On some determinants involving Jacobi symbols

TL;DR

This paper investigates conjectures on determinants with Jacobi symbol entries, proving several cases for specific moduli and primes, using character sums over finite fields to establish the results.

Contribution

The paper proves new conjectures by Sun regarding determinants with Jacobi symbol entries for specific moduli and primes, expanding understanding of their properties.

Findings

Certain determinants are zero for specific moduli, confirming conjectures.

Proved that $(10,9)_p=0$ for primes $p ot ext{divisible by }12$.

Established that $[5,5]_p=0$ for primes $p ot ext{divisible by }20$.

Abstract

In this paper we study some conjectures on determinants with Jacobi symbol entries posed by Z.-W. Sun. For any positive integer $n\equiv3\pmod4$, we show that $$(6,1)_n=[6,1]_n=(3,2)_n=[3,2]_n=0$$ and $$(4,2)_n=(8,8)_n=(3,3)_n=(21,112)_n=0$$ as conjectured by Sun, where $$(c,d)_n=\bigg|\left(\frac{i^2+cij+dj^2}n\right)\bigg|_{1\le i,j\le n-1}$$ and $$[c,d]_n=\bigg|\left(\frac{i^2+cij+dj^2}n\right)\bigg|_{0\le i,j\le n-1}$$ with $(\frac{\cdot}n)$ the Jacobi symbol. We also prove that $(10,9)_p=0$ for any prime $p\equiv5\pmod{12}$, and $[5,5]_p=0$ for any prime $p\equiv 13,17\pmod{20}$, which were also conjectured by Sun. Our proofs involve character sums over finite fields.

Equations

$$(6,1{)}_n=[6,1{]}_n=(3,2{)}_n=[3,2{]}_n=0$$

$$(4,2{)}_n=(8,8{)}_n=(3,3{)}_n=(21,112{)}_n=0$$

$$(c,d)_{n}=\bigg{|}\left(\frac{i^{2}+cij+dj^{2}}{n}\right)\bigg{|}_{1\leqslant i,j\leqslant n-1}$$

$$[c,d]_{n}=\bigg{|}\left(\frac{i^{2}+cij+dj^{2}}{n}\right)\bigg{|}_{0\leqslant i,j\leqslant n-1}$$

$$\bigg{|}\left(\frac{i+j-1}{p}\right)\bigg{|}_{1\leqslant i,j\leqslant(p-1)/2}=\left(\frac{-1}{p}\right)\bigg{|}\left(\frac{i+j}{p}\right)\bigg{|}_{1\leqslant i,j\leqslant(p-1)/2}$$

$$\bigg{|}\left(\frac{i+j-1}{p}\right)\bigg{|}_{1\leqslant i,j\leqslant(p+1)/2}=\bigg{|}\left(\frac{i+j}{p}\right)\bigg{|}_{0\leqslant i,j\leqslant(p-1)/2},$$

$$\bigg{|}\left(\frac{j-i}{p}\right)\bigg{|}_{0\leqslant i,j\leqslant(p-1)/2}$$

$$(c,d)_{n}:=\bigg{|}\left(\frac{i^{2}+cij+dj^{2}}{n}\right)\bigg{|}_{1\leqslant i,j\leqslant n-1}$$

$$[c,d]_{n}:=\bigg{|}\left(\frac{i^{2}+cij+dj^{2}}{n}\right)\bigg{|}_{0\leqslant i,j\leqslant n-1},$$

$$\left(\frac{d}{n}\right)=-1\Rightarrow (c,d{)}_n=0,$$

$$\left(\frac{d}{p}\right)=1\Rightarrow [c,d{]}_p=\{\begin{array}{ll}\frac{p-1}{2}(c,d{)}_p& \text{if}p\nmid c^2-4d,\\ \frac{1-p}{p-2}(c,d{)}_p& \text{if}p\mid c^2-4d.\end{array}$$

$$(6,1{)}_n=(3,2{)}_n=0\text{and}[6,1{]}_n=[3,2{]}_n=0.$$

$$(4,2{)}_n=(8,8{)}_n=0\text{and}[4,2{]}_n=[8,8{]}_n=0.$$

$$(3,3{)}_n=(6,-3{)}_n=0\text{and}[3,3{]}_n=[6,-3{]}_n=0.$$

$$(21,112{)}_n=(42,-7{)}_n=0\text{and}[21,112{]}_n=[42,-7{]}_n=0.$$

$$(6,1{)}_n=[6,1{]}_n=(3,2{)}_n=[3,2{]}_n=0$$

$$(4,2{)}_n=(8,8{)}_n=(3,3{)}_n=(21,112{)}_n=0.$$

$$\sum _{j=0}^{n-1}\left(\frac{j}{n}\right)\left(\frac{i^2+cij+d{j}^2}{n}\right)=\sum _{j=0}^{n-1}\left(\frac{-j}{n}\right)\left(\frac{i^2+2cij+(c^2-4d)j^2}{n}\right).$$

$$\sum _{j=0}^{n-1}\left(\frac{j}{n}\right)\left(\frac{i^2+3ij+2j^2}{n}\right)=0$$

$$\sum _{j=0}^{n-1}\left(\frac{j}{n}\right)\left(\frac{i^2+4ij+2j^2}{n}\right)=0$$

$$\sum _{j=0}^{n-1}\left(\frac{j}{n}\right)\left(\frac{i^2+3ij+3j^2}{n}\right)=0$$

$$\sum _{j=0}^{n-1}\left(\frac{j}{n}\right)\left(\frac{i^2+21ij+112j^2}{n}\right)=0$$

$$\sum _{x=0}^{p-1}\left(\frac{x^5+10x^{3}y+9x{y}^2}{p}\right)=0$$

$$\sum _{x=0}^{p-1}\left(\frac{x^5+5x^{3}y+5x{y}^2}{p}\right)=0$$

$$\sum _{x=0}^{p-1}\left(\frac{x^5+8x^{3}y+18x{y}^2}{p}\right)=0$$

$$\sum _{x=0}^{p-1}\left(\frac{a{x}^5+b{x}^3+cx}{p}\right)=2\sum _{x=1}^{(p-1)/2}\left(\frac{x}{p}\right)\left(\frac{a(x^2{)}^2+b{x}^2+c}{p}\right)=0.$$

$$\sum _{j=0}^{p-1}\left(\frac{j}{p}\right)\left(\frac{i^2+cij+d{j}^2}{p}\right)\equiv -\left(\frac{ci}{p}\right)\sum _{k=0}^{p-1}\left(\genfrac{}{}{0ex}{}{4k}{2k}\right)\left(\genfrac{}{}{0ex}{}{2k}{k}\right){\left(\frac{d}{16c^2}\right)}^k(\mathrm{mod}p).$$

$$L=$$

$$\equiv$$

$$\sum _{r=1}^{p-1}{r}^s\equiv \{\begin{array}{ll}-1(\mathrm{mod}p)& \text{if}p-1\mid s,\\ 0(\mathrm{mod}p)& \text{otherwise}.\end{array}$$

$$\sum _{r=1}^{p-1}{\left(r^{-1}+c+dr\right)}^{(p-1)/2}=\sum _{s=-(p-1)/2}^{(p-1)/2}{a}_s\sum _{r=1}^{p-1}{r}^s\equiv -a_0(\mathrm{mod}p).$$

$$a_0=$$

Full text

[TABLE]

and

[TABLE]

as conjectured by Sun, where

[TABLE]

and

[TABLE]

with $(\frac{\cdot}{n})$ the Jacobi symbol. We also prove that $(10,9)_{p}=0$ for any prime $p\equiv 5\ ({\rm{mod}}\ 12)$, and $[5,5]_{p}=0$ for any prime $p\equiv 13,17\ ({\rm{mod}}\ 20)$, which were also conjectured by Sun. Our proofs involve character sums over finite fields.

2020 Mathematics Subject Classification. Primary 11C20, 11T24; Secondary 11E16, 15A15.

The work is supported by the NSFC (Natural Science Foundation of China)-RFBR (Russian Foundation for Basic Research) Cooperation and Exchange Program (grants NSFC 11811530072 and RFBR 18-51-53020-GFEN-a). The third author is also supported by the Natural Science Foundation of China (grant 11971222).

1. Introduction

For an $n\times n$ matrix $[a_{ij}]_{1\leqslant i,j\leqslant n}$ over a field, we simply denote its determinant by $|a_{ij}|_{1\leqslant i,j\leqslant n}$. In this paper we study some conjectures on determinants with Jacobi symbol entries posed by Z.-W. Sun [11].

Let $p$ be an odd prime. In 2004, R. Chapman [2] determined the values of

[TABLE]

and

[TABLE]

where $(\frac{\cdot}{p})$ denotes the Legendre symbol. Chapman’s conjecture on the evaluation of

[TABLE]

was confirmed by M. Vsemirnov [12, 13] via matrix decomposition. With this background, Z.-W. Sun [11] studied some new kinds of determinants with Legendre symbol or Jacobi symbol entries.

For any odd integer $n>1$ and integers $c$ and $d$, Sun [11] introduced the notations

[TABLE]

and

[TABLE]

where $(\frac{\cdot}{n})$ denotes the Jacobi symbol. He showed that

[TABLE]

and that for any odd prime $p$ we have

[TABLE]

For $a\in\mathbb{Z}$ and $n\in\mathbb{Z}^{+}=\{1,2,3,\ldots\}$ , if $a$ is relatively prime to $n$ and $x^{2}\equiv a\ ({\rm{mod}}\ n)$ for some $x\in\mathbb{Z}$, then $a$ is called a quadratic residue modulo $n$. If $n$ is odd and $a$ is a quadratic residue modulo $n$, then $(\frac{a}{n})=1$ since $a$ is a quadratic residue modulo any prime divisor of $n$.

Now we state our first theorem.

Theorem 1.1**.**

Let $n>1$ be an odd integer.

(i)* If $-1$ is not a quadratic residue modulo $n$, then*

[TABLE]

(ii)* If $-2$ is not a quadratic residue modulo $n$, then*

[TABLE]

(iii)* If $-3$ is not a quadratic residue modulo $n$, then*

[TABLE]

(iv)* If $-7$ is not a quadratic residue modulo $n$, then*

[TABLE]

Combining Theorem 1.1 with (1.3), we immediately obtain the following consequence which was conjectured by Sun [11, Conjecture 4.8(ii)].

Corollary 1.1**.**

For any positive integer $n\equiv 3\ ({\rm{mod}}\ 4)$, we have

[TABLE]

and

[TABLE]

Actually we deduce Theorem 1.1 from the following theorems.

Theorem 1.2**.**

Let $n$ be a positive odd integer which is squarefree. For any $c,d,i\in\mathbb{Z}$, we have

[TABLE]

Theorem 1.3**.**

Let $n$ be a positive odd integer which is squarefree, and let $i\in\mathbb{Z}$. Then

[TABLE]

where the notation $m\ R\ n$ means that $m$ is a quadratic residue modulo $n$.

Our following result was originally conjectured by Sun [11, Conjecture 4.8(iv)].

Theorem 1.4**.**

(i)* $(10,9)_{p}=0$ for any prime $p\equiv 5\ ({\rm{mod}}\ 12)$.*

(ii)* $[5,5]_{p}=0$ for any prime $p\equiv 13,17\ ({\rm{mod}}\ 20)$.*

In fact, our proof of Theorem 1.4 yields a stronger result: For each integer $y$, we have

[TABLE]

for any prime $p\equiv 5\ ({\rm{mod}}\ 12)$, and

[TABLE]

for any prime $p\equiv 13,17\ ({\rm{mod}}\ 20)$.

We will prove Theorem 1.2, Theorems 1.3 and 1.1, and Theorem 1.4 in Sections 2-4 respectively.

Sun [11, Conjecture 4.8(iv)] also conjectured that $(8,18)_{p}=[8,18]_{p}=0$ for any prime $p\equiv 13,17\ ({\rm{mod}}\ 24)$. Moreover, Sun [10] conjectured that

[TABLE]

for any prime $p\equiv 13,17\ ({\rm{mod}}\ 24)$ and integer $y$, and this was confirmed by M. Stoll via two elliptic curves with complex multiplication by $\mathbb{Z}[\sqrt{-6}]$ (see the answer in [10]).

For any prime $p\equiv 1\ ({\rm{mod}}\ 4)$ and $a,b,c\in\mathbb{Z}$, we provide in Section 5 a sufficient condition for

[TABLE]

2. Proof of Theorem 1.2

Lemma 2.1**.**

Let $p$ be an odd prime and let $c,d,i\in\mathbb{Z}$ with $p\nmid c$. Then

[TABLE]

Proof. If $p\mid i$, then both sides of the congruence (2.1) are zero.

Below we assume $p\nmid i$ and let $L$ denote the left-hand side of the congruence (2.1). As $\{ir:\ r=0,\ldots,p-1\}$ is a complete system of residues modulo $p$, we have

[TABLE]

We may write $(x^{-1}+c+dx)^{(p-1)/2}=\sum_{s=-(p-1)/2}^{(p-1)/2}a_{s}x^{s}$ with $a_{s}\in\mathbb{Z}$. For any integer $s$, it is well known (cf. [4, p. 235]) that

[TABLE]

Therefore,

[TABLE]

Clearly,

[TABLE]

So, by the above, we finally obtain (2.1). ∎

Lemma 2.2**.**

Let $p$ be any odd prime. Then we have the congruence

[TABLE]

in the ring $\mathbb{Z}_{p}[x]$, where $\mathbb{Z}_{p}$ is the ring of all $p$-adic integers, and $\delta_{p,3}$ is $1$ or [math] according as $p=3$ or not.

Remark 2.1*.*

For any prime $p>3$, the congruence (2.3) is due to Sun [9, (1.15)]. We can easily verify that (2.3) also holds for $p=3$.

Proof of Theorem 1.2. Clearly both sides of (1.5) vanish if $n=1$. Below we assume $n>1$ and distinguish three cases.

Case 1. $n$ is an odd prime $p$.

Define

[TABLE]

If $p\mid c$ and $p\equiv 3\ ({\rm{mod}}\ 4)$, then

[TABLE]

When $p\mid c$ and $p\equiv 1\ ({\rm{mod}}\ 4)$, for $q=((p-1)/2)!$ we have $q^{2}\equiv-1\ ({\rm{mod}}\ p)$ and $(\frac{2q}{p})=1$ (cf. [11, Remark 1.1 and Lemma 2.3]), thus

[TABLE]

Now suppose that $p\nmid c$. By Lemma 2.2,

[TABLE]

Combining this with Lemma 2.1, we obtain that

[TABLE]

Thus $D\equiv 0\ ({\rm{mod}}\ p)$. Clearly $|D|<2p$.

If $p\mid i$, then

[TABLE]

Now assume that $p\nmid i$. If neither $c^{2}-4d$ nor $(2c)^{2}-4(c^{2}-4d)=16d$ is divisible by $p$, then

[TABLE]

and

[TABLE]

hence $D$ is even. When $p\mid d$, we also have $2\mid D$ since

[TABLE]

If $p\mid c^{2}-4d$, then $2\mid D$ since

[TABLE]

and

[TABLE]

So $D$ is always even, and hence $D=0$ as $p\mid D$ and $|D|<2p$.

Case 2. $n=p_{1}\ldots p_{r}$ with $r\geqslant 2$, where $p_{1},\ldots,p_{r}$ are distinct primes.

By the Chinese Remainder Theorem,

[TABLE]

and hence

[TABLE]

Similarly,

[TABLE]

Thus, (1.5) holds in view of Case 1. This concludes the proof. ∎

3. Proofs of Theorems 1.3 and 1.1

Lemma 3.1**.**

Let $p>3$ be a prime. If $p\equiv 1,3\ ({\rm{mod}}\ 8)$ and $p=x^{2}+2y^{2}$ with $x,y\in\mathbb{Z}$ and $x\equiv 1\ ({\rm{mod}}\ 4)$, then

[TABLE]

If $(\frac{-2}{p})=-1$, i.e., $p\equiv 5,7\ ({\rm{mod}}\ 8)$, then

[TABLE]

Remark 3.1*.*

The first assertion in Lemma 3.1 was conjectured by Z.-W. Sun [8] and confirmed by his twin brother Z.-H. Sun [7, Theorem 4.3]. The second assertion was proved by Z.-W. Sun [9, Corollary 1.3] as a consequence of (2.3) with $x=1/2$.

Lemma 3.2**.**

Let $p$ be an odd prime and let $c,d,i\in\mathbb{Z}$ with $p\nmid d$. Then

[TABLE]

Proof. Both sides of (3.1) vanish if $p\mid i$. Below we assume $p\nmid i$.

Clearly,

[TABLE]

and

[TABLE]

So (3.1) holds. ∎

Lemma 3.3**.**

Let $p$ be any odd prime and let $i\in\mathbb{Z}$.

(i)* We have*

[TABLE]

(ii)* We have*

[TABLE]

Also,

[TABLE]

and

[TABLE]

Remark 3.2*.*

It is well known that any prime $p\equiv 1\ ({\rm{mod}}\ 4)$ can be written as $x^{2}+4y^{2}$ with $x,y\in\mathbb{Z}$. Also, for each $m\in\{2,3,7\}$ any odd prime $p$ with $(\frac{-m}{p})=1$ can be written $x^{2}+my^{2}$ with $x,y\in\mathbb{Z}$ (cf. [3]).

Proof of Lemma 3.3. It is easy to verify that (3.2)-(3.5) hold for $p=3$. Below we assume $p>3$.

(i) As $16\times 4^{2}/2=128$, combining Lemma 2.1 and Lemma 3.1 we find that

[TABLE]

Observe that

[TABLE]

is even (since $|\{1\leqslant j\leqslant p-1:\ i^{2}+4ij+2j^{2}\equiv 0\ ({\rm{mod}}\ p)\}|\in\{0,2\}$), and its absolute value is smaller than $p$. If $p=x^{2}+2y^{2}$ with $x,y\in\mathbb{Z}$ and $x\equiv 1\ ({\rm{mod}}\ 4)$, then $|2x|<2\sqrt{p}<p$. So (3.2) holds.

(ii) In light of Lemma 3.2,

[TABLE]

and

[TABLE]

On the other hand, by [1, Theorem 6.2.9] and [1, pp. 195-196],

[TABLE]

and

[TABLE]

So we have (3.3) and (3.4).

Now we prove (3.5). Clearly, (3.5) is valid if $p\mid i$ or $p=7$. Below we assume that $p\nmid i$ and $p\not=7$. Observe that

[TABLE]

By a result of Rajwade [6],

[TABLE]

Therefore (3.5) holds.

The proof of Lemma 3.3 is now complete. ∎

Proof of Theorem 1.3. Write $n=p_{1}\ldots p_{r}$ with $p_{1},\ldots,p_{r}$ distinct primes. In light of (2.4) and Lemma 3.3(i), if $-2\ R\ n$ fails (i.e., $(\frac{-2}{p_{s}})=-1$ for some $s=1,\ldots,r$) then

[TABLE]

Thus (1.7) holds. Note that if $-2\ R\ n$ then for each $s=1,\ldots,r$ we may write $p_{s}=x_{s}^{2}+2y_{s}^{2}$ with $x_{s},y_{s}\in\mathbb{Z}$ and $x_{s}\equiv 1\ ({\rm{mod}}\ 4)$ and hence

[TABLE]

Similarly, (1.6), (1.8) and (1.9) also hold in view of (2.4) and Lemma 3.3(ii). This concludes our proof of Theorem 1.3. ∎

Proof of Theorem 1.1. Suppose that $n=\prod_{s=1}^{r}p_{s}^{a_{s}}$, where $p_{1},\ldots,p_{r}$ are distinct primes and $a_{1},\ldots,a_{r}$ are positive integers. If $a_{t}>1$ with $1\leqslant t\leqslant r$, then $n/p_{t}\equiv 0\ ({\rm{mod}}\ p_{1}\ldots p_{r})$ and hence for any $i\in\mathbb{Z}$ we have

[TABLE]

for all $j=0,\ldots,n-1$. Therefore

[TABLE]

Below we assume that $n$ is squarefree. If $-1\ R\ n$ fails, then by Theorems 1.2 and 1.3 we have

[TABLE]

for all $i=0,\ldots,n-1$, hence $(3,2)_{n}=(6,1)_{n}=0$ and $[3,2]_{n}=[6,1]_{n}=0$. This proves part (i) of Theorem 1.1. Similarly, parts (ii)-(iv) of Theorem 1.1 follow from Theorems 1.2 and 1.3. This ends the proof. ∎

4. Proof of Theorem 1.4

Let $q>1$ be a prime power and let $\mathbb{F}_{q}$ be the finite field of order $q$. A multiplicative character $\chi$ on $\mathbb{F}_{q}$ is called trivial (or principal) if $\chi(a)=1$ for all $a\in\mathbb{F}_{q}^{*}=\mathbb{F}_{q}\setminus\{0\}$. For a polynomial $P(x)=\sum_{s=0}^{n}c_{s}x^{s}\in\mathbb{F}_{q}[x]$, we define the homogenous polynomial

[TABLE]

Fix a list of the elements of $\mathbb{F}_{q}$. For a multiplicative character $\chi$ on $\mathbb{F}_{q}$, we introduce the matrices

[TABLE]

Lemma 4.1**.**

Let $q>1$ be a prime power and let $\chi$ be a nontrivial multiplicative character on $\mathbb{F}_{q}$. Suppose that $P(x)\in\mathbb{F}_{q}[x]$ and $\sum_{x\in\mathbb{F}_{q}}\chi(xP(x))=0$. Then $M(P,\chi)$ is singular $($i.e., $\det M(P,\chi)=0)$. If the character $\chi^{n+1}$ is nontrivial with $n=\deg P$, then the matrix $M_{0}(P,\chi)$ is singular too.

Proof.

We introduce the column vector $v$ whose coordinates are $v_{b}=\chi(b)$ for $b\in\mathbb{F}_{q}^{*}$. Let $M=M(P,\chi)$. Then, for any $a\in\mathbb{F}_{q}^{*}$ we have

[TABLE]

Since $v$ is a nonzero vector, the matrix $M$ is singular.

Now suppose that the degree of $P$ is $n$ and the character $\chi^{n+1}$ is nontrivial. Let $M_{0}=M_{0}(P,\chi)$ and introduce the vector $v$ with coordinates $v_{b}=\chi(b)$ for $b\in\mathbb{F}_{q}$. Then $(M_{0}v)_{a}=0$ for all $a\in\mathbb{F}_{q}^{*}$ as before. Let $c_{n}$ be the leading coefficient of the polynomial $P(x)$. Then

[TABLE]

Therefore $M_{0}v$ is the zero vector and hence $M_{0}$ is singular. ∎

Motivated by Lemma 4.1, we give the following more sophisticated lemma.

Lemma 4.2**.**

Let $q>1$ be an odd prime power. Suppose that $g\in\mathbb{F}_{q}$ is not a square and $\chi$ is a nontrivial multiplicative character on $\mathbb{F}_{q}$ with $\chi(-1)=1$. Assume that $P(x)\in\mathbb{F}_{q}[x]$ and

[TABLE]

(i)* We have $\operatorname{dim}(\operatorname{Ker}(M(P,\chi)))\geqslant 2$, in particular $M(P,\chi)$ is singular.*

(ii)* Assume that the character $\chi^{2n+1}$ with $n=\deg P$ is nontrivial. Then $\operatorname{dim}(\operatorname{Ker}(M_{0}(P,\chi)))\geqslant 2$.*

Proof.

For $a,b\in\mathbb{F}_{q}$, set

[TABLE]

This is well defined since $\chi(\pm 1)=1$, The matrix $V=[v_{a,b}]_{a,b\in\mathbb{F}_{q}^{*}}$ has rank 2; in fact, if $b^{\prime}=bc^{2}$ for some $c\in\mathbb{F}_{q}$ then columns $b$ and $b^{\prime}$ in $V$ are proportional, but columns 1 and $g$ are not proportional.

(i) Write $M$ for $M(P,\chi)$. It suffices to show that $MV$ is the zero matrix. For $a,b\in\mathbb{F}_{q}^{*}$, the $(a,b)$-entry of the matric $MV$ is

[TABLE]

where $P_{c}(x)=xP(cx^{2})$ for any $c\in\mathbb{F}_{q}$.

Now it remains to show for any $c\in\mathbb{F}_{q}^{*}$ the identity

[TABLE]

Clearly, $c=c_{0}d^{2}$ for some $c_{0}\in\{1,g\}$ and $d\in\mathbb{F}_{q}^{*}$. Thus

[TABLE]

This proves part (i) of Lemma 4.2.

(ii) Write $M_{0}$ for $M_{0}(P,\chi)$, and define $V_{0}=[v_{a,b}]_{a,b\in\mathbb{F}_{q}}$. (Note the slight difference between $V_{0}$ and $V$.) The rank of $V_{0}$ is still equal to $2$, so it suffices to show that $M_{0}V_{0}$ is the zero matrix. Note that the $(a,b)$-entry of $M_{0}V_{0}$ is trivially zero if $b=0$ since $v_{c,0}=0$ for all $c\in\mathbb{F}_{q}$. For $a,b\not=0$ we can repeat the computation for $MV$ verbatim. Let $c_{n}$ denote the leading coefficient of $P(x)$. If $a=0$ and $b\not=0$, then the $(a,b)$-entry of $M_{0}V_{0}$ is

[TABLE]

This is zero since $\chi^{2n+1}$ is nontrivial. We are done. ∎

Theorem 4.1**.**

Let $q>1$ be an odd prime power and let $m\in\mathbb{Z}^{+}$ with $\gcd(m,q-1)=1$. Let $\chi$ be a nontrivial quadratic character on $\mathbb{F}_{q}$, and let

[TABLE]

with $a\in\mathbb{F}_{q}^{*}=\mathbb{F}_{q}\setminus\{0\}$. Then

[TABLE]

If $\chi(-1)=1$, then both $M(P_{m}(x,a),\chi)$ and $M_{0}(P_{m}(x,a),\chi)$ are singular, and moreover either of them has a kernel of dimension at least two.

Proof.

In view of Lemma 4.2, we only need to prove (4.5). As $P_{m}(gx^{2},a)=g^{m-1}P_{m}(x^{2},ag^{-1})$ for all $g\in\mathbb{F}_{q}^{*}$, it suffices to show that

[TABLE]

for any $a\in\mathbb{F}_{q}^{*}$.

Clearly, $m$ is odd since $\gcd(m,q-1)=1$. Recall that $\chi^{2}$ is the trivial character, and note that

[TABLE]

If $a$ is not a square in $\mathbb{F}_{q}$, then $\chi(a)^{m}=(-1)^{m}=-1$ and hence (4.6) holds by the above.

Now assume that $a=b^{2}$ with $b\in\mathbb{F}_{q}^{*}$. Since

[TABLE]

it remains to show that $\sum_{x\in\mathbb{F}_{q}}\chi(xP_{m}(x^{2},1))=0$. Since $\chi=\chi^{-1}$ and

[TABLE]

we have

[TABLE]

Thus $\sum_{x\in\mathbb{F}_{q}}\chi(xP_{m}(x^{2},1))=0$ as desired.

The proof of Theorem 4.1 is now complete. ∎

Proof of Theorem 1.4(i). Let $p$ be any prime with $p\equiv 5\ ({\rm{mod}}\ 12)$, and let $\chi$ be the quadratic character of $\mathbb{F}_{p}=\mathbb{Z}/p\mathbb{Z}$ with $\chi(x+p\mathbb{Z})=(\frac{x}{p})$ for all $x\in\mathbb{Z}$. Note that $\chi(-1)=1$ since $p\equiv 1\ ({\rm{mod}}\ 4)$. Clearly,

[TABLE]

Applying Theorem 4.1, we obtain that

[TABLE]

and

[TABLE]

Note that Sun stated in [S19, Remark 4.9] that $(10,9)_{p}=0$ if and only if $[10,9]_{p}=0$. ∎

Let $\mathbb{F}_{q}$ be a finite field of order $q$. A polynomial $P(x)\in\mathbb{F}_{q}[x]$ is called a permutation polynomial if $P$ is bijective as a function on $\mathbb{F}_{q}$. If $\chi$ is a nontrivial multiplicative character on $\mathbb{F}_{q}$ and $P(x)\in\mathbb{F}_{q}[x]$ is a permutation polynomial, then

[TABLE]

and also

[TABLE]

provided that $P(0)=0$.

Theorem 4.2**.**

Let $q>1$ be an odd prime power and let $m\in\mathbb{Z}^{+}$ with $\gcd(m,q^{2}-1)=1$. Let $\chi$ be a nontrivial multiplicative character on $\mathbb{F}_{q}$ with $\chi(-1)=1$. For the polynomial

[TABLE]

with $a\in\mathbb{F}_{q}^{*}$, we have

[TABLE]

Moreover, if the character $\chi^{m}$ is nontrivial, then

[TABLE]

Proof.

Let $a\in\mathbb{F}_{q}$. It is a classical result (cf. [5, pp. 355-357]) that the Dickson polynomial $D_{m}(x,a):=xQ_{m}(x^{2},a)$ is a permutation polynomial on $\mathbb{F}_{q}$. For any $g\in\mathbb{F}_{q}^{*}$, as $Q_{m}(gx^{2},a)=g^{(m-1)/2}Q_{m}(x^{2},ag^{-1})$, the polynomial $xQ_{m}(gx^{2},a)$ is also a permutation polynomial on $\mathbb{F}_{q}$. Thus

[TABLE]

Combining this with Lemma 4.2, we immediately obtain the desired results. ∎

Proof of Theorem 1.4(ii). Let $p$ be any prime with $p\equiv 13,17\ ({\rm{mod}}\ 20)$. Then $\gcd(5,p^{2}-1)=1$. Let $\chi$ be the quadratic character of $\mathbb{F}_{p}=\mathbb{Z}/p\mathbb{Z}$ with $\chi(x+p\mathbb{Z})=(\frac{x}{p})$ for all $x\in\mathbb{Z}$. Then $\chi(-1)=1$ since $p\equiv 1\ ({\rm{mod}}\ 4)$. Clearly $\chi^{5}=\chi$ is nontrivial and $Q_{5}(x,-1)=x^{2}+5x+5$. Applying Theorem 4.2, we get that

[TABLE]

This concludes the proof. ∎

Note that actually our method to prove Theorem 1.4 yields a stronger result stated after Theorem 1.4 in Section 1.

5. A sufficient condition for $\sum_{x=0}^{p-1}(\frac{ax^{5}+bx^{3}+cx}{p})=0$

For an odd prime power $q>1$, we let $\chi_{q}$ denote the quadratic multiplicative character on the finite field $\mathbb{F}_{q}$.

Let $p\equiv 1\ ({\rm{mod}}\ 4)$ be a prime and let $a$ be a nonzero element of $\mathbb{F}_{p}$. If $\chi_{p}(a)=1$, then we define $\sqrt{a}$ as an element $\alpha\in\mathbb{F}_{p}$ with $\alpha^{2}=a$. When $\chi_{p}(a)=-1$, the finite field $\mathbb{F}_{p^{2}}\cong\mathbb{F}_{p}[x]/(x^{2}-a)$ contains an element $\alpha$ with $\alpha^{2}=a$, and we denote such an $\alpha\in\mathbb{F}_{p^{2}}$ by $\sqrt{a}$.

Theorem 5.1**.**

Let $p\equiv 1\ ({\rm{mod}}\ 4)$ be a prime and let $a,b,c$ be nonnzero elements of the field $\mathbb{F}_{p}$. Let $q$ be $p$ or $p^{2}$ according as $\chi_{p}(ac)$ is $1$ or $-1$, and set

[TABLE]

Let $N$ be the number of $\mathbb{F}_{q}$-points on the affine curve $y^{2}=x^{4}+x^{2}+\gamma$. If $N\equiv-1\ ({\rm{mod}}\ p)$, then

[TABLE]

For the sake of convenience, for an odd prime $p$ we introduce the following two polynomials over $\mathbb{F}_{p}$:

[TABLE]

and

[TABLE]

Lemma 5.1**.**

Let $p$ be a prime with $p\equiv 1\ ({\rm{mod}}\ 4)$, and let $a,b,c\in\mathbb{F}_{p}\setminus\{0\}$. Define

[TABLE]

Viewing $A_{p}\ ({\rm{mod}}\ p)$ as an element of $\mathbb{F}_{p}$, we have

[TABLE]

Consequently, $A_{p}=0$ if $(a^{-1}c)^{(p-1)/4}=-1$ or $f(ac/b^{2})=0$.

Proof.

As $A_{p}=\sum_{x\in\mathbb{F}_{p}\setminus\{0\}}\chi_{p}(ax^{5}+bx^{3}+cx)$, we have $|A_{p}|<p$. So, the second assertion in Lemma 5.1 follows from the first one.

Now we come to prove the first assertion. With the help of (2.2), in $\mathbb{F}_{p}$ we have

[TABLE]

(Note that if $k_{1},k_{3},k_{5}$ are nonnegative integers with $k_{1}+k_{3}+k_{5}=(p-1)/2$ then $k_{1}+3k_{3}+5k_{5}\leqslant 5(k_{1}+k_{3}+k_{5})<3(p-1).$) Thus,

[TABLE]

as desired.

∎

Lemma 5.2**.**

Let $p$ be an odd prime and let $q=p^{n}$ with $n\in\mathbb{Z}^{+}$. For any polynomial

[TABLE]

we have

[TABLE]

Proof.

As the multiplicative group $\mathbb{F}_{q}\setminus\{0\}$ is cyclic, similar to (2.2), for each $s=0,1,2,\ldots$ we have

[TABLE]

where we treat $0^{0}$ as $1$ when $s=0$. Note also that

[TABLE]

for all integers $i\geqslant 0$. Thus

[TABLE]

where $\sum^{\ast}$ means that the sum is taken over all $k_{0},\dots,k_{n-1}\in\{0,1,\ldots,2p-2\}$ subject to the condition

[TABLE]

Write $k_{i}=p-1+t_{i}$, where $-(p-1)\leqslant t_{i}\leqslant p-1$. Then

[TABLE]

Note that

[TABLE]

and the equality is possible only if $t_{0}=\cdots=t_{n-1}=p-1$ (i.e., $k_{0}=\cdots=k_{n-1}=2(p-1)$) or $t_{0}=\cdots=t_{n-1}=-(p-1)$ (i.e., $k_{0}+k_{1}p+\cdots+k_{n-1}p^{n-1}=0$). Since $|t_{i}|<p$, if $\sum_{i=0}^{n-1}t_{i}p^{i}=0$ then we obtain step by step that $t_{0}=\cdots=t_{n-1}=0$ (i.e., $k_{0}=\cdots=k_{n-1}=p-1$).

Combining the above, we finally obtain (5.3). ∎

Lemma 5.3**.**

Let $p$ be an odd prime and let $q=p^{n}$ with $n\in\mathbb{Z}^{+}$. Let $\alpha,\beta,\gamma\in\mathbb{F}_{q}\setminus\{0\}$ and set

[TABLE]

Viewing $B_{q}\ ({\rm{mod}}\ p)$ as an element of $\mathbb{F}_{q}$, we have

[TABLE]

Proof.

Write

[TABLE]

In view of Lemma 5.2, we have

[TABLE]

Clearly,

[TABLE]

Note also that

[TABLE]

and hence

[TABLE]

Combining (5.6) with (5.7) and (5.8), we immediately obtain the desired (5.5). ∎

Now we study further properties of the polynomials $f$ and $g$ defined by (5.1) and (5.2). They may be viewed as truncated versions of certain hypergeometric series.

Lemma 5.4**.**

Let $p$ be an odd prime and let $q=p^{n}$ with $n\in\mathbb{Z}^{+}$.

(i)* A polynomial $u\in\mathbb{F}_{q}[z]$ with $\deg u\leqslant\lfloor(p-1)/4\rfloor$ satisfies the differential equation*

[TABLE]

if and only if $u=cg$ for some $c\in\mathbb{F}_{q}$.

(ii)* Suppose that $p\equiv 1\ ({\rm{mod}}\ 4)$. Then a polynomial $v\in\mathbb{F}_{q}[z]$ with $\deg v\leqslant\lfloor(p-1)/8\rfloor$ satisfies the differential equation*

[TABLE]

if and only if $v=cf$ for some $c\in\mathbb{F}_{q}$.

Proof.

It is straightforward to verify that $u=g$ and $v=f$ satisfy (5.9) and (5.10) respectively. So, the “if” parts of (i) and (ii) are easy.

Now we prove the “only if” part of (i). If a polynomial $u\in\mathbb{F}_{q}[z]$ with $\deg u\leq\lfloor(p-1)/4\rfloor$ satisfies (5.9), then there is a constant $c\in\mathbb{F}_{q}$ such that $\tilde{u}=u-cg$ is a solution of (5.9) with $\deg\tilde{u}<\lfloor(p-1)/4\rfloor$. Thus, it suffices to show that (5.9) has no nonzero solution $u=c_{d}z^{d}+\cdots+c_{0}$ with $\deg u=d<\lfloor(p-1)/4\rfloor$. In fact, the coefficient of $z^{d}$ in $(4z-16z^{2})u^{\prime\prime}+(4-32z)u^{\prime}-3u$ is $-(4d+1)(4d+3)c_{d}\neq 0$ provided $d<\lfloor(p-1)/4\rfloor$.

Similarly, we can show the “only if” part of (ii). ∎

Lemma 5.5**.**

Let $p=4n+1$ be a prime with $n\in\mathbb{Z}^{+}$. Then

[TABLE]

Proof.

Clearly, $u=(16z-2)^{n}f((16z-2)^{-2})$ is a polynomial of degree $n=(p-1)/4$ with the leading coefficient $1$. A direct computation based on (5.10) shows that $u$ satisfies (5.9). Now we apply Lemma 5.4 and compare the leading terms of both sides. Since

[TABLE]

we immediately get the desired result. ∎

Proof of Theorem 5.1.

Since

[TABLE]

the assumption $N\equiv-1\ ({\rm{mod}}\ p)$, together with Lemma 5.3 in the case $\alpha=\beta=1$, implies that $g(\gamma)=0$. As $(16\gamma-2)^{-2}=ac/b^{2}$, we have $f(ac/b^{2})=0$ by Lemma 5.5. Applying Lemma 5.1 we obtain the desired result. ∎