Cyclic codes of length 5p with MDS symbol-pair

Abstract

Let p be a prime with $5 |(p-1)$. Let S be a set of all repeated-root cyclic codes ${\mathcal {C}}=\langle g(x)\rangle$, $(x^5-1)|g(x)$, of length 5p over a field field ${\mathbb {F}}_p$, whose Hamming distances are at most 7. In this paper, we present a method to find all maximum distance separable (MDS) symbol-pair codes in S. By this method we can easily obtain the results in Ma and Luo (Des Codes Cryptogr 90:121–137, 2022) and new MDS symbol-pair codes, so we remain two possible MDS symbol-pair codes for readers.

1 Introduction

Symbol-pair codes introduced by Cassuto and Blaum [1] are designed to protect against pair errors in symbol-pair read channels. Cassuto and Litsyn [3] constructed cyclic symbol-pair codes using algebraic methods and showed that there exist symbol-pair codes whose rates are strictly higher, compared to codes for the Hamming metric with the same relative distance. Yaakobi et al. [16] studied b-symbol read channels and generalized some of the known results for symbol-pair codes to those for b-symbol read channels. Dinh et al. [9,10,11] investigated the symbol-pair weight distributions of repeated-root constacyclic codes etc.

The minimum symbol-pair distance plays an important role in determining the error correcting capability of a symbol-pair code. In general, a code over ${\mathbb {F}}_q$ of length n with size M and minimum pair-distance $d_p$ is called an $(n, M, d_p)$ symbol-pair code. An $(n, M, d_p)$ symbol-pair code can correct up to $\lfloor (d_p-1)/2\rfloor$ pair errors (see [1, Proposition 3]). Chee et al. [4] gave the Singleton-type bound for symbol-pair codes relates the parameters n, M and $d_p$.

Lemma 1.1

[4] (Singleton Bound) Let q be a prime power and $2 \le d_p \le n$. If ${\mathcal {C}}$ is an $(n, M, d_p)$ symbol-pair code over ${\mathbb {F}}_q$, then $M \le q^{n-d_p+2}$. If $M = q^{n-d_p+2}$, then it is called an maximum distance separable (MDS) symbol-pair code.

A q-ary MDS symbol-pair code with parameters $(n, M, d_p)$ is simply called an MDS $(n, d_p)$ symbol-pair code.

There are several works that have contributed to the constructions of MDS symbol-pair codes. Chee et al. [4, 5] obtained many classes of MDS symbol-pair codes from classical MDS codes and interleaving method of Cassuto and Blaum [1]. Moreover, they obtained nontrivial MDS symbol-pair codes with length $(q^2+2q)/2$ by employing classical MDS codes and Eulerian graphs of certain girth. Kai et al. [12] constructed MDS symbol-pair codes with $d_p=5$ based on constacyclic codes. Later Kai et al. [13] derived three families of MDS symbol-pair codes by using repeated-root constacyclic codes. Ding et al. [7] obtained MDS symbol-pair codes with $d_p=6$, whose lengths from 6 to $q^2 + 1$, moreover, they found some MDS symbol-pair codes with $d_p\ge 7$ utilizing elliptic curves. Then they investigated MDS b-symbol codes [8]. Li et al. [14] gave a number of MDS symbol-pair codes with $d_p=7$ by analyzing some linear fractional transformations. Chen et al. [6] obtained MDS symbol-pair codes with $d_p=8$ of length 3p from repeated-root cyclic codes. Recently, Ma and Luo [15] constructed two classes of MDS symbol-pair codes with $d_p=10$ and $d_p=12$ from repeated-root cyclic codes of length 3p over ${\mathbb {F}}_p$. However, it becomes difficult to find MDS symbol-pair codes possessing comparatively large length and minimum pair-distance.

In this paper, let p be a prime with $5 |(p-1)$. Let S be a set of all repeated-root cyclic codes ${\mathcal {C}}=\langle g(x)\rangle$, $(x^5-1)|g(x)$, we present a method to find MDS symbol-pair codes of length 5p over ${\mathbb {F}}_p$. Moreover, by the method we can easily obtain the results in [15]. This paper is organized as follows. In Sect. 2, basic notations and results about cyclic codes and symbol-pair codes are provided. In Sect. 3, an unique class of MDS symbol-pair codes with $d_p=12$ among all repeated-root cyclic codes whose Hamming distance is equal to 6 are investigated. In Sect. 4, we conclude this paper with remarks.

2 Preliminaries

In this section, we review some basic notations, results on cyclic codes, and symbol-pair codes over a finite field, which will be used to prove our main results in the sequel.

2.1 Cyclic code

Let ${\mathbb {F}}_q$ be a finite field with q elements, where $q=p^s$, p is a prime and s is a positive integer. Let ${\mathcal {C}}$ be an [n, l] linear code over ${\mathbb {F}}_q$, i.e., it is an l-dimensional subspace of ${\mathbb {F}}_{q}^n$. If for each codeword $(c_0,c_1,\ldots ,c_{n-1})\in {\mathcal {C}}$, $( c_{n-1},c_0,\ldots ,c_{n-2})$ is also in $\mathcal {C}$, then we call $\mathcal {C}$ a cyclic code. We identify a codeword $\textbf{c} =(c_0, c_1, \ldots , c_{n-1})$ in ${\mathcal {C}}$ with the polynomial $c(x)=c_0+c_1x+c_2x^2+\cdots +c_{n-1}x^{n-1}$ in ${\mathbb {F}}_q[x]/\langle x^n-1\rangle$. A code ${\mathcal {C}}$ of length n over ${\mathbb {F}}_q$ corresponds to a subset of ${\mathbb {F}}_q[x]/\langle x^n-1\rangle$. Then ${\mathcal {C}}$ is a cyclic code if and only if the corresponding subset is an ideal of ${\mathbb {F}}_q[x]/\langle x^n-1\rangle$. Hence there exists a monic divisor g(x) of $x^n-1\in {\mathbb {F}}_q[x]$ such that

$$\begin{aligned} {\mathcal {C}} =\langle g(x) \rangle =\{ f(x)g(x)\pmod {x^n-1}: f(x)\in {\mathbb {F}}_q[x]\}. \end{aligned}$$

The g(x) is called the generator polynomial of ${\mathcal {C}}$.

A cyclic code is called simple-root cyclic code if $\gcd (n,p)=1$ and a repeated-root cyclic code if p|n. Castagnoli et al. in [2] studied the Hamming distance of repeated-root cyclic codes by using polynomial algebra, they showed that the Hamming distance of a repeated-root cyclic code ${\mathcal {C}}$ can be expressed in terms of $d_H(\overline{{\mathcal {C}}}_t)$, where $\overline{{\mathcal {C}}}_t$ are simple-root cyclic codes fully determined by ${\mathcal {C}}$.

Let ${\mathcal {C}}=\langle g(x)\rangle$ be a repeated-root cyclic code of length $\ell p^s$ over ${\mathbb {F}}_q$, where $\ell >1$ is a positive integer such that $\gcd (\ell ,p)=1$ and s is a positive integer. Suppose that $g(x)=\Pi _{i=1}^s m_i(x)^{e_i}$ is the factorization of g(x) over ${\mathbb {F}}_q$, where $m_i(x), i=1,\ldots ,s$ are distinct monic irreducible polynomials of multiplicity $e_i$. Fixing an integer t, $0\le t\le p^s-1$, we define $\overline{{\mathcal {C}}}_t=\langle {\overline{g}}_t(x)\rangle$ a simple-root cyclic code of length $\ell$ over ${\mathbb {F}}_q$, where ${\overline{g}}_t(x)$ is the product of those irreducible factors $m_i(x)$ with $e_i>t$. If this product is equal to $x^{\ell }-1$, i.e., $\overline{{\mathcal {C}}}_t$ contains only the zero codeword, then $d_H(\overline{{\mathcal {C}}}_t)=\infty$. If all $e_i$ satisfy $e_i\le t$, then ${\overline{g}}_{t}(x)=1$ and $d_H(\overline{{\mathcal {C}}}_t)=1$.

The following lemma will be used to determine the Hamming distance of repeated-root cyclic codes ${\mathcal {C}}$, which obtained from [2].

Lemma 2.1

[2] Let ${\mathcal {C}}=\langle g(x)\rangle$ be a repeated-root cyclic code of length $\ell p^s$ over ${\mathbb {F}}_q$, where p is a prime with $\gcd (\ell ,p)=1$ and s is a positive integer. Then

$$\begin{aligned} d_H({\mathcal {C}}) = \min \{P_t\cdot d_H(\overline{{\mathcal {C}}}_t) : t \in T\}, \end{aligned}$$

where for each $t\in T=\{t:0\le t\le p^s-1\}$, $t=t_{0}+t_{1}p+\cdots +t_{s-1}p^{s-1}$ is the p-adic representation and $P_t=\prod _{m=0}^{s-1}(t_m+1)=w_{H}((x-1)^t)$.

2.2 Symbol-pair codes

For $x=(x_0,x_1,\ldots ,x_{n-1})\in {\mathbb {F}}_q^n$, the symbol-pair read vector of x is

$$\begin{aligned} \pi _p(x) =((x_0, x_1), (x_1, x_2), \ldots , (x_{n-1}, x_{0} )). \end{aligned}$$

For a code ${\mathcal {C}}\subset {\mathbb {F}}_q^n$, there is the symbol-pair code generated by ${\mathcal {C}}$:

$$\begin{aligned} \pi _p({\mathcal {C}}):=\{\pi _p(x): x\in {\mathcal {C}}\}. \end{aligned}$$

Let $x=(x_0, x_1, \ldots , x_{n-1})$ and $y=(y_0, y_1, \ldots , y_{n-1})\in {\mathbb {F}}_q^n$. Recall that the Hamming weight of the vector x is defined as $w_H (x) =|\{i: x_i \ne 0, 0\le i\le n-1\}|$ and the Hamming distance between x and y is defined as $d_H (x,y) =|\{i:x_i \ne y_i, 0\le i\le n-1 \}|$. Define the symbol-pair weight of x as

$$\begin{aligned} w_p(x)=w_H(\pi _p(x))=|\{(x_i,x_{i+1}): (x_i,x_{i+1})\ne (0,0), 0\le i\le n-1\}|, \end{aligned}$$

define the symbol-pair distance between x and y as

$$\begin{aligned} d_p(x, y)= & {} d(\pi _p(x), \pi _p(y))\\ {}= & {} |\{i: (x_i, x_{i+1})\ne (y_i, y_{i+1}),0\le i\le n-1\}|, \end{aligned}$$

where the subscripts $i+1$ are reduced modulo n.

An $(n, M, d_p)$ symbol-pair code $\pi _p({\mathcal {C}})$ generated by ${\mathcal {C}}\subset {\mathbb {F}}_q^n$ has size M and minimum symbol-pair distance $d_p$, where $d_p = \min \{d_p(x, y):x, y \in {\mathcal {C}}, x\ne y\}$. Similar to the classical case, if ${\mathcal {C}}$ is a linear code, then the minimum symbol-pair distance of $\pi _p({\mathcal {C}})$ is the smallest symbol-pair weight of nonzero codewords of $\pi _p({\mathcal {C}})$, that is

$$\begin{aligned} d_p({\mathcal {C}})=\min \{w_p(x):x \in {\mathcal {C}}, x\ne 0\}. \end{aligned}$$

It is known in [1] that for any $0< d_H({\mathcal {C}}) < n$,

$$\begin{aligned} d_H({\mathcal {C}}) +1\le d_p({\mathcal {C}}) \le 2d_H({\mathcal {C}}). \end{aligned}$$

Let $S=\{(x_i, x_{i+1}): 0\le i\le n-1\}$ be the set from the vector x. There are two subsets of S:

$$\begin{aligned} S_0=\{(x_i,x_{i+1} )\in S: x_i\ne 0\} \end{aligned}$$

and

$$\begin{aligned} S_1=\{(x_i,x_{i+1} )\in S: x_i=0, x_{i+1}\ne 0\}. \end{aligned}$$

It is obvious that $w_H(x)=|S_0|$ and

$$\begin{aligned} w_p(x)=|S_0|+L, \end{aligned}$$

(2.1)

where $L= |S_1|$. In fact if $x=(x_0,x_1,\ldots , x_{n-1})\in {\mathbb {F}}_q^n$ is viewed as a cycle of length n, then L is the number of a sequence of 0’s in the cyclic of x. For example, in $x=(1,0,0,1,0,0,0,1,0,1)$ and $y=(0,1,0,0,1,0,1,0,1,0)\in {\mathbb {F}}_{2}^{10}$, we have $L=3$ and $L=4$, respectively.

In this paper, we will utilize repeated-root cyclic codes to obtain a class of new MDS symbol-pair codes. A simple notation is given below.

Definition 2.2

The support of a polynomial $f(y)=\sum _{i=0}^{\ell -1}a_iy^i$ is the set

$$\begin{aligned} supp(f)=\{ i: a_i\ne 0, 0\le i\le \ell -1\}, \end{aligned}$$

and denote the number of elements in supp(f) by N.

3 MDS symbol-pair codes

In this section, we always assume that p is a prime number and $5|(p-1)$. There is an irreducible factorization over ${\mathbb {F}}_p$:

$$\begin{aligned} x^{5 p}-1=\prod _{i=0}^{4}(x-\zeta ^i)^{p }, \end{aligned}$$

where $\zeta$ ia a primitive 5-th root of unity in ${\mathbb {F}}_p$.

Let

$$\begin{aligned} S=\left\{ {\mathcal {C}}=\langle g(x)\rangle : g(x)=\prod _{i=0}^{4}(x-\zeta ^{i})^{j_i}, p\ge j_0\ge j_1\ge j_2\ge j_3 \ge j_4\ge 1\right\} \end{aligned}$$

(3.1)

be a set of nontrivial cyclic codes with length 5p over ${\mathbb {F}}_p$.

First, we shall find MDS symbol-pair codes from all repeated-root cyclic codes of length 5p with $d_H({\mathcal {C}})\le 7$ defined as (3.1).

Theorem 3.1

If ${\mathcal {C}}=\langle g(x)\rangle \in S, d_H({\mathcal {C}})\le 7$, and ${\mathcal {C}}$ is an MDS symbol-pair code. Then there is a unique possible code as follows: $d_H({\mathcal {C}})=6$ and

$$\begin{aligned} g(x)=(x-1)^5(x-\zeta )^2(x-\zeta ^2)(x-\zeta ^3)(x-\zeta ^4). \end{aligned}$$

(3.2)

Proof

Suppose that ${\mathcal {C}}=\langle g(x)\rangle$ is a $[5p,l,d_H({\mathcal {C}})]$ cyclic code with MDS symbol-pair. Then $d_p({\mathcal {C}})=5p-l+2$ with $l=5p-\deg (g(x))$, so

$$\begin{aligned} d_p( {{\mathcal {C}}} )=\deg (g(x))+2. \end{aligned}$$

(3.3)

In (3.1), $(x^5-1)| {g}(x)$ and $\deg (g(x))\ge 5$. Recall that $d_p( {{\mathcal {C}}} )\le 2d_H( {{\mathcal {C}}} )$. Then $d_H({\mathcal {C}})\ge 4$.

By Lemma 2.1, $d_H({\mathcal {C}})=\min \{P_t\cdot d_H(\overline{{\mathcal {C}}}_t): t=1,2,\ldots ,p-1\}$, where $\overline{ {\mathcal {C}} }_t=\langle {g}_t(x)\rangle$, it is clear that ${g}_0 (x)= x^5-1$ and $P_0\cdot d_H(\overline{{\mathcal {C}}}_0)=\infty$. So we only consider $1\le t\le p-1$ and $P_t=t+1$.

(1) Suppose that $d_H({\mathcal {C}})=4$. Then $d_p( {{\mathcal {C}}} )\le 8$.

If $t=1$, then $d_H(\overline{{\mathcal {C}}}_1)\ge 2$ and ${g}_1 (x)$ has at least one factor: $x-1$, this means $j_0\ge 2$.

If $t=2$, then $d_H(\overline{{\mathcal {C}}}_2)\ge 2$ and ${g}_2 (x)$ has at least one factor: $x-1$, this means $j_0\ge 3$.

Thus $j_0\ge 3$ and $j_1\ge j_2\ge j_3\ge j_4 \ge 1$ and $\deg (g(x))\ge 7$, which is a contradiction.

(2) Suppose that $d_H({\mathcal {C}})=5$. Then $d_p({\mathcal {C}})\le 10$.

If $t=1$, then $d_H(\overline{{\mathcal {C}}}_1)\ge 3$ and ${g}_1 (x)$ has at least two factors: $x-1$ and $x-\zeta$, this means $j_0\ge 2$ and $j_1\ge 2$.

If $t=2$, then $d_H(\overline{{\mathcal {C}}}_2)\ge 2$ and ${g}_2 (x)$ has at least one factor: $x-1$, this means $j_0\ge 3$.

If $t=3$, then $d_H(\overline{{\mathcal {C}}}_3)\ge 2$ and ${g}_2 (x)$ has at least one factor: $x-1$, this means $j_0\ge 4$.

Thus $j_0\ge 4$, $j_1\ge 2$, and $j_2\ge j_3\ge j_4 \ge 1$, and $\deg (g(x))\ge 9$, which is a contradiction.

(3) Suppose that $d_H({\mathcal {C}})=7$. Then $d_p({\mathcal {C}})\le 14$.

If $t=1$, then $d_H(\overline{{\mathcal {C}}}_1)\ge 4$ and ${g}_1 (x)$ has at least three factors: $x-1$, $x-\zeta$, and $x-\zeta ^2$, this means $j_0\ge 2$, $j_1\ge 2$, $j_2\ge 2$.

If $t=2$, then $d_H(\overline{{\mathcal {C}}}_2)\ge 3$ and ${g}_2 (x)$ has at least two factors: $x-1$ and $x-\zeta$, this means $j_0\ge 3$ and $j_1\ge 3$.

If $t=3$, then $d_H(\overline{{\mathcal {C}}}_3)\ge 2$ and ${g}_2 (x)$ has at least one factor: $x-1$, this means $j_0\ge 4$.

If $t=4$, then $d_H(\overline{{\mathcal {C}}}_4)\ge 2$ and ${g}_2 (x)$ has at least one factor: $x-1$, this means $j_0\ge 5$.

If $t=5$, then $d_H(\overline{{\mathcal {C}}}_5)\ge 2$ and ${g}_2 (x)$ has at least one factor: $x-1$, this means $j_0\ge 6$.

Thus $j_0\ge 6$, $j_1\ge 3$, $j_2\ge 2$, and $j_3\ge j_4 \ge 1$, and $\deg (g(x))\ge 13$, which is a contradiction.

(4) Suppose that $d_H({\mathcal {C}})=6$. Then $d_p({\mathcal {C}})\le 12$.

If $t=1$, then $d_H(\overline{{\mathcal {C}}}_1)\ge 3$ and ${g}_1 (x)$ has at least two factors: $x-1$ and $x-\zeta$, this means $j_0\ge 2$ and $j_1\ge 2$.

If $t=2$, then $d_H(\overline{{\mathcal {C}}}_2)\ge 2$ and ${g}_2 (x)$ at least one factor: $x-1$, this means $j_0\ge 3$.

If $t=3$ and $t=4$, then either ${g}_3 (x)$ or ${g}_4 (x)$ has at least one factor: $x-1$, this means $j_0\ge 5$.

Thus $j_0\ge 5$, $j_1\ge 2$, and $j_2\ge j_3\ge j_4\ge 1$. Then

$$\begin{aligned} {g}(x)=(x-1)^{5+j_0^\prime }(x-\zeta )^{2+j_1^\prime }(x-\zeta ^2)^{1+j_2^\prime } (x-\zeta ^3)^{1+j_3^\prime }(x-\zeta ^4)^{1+j_4^\prime }, \end{aligned}$$

where for $0\le i\le 4$, $j_i^\prime$ is a positive integer, and $\deg (g(x))=10+\sum _{i=0}^4j_i^\prime$.

By (3.3), we have

$$\begin{aligned} d_p({\mathcal {C}})=10+\sum _{i=0}^4j_i^\prime +2 \le 12, \end{aligned}$$

it can only have

$$\begin{aligned} j_0^\prime =j_1^\prime = j_2^\prime = j_3^\prime =j_4^\prime =0. \end{aligned}$$

Hence if ${\mathcal {C}}=\langle g(x)\rangle \in S$ and ${\mathcal {C}}$ is an MDS symbol-pair code, then there is a unique possible code as follows: $d_H({\mathcal {C}})=6$ and

$$\begin{aligned} g(x)=(x-1)^5(x-\zeta )^2(x-\zeta ^2)(x-\zeta ^3)(x-\zeta ^4). \end{aligned}$$

This is completed the proof. $\square$

Next, we shall verify that the code in Theorem 3.1 is just MDS symbol-pair with $d_H({\mathcal {C}})=6$.

Suppose that c(x) is a nonzero code polynomial of ${\mathcal {C}}=\langle g(x)\rangle \in S$. Then g(x)|c(x) and c(x) can be written as the form $c(x)=\sum _{i=0}^{4} x^iV_i(x^{5})$, for convenience, we write

$$\begin{aligned} c(x)=(V_0(x^5),V_1(x^5),V_2(x^5),V_3(x^5),V_4(x^5) ), \end{aligned}$$

where $V_i(x^{5})$ is a polynomial of $x^{5}$. Let $N_i=|supp(V_i(x^5))|, 0\le i\le 4$, where each $supp(V_i(x^5)$ is in Definition 2.2.

By $c(1)=c(\zeta )=\cdots =c(\zeta ^{4})=0$, we obtain a system of 5 equations over ${\mathbb {F}}_p$ as follows:

$$\begin{aligned} \left( \begin{array}{cccc} (\zeta ^0)^0 &{} (\zeta ^0)^1 &{} \ldots &{} (\zeta ^0)^{4}\\ (\zeta ^{1})^0 &{} (\zeta ^{1})^1 &{} \ldots &{} (\zeta ^{1})^{4}\\ \vdots &{} \vdots &{} &{}\vdots \\ (\zeta ^{4})^0&{} (\zeta ^{4})^1&{} \ldots &{}(\zeta ^{4})^{4} \end{array} \right) \left( \begin{array}{c} V_0(1)\\ V_1(1)\\ \vdots \\ V_4(1) \end{array} \right) =0. \end{aligned}$$

(3.4)

It is easy to check that the coefficient matrix of (3.4) is nonsingular. Then

$$\begin{aligned} V_0(1)=V_1(1)=\cdots =V_4(1)=0, \end{aligned}$$

it is implied that $(x^{5}-1)|V_{i}(x^{5})$ for each $0\le i\le 4$. Suppose that $V_i(x^{5})=\sum _{j=0}^n a_j{(x^5)}^j$, it follows from $V_{i}(1)=0$ that $a_0=-(a_1+\ldots +a_n)$.

Theorem 3.2

Let g(x) be defined as (3.2) and ${\mathcal {C}}=\langle g(x)\rangle$. Then ${\mathcal {C}}$ is an MDS symbol-pair codes with $d_H({\mathcal {C}})=6$.

Now we give some lemmas to prove Theorem 3.2.

Lemma 3.3

If $w_H(c(x))=6$, then $w_p(c(x))=12$.

Proof

We divide into three cases to investigate $w_p(c(x))$ with $w_H(c(x))=6$.

Case 1: If $c(x)=(V_i(x^5), V_j(x^5))$ with $(N_i, N_j)=(4,2)$ and $0\le i< j\le 4$. Since $w_H(x^iV_i(x^5))=w_H(V_i(x^5))$, without loss of generality, we consider $c(x)=(V_0(x^5), V_k(x^5))$ with $1\le k\le 4$.

Suppose that $k\in \{2,3\}$. Then $L=6$ and $w_p(c(x))=12$.

Suppose that $k=1$. Let $V_0(x^5)=a_0+ a_1x^{5r_1}+a_2x^{5r_2}+a_3x^{5r_3}$ with $1\le r_1<r_2<r_3<p$ and $V_1(x^5)=b_1 (x^{5r_4}-1), 1\le r_4<p$. Then

$$\begin{aligned} c(x)= & {} a_0+ a_1x^{5r_1}+a_2x^{5r_2}+a_3x^{5r_3}+x(-b_1+b_1x^{5r_4})\nonumber \\= & {} a_0-b_1x+ a_1x^{5r_1}+a_2x^{5r_2}+a_3x^{5r_3}+ b_1x^{5r_4+1}\in {\mathbb {F}}_p^*[x]. \end{aligned}$$

The first, the second and the third formal derivative of c(x) respectively gives

$$\begin{aligned} c^{(1)}(x)= & {} -b_1 +5r_1a_1x^{5r_1-1}+5r_2a_2x^{5r_2-1}+5r_3a_3x^{5r_3-1}\\ {}{} & {} +(5r_4+1)b_1x^{5r_4},\\ c^{(2)}(x)= & {} 5r_1(5r_1-1)a_1x^{5r_1-2}+5r_2(5r_2-1)a_1x^{5r_2-2}\\{} & {} +5r_3(5r_3-1)a_1x^{5r_3-2}+5(5r_4+1)r_4b_1x^{5r_4-1}, \end{aligned}$$

and

$$\begin{aligned} c^{(3)}(x)= & {} 5r_1(5r_1-1)(5r_1-2)a_1x^{5r_1-3}+5r_2(5r_2-1)(5r_2-2)a_1x^{5r_2-3}\\{} & {} +5r_3(5r_3-1)(5r_3-2)a_1x^{5r_3-3}+5(5r_4+1)(5r_4-1)r_4b_1x^{5r_4-2}. \end{aligned}$$

Since $(x -1)^5$ and $(x-\zeta )^2$ are divisors of c(x), it follows from $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=0$, note that $a_0=-(a_1+a_2+a_3)$, that

$$\begin{aligned} B( a_1,a_2,a_3,b_1)^\top =0, \end{aligned}$$

(3.5)

where $B=(B_1,B_2,B_3,B_4)$, and for $1\le i\le 3$,

$$\begin{aligned} B_i=\left( \begin{array}{cccc} r_i\\ r_i\zeta ^{-1} \\ r_i(5r_i-1) \\ r_i(5r_i-1)(5r_i-2) \\ \end{array} \right) \end{aligned}$$

(3.6)

and

$$\begin{aligned} B_4=\left( \begin{array}{cccc} r_4\\ r_4 \\ r_4(5r_4+1)\\ r_4(25r_4^2-1)\\ \end{array} \right) . \end{aligned}$$

(3.7)

We make some elementary transformations:

$$\begin{aligned} B \sim \left( \begin{array}{cccc} 1&{} 0&{}0&{}0\\ 0&{}0&{}0&{}1\\ 0&{}5(r_2-r_1) &{}5(r_3-r_1)&{}5r_4+1\\ 0&{}0&{}25(r_3-r_2)(r_3-r_1)&{}\lambda \end{array}\right) , \end{aligned}$$

where $\lambda =(5r_4+1)(5r_4-5r_1-5r_2+2)$. Since $1\le r_1<r_2<r_3<p$, we can verfy that the matrix B is nonsingular, thus $a_1=a_2=a_3=b_1=0$, which contradicts with that $b_1, a_j\in {\mathbb {F}}_p^*, 0\le j\le 3$.

Suppose that $k=4$, that is

$$\begin{aligned} c(x)=a_0-b_1x^4+ a_1x^{5r_1}+a_2x^{5r_2}+a_3x^{5r_3}+ b_1x^{5r_4+4}\in {\mathbb {F}}_p^*[x], \end{aligned}$$

similarly, by $c(1)=0$ and $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=0$, then we derive a contradiction.

Hence if $c(x)=(V_i(x^5), V_j(x^5))$ with $(N_i, N_j)=(4,2)$, then $w_p(c(x))=12$.

Case 2: If $c(x)=(V_0(x^5), V_k(x^5)), 1\le k\le 4$, with $(N_0, N_k)=(3,3)$.

Let $V_0(x^5)=a_0+ a_1x^{5r_1}+a_2x^{5r_2}$ with $1\le r_1<r_2<p$ and $V_k(x^5)=b_0+ b_1x^{5r_3}+b_2x^{5r_4}$ with $1\le r_3<r_4 <p$, where $a_0=-a_1-a_2$ and $b_0=-b_1-b_2$. Then

$$\begin{aligned} c(x)= & {} a_0+ a_1x^{5r_1}+a_2x^{5r_2} +x^k(b_0+ b_1x^{5r_3}+b_2x^{5r_4})\\= & {} a_0+b_0x^k+ a_1x^{5r_1}+a_2x^{5r_2}+b_1x^{5r_3+k}+ b_2x^{5r_4+k}\in {\mathbb {F}}_p^*[x]. \end{aligned}$$

It is obvious that if $k\in \{2,3\}$, then $L=6$ and $w_p(c(x))=12$.

Suppose that $k=1$. Then

$$\begin{aligned} c(x)=-(a_1+a_2)-(b_1+b_2)x+ a_1x^{5r_1}+a_2x^{5r_2}+b_1x^{5r_3+1}+ b_2x^{5r_4+1}, \end{aligned}$$

by $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=0$, we have

$$\begin{aligned} (B_1,B_2,B_3^\prime ,B_4)\left( \begin{array}{cccc} a_1\\ a_2 \\ b_1\\ b_2\\ \end{array} \right) =0, \end{aligned}$$

where $B_1,B_2,B_4$ are defined as (3.6), (3.7) and $B_3^\prime$ is given by changing $r_4$ into $r_3$ in $B_4$. Note that $1\le r_1<r_2$ and $1\le r_3<r_4$, we can obtain that the determinant of $B=(B_1,B_2,B_3^\prime ,B_4)$ is not equal to 0. Hence $k=1$ is impossible.

Suppose that $k=4$, then

$$\begin{aligned} c(x)=-(a_1+a_2)-(b_1+b_2)x^4+ a_1x^{5r_1}+a_2x^{5r_2}+b_1x^{5r_3+4}+ b_2x^{5r_4+4}, \end{aligned}$$

similar to the argument with $k=1$, we know that $k=4$ is also impossible.

Case 3: If $c(x)= (V_0(x^5), V_i(x^5), V_j(x^5))$ with $(N_0, N_i, N_j)=(2,2,2)$ and $1\le i<j\le 4$.

Let $V_0(x^5)=a_1(x^{5r_1}-1)$, $V_i(x^5)=a_2(x^{5r_2}-1)$, and $V_j(x^5)=a_3(x^{5r_3}-1)$. Then

$$\begin{aligned} c(x)= & {} a_1(x^{5r_1}-1)+x^ia_2(x^{5r_2}-1)+x^ja_3(x^{5r_3}-1)\\= & {} -a_1-a_2x^{i}-a_3x^j+a_1x^{5r_1}+a_2x^{5r_2+i}+a_3x^{5r_3+j}\in {\mathbb {F}}_p^*[x]. \end{aligned}$$

Note that $1\le i<j\le 4$, then

$$\begin{aligned} (i,j)\in \{(1,2), (1,3), (1,4), (2,3), (2,4), (3,4)\}. \end{aligned}$$

The first and the second formal derivative of c(x) respectively gives

$$\begin{aligned} c^{(1)}(x)= & {} -ia_2x^{i-1}-j a_3x^{j-1} +5r_1a_1x^{5r_1-1}+(5r_2+i)a_2x^{5r_2+i-1}\\{} & {} +(5r_3+j)a_3x^{5r_3+j-1}, \end{aligned}$$

and

$$\begin{aligned} c^{(2)}(x)= & {} -i(i-1)a_2x^{i-2}-j(j-1) a_3x^{j-2} +5r_1(5r_1-1)a_1x^{5r_1-2}\\{} & {} + (5r_2+i)(5r_2+i-1)a_2x^{5r_2+i-2}+(5r_3+j)(5r_3+j-1)a_3x^{5r_3+j-2}. \end{aligned}$$

(1) Suppose that $(i,j)=(1,2)$. Since $(x -1)^5$ and $(x-\zeta )^2$ are divisors of c(x), $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=0$. Then

$$\begin{aligned} (B_1,B_2,B_3)\left( \begin{array}{c} a_1\\ a_2 \\ a_3\\ \end{array} \right) =0, \end{aligned}$$

(3.8)

where

$$\begin{aligned} B_1=\left( \begin{array}{c } r_{1} \\ r_1\zeta ^{-1} \\ r_1(5r_1-1) \\ \end{array} \right) , B_2=\left( \begin{array}{c} r_{2} \\ r_2 \\ r_2(5r_2+1) \\ \end{array} \right) , B_3=\left( \begin{array}{c} r_{3} \\ r_3\zeta \\ r_3(5r_3+3) \\ \end{array} \right) . \end{aligned}$$

(3.9)

We make some elementary transformations:

$$\begin{aligned} (B_1,B_2,B_3) \sim \left( \begin{array}{cccc} 1 &{} 0&{}0 \\ 0 &{} 1-\zeta ^{-1} &{} \zeta -\zeta ^{-1} \\ 0&{}5(r_2-r_1)+2&{}5(r_3-r_1)+4 \\ \end{array} \right) . \end{aligned}$$

Note that $1\le r_1,r_2,r_3<p$ are positive integers, we conclude that

$$\begin{aligned}{} & {} \left| \begin{array}{ll}1-\zeta ^{-1} &{} \zeta -\zeta ^{-1}\\ 5(r_2-r_1)+2 &{} 5(r_3-r_1)+4\end{array}\right| \\= & {} 5r_3-5r_1+4-(5r_2-5r_1+2)\zeta +(5r_2-5r_3-2)\zeta ^{-1}\ne 0. \end{aligned}$$

The solution of Eq. (3.8) has only zero, which is a contradiction.

(2) Suppose that $(i,j)=(1,3)$. By $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=0$, then

$$\begin{aligned} (B_1,B_2,B_3^\prime )\left( \begin{array}{cccc} a_1\\ a_2 \\ a_3\\ \end{array} \right) =0, \end{aligned}$$

(3.10)

where $B_1,B_2$ are defined as (3.9) and

$$\begin{aligned} B_3^\prime =\left( \begin{array}{c} r_{3} \\ r_3\zeta ^2 \\ r_3(5r_3+5) \\ \end{array} \right) . \end{aligned}$$

(3.11)

(3) Suppose that $(i,j)=(1,4)$. By $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=0$, then

$$\begin{aligned} (B_1,B_2,B_4 )\left( \begin{array}{cccc} a_1\\ a_2 \\ a_3\\ \end{array} \right) =0, \end{aligned}$$

(3.12)

where $B_1,B_2$ are defined as (3.9) and

$$\begin{aligned} B_4=\left( \begin{array}{c} r_{3} \\ r_3\zeta ^3 \\ r_3(5r_3+7) \\ \end{array} \right) .\end{aligned}$$

(3.13)

(4) Suppose that $(i,j)=(2,3)$. By $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=0$, then

$$\begin{aligned} (B_1,B_2^\prime ,B_3^\prime )\left( \begin{array}{cccc} a_1\\ a_2 \\ a_3\\ \end{array} \right) =0, \end{aligned}$$

(3.14)

where $B_1,B_3^\prime$ is defined as (3.9), (3.11), respectively, and

$$\begin{aligned} B_2^\prime =\left( \begin{array}{c} r_{2} \\ r_2\zeta \\ r_2(5r_2+3) \\ \end{array} \right) . \end{aligned}$$

(3.15)

(5) Suppose that $(i,j)=(2,4)$. By $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=0$, then

$$\begin{aligned} (B_1,B_2^{\prime },B_4 )\left( \begin{array}{cccc} a_1\\ a_2 \\ a_3\\ \end{array} \right) =0, \end{aligned}$$

(3.16)

where $B_1$, $B_2^{\prime }$, and $B_4$ is defined as (3.9), (3.15), and (3.13), respectively.

(6) Suppose that $(i,j)=(3,4)$. By $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=0$, then

$$\begin{aligned} (B_1,B_2^{\prime \prime },B_4 )\left( \begin{array}{cccc} a_1\\ a_2 \\ a_3\\ \end{array} \right) =0, \end{aligned}$$

(3.17)

where $B_1$ and $B_4$ is defined as (3.9) and (3.13), respectively, $B_2^{\prime \prime }$ is replaced $r_3$ by $r_2$ in $B_3^\prime$ defined as (3.11).

Similar to the case $i=1$ and $j=2$, the solutions of (3.10), (3.12), (3.14), (3.16) and (3.17) are zero, which are contradictions.

Hence if $w_H( c(x))=6$, then $w_p(c(x))=12$. $\square$

Lemma 3.4

If $w_H(c(x))=7$, then $w_p(c(x))\ge 12$.

Proof

We divide into three cases to investigate $w_p(c(x))$ with $w_H(c(x))=7$.

Case 1: If $c(x)=(V_0(x^5), V_k(x^5)), 1\le k\le 4$, with $(N_0, N_k)=(4,3)$.

Let $V_0(x^5)=a_0+ a_1x^{5r_1}+a_2x^{5r_2}+a_3x^{5r_3}$ with $1\le r_1<r_2<r_3<p$ and $V_k(x^5)=b_0+ b_1x^{5r_4}+b_2x^{5r_5}$ with $1\le r_4<r_5<p$. Then

$$\begin{aligned} c(x)= & {} a_0+ a_1x^{5r_1}+a_2x^{5r_2}+a_3x^{5r_3}+x^k(b_0+ b_1x^{5r_4}+b_2x^{5r_5})\\= & {} a_0+b_0x^k+ a_1x^{5r_1}+a_2x^{5r_2}+a_3x^{5r_3}+ b_1x^{5r_4+k}+ b_2x^{5r_5+k}\in {\mathbb {F}}_p^*[x]. \end{aligned}$$

It is obvious that if $k\in \{2,3\}$, then $L=7$ and $w_p(c(x))=14$.

Suppose that $k=1$. Then

$$\begin{aligned} c(x)=a_0+b_0x+ a_1x^{5r_1}+a_2x^{5r_2}+a_3x^{5r_3}+ b_1x^{5r_4+1}+ b_2x^{5r_5+1}, \end{aligned}$$

then $w_p(c(x))\ge 12$ except $(r_4,r_5)\in \{ (r_1,r_2), (r_1,r_3), (r_2,r_3)\}$. Without loss of generality, we assume that $r_4=r_1$ and $r_5=r_2$. That is

$$\begin{aligned} c(x)=a_0+b_0x+ a_1x^{5r_1}+ b_1x^{5r_1+1}+a_2x^{5r_2}+b_2x^{5r_2+1}+a_3x^{5r_3}, \end{aligned}$$

in this case $L=4$ and $w_p(c(x))=11$. But, this is impossible. The details are the below.

The i-th $1\le i\le 4$, formal derivative of c(x) respectively gives

$$\begin{aligned} c^{(1)}(x)= & {} b_0 +5r_1a_1x^{5r_1-1}+5r_2a_2x^{5r_2-1}+5r_3a_3x^{5r_3-1}\\{} & {} +(5r_1+1)b_1x^{5r_1}+(5r_2+1)b_2x^{5r_2}, \\ c^{(2)}(x)= & {} 5r_1(5r_1-1)a_1x^{5r_1-2}+5r_2(5r_2-1)a_1x^{5r_2-2}+5r_3(5r_3-1)a_1x^{5r_3-2}\\{} & {} +5(5r_1+1)r_1b_1x^{5r_1-1}+5(5r_2+1)r_2b_2x^{5r_2-1},\\ c^{(3)}(x)= & {} 5r_1(5r_1-1)(5r_1-2)a_1x^{5r_1-3}+5r_2(5r_2-1)(5r_2-2)a_1x^{5r_2-3}\\{} & {} +5r_3(5r_3-1)(5r_3-2)a_1x^{5r_3-3}+5(5r_1+1)(5r_1-1)r_1b_1x^{5r_1-2}\\{} & {} +5(5r_2+1)(5r_2-1)r_2b_2x^{5r_2-2}, \end{aligned}$$

and

$$\begin{aligned} c^{(4)}(x)= & {} 5r_1(5r_1-1)(5r_1-2)( 5r_1-3)a_1x^{5r_1-4}+5r_2(5r_2-1)(5r_2-2)( 5r_2-3)a_1x^{5r_2-4}\\{} & {} +5r_3(5r_3-1)(5r_3-2)( 5r_3-3)a_1x^{5r_3-4}+5(5r_1+1)(5r_1-1)( 5r_1-2)r_1b_1x^{5r_1-3}\\{} & {} +5(5r_2+1)(5r_2-1)( 5r_2-2)r_2b_2x^{5r_2-3}, \end{aligned}$$

Since $(x -1)^5$ and $(x-\zeta )^2$ are divisors of c(x), it follows from $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=c^{(4)}(1)=0$, note that $a_0=-(a_1+a_2+a_3)$ and $b_0=-(b_1+b_2)$, that

$$\begin{aligned} \left( \begin{array}{cc} B&{} \alpha \\ \beta &{} a_{55}\end{array}\right) (a_1,a_2,a_3,b_1,b_2)^\top =0, \end{aligned}$$

(3.18)

where B is defined as (3.5), $\alpha =( r_2, r_2, r_2(5r_2+1), r_2(25r_2^2-1))^\top$, $\beta = (r_1(5r_1-1)(5r_1-2)(5r_1-3),r_2(5r_2-1)(5r_2-2)(5r_2-3),r_3(5r_3-1)(5r_3-2)(5r_3-3),r_1(25r_1^2-1)(5r_1-2))$, and $a_{55}=r_2(25r_2^2-1)(5r_2-2)$. By make some elementary transformations, note that $1\le r_1<r_2<r_3<p$ and $1\le r_4<r_5<p$, we can check that the matrix $\left( \begin{array}{cc} B&{} \alpha \\ \beta &{} a_{55}\end{array}\right)$ is nonsingular, hence the solution of (3.18) is zero, which is a contradiction.

Suppose that $k=4$. Then

$$\begin{aligned} c(x)=a_0+b_0x^4+ a_1x^{5r_1}+a_2x^{5r_2}+a_3x^{5r_3}+ b_1x^{5r_4+4}+ b_2x^{5r_5+4}, \end{aligned}$$

then $w_p(c(x))\ge 12$ except $r_1=1$ and $(r_4,r_5)=(r_2-1,r_3-1)$. That is

$$\begin{aligned} c(x)=a_0+b_0x^4+ a_1x^{5}+ b_1x^{5r_2-1}+a_2x^{5r_2}+ b_2x^{5r_3-1}+a_3x^{5r_3}, \end{aligned}$$

using arguments similar to the above, $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=c^{(4)}(1)=0$, we derive a contradiction.

Hence if $c(x)=( V_0(x^5), V_k(x^5)), 1\le k\le 4$ with $(N_0, N_k)=(4,3)$, then $w_p(c(x))\ge 12$.

Case 2: If $c(x)=(V_0(x^5),V_k(x^5)), 1\le k\le 4$, with $(N_0, N_k)=(5,2)$. It is easy to see that $L\ge 5$ and $w_p(c(x))\ge 12$.

Case 3: If $c(x)= (V_0(x^5), V_i(x^5), V_j(x^5))$ with $(N_0, N_i, N_j)=(2,2,3)$ and $1\le i<j\le 4$.

Let $V_0(x^5)=a_1(x^{5r_1}-1)$, $V_i(x^5)=a_2(x^{5r_2}-1)$, and $V_j(x^5)= b_0+b_1x^{5r_3}+b_2x^{5r_4}$, where $1\le r_3<r_4<p$. Then

$$\begin{aligned} c(x)= & {} a_1(x^{5r_1}-1)+x^ia_2(x^{5r_2}-1)+x^j(b_0+b_1x^{5r_3}+b_2x^{5r_4})\\= & {} -a_1-a_2x^{i}+b_0x^j+a_1x^{5r_1}+a_2x^{5r_2+i}+b_1x^{5r_3+j}+b_2x^{5r_4+j}\in {\mathbb {F}}_p^*[x]. \end{aligned}$$

Note that $1\le i<j\le 4$, it is easy to check that $w_p(c(x))\ge 12$ except

$$\begin{aligned} (i,j)\in \{(1,2),(1,4),(3,4)\}. \end{aligned}$$

In the following, we discuss the subcases: (1) $i=1$ and $j=2$; (2) $i=1$ and $j=4$; (3) $i=3$ and $j=4$.

The first, the second, and the third formal derivative of c(x) respectively gives

$$\begin{aligned} c^{(1)}(x)= & {} -ia_2x^{i-1}+j b_0x^{j-1} +5r_1a_1x^{5r_1-1}+(5r_2+i)a_2x^{5r_2+i-1}\\{} & {} +(5r_3+j)b_1x^{5r_3+j-1}+(5r_4+j)b_2x^{5r_4+j-1},\\ c^{(2)}(x)= & {} -i(i-1)a_2x^{i-2}+j(j-1) b_0x^{j-2} +5r_1(5r_1-1)a_1x^{5r_1-2}\\{} & {} + (5r_2+i)(5r_2+i-1)a_2x^{5r_2+i-2}+(5r_3+j)(5r_3+j-1)b_1x^{5r_3+j-2}\\ {}{} & {} +(5r_4+j)(5r_4+j-1)b_2x^{5r_4+j-2}.\\ c^{(3)}(x)= & {} -i(i-1)(i-2)a_2x^{i-3}+j(j-1)(j-2) b_0x^{j-3} +5r_1(5r_1-1)(5r_1-2)a_1x^{5r_1-3}\\{} & {} + (5r_2+i)(5r_2+i-1)(5r_2+i-2)a_2x^{5r_2+i-3}\\ {}{} & {} +(5r_3+j)(5r_3+j-1)(5r_3+j-2)b_1x^{5r_3+j-3}\\{} & {} +(5r_4+j)(5r_4+j-1)(5r_4+j-2)b_2x^{5r_4+j-3}.\end{aligned}$$

(1) Suppose that $(i,j)=(1,2)$. Since $(x -1)^5$ and $(x-\zeta )^2$ are divisors of c(x), $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=0$. Then

$$\begin{aligned} (B_1,B_2,B_3, B_4)\left( \begin{array}{c} a_1\\ a_2 \\ b_1\\ b_2\\ \end{array} \right) =0,\end{aligned}$$

(3.19)

where

$$\begin{aligned} B_1=\left( \begin{array}{c } r_{1} \\ r_1\zeta ^{-1} \\ r_1(5r_1-1) \\ r_1(5r_1-1)(5r_1-2) \end{array} \right) , B_2=\left( \begin{array}{c} r_{2} \\ r_2 \\ r_2(5r_2+1) \\ r_2(5r_2+1)(5r_2-1)\\ \end{array} \right) , \end{aligned}$$

(3.20)

and

$$\begin{aligned} B_3=\left( \begin{array}{c} r_{3} \\ r_3\zeta \\ r_3(5r_3+3) \\ r_3(5r_3+2)(5r_3+1)\\ \end{array} \right) , B_4=\left( \begin{array}{c} r_{4} \\ r_4\zeta \\ r_4(5r_4+3) \\ r_4(5r_4+2)(5r_4+1)\\ \end{array} \right) . \end{aligned}$$

(3.21)

Note that $r_1,r_2,r_3<r_4<p$ are positive integers and $\zeta$ is a primitive 5-th root of unity in ${\mathbb {F}}_p$, by making some elementary transformations, we obtain $(B_1,B_2,B_3,B_4)$ is nonsingular. The solution of Eq. (3.19) has only zero, which is a contradiction.

(2) Suppose that $(i,j)=(1,4)$. By $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=0$, then

$$\begin{aligned} (B_1,B_2,B_3^\prime ,B_4^\prime )\left( \begin{array}{cccc} a_1\\ a_2 \\ b_1\\ b_2\\ \end{array} \right) =0, \end{aligned}$$

(3.22)

where $B_1,B_2$ are defined as (3.20) and

$$\begin{aligned} B_3^\prime =\left( \begin{array}{c} r_{3} \\ r_3\zeta ^3 \\ r_3(5r_3+7) \\ r_3((5r_3+5)(5r_3+4)+6)\\ \end{array} \right) , B_4^\prime =\left( \begin{array}{c} r_{4} \\ r_4\zeta ^3 \\ r_4(5r_4+7) \\ r_4((5r_4+5)(5r_4+4)+6)\\ \end{array} \right) . \end{aligned}$$

(3.23)

(3) Suppose that $(i,j)=(3,4)$. By $c^{(1)}(1)=c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=0$, then

$$\begin{aligned} (B_1,B_2^{\prime },B_3^\prime ,B_4^\prime )\left( \begin{array}{cccc} a_1\\ a_2 \\ b_1\\ b_2\\ \end{array} \right) =0, \end{aligned}$$

(3.24)

where $B_1$ and $B_3^\prime , B_4^\prime$ is defined as (3.20) and (3.23), respectively, and

$$\begin{aligned} B_2^\prime =\left( \begin{array}{c} r_{2} \\ r_2\zeta ^2 \\ r_2(5r_2+5) \\ r_2((5r_2+1)(5r_2+5)+6)\\ \end{array} \right) . \end{aligned}$$

Similar to the case $i=1$ and $j=2$, the solutions of (3.22) and (3.24) have zero, a contradiction.

Hence if $w_H( c(x))=7$, then $w_p(c(x))\ge 12$. $\square$

Lemma 3.5

If $w_H(c(x))=8$, then $w_p(c(x))\ge 12$.

Proof

If $w(c(x))=8$. Suppose that $c(x)= (V_0(x^5),V_k(x^5)),1\le k\le 4,$ with $(N_0, N_k)\in \{(2,6),(3,5),(4,4)\}$. Then $w_p(c(x))\ge 12$. We only need to consider the following two cases.

Case 1: If $c(x)= (V_0(x^5),V_i(x^5),V_j(x^5))$ with $(N_0, N_i,N_j)=(3, 3, 2)$ and $1\le i<j\le 4$.

Let $V_0(x^5)=a_0+ a_1x^{5r_1}+a_2x^{5r_2}$ with $1\le r_1<r_2<p$, $V_i(x^5)=b_0+ b_1x^{5r_3}+b_2x^{5r_4}$ with $1\le r_3<r_4 <p$, and $V_j(x^5)=b_3(x^{5r_5}-1)$, where $a_0=-a_1-a_2$ and $b_0=-b_1-b_2$. Then

$$\begin{aligned} c(x)= & {} a_0+ a_1x^{5r_1}+a_2x^{5r_2} +x^i(b_0+ b_1x^{5r_3}+b_2x^{5r_4})+x^j b_3(x^{5r_5}-1)\\= & {} a_0+b_0x^i-b_3x^j+ a_1x^{5r_1}+b_1x^{5r_3+i}+b_3x^{5r_3+j}+a_2x^{5r_2}+ b_2x^{5r_4+i}\in {\mathbb {F}}_p^*[x]. \end{aligned}$$

Note that $1\le i<j \le 4$, then

$$\begin{aligned} (i,j)\in \{(1,2), (1,3), (1,4), (2,3), (2,4), (3,4)\}. \end{aligned}$$

We can quickly check that $d_p(c(x))\ge 12$ except $(i,j)\in \{ (1,2), (1,4)\}$.

(1) Suppose that $(i,j)=(1,2)$. We can now see that if $r_1=r_3$ and $r_2=r_4$, then $w_p(c(x))=8+3=11$; otherwise, $w_p(c(x))\ge 12$. Without loss of generality, we assume that $r_1=r_3=1$ and $r_2=r_4=2$. Then

$$\begin{aligned} c(x)=a_0+b_0x -b_3x^2+ a_1x^{5 }+b_1x^{6}+b_3x^{7}+a_2x^{10}+ b_2x^{11}\in {\mathbb {F}}_p^*[x]. \end{aligned}$$

By $c^{(1)}(1) =c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=c^{(4)}(1)=0$, note that $a_0=-(a_1+a_2)$ and $b_0=-(b_1+b_2)$, we have

$$\begin{aligned} \left( \begin{array}{ccccc} 1 &{} 2&{}1&{}2&{}1 \\ \zeta ^4 &{} 2\zeta ^4 &{}1 &{} 2&{} \zeta \\ 2&{}9&{}3&{} 11&{}4 \\ 2&{}24&{}4 &{}33&{}7\\ 2&{}126&{}9 &{}198&{}21\\ \end{array} \right) \left( \begin{array}{cccc} a_1\\ a_2 \\ b_1\\ b_2\\ b_3\\ \end{array} \right) =0, \end{aligned}$$

it is easy to verify the solution of the above equation is zero, which is a contradiction.

(2) Suppose that $(i,j)=(1,4)$. We can easily observe that $w_p(c(x))\ge 12$ except the case $r_1=r_3=1$ and $r_2=r_4=2$. In a similar way, by $c^{(1)}(1) =c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=c^{(4)}(1)=0$, this is also a contradiction.

Case 2: If $c(x)= (V_0(x^5),V_i(x^5),V_j(x^5),V_k(x^5))$ with $(N_0, N_i,N_j,N_k)=(2,2,2,2)$ and $1\le i<j<k\le 4$.

Let $V_0(x^5)=a_1(x^{5r_1}-1)$, $V_i(x^5)=a_2(x^{5r_2}-1)$, $V_j(x^5)=a_3(x^{5r_3}-1)$, and $V_k(x^5)=a_4(x^{5r_4}-1)$. Then

$$\begin{aligned} c(x)= & {} a_1(x^{5r_1}-1)+a_2x^i(x^{5r_2}-1)+a_3x^j(x^{5r_3}-1)+a_4x^k(x^{5r_4}-1)\nonumber \\= & {} -a_1-a_2x^i-a_3x^j-a_4x^k+a_1x^{5r_1}+a_2x^{5r_2+i} +a_3x^{5r_3+j}+a_4x^{5r_4+k}.\nonumber \\ \end{aligned}$$

(3.25)

Note that $1\le i<j<k\le 4$. Then

$$\begin{aligned} (i,j,k)\in \{ (1,2,4),(1,3,4),(2,3,4), (1,2,3)\}. \end{aligned}$$

(1) Suppose that $(i,j,k)=(1,2,4)$. It follows from (3.25) that $w_p(c(x))\ge 12$ except $r_1=r_2=r_3=1$.

If $(i,j,k)=(1,2,4)$ and $r_1=r_2=r_3=1$, then

$$\begin{aligned} c(x)=-a_1-a_2x -a_3x^2-a_4x^4+a_1x^{5 }+a_2x^{6}+a_3x^{7}+a_4x^{5r_4+4}\in {\mathbb {F}}_p^*[x]. \end{aligned}$$

By $c^{(1)}(1) =c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=0$, we have

$$\begin{aligned} \left( \begin{array}{cccc} 1 &{} 1&{}1&{}r_4 \\ \zeta ^4 &{} 1 &{} \zeta &{}r_4\zeta ^{3} \\ 4&{}6&{}8&{} r_4(5r_4+7) \\ 12&{}24&{}42&{}\mu \\ \end{array} \right) \left( \begin{array}{cccc} a_1\\ a_2 \\ a_3\\ a_4\\ \end{array} \right) =0, \end{aligned}$$

where $\mu =r_4((5r_4+4)(5r_4+5)+6)$. The solution of the above equation has only zero, which is a contradiction.

(2) Suppose that $(i,j,k)=(1,3,4)$. It follows from (3.25) that $w_p(c(x))\ge 12$ except $r_1=r_2=1$ and $r_3=r_4<p$.

If $(i,j,k)=(1,3,4)$, $r_1=r_2= 1$, and $r_3=r_4$, then

$$\begin{aligned} c(x)=-a_1-a_2x -a_3x^3-a_4x^4+a_1x^{5 }+a_2x^{6}+a_3x^{5r_3+3}+a_4x^{5r_3+4}\in {\mathbb {F}}_p^*[x]. \end{aligned}$$

A similar argument to the above, by $c^{(1)}(1) =c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=0$, we conclude that if $(i,j,k)=(1,3,4)$ and $r_1=r_2=1$ and $r_3=r_4<p$ is impossible.

(3) Suppose that $(i,j,k)=(2,3,4)$. It follows from (3.25) that $w_p(c(x))\ge 12$ except $r_1=1$ and $r_2=r_3=r_4<p$.

If $(i,j,k)=(2,3,4)$, $r_1= 1$, and $r_2=r_3=r_4$, then

$$\begin{aligned} c(x)=-a_1-a_2x^2 -a_3x^3-a_4x^4+a_1x^{5 }+a_2x^{5r_2+2}+a_3x^{5r_2+3}+a_4x^{5r_2+4}\in {\mathbb {F}}_p^*[x]. \end{aligned}$$

A similar argument to the above, by $c^{(1)}(1) =c^{(1)}(\zeta )=c^{(2)}(1)=c^{(3)}(1)=0$, we know that if $(i,j,k)=(2,3,4)$ and $r_1 =1$ and $r_2=r_3=r_4$ is impossible.

(4) Suppose that $(i,j,k)=(1,2,3)$. It follows from (3.25) that $w_p(c(x))\ge 12$ except $r_1=r_2=r_3<p$ or $r_2=r_3=r_4<p$ or $r_1=r_2<p, r_3=r_4<p$. But the three cases are not happen.

Hence if $w_H( c(x))=8$ then $w_p(c(x))\ge 12$. $\square$

Now we are ready to complete the proof of Theorem 3.2.

Proof

From Lemmas 3.3, 3.4, 3.5, we know that for $0\ne c(x)\in {\mathcal {C}}$, if $6\le w_H(c(x))\le 8$, then $w_p(c(x))\ge 12$. Furthermore, if $w_H(c(x))\ge 9$, then by (2.1), it is easy to verify that no such codeword c(x) in ${\mathcal {C}}$ exists such that $w_p(c(x))< 12$. Hence we conclude that $d_p({\mathcal {C}})=12$.

Therefore, if $g(x)=(x-1)^5(x-\zeta )^2(x-\zeta ^2)(x-\zeta ^3)(x-\zeta ^4)$, then ${\mathcal {C}}=\langle g(x)\rangle$ is a (5p, 12) MDS symbol-pair code. This completes the proof of Theorem 3.2. $\square$

Example 3.6

Let $p=11$ and $g(x)=(x-1)^5(x-3)^2(x-9)(x-5)(x-4)$. Then ${\mathcal {C}}=\langle g(x)\rangle$ is a [55, 45, 6] cyclic code. By Theorem 3.2, its minimum symbol-pair distance is 12. The code ${\mathcal {C}}$ is an MDS symbol-pair code.

Example 3.7

Let $p=31$ and $g(x)=(x-1)^5(x-4)^2(x-16)(x-2)(x-8)$. Then ${\mathcal {C}}=\langle g(x)\rangle$ is a [155, 45, 6] cyclic code. By Theorem 3.2, its minimum symbol-pair distance is 12. The code ${\mathcal {C}}$ is an MDS symbol-pair code.

Suppose that $3|(p-1)$. Let

$$\begin{aligned} S^\prime =\{{\mathcal {C}}=\langle g(x)\rangle : g(x)=(x-1)^{j_0}(x-\omega )^{j_1}(x-\omega ^2)^{j_2}, p\ge j_0\ge j_1\ge j_2\ge 1\} \end{aligned}$$

be a set of nontrivial cyclic codes of length 3p over ${\mathbb {F}}_p$, where $\omega$ is a primitive 3-th root of unity in ${\mathbb {F}}_p$. From the proof of Theorem 3.1 and the results in [15], we have the following results.

Theorem 3.8

Let ${\mathcal {C}}=\langle g(x)\rangle \in S^\prime$ and $d_H({\mathcal {C}})= 5$. Then there is a unique MDS symbol-pair code of length 3p over ${\mathbb {F}}_p$ as follows:

$$\begin{aligned} g(x)=(x-1)^4(x-\omega )^2(x-\omega ^2)^2. \end{aligned}$$

Let ${\mathcal {C}}=\langle g(x)\rangle \in S^\prime$ and $d_H({\mathcal {C}})= 6$. Then there is a unique MDS symbol-pair code of length 3p over ${\mathbb {F}}_p$ as follows:

$$\begin{aligned} g(x)=(x-1)^5(x-\omega )^3(x-\omega ^2)^2. \end{aligned}$$

Furthermore, by the proof of Theorem 3.1, we know the following results.

Proposition 3.9

(1) If ${\mathcal {C}}=\langle g(x)\rangle \in S, d_H({\mathcal {C}})=8$, and ${\mathcal {C}}$ is an MDS symbol-pair code of length 5p over ${\mathbb {F}}_p$. Then there is a unique possible code as follows:

$$\begin{aligned} g(x)=(x-1)^7(x-\zeta )^3(x-\zeta ^2)^2(x-\zeta ^3)(x-\zeta ^4). \end{aligned}$$

(2) If ${\mathcal {C}}=\langle g(x)\rangle \in S^\prime , d_H({\mathcal {C}})=7$, and ${\mathcal {C}}$ is an MDS symbol-pair code of length 3p over ${\mathbb {F}}_p$. Then there is a unique possible code as follows:

$$\begin{aligned} g(x)=(x-1)^6(x-\omega )^3(x-\omega ^2)^3. \end{aligned}$$

Question 3.10

In Proposition 3.9, are two codes MDS symbol-pair codes?

4 Concluding remarks

Let p be a prime and $5|(p-1)$. Let S be a set of all repeated-root cyclic codes ${\mathcal {C}}=\langle g(x)\rangle$, $(x^5-1)|g(x)$, of length 5p over a field field ${\mathbb {F}}_p$. In this paper, we provided a method to find MDS symbol-pair codes in S whose Hamming distance is 6. By the method we can easily obtain the results in [15] and new MDS symbol-pair codes of length $\ell p$ over ${\mathbb {F}}_p$, where $\ell$ is a positive integer with $\ell |(p-1)$ and $(x^{\ell }-1)|g(x)$.