)}{p}\Big)$ and Lemma 2.1 implying that $G(\varphi, \psi_1)\overline{G(\varphi, \psi_1)} = q$, it follows that

By using the following simple facts

we deduce that

Since $-xc$ runs over $\mathbb{F}_q^*$ as $x$ runs through $\mathbb{F}_q^*$, it follows from (3.1), (3.2) and the definition of Gauss sum that

Note that $\frac{1}{\varphi(c)} = \overline{\varphi}(c), \quad \overline{\varphi}^2(c) = \varphi^2(c)$. Then from Lemmas 2.1 and 2.2, we derive that

Since $\varphi(1) = \overline{\varphi}(1) = 1$, it follows from (3.3) that

From Lemma 2.9 and letting $a+b{\rm i}: = (a'+b'{\rm i})^s$, we know that

Thus

Then by Corollary 2.1 we get

From (3.3), (3.4) and noticing that $\varphi(g) = {\rm i}$, we obtain that

So applying Corollary 2.1 gives that

Similarly, from $\varphi(g^2) = -1$ and $\varphi(g^3) = -{\rm i}$, one can deduce that

and

Hence part $\rm(i)$ of Theorem 1.2 is proved.

(ⅱ). Let $p\equiv3\pmod4$ and $q\equiv1\pmod4.$ Then $s$ is even. From Corollary 2.1 and (3.3), one has

Then from Lemma 2.11, Remark 2.1 and (3.9), it follows that

and

as required. So part (ⅱ) of Theorem 1.2 is proved.

This concludes the proof of Theorem 1.2.

Now we turn our attention to the proof of Theorem 1.3.

Proof of Theorem 1.3. (ⅰ). Let $p\equiv 1\pmod 4.$ Then applying Lemma 2.10 to $n = 4$ gives us that

Let

Since

it follows from Lemma 2.1 that

Then from the following identities

we deduce that

Since $-xc$ run through $\mathbb{F}_q^*$ whenever $x$ run through $\mathbb{F}_q^*$ and

if follows from (3.14), (3.15), Lemmas 2.1 and Lemma 2.2 that

But $\varphi(1) = \overline{\varphi(1)} = 1$. By (3.16) we have

and by (3.4), one has

Thus using Corollary 2.1 gives us that

From (3.16) and $\varphi(g) = {\rm i}$ we know that

Also (3.4) gives that

Thus using Corollary 2.1 tells us that

In the similar way, noting that $\varphi(g^2) = -1$ and $\varphi(g^3) = -{\rm i}$, one can easily deduce that

Combining (3.17) to (3.20) concludes the proof of part $\rm(i)$ of Theorem 1.3.

$\rm(ii).$ Let $p\equiv 3\pmod 4$ and $q\equiv 1\pmod 4.$ Then $s$ is even. From Corollary 2.1, Lemma 2.11 and (3.16), one derives that

From Lemma 2.11 and Remark 2.1, we have

and

Finally, combining (3.22) to (3.25), we arrive at the results given in part $\rm(ii)$ of Theorem 1.3. This finishes the proof of Theorem 1.3.

4.   Two Examples

In this section, we give two examples to demonstrate the validity of our main results.

Example 4.1. Let $N(c)$ be the number of $\mathbb{F}_5$-rational points of the affine hypersurface $x_1^4+x_2^4+x_3^4+x_4^4 = c$ with $c\in \mathbb{F}_5^*$. It is easy to see that $2$ is a generator of $\mathbb{F}_5^*$, ${\rm ind}_2(1)\equiv 0\pmod 4$, ${\rm ind}_2(2)\equiv 1\pmod 4$ and $s = 1$. From Theorem 1.3, we can deduce that $a = -1$, $b = -2$, $N(1) = 16$ and $N(2) = 96$.

On the other hand, by Matlab, one can calculate that

and

which coincide with the results in Theorem 1.3.

Example 4.2. Let $N(c)$ be the number of $\mathbb{F}_{25}$-rational points of the affine hypersurface $x_1^4+x_2^4+x_3^4+x_4^4 = c$ with $c\in \mathbb{F}_{25}^*$. Observe that $x^2-2$ is irreducible over $\mathbb{F}_5$. Let $\alpha$ be a root of $x^2-2$ over its split field. Then $\mathbb{F}_5(\alpha)$ is a extensive field of $\mathbb{F}_5$ with $25$ elements and denoted by $\mathbb{F}_{25}$. One can write

The additivity and multiplicity of $\mathbb{F}_{25}$ are defined as follows: $\forall$ ${x_i+y_i}\in \mathbb{F}_{25}$ with $i = 1, 2$, define

and

By Matlab, we get that $2+4\alpha$ is a generator of $\mathbb{F}_{25}$, ${\rm ind}_{2+4\alpha}(1)\equiv 0\pmod 4$, ${\rm ind}_{2+4\alpha}(\alpha)\equiv 3\pmod 4$, and $(2+4\alpha)^6 = 2$.

Now letting $s = 2$, $a' = -1$ and $b' = -2$ in Theorem 1.3, one obtains that $a = -3$, $b = 4$, $N(1) = 19664$ and $N(\alpha) = 16064$. This coincide with the results for $N(1)$ and $N(2)$ gotten by Matlab.

Acknowledgments

The authors thank the anonymous referees for helpful comments and suggestions. S. F. Hong was supported partially by National Science Foundation of China Grant #11771304.

Conflict of interest

We declare that we have no conflict of interest.