1 Introduction

For an \(n\times n\) matrix \([a_{ij}]_{1\le i,j\le n}\) over a commutative ring with identity, we use \(\det |a_{ij}|_{1\le i,j\le n}\) or \(|a_{ij}|_{1\le i,j\le n}\) to denote its determinant.

Let p be an odd prime, and let \(b,c\in {\mathbb {Z}}\). Sun [4] introduced

$$\begin{aligned} (b,c)_p=\det \left[ \left( \frac{i^2+bij+cj^2}{p}\right) \right] _{1\le i,j\le p-1} \end{aligned}$$

and

$$\begin{aligned}{}[b,c]_p=\det \left[ \left( \frac{i^2+bij+cj^2}{p}\right) \right] _{0\le i,j\le p-1}, \end{aligned}$$

and proved the following results:

$$\begin{aligned} \left( \frac{c}{p}\right) =-1\Longrightarrow (b,c)_p=0, \end{aligned}$$
(1.1)

and

$$\begin{aligned} \left( \frac{c}{p}\right) =1\Longrightarrow [b,c]_p={\left\{ \begin{array}{ll}\frac{p-1}{2}(b,c)_p&{}\text{ if }\ p\not \mid b^2-4c, \\ \frac{1-p}{p-2}(b,c)_p&{}\text{ if }\ p\mid b^2-4c,\end{array}\right. } \end{aligned}$$

where \((\frac{\cdot }{p})\) denotes the Legendre symbol. Grinberg, Sun and Zhao [1, Theorem 1.3] determined \((\frac{S_c(b,p)}{p})\) in the case \(p\not \mid bc\), where

$$\begin{aligned} S_c(b,p)=\det \left[ \left( \frac{i^2+bj^2+c}{p}\right) \right] _{1\le i,j\le (p-1)/2}. \end{aligned}$$

For each prime \(p\equiv 5\ (\mathrm{{mod}}\ 6)\), Sun [4] conjectured that

$$\begin{aligned} 2\det \left[ \frac{1}{i^2-ij+j^2}\right] _{1\le i,j \le p-1} \end{aligned}$$

is a quadratic residue modulo p. This was confirmed by Wu et al. [8].

For any odd prime p and a p-adic integer \(x\not \equiv 0\ (\mathrm{{mod}}\ p)\), clearly

$$\begin{aligned} \frac{1}{x}\equiv x^{p-2}\ (\mathrm{{mod}}\ p)\quad \text{ and }\quad \frac{1}{x^2}\equiv x^{p-3}\ (\mathrm{{mod}}\ p). \end{aligned}$$

Let p be an odd prime, and let \(b,c\in {\mathbb {Z}}\). Sun [5] showed that for any integer n with \((p-1)/2<n<p-1\) we have

$$\begin{aligned} \det [(i^2+bij+cj^2)^n]_{0\le i,j\le p-1}\equiv 0\ (\mathrm{{mod}}\ p). \end{aligned}$$

Sun [5] also introduced

$$\begin{aligned} D_p(b,c)=\det [(i^2+bij+cj^2)^{p-2}]_{1\le i,j\le p-1}, \end{aligned}$$

and proved that for any prime \(p>3\) with \(p\equiv 3\ (\mathrm{{mod}}\ 4)\) we have

$$\begin{aligned} D_p(b,-1)\equiv D_p(2,2)\equiv 0\ (\mathrm{{mod}}\ p). \end{aligned}$$

By Wu et al. [8], we actually have \(\left( \frac{D_p(1,1)}{p}\right) =\left( \frac{-2}{p}\right) \) if \(p\equiv 2\ (\mathrm{{mod}}\ 3)\). Recently, Luo and Sun [3] have proved that

$$\begin{aligned} \left( \frac{D_p(1,1)}{p}\right) ={\left\{ \begin{array}{ll}0&{}\text{ if }\ p\equiv 7\ (\mathrm{{mod}}\ 9), \\ 1&{}\text{ if }\ p\equiv 1,4\ (\mathrm{{mod}}\ 9),\end{array}\right. } \end{aligned}$$
(1.2)

and that

$$\begin{aligned} \left( \frac{D_p(2,2)}{p}\right) ={\left\{ \begin{array}{ll}1&{}\text{ if }\ p\equiv 1\ (\mathrm{{mod}}\ 8),\\ 0&{}\text{ if }\ p\equiv 5\ (\mathrm{{mod}}\ 8). \end{array}\right. }\end{aligned}$$
(1.3)

Their tools include generalized trinomial coefficients and Lucas sequences. Similar to (1.1), Wu and She [7] extended a result of Sun [5] by proving that \(D_p(b,c)\equiv 0\ (\mathrm{{mod}}\ p)\) if \((\frac{c}{p})=-1\).

We first present a basic result which is similar to (1.1) and Sun [5, Theorem 1.2].

Theorem 1.1

Let p be an odd prime, and let \(b,c\in {\mathbb {Z}}\) with \((\frac{c}{p})=-1\). For any integer n in the interval \([1,p-1]\), we have

$$\begin{aligned} \det [(i^2+bij+cj^2)^n]_{1\le i,j\le p-1}\equiv 0\ (\mathrm{{mod}}\ p). \end{aligned}$$
(1.4)

Proof

For \(j=1,\ldots ,p-1\), let \(\pi _c(j)=\{cj\}_p\), the least nonnegative residue of cj modulo p. By Zolotarev’s Lemma (cf. [9]), the sign of \(\pi _c \in S_{p-1}\) is exactly the Legendre symbol \((\frac{c}{p})\). Observe that

$$\begin{aligned}&c^{n(p-1)}\det [(i^2+bij+cj^2)^n]_{1\le i,j\le p-1}\\&\quad =\det \left[ (ci^2+bi(cj)+(cj)^2)^{n}\right] _{1\le i,j\le p-1}\\&\quad =\det \left[ (ci^2+bi\pi _c(j)+\pi _c(j)^2)^{n}\right] _{1\le i,j\le p-1}\\&\quad =\sum _{\sigma \in S_{p-1}}\textrm{sign}(\sigma )\prod _{i=1}^{p-1}(ci^2+bi\pi _c(\sigma (i))+\pi _c(\sigma (i))^2)^{n}\\&\quad =\textrm{sign}(\pi _c)\sum _{\tau \in S_{p-1}}\textrm{sign}(\tau )\prod _{i=1}^{p-1}(ci^2+bi\tau (i)+\tau (i)^2)^{n}\\&\quad =\left( \frac{c}{p}\right) \det [(i^2+bij+cj^2)^n]_{1\le i,j\le p-1}=-\det [(i^2+bij+cj^2)^n]_{1\le i,j\le p-1}. \end{aligned}$$

Thus, with the aid of Fermat’s little theorem, we obtain (1.4). \(\square \)

Let p be an odd prime, and let \(b,c\in {\mathbb {Z}}\). In contrast to the notation \(D_p(b,c)\), we introduce

$$\begin{aligned} D_p^*(b,c)=\det [(i^2+bij+cj^2)^{p-3}]_{1\le i,j\le p-1}. \end{aligned}$$
(1.5)

If \((\frac{b^2-4c}{p})=-1\), then \(i^2+bij+cj^2\not \equiv 0\ (\mathrm{{mod}}\ p)\) for all \(i,j=1,\ldots ,p-1\), and hence,

$$\begin{aligned} D_p^*(b,c)\equiv \det \left[ \frac{1}{(i^2+bij+cj^2)^2}\right] _{1\le i,j\le p-1}\ (\mathrm{{mod}}\ p). \end{aligned}$$

The notations \(D_p(b,c)\) and \(D_p^*(b,c)\) are motivated by Wolstenholme’s congruences (cf. [6])

$$\begin{aligned} \sum _{k=1}^{p-1}\frac{1}{k}\equiv 0\ (\mathrm{{mod}}\ p^2)\quad \text{ and }\quad \sum _{k=1}^{p-1}\frac{1}{k^2}\equiv 0\ (\mathrm{{mod}}\ p) \end{aligned}$$

provided \(p>3\).

Now we state our main results.

Theorem 1.2

Let p be an odd prime. Then

$$\begin{aligned} \left( \frac{D_p^*(1,1)}{p}\right) ={\left\{ \begin{array}{ll} \genfrac(){}{}{-1}{p}&{}\text{ if }\ p\equiv 2\ (\mathrm{{mod}}\ 3), \\ \left( \frac{p}{5}\right) \text{ or }\ 0&{}\text{ if }\ p\equiv 1\ (\mathrm{{mod}}\ 3). \end{array}\right. } \end{aligned}$$
(1.6)

Consequently, when \(p\equiv 2\ (\mathrm{{mod}}\ 3)\) the p-adic integer

$$\begin{aligned} -\det \left[ \frac{1}{(i^2+ij+j^2)^{2}}\right] _{1\le i,j\le p-1} \end{aligned}$$

is a quadratic residue modulo p.

Theorem 1.3

Let p be an odd prime. Then

$$\begin{aligned} \left( \frac{D_p^*(2,2)}{p}\right) ={\left\{ \begin{array}{ll} 0\ \text{ or }\ 1&{}\text{ if }\ p\equiv 1\ (\mathrm{{mod}}\ 8)\\ \left( \frac{p}{3}\right) (-1)^\frac{p+1}{8}&{}\text{ if }\ p\equiv 7\ (\mathrm{{mod}}\ 8), \\ 0&{}\text{ if }\ p\equiv {\pm 3}\ (\mathrm{{mod}}\ 8). \end{array}\right. } \end{aligned}$$
(1.7)

Remark 1.1

Note that for any prime \(p\equiv 3\ (\mathrm{{mod}}\ 4)\) we have

$$\begin{aligned} D_p^*(2,2)\equiv \det \left[ \frac{1}{((i+j)^2+j^2)^2}\right] _{1\le i,j\le p-1}\ (\mathrm{{mod}}\ p). \end{aligned}$$

Let \(n\in {\mathbb {N}}=\{0,1,2,\ldots \}\) and \(b,c\in {\mathbb {Z}}\). The generalized trinomial coefficients

$$\begin{aligned} \left( {\begin{array}{c}n\\ k\end{array}}\right) _{b,c}\quad (k\in {\mathbb {Z}}) \end{aligned}$$

are given by

$$\begin{aligned} \left( x+b+\frac{c}{x}\right) ^n=\sum _{k\in {\mathbb {Z}}}\left( {\begin{array}{c}n\\ k\end{array}}\right) _{b,c}x^k. \end{aligned}$$
(1.8)

We will make use of generalized trinomial coefficients to prove Theorems 1.2 and 1.3 in Sects. 2 and 3, respectively. Note that Theorems 1.2 and 1.3 cannot be deduced from Luo and Sun’s results (1.2) and (1.3), and their proofs are somewhat sophisticated.

2 Proof of Theorem  1.2

Lemma 2.1

([2, Lemma 10]) Let R be a commutative ring with identity, and let \(P(x)=\sum _{i=0}^{n-1}a_ix^i\in R[x]\). Then

$$\begin{aligned} \det [P(X_iY_j)]_{1\le i, j\le n}=a_0a_1\cdots a_{n-1}\prod _{1\le i, j\le n}(X_i-X_j)(Y_i-Y_j). \end{aligned}$$

Lemma 2.2

(Luo and Sun [3, (3.2)]) For any odd prime p, we have

$$\begin{aligned} \prod _{1\le i,j\le p-1}(i-j)\left( \frac{1}{i}-\frac{1}{j}\right) =(-1)^{(p+1)/2}\prod _{j=1}^{p-2}(j!)^2. \end{aligned}$$
(2.1)

Lemma 2.3

([3, Lemma 2.1]) Let p be an odd prime, and let \(b,c\in {\mathbb {Z}}\). For \(k\in \{-p+2,\ldots ,p-2\}\), we have

$$\begin{aligned} (4c^2-b)\left( {\begin{array}{c}p-2\\ k\end{array}}\right) _{b,c}\equiv {\left\{ \begin{array}{ll} \left( {\begin{array}{c}p-1\\ -1\end{array}}\right) _{b,c}+c\left( {\begin{array}{c}p-1\\ 1\end{array}}\right) _{b,c}-b\ (\mathrm{{mod}}\ p)&{}\text{ if }\ k=0,\\ (k+1)\left( {\begin{array}{c}p-1\\ k-1\end{array}}\right) _{b,c}-(k-1)c\left( {\begin{array}{c}p-1\\ k+1\end{array}}\right) _{b,c}\ (\mathrm{{mod}}\ p)&\text{ otherwise }. \end{array}\right. }\nonumber \\ \end{aligned}$$
(2.2)

Lemma 2.4

Let p be an odd prime, and let \(b,c\in {\mathbb {Z}}\). For \(k\in \{-p+3,\ldots ,p-3\}\), we have

$$\begin{aligned}{} & {} (k+3)\left( {\begin{array}{c}p-2\\ k-1\end{array}}\right) _{b,c}-(k-3)c\left( {\begin{array}{c}p-2\\ k+1\end{array}}\right) _{b,c}-2\left( {\begin{array}{c}p-1\\ k\end{array}}\right) _{b,c}\nonumber \\{} & {} \quad \equiv 2(4c-b^2)\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{b,c}\ (\mathrm{{mod}}\ p). \end{aligned}$$
(2.3)

Proof

For \(n\in {\mathbb {N}}\) and \(k\in {\mathbb {Z}}\), we simply write \({n\brack k}\) for \(\left( {\begin{array}{c}n\\ k\end{array}}\right) _{b,c}\).

Taking derivatives of both sides of the following identity

$$\begin{aligned} \sum _{k=-p+1}^{p-1}{p-1\brack k}x^k=\left( x+b+\frac{c}{x}\right) ^{p-1}, \end{aligned}$$

we get

$$\begin{aligned} \sum _{k=-p+1}^{p-1}k{p-1\brack k}x^{k-1}=(p-1)\left( x+b+\frac{c}{x}\right) ^{p-2}\left( 1-\frac{c}{x^2}\right) . \end{aligned}$$
(2.4)

Comparing the coefficients of \(x^{k-1}\) on both sides of (2.4), we obtain

$$\begin{aligned} k{p-1\brack k}=(p-1)\left( {p-2\brack k-1}-c{p-2\brack k+1}\right) . \end{aligned}$$
(2.5)

Taking derivatives of both sides of (2.4), we get

$$\begin{aligned} \sum _{k=-p+1}^{p-1}k(k-1){p-1\brack k}x^{k-2}= & {} (p-1)(p-2)\left( x+b+\frac{c}{x}\right) ^{p-3} \left( 1-\frac{c}{x^2}\right) ^2\nonumber \\{} & {} +\frac{2c}{x^3}(p-1)\left( x+b+\frac{c}{x}\right) ^{p-2}. \end{aligned}$$
(2.6)

Comparing the coefficients of \(x^{k-2}\) on both sides of (2.6), we deduce that

$$\begin{aligned} k(k-1){p-1\brack k}= & {} (p-1)(p-2)\left( {p-3\brack k-2}-2c{p-3\brack k}+c^2{p-3\brack k+2}\right) \nonumber \\{} & {} +\,2c(p-1){p-2\brack k+1}. \end{aligned}$$
(2.7)

For \(n\in {\mathbb {Z}}^+=\{1,2,3,\ldots \}\) and \(k\in {\mathbb {Z}}\), we have the recurrence

$$\begin{aligned} {n\brack k}={n-1\brack k-1}+b{n-1\brack k}+c{n-1\brack k+1} \end{aligned}$$

by Luo and Sun [3, (2.3)]. With the aid of this, we have

$$\begin{aligned}&{p-3\brack k-2}-2c{p-3\brack k}+c^2{p-3\brack k+2}\\&\quad = {p-2\brack k-1}-b{p-3\brack k-1}-c{p-3\brack k}-2c{p-3\brack k}\\&\qquad +c\left( {p-2\brack k+1} -b{p-3\brack k+1}-{p-3\brack k}\right) \\&\quad = {p-2\brack k-1}+c{p-2\brack k+1}-4c{p-3\brack k}-b\left( {p-3\brack k-1}+c{p-3\brack k+1}\right) \\&\quad = {p-2\brack k-1}+c{p-2\brack k+1}-4c{p-3\brack k}-b\left( {p-2\brack k}-b{p-3\brack k}\right) \\&\quad = {p-2\brack k-1}-b{p-2\brack k}+c{p-2\brack k+1}+(b^2-4c){p-3\brack k}\\&\quad = {p-2\brack k-1}-\left( {p-1\brack k}-{p-2\brack k-1}-c{p-2\brack k+1}\right) \\&\qquad +c{p-2\brack k+1}+(b^2-4c){p-3\brack k}, \end{aligned}$$

and hence,

$$\begin{aligned}{} & {} {p-3\brack k-2}-2c{p-3\brack k}+c^2{p-3\brack k+2}\nonumber \\{} & {} \quad = -{p-1\brack k}+2{p-2\brack k-1}+2c{p-2\brack k+1}+(b^2-4c){p-3\brack k}. \end{aligned}$$
(2.8)

Combining (2.5), (2.7) and (2.8), we get

$$\begin{aligned} \begin{aligned}&(k-1)\left( {p-2\brack k-1}-c{p-2\brack k+1}\right) -2c(p-1){p-2\brack k+1}\\&\quad = \frac{k(k-1)}{p-1}{p-1\brack k}-2c(p-1){p-2\brack k+1}\\&\quad = (p-2)\left( {p-3\brack k-2}-2c{p-3\brack k}+c^2{p-3\brack k+2}\right) -2c(p-2){p-2\brack k+1}\\&\quad = (p-2)\left( -{p-1\brack k}+2{p-2\brack k-1}+(b^2-4c){p-3\brack k}\right) . \end{aligned} \end{aligned}$$

Hence, we have

$$\begin{aligned}&(k-1)\left( {p-2\brack k-1}-c{p-2\brack k+1}\right) +2c{p-2\brack k+1}\\&\quad \equiv 2\left( {p-1\brack k}-2{p-2\brack k-1}-(b^2-4c){p-3\brack k}\right) \ (\mathrm{{mod}}\ p), \end{aligned}$$

that is,

$$\begin{aligned} (k+3){p-2\brack k-1}-(k-3)c{p-2\brack k+1}-2{p-1\brack k}\equiv 2(4c-b^2){p-3\brack k}\ (\mathrm{{mod}}\ p).\nonumber \\ \end{aligned}$$
(2.9)

This concludes the proof. \(\square \)

Proof of Theorem 1.2

Let \(b,c\in {\mathbb {Z}}\). By Luo and Sun [3, (2.2)], we have

$$\begin{aligned} \left( {\begin{array}{c}n\\ -k\end{array}}\right) _{b,c}=c^k\left( {\begin{array}{c}n\\ k\end{array}}\right) _{b,c}\quad \text{ for } \text{ all }\ n\in {\mathbb {N}}\quad \text{ and }\quad k\in {\mathbb {Z}}. \end{aligned}$$
(2.10)

Thus,

$$\begin{aligned}{} & {} (x^2+bx+c)^{p-3}-\left( {\begin{array}{c}p-3\\ 0\end{array}}\right) _{b,c}x^{p-3}\nonumber \\{} & {} \quad = \sum _{\begin{array}{c} k=-p-3\\ k\ne 0 \end{array}}^{p-3}\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{b,c}x^{p-3+k}\nonumber \\{} & {} \quad =\sum _{k=1}^{p-3}\left( \left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{b,c}x^{p-3+k}+\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{b,c}c^kx^{p-3-k}\right) \nonumber \\{} & {} \quad =\sum _{k=2}^{p-3}\left( \left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{b,c}x^{p-1}+\left( {\begin{array}{c}p-3\\ p-1-k\end{array}}\right) _{b,c}c^{p-1-k}\right) x^{k-2}\nonumber \\{} & {} \qquad +\,\left( {\begin{array}{c}p-3\\ 1\end{array}}\right) _{b,c}x^{p-2}+\left( {\begin{array}{c}p-3\\ 1\end{array}}\right) _{b,c}cx^{p-4}. \end{aligned}$$
(2.11)

Let \(k\in \{-p+3,\ldots ,p-3\}\). Taking \(b=c=1\) in Lemma 2.4, we get

$$\begin{aligned} 6\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{1,1}\equiv (k+3)\left( {\begin{array}{c}p-2\\ k-1\end{array}}\right) _{1,1}-(k-3)\left( {\begin{array}{c}p-2\\ k+1\end{array}}\right) _{1,1}-2\left( {\begin{array}{c}p-1\\ k\end{array}}\right) _{1,1}\ (\mathrm{{mod}}\ p).\nonumber \\ \end{aligned}$$
(2.12)

Putting \(b=c=1\) in (2.2) and noting (2.10), we obtain

$$\begin{aligned} 3\left( {\begin{array}{c}p-2\\ k\end{array}}\right) _{1,1}\equiv {\left\{ \begin{array}{ll} 2\left( {\begin{array}{c}p-1\\ 1\end{array}}\right) _{1,1}-1\ (\mathrm{{mod}}\ p)&{}\text{ if }\ k=0,\\ (k+1)\left( {\begin{array}{c}p-1\\ k-1\end{array}}\right) _{1,1}-(k-1)\left( {\begin{array}{c}p-1\\ k+1\end{array}}\right) _{1,1}\ (\mathrm{{mod}}\ p)&\text{ if }\ k\not =0. \end{array}\right. }\qquad \end{aligned}$$
(2.13)

Combining (2.12) with (2.13), we see that

$$\begin{aligned}&18\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{1,1}\\&\quad \equiv 3(k+3)\left( {\begin{array}{c}p-2\\ k-1\end{array}}\right) _{1,1}-3(k-3)\left( {\begin{array}{c}p-2\\ k+1\end{array}}\right) _{1,1}-6\left( {\begin{array}{c}p-1\\ k\end{array}}\right) _{1,1}\\&\quad \equiv {\left\{ \begin{array}{ll} -2\left( {\begin{array}{c}p-1\\ -3\end{array}}\right) _{1,1}+8\left( {\begin{array}{c}p-1\\ 1\end{array}}\right) _{1,1}-4\ (\mathrm{{mod}}\ p)&{}\text{ if }\ k=-1,\\ -2\left( {\begin{array}{c}p-1\\ 3\end{array}}\right) _{1,1}+8\left( {\begin{array}{c}p-1\\ 1\end{array}}\right) _{1,1}-4\ (\mathrm{{mod}}\ p)&{}\text{ if }\ k=1,\\ k(k+3)\left( {\begin{array}{c}p-1\\ k-2\end{array}}\right) _{1,1}-2(k^2-3)\left( {\begin{array}{c}p-1\\ k\end{array}}\right) _{1,1}+k(k-3)\left( {\begin{array}{c}p-1\\ k+2\end{array}}\right) _{1,1}\ (\mathrm{{mod}}\ p)&\text{ if }\ k\not =\pm 1. \end{array}\right. } \end{aligned}$$

For each \(k\in \{0,\ldots ,p-3\}\), we have

$$\begin{aligned} \left( {\begin{array}{c}p-1\\ p-k\end{array}}\right) _{1,1}\equiv \genfrac(){}{}{k}{3}\ (\mathrm{{mod}}\ p)\end{aligned}$$

by Luo and Sun [3, (2.14)], and hence,

$$\begin{aligned}{} & {} 18\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{1,1}\equiv 3(k+3)\left( {\begin{array}{c}p-2\\ k-1\end{array}}\right) _{1,1}-3(k-3)\left( {\begin{array}{c}p-2\\ k+1\end{array}}\right) _{1,1}-6\left( {\begin{array}{c}p-1\\ k\end{array}}\right) _{1,1}\nonumber \\{} & {} \quad \equiv {\left\{ \begin{array}{ll} -2\genfrac(){}{}{p}{3}+8\genfrac(){}{}{p-1}{3}-4\ (\mathrm{{mod}}\ p)&{}\text{ if }\ k=1,\\ k(k+3)\genfrac(){}{}{p-k+2}{3}-2(k^2-3)\genfrac(){}{}{p-k}{3}+k(k-3)\genfrac(){}{}{p-k-2}{3}\ (\mathrm{{mod}}\ p)&{}\text{ if }\ 2\le k\le p-3. \end{array}\right. }\qquad \end{aligned}$$
(2.14)

When \(2\le k\le p-3\), we have

$$\begin{aligned}&18\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{1,1}+18\left( {\begin{array}{c}p-3\\ p-1-k\end{array}}\right) _{1,1}\\&\quad \equiv k(k+3)\genfrac(){}{}{p-k+2}{3}+(p-1-k)(p+2-k)\genfrac(){}{}{k+3}{3}\\&\qquad -\,2(k^2-3)\genfrac(){}{}{p-k}{3}-2((k+1)^2-3)\genfrac(){}{}{k+1}{3}\\&\qquad +\,k(k-3)\genfrac(){}{}{p-k-2}{3}+(p-1-k)(p-4-k)\genfrac(){}{}{k-1}{3}\ (\mathrm{{mod}}\ p), \end{aligned}$$

and hence,

$$\begin{aligned}{} & {} \quad 18\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{1,1}+18\left( {\begin{array}{c}p-3\\ p-1-k\end{array}}\right) _{1,1}\nonumber \\{} & {} \equiv {\left\{ \begin{array}{ll} (-3k^2+3k+6)\genfrac(){}{}{k+1}{3}+(3k^2+9k)\genfrac(){}{}{k+2}{3}\ (\mathrm{{mod}}\ p)&{}\text{ if }\ p\equiv 1\ (\mathrm{{mod}}\ 3),\\ -6k\genfrac(){}{}{k+1}{3}+6\genfrac(){}{}{k+2}{3}\ (\mathrm{{mod}}\ p)&{}\text{ if }\ p\equiv 2\ (\mathrm{{mod}}\ 3). \end{array}\right. }\qquad \quad \end{aligned}$$
(2.15)

In view of (2.11) and (2.14), we obtain

$$\begin{aligned}&18(x^2+x+1)^{p-3}\\&\quad \equiv 18\left( {\begin{array}{c}p-3\\ 0\end{array}}\right) _{1,1}x^{p-3}+\sum _{k=2}^{p-3}\left( 18\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{1,1}+18\left( {\begin{array}{c}p-3\\ p-1-k\end{array}}\right) _{1,1}\right) x^{k-2}\\&\qquad +\,18\left( {\begin{array}{c}p-3\\ 1\end{array}}\right) _{1,1}(x^{p-2}+x^{p-4})\\&\quad \equiv 6\genfrac(){}{}{p}{3}x^{p-3}+\sum _{k=2}^{p-3}\left( 18\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{1,1}+18\left( {\begin{array}{c}p-3\\ p-1-k\end{array}}\right) _{1,1}\right) x^{k-2}\\&\qquad +\,\left( -2\genfrac(){}{}{p}{3}+8\genfrac(){}{}{p+2}{3}-4\right) (x^{p-2}+x^{p-4})\ (\mathrm{{mod}}\ p). \end{aligned}$$

Thus, with the aid of (2.15), we have

$$\begin{aligned}{} & {} 18(x^2+x+1)^{p-3}\nonumber \\{} & {} \quad \equiv 6\left( \frac{p}{3}\right) (x^{p-3}-x^{p-2}-x^{p-4})\nonumber \\{} & {} \qquad +\, {\left\{ \begin{array}{ll} 18\sum _{k=2}^{p-3}\left( -\frac{(k+1)(k-2)}{6}\genfrac(){}{}{k+1}{3}+\frac{k(k+3)}{6}\genfrac(){}{}{k+2}{3}\right) x^{k-2}\ (\mathrm{{mod}}\ p)&{}\text{ if }\ p\equiv 1\ (\mathrm{{mod}}\ 3),\\ 18\sum _{k=2}^{p-3}\left( -\frac{k}{3}\genfrac(){}{}{k+1}{3}+\frac{1}{3}\genfrac(){}{}{k+2}{3}\right) x^{k-2}\ (\mathrm{{mod}}\ p)&{}\text{ if }\ p\equiv 2\ (\mathrm{{mod}}\ 3). \end{array}\right. }\nonumber \\ \end{aligned}$$
(2.16)

Let \(F(x)=(x^2+x+1)^{p-3}\). For \(1\le i, j\le p-1\), we have

$$\begin{aligned} \frac{(i^2+ij+j^2)^{p-3}}{j^{2(p-3)}}=\left( \frac{i^2}{j^2}+\frac{i}{j}+1\right) ^{p-3}=F\left( \frac{i}{j}\right) , \end{aligned}$$

and hence,

$$\begin{aligned} \left( \frac{D_p^*(1,1)}{p}\right) =\left( \frac{|F(i/j)|_{1\le i, j\le p-1}}{p}\right) \end{aligned}$$
(2.17)

by Fermat’s little theorem.

Case 1. \(p\equiv 1\ (\mathrm{{mod}}\ 3)\).

Applying Lemma 2.1 with \(P(x)=F(x)\), \(X_i=i\) and \(Y_j=1/j\), and noting the identity (2.1), we get

$$\begin{aligned}&\bigg |F\left( \frac{i}{j}\right) \bigg |_{1\le i,j\le p-1}\\&\quad \equiv \frac{1}{27}\prod _{k=2}^{p-3}\left( -\frac{(k+1)(k-2)}{6}\genfrac(){}{}{k+1}{3}+\frac{k(k+3)}{6}\genfrac(){}{}{k+2}{3}\right) \\&\qquad \times \,\prod _{1\le i<j\le p-1}(i-j)\left( \frac{1}{i}-\frac{1}{j}\right) \\&\quad \equiv \frac{1}{27}(-1)^{\frac{p+1}{2}}(-1)^{\frac{p-4}{3}}\prod _{j=1}^{p-2}(j!)^2\prod _{k\in \{2+3a:\ 0\le a\le \frac{p-7}{3}\}}\frac{k(k+3)}{6}\\&\qquad \times \,\prod _{k\in \{4+3a:\ 0\le a\le \frac{p-7}{3}\}}\frac{(k+1)(k-2)}{6} \times \prod _{k\in \{3+3a:\ 0\le a\le \frac{p-7}{3}\}}\frac{k^2+k-1}{3}\\&\quad \equiv \frac{1}{27}(-1)^{\frac{p+1}{2}+\frac{p-4}{3}}\prod _{j=1}^{p-2}(j!)^2\times \prod _{k\in \{2+3a:\ 0\le a\le \frac{p-7}{3}\}}\left( \frac{k(k+3)}{3}\right) ^2\\\&\qquad \times \, 3^{-\frac{p-4}{3}}\prod _{\begin{array}{c} 3\le k\le p-4\\ 3\mid k \end{array}}(k^2+k-1)\\&\quad \equiv 3^{-\frac{p+5}{3}}(-1)^{\frac{p+1}{2}+\frac{p-4}{3}}\prod _{j=1}^{p-2}(j!)^2 \times \prod _{k\in \{2+3a:\ 0\le a\le \frac{p-7}{3}\}}\left( \frac{k(k+3)}{3}\right) ^2\\&\qquad \times \, \prod _{\begin{array}{c} 3\le k<\frac{p-1}{2}\\ 3\mid k \end{array}}(k^2+k-1)((p-1-k)^2+(p-1-k)-1)\\&\qquad \times \,\left( \left( \frac{p-1}{2}\right) ^2+\frac{p-1}{2}-1\right) , \end{aligned}$$

and hence,

$$\begin{aligned} \bigg |F\left( \frac{i}{j}\right) \bigg |_{1\le i,j\le p-1}\equiv & {} 3^{-\frac{p+5}{3}}(-1)^{\frac{p+1}{2}+\frac{p-4}{3}}\frac{p^2-5}{4}\prod _{j=1}^{p-2}(j!)^2\nonumber \\{} & {} \times \prod _{k\in \{2+3a:\ 0\le a\le \frac{p-7}{3}\}}\left( \frac{k(k+3)}{3}\right) ^2\nonumber \\{} & {} \times \prod _{\begin{array}{c} 3\le k<\frac{p-1}{2}\\ 3\mid k \end{array}}(k^2+k-1)^2\ (\mathrm{{mod}}\ p). \end{aligned}$$
(2.18)

Observe that

$$\begin{aligned} \genfrac(){}{}{3}{p}^{-\frac{p+5}{3}}\genfrac(){}{}{-1}{p}^{\frac{p+1}{2}+\frac{p-4}{3}}\genfrac(){}{}{-5}{p}= & {} \genfrac(){}{}{-1}{p}^{\frac{p+1}{2}+\frac{p-4}{3}+1}\genfrac(){}{}{5}{p}\nonumber \\= & {} (-1)^{\frac{p-1}{2}\cdot \frac{p+1}{2}}\left( \frac{5}{p}\right) =\genfrac(){}{}{5}{p}. \end{aligned}$$
(2.19)

Combining (2.17)–(2.19), we obtain

$$\begin{aligned} \left( \frac{D_p^*(1,1)}{p}\right) =\left( \frac{|F(i/j)|_{1\le i, j\le p-1}}{p}\right) =\genfrac(){}{}{5}{p}\ \text{ or }\ 0. \end{aligned}$$

Case 2. \(p\equiv 2\ (\mathrm{{mod}}\ 3)\).

By Lemma 2.1 and the identity (2.1), we have

$$\begin{aligned} \bigg |F\left( \frac{i}{j}\right) \bigg |_{1\le i,j\le p-1}&\equiv -\frac{1}{27}\prod _{k=2}^{p-3}\left( -\frac{k}{3}\genfrac(){}{}{k+1}{3}+\frac{1}{3}\genfrac(){}{}{k+2}{3}\right) \prod _{1\le i<j\le p-1}(i-j)\left( \frac{1}{i}-\frac{1}{j}\right) \\&\equiv -\frac{1}{27}(-1)^{\frac{p+1}{2}}(-1)^{\frac{p-5}{3}}\prod _{j=1}^{p-2}(j!)^2\times \prod _{k\in \{2+3a:\ 0\le a\le \frac{p-5}{3}\}}\frac{1}{3}\\&\quad \times \prod _{k\in \{4+3a:\ 0\le a\le \frac{p-8}{3}\}}\frac{k}{3}\times \prod _{k\in \{3+3a:\ 0\le a\le \frac{p-8}{3}\}}\frac{k+1}{3}\\&\equiv 3^{-(p-4)-3}(-1)^{\frac{p+1}{2}+\frac{p-5}{3}+1}\prod _{j=1}^{p-2}(j!)^2\times \prod _{k\in \{4+3a:\ 0\le a\le \frac{p-8}{3}\}}k^2. \end{aligned}$$

Combining this with (2.17), we finally obtain

$$\begin{aligned} \begin{aligned} \left( \frac{D_p^*(1,1)}{p}\right)&=\left( \frac{|F(i/j)|_{1\le i, j\le p-1}}{p}\right) =\genfrac(){}{}{3}{p}^{-p+1}\genfrac(){}{}{-1}{p}^{\frac{p+1}{2}+\frac{p-5}{3}+1}\\ =&\genfrac(){}{}{-1}{p}^{\frac{p+1}{2}+\frac{p-5}{3}+1}=\genfrac(){}{}{-1}{p}. \end{aligned} \end{aligned}$$

In view of the above, we have completed our proof of Theorem 1.2. \(\square \)

3 Proof of Theorem 1.3

Lemma 3.1

Let \(p>5\) be a prime, and let \(b,c\in {\mathbb {Z}}\). Then

$$\begin{aligned} \left( \frac{D_p^*(b,c)}{p}\right) =\left( \frac{c}{p}\right) ^{\frac{(p-1)(p-3)}{8}}\left( \frac{\left( {\begin{array}{c}p-3\\ 0\end{array}}\right) _{b,c}\left( {\begin{array}{c}p-3\\ 1\end{array}}\right) _{b,c}^2}{p}\right) \left( \frac{W(\frac{p-1}{2})\prod _{k=2}^{(p-3)/2}W(k)^2}{p}\right) ,\nonumber \\ \end{aligned}$$
(3.1)

where

$$\begin{aligned} W(k)=\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{b,c}+\left( {\begin{array}{c}p-3\\ p-1-k\end{array}}\right) _{b,c}c^{p-1-k}. \end{aligned}$$

Proof

Let \(G(x)=(x^2+bx+c)^{p-3}\). For \(1\le i, j\le p-1\), we have

$$\begin{aligned} \frac{(i^2+bij+cj^2)^{p-3}}{j^{2(p-3)}}=\left( \frac{i^2}{j^2}+b\frac{i}{j}+c\right) ^{p-3}=G\left( \frac{i}{j}\right) , \end{aligned}$$

and hence,

$$\begin{aligned} \left( \frac{D_p^*(b,c)}{p}\right) =\left( \frac{|G(i/j)|_{1\le i, j\le p-1}}{p}\right) \end{aligned}$$

by Fermat’s little theorem. In view of (2.11), and Lemmas 2.1 and 2.2, we see that

$$\begin{aligned} \begin{aligned} \bigg |G\left( \frac{i}{j}\right) \bigg |_{1\le i,j\le p-1}&= \left( {\begin{array}{c}p-3\\ 0\end{array}}\right) _{b,c}\left( {\begin{array}{c}p-3\\ 1\end{array}}\right) _{b,c}^2\\&\quad \times c\prod _{k=2}^{p-3}\left( \left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{b,c}+\left( {\begin{array}{c}p-3\\ p-1-k\end{array}}\right) _{b,c}c^{p-1-k}\right) \\&\quad \times \prod _{1\le i<j\le p-1}(i-j)\left( \frac{1}{i}-\frac{1}{j}\right) \\&= \left( {\begin{array}{c}p-3\\ 0\end{array}}\right) _{b,c}\left( {\begin{array}{c}p-3\\ 1\end{array}}\right) _{b,c}^2c\prod _{k=2}^{p-3}W(k)\times (-1)^{\frac{p+1}{2}}\prod _{j=1}^{p-2}(j!)^2\\&=\left( {\begin{array}{c}p-3\\ 0\end{array}}\right) _{b,c}\left( {\begin{array}{c}p-3\\ 1\end{array}}\right) _{b,c}^2cW\left( \frac{p-1}{2}\right) \prod _{k=2}^{\frac{p-3}{2}}(W(k)W(p-1-k))\\&\quad \times (-1)^{\frac{p+1}{2}}\prod _{j=1}^{p-2}(j!)^2. \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} W(p-1-k)=\left( {\begin{array}{c}p-3\\ p-1-k\end{array}}\right) _{b,c}+c^k\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{b,c}\equiv c^kW(k)\ (\mathrm{{mod}}\ p) \end{aligned}$$

for all \(k=2,\ldots ,(p-3)/2\), we have

$$\begin{aligned} \prod _{k=2}^{\frac{p-3}{2}}W(k)W(p-1-k)&=\prod _{k=2}^{\frac{p-3}{2}}\genfrac(){}{}{c^kW(k)^2}{p}=\genfrac(){}{}{c}{p}^{\sum _{k=2}^{\frac{p-3}{2}}k} \left( \frac{\prod _{k=2}^{(p-3)/2}W(k)^2}{p}\right) \\&=\genfrac(){}{}{c}{p}^{\frac{p^2-4p-5}{8}}\left( \frac{\prod _{k=2}^{(p-3)/2}W(k)^2}{p}\right) . \end{aligned}$$

Combining the above, we immediately obtain the desired identity (3.1). \(\square \)

Proof of Theorem 1.3

Applying Theorem 1.1 with \(c=2\), we see that \((\frac{D_p^*(2,2)}{p})=0\) if \(p\equiv \pm 3\ (\mathrm{{mod}}\ 8)\). Below we assume that \(p\equiv \pm 1\ (\mathrm{{mod}}\ 8)\).

Let \(k\in \{0,\ldots ,p-3\}\). Taking \(b=c=2\) in Lemma 2.4 and (2.2), we get

$$\begin{aligned} 8\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{2,2}\equiv (k+3)\left( {\begin{array}{c}p-2\\ k-1\end{array}}\right) _{2,2}-2(k-3)\left( {\begin{array}{c}p-2\\ k+1\end{array}}\right) _{2,2}-2\left( {\begin{array}{c}p-1\\ k\end{array}}\right) _{2,2}\ (\mathrm{{mod}}\ p)\nonumber \\ \end{aligned}$$
(3.2)

and

$$\begin{aligned} 4\left( {\begin{array}{c}p-2\\ k\end{array}}\right) _{2,2}\equiv {\left\{ \begin{array}{ll} \left( {\begin{array}{c}p-1\\ -1\end{array}}\right) _{2,2}+2\left( {\begin{array}{c}p-1\\ 1\end{array}}\right) _{2,2}-2\ (\mathrm{{mod}}\ p)&{}\text{ if }\ k=0,\\ (k+1)\left( {\begin{array}{c}p-1\\ k-1\end{array}}\right) _{2,2}-2(k-1)\left( {\begin{array}{c}p-1\\ k+1\end{array}}\right) _{2,2}\ (\mathrm{{mod}}\ p)&\text{ otherwise }. \end{array}\right. }\qquad \end{aligned}$$
(3.3)

Combining (3.2) and (3.3), we have

$$\begin{aligned}{} & {} 32\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{2,2}\equiv 4(k+3)\left( {\begin{array}{c}p-2\\ k-1\end{array}}\right) _{2,2}-8(k-3)\left( {\begin{array}{c}p-2\\ k+1\end{array}}\right) _{2,2}-8\left( {\begin{array}{c}p-1\\ k\end{array}}\right) _{2,2}\nonumber \\{} & {} \equiv {\left\{ \begin{array}{ll} 20\left( {\begin{array}{c}p-1\\ 1\end{array}}\right) _{2,2}-8\left( {\begin{array}{c}p-1\\ 3\end{array}}\right) -8\ (\mathrm{{mod}}\ p)&{}\text{ if }\ k=1,\\ k(k+3)\left( {\begin{array}{c}p-1\\ k-2\end{array}}\right) _{2,2}-4(k+2)(k-2)\left( {\begin{array}{c}p-1\\ k\end{array}}\right) _{2,2}+4k(k-3)\left( {\begin{array}{c}p-1\\ k+2\end{array}}\right) _{2,2}\ (\mathrm{{mod}}\ p)&\text{ otherwise }, \end{array}\right. }\nonumber \\ \end{aligned}$$
(3.4)

and also

$$\begin{aligned} 32\left( {\begin{array}{c}p-3\\ p-1-k\end{array}}\right) _{2,2}\equiv & {} (k^2-k-2)\left( {\begin{array}{c}p-1\\ p-3-k\end{array}}\right) _{2,2}-4(k^2+2k-3)\left( {\begin{array}{c}p-1\\ p-1-k\end{array}}\right) _{2,2}\nonumber \\{} & {} +\,4(k^2+5k+4)\left( {\begin{array}{c}p-1\\ p+1-k\end{array}}\right) _{2,2}\ (\mathrm{{mod}}\ p) \end{aligned}$$
(3.5)

if \(2\le k\le p-3\).

For any \(2\le k\le p-3\), define

$$\begin{aligned} W(k)=\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{2,2}+\left( {\begin{array}{c}p-3\\ p-1-k\end{array}}\right) _{2,2}2^{p-1-k}. \end{aligned}$$

Then

$$\begin{aligned} 32W(k)&= 32\left( {\begin{array}{c}p-3\\ k\end{array}}\right) _{2,2}+2^{p-1-k}\times 32\left( {\begin{array}{c}p-3\\ p-1-k\end{array}}\right) _{2,2}\\&= k(k+3)\left( {\begin{array}{c}p-1\\ k-2\end{array}}\right) _{2,2}-4(k+2)(k-2)\left( {\begin{array}{c}p-1\\ k\end{array}}\right) _{2,2}+4k(k-3)\left( {\begin{array}{c}p-1\\ k+2\end{array}}\right) _{2,2}\\&\quad +\,2^{p-1-k}\bigg ((k^2-k-2)\left( {\begin{array}{c}p-1\\ p-3-k\end{array}}\right) _{2,2}-4(k^2+2k-3)\left( {\begin{array}{c}p-1\\ p-1-k\end{array}}\right) _{2,2}\\&\quad +\,4(k^2+5k+4)\left( {\begin{array}{c}p-1\\ p+1-k\end{array}}\right) _{2,2}\bigg ). \end{aligned}$$

Define the sequence \((u_n)_{n\ge 0}\) by

$$\begin{aligned} u_0=0,\ u_1=1,\quad \text{ and }\quad u_{n+1}=-2u_n-2u_{n-1}\quad \text{ for }\ n=1,2,3,\ldots . \end{aligned}$$

By Luo and Sun [3, (4.3)],

$$\begin{aligned} \left( {\begin{array}{c}p-1\\ p-k\end{array}}\right) _{2,2}\equiv u_k\ (\mathrm{{mod}}\ p) \end{aligned}$$
(3.6)

for all \(k=0,1,\ldots ,p-1\). Thus,

$$\begin{aligned}{} & {} 32W(k)= k(k+3)u_{p-k+2}-4(k+2)(k-2)u_{p-k}+4k(k-3)u_{p-k-2}\nonumber \\{} & {} +\,2^{-k}\left( (k+1)(k-2)u_{k+3}-4(k-1)(k+3)u_{k+1}+4(k+1)(k+4)u_{k-1}\right) \nonumber \\ \end{aligned}$$
(3.7)

for all \(k=2,\ldots ,p-3\).

Clearly, (3.5) with \(k=(p-1)/2\) yields that

$$\begin{aligned} 32W\left( \frac{p-1}{2}\right)= & {} \frac{p-1}{2}\cdot \frac{p+5}{2}u_{\frac{p+5}{2}}-4\frac{p+3}{2} \cdot \frac{p-5}{2}u_{\frac{p+1}{2}}+4\frac{p-1}{2}\cdot \frac{p-7}{2}u_{\frac{p-3}{2}}\nonumber \\{} & {} +\,\genfrac(){}{}{2}{p}\left( \frac{p+1}{2}\cdot \frac{p-5}{2}u_{\frac{p+5}{2}}-4\frac{p-3}{2}\cdot \frac{p+5}{2}u_{\frac{p+1}{2}}\right. \nonumber \\{} & {} \left. +4\frac{p+1}{2}\cdot \frac{p+7}{2}u_{\frac{p-3}{2}}\right) . \end{aligned}$$
(3.8)

By Luo and Sun [3, (4.9)], for any \(k\in {\mathbb {N}}\) we have

$$\begin{aligned} u_k=(-4)^{\lfloor {\frac{k}{4}}\rfloor }\times {\left\{ \begin{array}{ll} 0&{}\text{ if }\ k\equiv 0\ (\mathrm{{mod}}\ 4),\\ 1&{}\text{ if }\ k\equiv 1\ (\mathrm{{mod}}\ 4),\\ -2&{}\text{ if }\ k\equiv 2\ (\mathrm{{mod}}\ 4),\\ 2&{}\text{ if }\ k\equiv 3\ (\mathrm{{mod}}\ 4).\end{array}\right. } \end{aligned}$$
(3.9)

Case 1. \(p\equiv 1\ (\mathrm{{mod}}\ 8)\).

By (3.4) and (3.6), we have

$$\begin{aligned} \left( {\begin{array}{c}p-3\\ 0\end{array}}\right) _{2,2}= & {} \frac{1}{2}\left( {\begin{array}{c}p-1\\ 0\end{array}}\right) _{2,2}=\frac{u_p}{2} =\frac{1}{2}(-4)^{\lfloor {\frac{p}{4}}\rfloor }\nonumber \\\equiv & {} \frac{1}{2}\cdot 2^{\frac{p-1}{2}}=\frac{1}{2}\genfrac(){}{}{2}{p}=\frac{1}{2}\ (\mathrm{{mod}}\ p). \end{aligned}$$
(3.10)

Write \(p=8q+1\) with \(q\in {\mathbb {N}}\). In view of (3.8) and (3.9), we have

$$\begin{aligned} \begin{aligned} 32W\left( \frac{p-1}{2}\right)&= 2\left( 4q(4q+3)u_{4q+3}-4(4q+2)(4q-2)u_{4q+1}+4(4q)(4q-3)u_{4q-1}\right) \\&= 8q(4q+3)\times 2(-4)^q-8(4q+2)(4q-2)(-4)^q\\&\quad +32q(4q-3)\times 2(-4)^{q-1} \end{aligned} \end{aligned}$$

and hence,

$$\begin{aligned} \begin{aligned} W\left( \frac{p-1}{2}\right)&=\ (-1)^q\left( 2q(4q+3)2^{2q-2}-(4q+2)(4q-2)2^{2q-2}-2q(4q-3)2^{2q-2}\right) \\&= (-1)^q2^{2q-2}\left( 8q^2+6q-16q^2+4-8q^2+6q\right) \\&= (-1)^q2^{2q}(-q+1)(4q+1)=(-1)^{\frac{p+7}{8}}2^{2q}\cdot \frac{p-9}{8}\cdot \frac{p+1}{2}. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \genfrac(){}{}{W(\frac{p-1}{2})}{p}=\genfrac(){}{}{-1}{p}^{\frac{p+7}{8}}\genfrac(){}{}{-1}{p}=1. \end{aligned}$$
(3.11)

Note that

$$\begin{aligned} \genfrac(){}{}{\left( {\begin{array}{c}p-3\\ 0\end{array}}\right) _{2,2}}{p}=\left( \frac{2}{p}\right) =1 \end{aligned}$$

by (3.10). Combining this with Lemma 3.1, (3.12) and (3.11), we obtain

$$\begin{aligned} \genfrac(){}{}{D_p^*(2,2)}{p}\ne -1. \end{aligned}$$

Case 2. \(p\equiv 7\ (\mathrm{{mod}}\ 8)\).

In this case, we write \(p=8q+7\) with \(q\in {\mathbb {N}}\). For \(2\le k\le p-3\), write \(k=4s+r\) with \(s\in {\mathbb {N}}\) and \(r\in \{0,1,2,3\}.\) We will first show that \(W(k)\not \equiv 0\ (\mathrm{{mod}}\ p)\) for any \(k\in \{2,3,\ldots ,p-3\}\).

Subcase 2.1. \(r=0\).

In this subcase, by (3.7), (3.9) and Fermat’s little theorem, we have

$$\begin{aligned}&32W(k)\\&\quad \equiv k(k+3)u_{p-k+2}-4(k+2)(k-2)u_{p-k}+4k(k-3)u_{p-k-2}\\&\qquad +\,2^{-k}\left( (k+1)(k-2)u_{k+3}-4(k-1)(k+3)u_{k+1}+4(k+1)(k+4)u_{k-1}\right) \\&\quad \equiv k(k+3)(-4)^{\lfloor {\frac{p-k+2}{4}}\rfloor }-8(k+2)(k-2)(-4)^{\lfloor {\frac{p-k}{4}}\rfloor }+4k(k-3)(-4)^{\lfloor {\frac{p-k-2}{4}}\rfloor }\\&\qquad +\,2^{-k}\left( 2(k+1)(k-2)(-4)^{\lfloor {\frac{k+3}{4}}\rfloor }-4(k-1)(k+3)(-4)^{\lfloor {\frac{k+1}{4}}\rfloor }\right. \\&\quad \left. +\,8({k+1})(k+4)(-4)^{\lfloor {\frac{k-1}{4}}\rfloor }\right) \\&\quad \equiv 4s(4s+3)(-4)^{2q-s+2}-8(4s+2)(4s-2)(-4)^{2q-s+1}+16s(4s-3)(-4)^{2q-s+1}\\&\qquad +\,2^{-k}\left( 2(4s+1)(4s-2)(-4)^{s}-4(4s-1)(4s+3)(-4)^{s}+8(k+1)(k+4)(-4)^{s-1}\right) \\&\quad \equiv 2k(k+3)(-4)^{-s}\times \genfrac(){}{}{2}{p}+4(k+2)(k-2)(-4)^{-s}\genfrac(){}{}{2}{p}-2k(k-3)\genfrac(){}{}{2}{p} (-4)^{-s}\\&\qquad +\,2^{-4s+1}(k+1)(k-2)(-4)^{s}\\&\qquad -\,2^{-4s+2}(k-1)(k+3)(-4)^s-2^{-4s+1}(k+1)(k+4)(-4)^s\\&\quad = 2(-4)^{-s}(-4k-8)\not \equiv 0\ (\mathrm{{mod}}\ p). \end{aligned}$$

Subcase 2.2. \(r=1\).

In this subcase, by (3.7) and (3.9) we have

$$\begin{aligned} 32W(k)&\equiv k(k+3)u_{p-k+2}-4(k+2)(k-2)u_{p-k}+4k(k-3)u_{p-k-2}\\&\quad +\,2^{-k}\left( (k+1)(k-2)u_{k+3}-4(k-1)(k+3)u_{k+1}+4(k+1)(k+4)u_{k-1}\right) \\&\equiv 8(k+2)(k-2)(-4)^{\lfloor {\frac{p-k}{4}}\rfloor }+2^{-k}\times 8(k-1)(k+3)(-4)^{\lfloor {\frac{k+1}{4}}\rfloor }\\&\equiv 8(k+2)(k-2)(-4)^{2q-s+1}+2^{-k}\times 8(k-1)(k+3)(-4)^s\\&\equiv -4(k+2)(k-2)(-4)^{-s}\genfrac(){}{}{2}{p}+2^{-4s+2}(k-1)(k+3)(-4)^s\\&= (-4)^{-s+1}(-1-2k)\not \equiv 0\ (\mathrm{{mod}}\ p). \end{aligned}$$

Subcase 2.3. \(r=2\).

In view of (3.7) and (3.9), we have

$$\begin{aligned}&32W(k)\\&\quad \equiv k(k+3)u_{p-k+2}-4(k+2)(k-2)u_{p-k}+4k(k-3)u_{p-k-2}\\&\qquad +\,2^{-k}\left( (k+1)(k-2)u_{k+3}-4(k-1)(k+3)u_{k+1}+4(k+1)(k+4)u_{k-1}\right) \\&\quad \equiv 2k(k+3)(-4)^{\lfloor {\frac{p-k+2}{4}}\rfloor }-4(k+2)(k-2)(-4)^{\lfloor {\frac{p-k}{4}}\rfloor }+8k(k-3)(-4)^{\lfloor {\frac{p-k-2}{4}}\rfloor }\\&\qquad +\,2^{-k}\left( (k+1)(k-2)(-4)^{\lfloor {\frac{k+3}{4}}\rfloor }-8(k-1)(k+3)(-4)^{\lfloor {\frac{k+1}{4}}\rfloor }\right. \\&\qquad \left. +\,4(k+1)(k+4)(-4)^{\lfloor {\frac{k-1}{4}}\rfloor }\right) \\&\quad \equiv 2k(k+3)(-4)^{2q-s+1}-4(k+2)(k-2)(-4)^{2q-s+1}+8k(k-3)(-4)^{2q-s}\\&\qquad +\,2^{-k}\left( (k+1)(k-2)(-4)^{s+1}-8(k-1)(k+3)(-4)^s+4(k+1)(k+4)(-4)^s\right) \\&\quad \equiv -(-4)^{-s}k(k+3)+2(-4)^{-s}(k+2)(k-2)+(-4)^{-s}k(k-3)\\&\qquad +\,2^{-4s-2}\left( -4(k+1)(k-2)(-4)^s-8(k-1)(k+3)(-4)^s+4(k+1)(k+4)(-4)^s\right) \\&\quad = (-4)^{-s+1}(k-1)\not \equiv 0\ (\mathrm{{mod}}\ p). \end{aligned}$$

Subcase 2.4. \(r=3\).

In this subcase, by (3.7) and (3.9) we have

$$\begin{aligned} 32W(k)&\equiv k(k+3)u_{p-k+2}-4(k+2)(k-2)u_{p-k}+4k(k-3)u_{p-k-2}\\&\quad +\,2^{-k}\left( (k+1)(k-2)u_{k+3}-4(k-1)(k+3)u_{k+1}+4(k+1)(k+4)u_{k-1}\right) \\&\equiv -2k(k+3)(-4)^{\lfloor {\frac{p-k+2}{4}}\rfloor }-8k(k-3)(-4)^{\lfloor {\frac{p-k-2}{4}}\rfloor }\\&\quad +\,2^{-k}(-2(k+1)(k-2)(-4)^{\lfloor {\frac{k+3}{4}}\rfloor }-8(k+1)(k+4)(-4)^{\lfloor {\frac{k-1}{4}}\rfloor })\\&\equiv -2k(k+3)(-4)^{2q-s+1}-8k(k-3)(-4)^{2q-s}\\&\quad +\,2^{-4s}\left( (k+1)(k-2)(-4)^s-(k+1)(k+4)(-4)^s\right) \\&\equiv k(k+3)(-4)^{-s}-k(k-3)(-4)^{-s}+(k+1)(k-2)(-4)^{-s}\\&\quad -\,(k+1)(k+4)(-4)^{-s}\\&= -6(-4)^{-s}\not \equiv 0\ (\mathrm{{mod}}\ p). \end{aligned}$$

In view of the discussions for all the four subcases, we do have

$$\begin{aligned} W(k)\not \equiv 0\ (\mathrm{{mod}}\ p)\quad \text{ for } \text{ any }\ 2\le k\le p-3. \end{aligned}$$
(3.12)

By (3.4) and (3.6), we have

$$\begin{aligned} \begin{aligned} 32\left( {\begin{array}{c}p-3\\ 1\end{array}}\right) _{2,2}&= 20\left( {\begin{array}{c}p-1\\ 1\end{array}}\right) _{2,2}-8\left( {\begin{array}{c}p-1\\ 3\end{array}}\right) _{2,2}-8\\&\equiv 20u_{p-1}-8u_{p-3}-8=-40(-4)^{\lfloor {\frac{p-1}{4}}\rfloor }-8=10(-4)^{2q+2}-8\\&= 10\times 2^{4q+4}-8=20\times 2^{\frac{p-1}{2}}-8=20\genfrac(){}{}{2}{p}-8=12\ (\mathrm{{mod}}\ p) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \left( {\begin{array}{c}p-3\\ 0\end{array}}\right) _{2,2}= & {} \frac{1}{2}\left( {\begin{array}{c}p-1\\ 0\end{array}}\right) _{2,2} \\\equiv & {} \frac{u_p}{2}=(-4)^{\lfloor {\frac{p}{4}}\rfloor }=-\frac{1}{2}2^{\frac{p-1}{2}} \equiv \ -\frac{1}{2}\genfrac(){}{}{2}{p}=-\frac{1}{2}\ (\mathrm{{mod}}\ p). \end{aligned}$$

Therefore,

$$\begin{aligned} \genfrac(){}{}{\left( {\begin{array}{c}p-3\\ 1\end{array}}\right) _{2,2}}{p}\ne 0\quad \text{ and }\quad \genfrac(){}{}{\left( {\begin{array}{c}p-3\\ 0\end{array}}\right) _{2,2}}{p}=-1. \end{aligned}$$
(3.13)

By (3.8) and (3.9), we have

$$\begin{aligned} \begin{aligned} 32W\left( \frac{p-1}{2}\right)&\equiv 2(4q+3)(4q+6)u_{4q+6}-8(4q+5)(4q+1)u_{4q+4}+8(4q+3)4qu_{4q+2}\\&\equiv 16(4q+3)(4q+6)(-4)^q-16(4q+3)4q(-4)^q\ (\mathrm{{mod}}\ p), \end{aligned} \end{aligned}$$

and hence,

$$\begin{aligned} \begin{aligned} W\left( \frac{p-1}{2}\right)&\equiv (4q+3)(2q+3)(-4)^q-(4q+3)2q(-4)^q\\&= (-1)^q((4q+3)(2q+3)2^{2q}-(4q+3)2q2^{2q})=3(-1)^q2^{2q}\frac{p-1}{2}\ (\mathrm{{mod}}\ p). \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \genfrac(){}{}{W(\frac{p-1}{2})}{p}=\genfrac(){}{}{3}{p}(-1)^{\frac{p-7}{8}}\genfrac(){}{}{p-1}{p}\genfrac(){}{}{2}{p}=\genfrac(){}{}{3}{p}(-1)^{\frac{p+1}{8}}. \end{aligned}$$
(3.14)

Combining Lemma 3.1, (3.12)–(3.14), we finally obtain

$$\begin{aligned} \genfrac(){}{}{D_p^*(2,2)}{p}=\genfrac(){}{}{\left( {\begin{array}{c}p-3\\ 0\end{array}}\right) _{2,2}}{p}\genfrac(){}{}{W(\frac{p-1}{2})}{p}=-1\times \genfrac(){}{}{3}{p}(-1)^{\frac{p+1}{8}}=\genfrac(){}{}{p}{3}(-1)^{\frac{p+1}{8}}. \end{aligned}$$

In view of the above, we have finished our proof of Theorem 1.3. \(\square \)