Constructions of MDS symbol-pair codes with minimum distance seven or eight

Abstract

Symbol-pair codes are proposed to guard against pair-errors in symbol-pair read channels. The minimum symbol-pair distance plays a vital role in determining the error-correcting capability and the constructions of symbol-pair codes with largest possible minimum symbol-pair distance is of great importance. Maximum distance separable (MDS) symbol-pair codes are optimal in the sense that such codes can acheive the Singleton bound. In this paper, for length 5p, two new classes of MDS symbol-pair codes with minimum symbol-pair distance seven or eight are constructed by utilizing repeated-root cyclic codes over ${\mathbb {F}}_{p}$, where p is a prime. In addition, we derive a class of MDS symbol-pair codes with minimum symbol-pair distance seven and length 4p.

Appendix Proof of Theorem 2:

Recall that ${\mathcal {C}}$ is a repeated-root cyclic code of length 5p over ${\mathbb {F}}_{p}$ with the generator polynomial

$$\begin{aligned} g(x)=\left( x-1\right) ^{3}\left( x-\beta \right) \left( x-\beta ^{2}\right) ^{2} \end{aligned}$$

where $\beta$ is a primitive 5-th root of unity in ${\mathbb {F}}_{p}$. By Lemma 3, one can derive that the parameter of ${\mathcal {C}}$ is $\left[ 5p,\,5p-6,\,4\right]$. Since ${\mathcal {C}}$ is not MDS, Lemma 4 yields that $d_{p}({\mathcal {C}})\ge 6$. Similar as Theorem 1, one can derive that there does not exist a codeword c(x) in ${\mathcal {C}}$ with $(w_H(c(x)), \,w_p(c(x)))=(5,\,6)$ or $(6,\,7)$. To prove that ${\mathcal {C}}$ is an MDS $\left( 5p,\,8\right) _{p}$ symbol-pair code, it suffices to determine that there does not exist a codeword in ${\mathcal {C}}$ with $(w_H(c(x)), \,w_p(c(x)))=(4,\,6)$, $(4,\,7)$ or $(5,\,7)$.

Case I   $(w_H(c(x)), \,w_p(c(x)))=(4,\,6)$. Since ${\mathcal {C}}$ is the subcode of the code occurred in Theorem 1 and the proof of Theorem 1 indicates that there does not exist a codeword c(x) in ${\mathcal {C}}$ with $(w_H(c(x)), \,w_p(c(x)))=(4,\,6)$ unless $p=41$. Now it is sufficient to show that for $p=41$, there does not exist a codeword c(x) in ${\mathcal {C}}$ with $(w_H(c(x)), \,w_p(c(x)))=(4,\,6)$. More precisely, we just need to consider Case II in Theorem 1. There are two subcases to be discussed:

• Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^{2}+a_{3}\,x^{5i+3}$ with $1\le i\le p-1$ and $a_{1},\,a_{2},\,a_{3}\in {\mathbb {F}}_p^{*}$. Notice that $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$ and Lemma 6 induces

  $$\begin{aligned} a_{1}=-\frac{\beta ^2+\beta +1}{\beta ^2},\quad a_{2}=\frac{\beta ^2+\beta +1}{\beta ^3},\quad a_{3}=-\frac{1}{\beta ^3}. \end{aligned}$$

  (17)

  It follows from $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ that

  $$\begin{aligned} \left\{ \begin{array}{l} a_{1}+2\,a_{2}+\left( 5i+3\right) a_{3}=0,\\ a_{1}+2\,a_{2}\,\beta ^2+\left( 5i+3\right) a_{3}\,\beta ^4=0\\ \end{array} \right. \end{aligned}$$

  which yields

  $$\begin{aligned} a_1\left( \beta ^4-1\right) +2\,a_2\left( \beta ^4-\beta ^2\right) =0. \end{aligned}$$

  Combining with (17), one can get $\left( \beta -1\right) ^2=0$, a contradiction.

• For the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^{2}+a_{3}\,x^{5i+4}$ with $0\le i\le p-2$ and $a_{1},\,a_{2},\,a_{3}\in {\mathbb {F}}_p^{*}$, by $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$ and Lemma 6, one can obtain that

  $$\begin{aligned} a_{1}=-\frac{1}{\beta },\quad a_{2}=-\frac{\beta }{\beta +1},\quad a_{3}=\frac{1}{\beta \left( \beta +1\right) }. \end{aligned}$$

  (18)

  On the other hand, $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ that

  $$\begin{aligned} \left\{ \begin{array}{l} a_{1}+2\,a_{2}+\left( 5i+4\right) a_{3}=0,\\ a_{1}+2\,a_{2}\,\beta ^2+\left( 5i+4\right) a_{3}\,\beta =0\\ \end{array} \right. \end{aligned}$$

  which induces

  $$\begin{aligned} a_1\left( \beta -1\right) =2\,a_2\left( \beta ^2-\beta \right) . \end{aligned}$$

  Together with (18), one can immediately obtain that

  $$\begin{aligned} 2\,\beta ^3=\beta +1. \end{aligned}$$

  This leads to

  $$\begin{aligned} \left( \beta -1\right) \left( 2\,\beta ^2+2\,\beta +1\right) =0. \end{aligned}$$

  The fact $\beta$ is a primitive 5-th root of unity implies that $2\beta ^2+2\beta +1=0$ and then one has

  $$\begin{aligned} \beta ^2+\beta =-\left( \beta ^2+\beta +1\right) =\beta ^4+\beta ^3 \end{aligned}$$

  which is impossible.

Case II   $(w_H(c(x)), \,w_p(c(x)))=(4,\,7)$. For this case, Lemma 1 implies that the cyclic shift of c(x) must have the form

$$\begin{aligned} \left( \star ,\,\star ,\,\mathbf {0},\,\star ,\,\mathbf {0},\,\star ,\,\mathbf {0}\right) . \end{aligned}$$

Assume that $c(x)=\left( x^{5}-1\right) ^{t}v(x)$, where $0\le t\le p-1$, $\left( x^{5}-1\right) \not \mid v(x)$ and

$$\begin{aligned} v(x)=v_{0}(x^{5})+x\,v_{1}(x^{5})+x^{2}\,v_{2}(x^{5})+x^{3}\,v_{3}(x^{5})+x^{4}\,v_{4}(x^{5}). \end{aligned}$$

Recall that $N_{v}=w_{H}\left( v(x)\,\mathrm{mod}\left( x^{5}-1\right) \right)$. Then by Lemma 5, one can deduce that

$$\begin{aligned} 4=w_{H}\left( \left( x^{5}-1\right) ^{t}\right) \cdot w_{H}\left( v(x)\,\mathrm{mod}\left( x^{5}-1\right) \right) =\left( 1+t\right) N_{v}. \end{aligned}$$

If $\left( N_{v},\,t\right) =\left( 1,\,3\right)$, then it is easily seen that the symbol-pair weight of c(x) is greater than 7.

If $\left( N_{v},\,t\right) =\left( 2,\,1\right)$, then there are three subcases to be discussed:

(1) For the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^{5i}+a_{3}\,x^{5j}$ with $1\le i<j \le p-1$ and $a_{1},\,a_{2},\,a_{3}\in {\mathbb {F}}_p^{*}$, it can be verified that

$$\begin{aligned} \left\{ \begin{array}{l} 1+a_{1}\,\beta +a_{2}+a_{3}=0,\\ 1+a_{1}\,\beta ^2+a_{2}+a_{3}=0\\ \end{array} \right. \end{aligned}$$

since $c\left( \beta \right) =c\left( \beta ^2\right) =0$. Then one can obtain that $a_{1}=0$, a contradiction.

(2) For the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^{5i+1}+a_{3}\,x^{5j+1}$ with $1\le i<j \le p-1$ and $a_{1},\,a_{2},\,a_{3}\in {\mathbb {F}}_p^{*}$, by $c\left( 1\right) =c\left( \beta \right) =0$, one can get

$$\begin{aligned} \left\{ \begin{array}{l} 1+a_{1}+a_{2}+a_{3}=0,\\ 1+a_{1}\,\beta +a_{2}\,\beta +a_{3}\,\beta =0.\\ \end{array} \right. \end{aligned}$$

This implies that $\beta =1$, which is impossible.

(3) For the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^{5i}+a_{3}\,x^{5j+1}$ with $1\le i<j \le p-1$ and $a_{1},\,a_{2},\,a_{3}\in {\mathbb {F}}_p^{*}$, it follows from $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ that

$$\begin{aligned} \left\{ \begin{array}{l} a_{1}+5i\,a_{2}+\left( 5j+1\right) a_{3}=0,\\ a_{1}+5i\,a_{2}\,\beta ^3+\left( 5j+1\right) a_{3}=0.\\ \end{array} \right. \end{aligned}$$

This leads to $\beta ^3=1$, a contradiction.

If $\left( N_{v},\,t\right) =\left( 4,\,0\right)$, then there are also three subcases to be considered:

(1) For the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^{5i+2}+a_{3}\,x^{5j+3}$ with $1\le i<j\le p-1$ and $a_{1},\,a_{2},\,a_{3}\in {\mathbb {F}}_p^{*}$, by Lemma 6 and $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$, one can derive that

$$\begin{aligned} a_{1}=-\frac{\beta ^2+\beta +1}{\beta ^2},\quad a_{2}=\frac{\beta ^2+\beta +1}{\beta ^3},\quad a_{3}=-\frac{1}{\beta ^3}. \end{aligned}$$

(19)

It follows from $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ that

$$\begin{aligned} \left\{ \begin{array}{l} a_{1}+\left( 5i+2\right) a_{2}+\left( 5j+3\right) a_{3}=0,\\ a_{1}+\left( 5i+2\right) a_{2}\,\beta ^2+\left( 5j+3\right) a_{3}\,\beta ^4=0\\ \end{array} \right. \end{aligned}$$

which indicates

$$\begin{aligned} \left\{ \begin{array}{l} \left( \beta ^4-1\right) a_{1}+(\beta ^4-\beta ^2)\left( 5i+2\right) a_{2}=0,\\ \left( \beta ^2-1\right) \left( 5i+2\right) a_{2}+\left( 5j+3\right) \left( \beta ^4-1\right) a_{3}=0.\\ \end{array} \right. \end{aligned}$$

Together with (19), one can immediately obtain that

$$\begin{aligned} \left\{ \begin{array}{l} \beta ^2+1=\left( 5i+2\right) \beta ,\\ \left( 5i+2\right) \left( \beta ^2+\beta +1\right) =\left( 5j+3\right) \left( \beta ^2+1\right) .\\ \end{array} \right. \end{aligned}$$

(20)

By substituting the value of $\beta ^2+1$ in the first equality into the second equality of (20), we can get

$$\begin{aligned} \left( 5i+2\right) \left( 5i+3\right) \beta =\left( 5i+2\right) \left( 5j+3\right) \beta \end{aligned}$$

which yields $i=j$ due to $p\not \mid (5i+2)$. This contradicts with $i<j$.

(2) Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^{5i+2}+a_{3}\,x^{5j+4}$ with $1\le i\le j\le p-2$ and $a_{1},\,a_{2},\,a_{3}\in {\mathbb {F}}_p^{*}$. The fact $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$ and Lemma 6 leads to

$$\begin{aligned} a_{1}=-\frac{1}{\beta },\quad a_{2}=-\frac{\beta }{\beta +1},\quad a_{3}=\frac{1}{\beta \left( \beta +1\right) }. \end{aligned}$$

(21)

It follows from $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ that

$$\begin{aligned} \left\{ \begin{array}{l} a_{1}=\beta \left( 5i+2\right) a_2,\\ \left( 5i+2\right) \left( \beta +1\right) a_{2}+\left( 5j+4\right) a_{3}=0.\\ \end{array} \right. \end{aligned}$$

By substituting (21), one can immediately derive that

$$\begin{aligned} \left\{ \begin{array}{l} \beta +1=\left( 5i+2\right) \beta ^3,\\ \left( 5i+2\right) \beta ^2\left( \beta +1\right) =5j+4.\\ \end{array} \right. \end{aligned}$$

(22)

This leads to $\left( 5i+2\right) ^2=5j+4$. Since it can be verified that $p\not \mid \left( 5i+2\right)$, it follows from $c^{\left( 2\right) }\left( 1\right) =0$ that

$$\begin{aligned} \beta ^2=\left( 5i+2\right) \left( 5i+3\right) . \end{aligned}$$

(23)

Then (21) and $c^{\left( 1\right) }\left( 1\right) =0$ indicates that

$$\begin{aligned} \left( 5i+2\right) \beta ^2+\beta -\left( 5j+3\right) =0. \end{aligned}$$

(24)

Let $t=5i+2$. Then one has $\beta +1=t\beta ^3$ and $\beta ^2=t\left( t+1\right)$ due to the first equality of (22) and (23). It follows from (24) that

$$\begin{aligned} t^2(t+1)+\beta -(t^2-1)=0 \end{aligned}$$

which implies $\beta +1=-t^3$. Combining with $\beta +1=t\beta ^3$, we have $\beta ^3=-t^2$. Since $\beta$ is a primitive 5-th root of unity, one can derive

$$\begin{aligned} \begin{aligned} 0=&\,\beta ^4+\beta ^3+\beta ^2+\beta +1 \\ =&\left( \beta +1\right) \left( \beta ^3+1\right) +\beta ^2 \\ =&-t^3\left( -t^2+1\right) +t\left( t+1\right) \\ =&\,t\left( t+1\right) \left( t^3-t^2+1\right) . \end{aligned} \end{aligned}$$

It follows from $t\left( t+1\right) =\beta ^2\ne 0$ that $t^3-t^2+1=0$. Then we obtain

$$\begin{aligned} \beta =-t^3-1=-t^2=\beta ^3 \end{aligned}$$

which yields $\beta ^2-1=0$, a contradiction.

(3) For the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^{5i+3}+a_{3}\,x^{5j+4}$ with $0\le i<j\le p-2$ and $a_{1},\,a_{2},\,a_{3}\in {\mathbb {F}}_p^{*}$, it follows from $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$ and Lemma 6 that

$$\begin{aligned} a_{1}=\frac{\beta ^2}{\beta +1},\quad a_{2}=-\frac{1}{\beta +1},\quad a_{3}=-\beta . \end{aligned}$$

(25)

Since $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$, one can immediately get

$$\begin{aligned} \left\{ \begin{array}{l} \left( 5i+3\right) \left( \beta ^4-1\right) a_{2}+\left( 5j+4\right) \left( \beta -1\right) a_{3}=0,\\ a_{1}(\beta -1)=\left( 5i+3\right) \left( \beta ^4-\beta \right) a_{2}.\\ \end{array} \right. \end{aligned}$$

Together with (25), one can conclude that

$$\begin{aligned} \left\{ \begin{array}{l} \left( 5i+3\right) \left( \beta ^2+1\right) +(5j+4)\beta =0,\\ \left( 5i+3\right) \left( \beta ^2+\beta +1\right) +\beta =0.\\ \end{array} \right. \end{aligned}$$

which indicates

$$\begin{aligned} \left\{ \begin{array}{l} \left( 5i+3\right) \beta ^2+(5j+4)\beta +5i+3=0,\\ \left( 5i+3\right) \beta ^2+(5i+4)\beta +5i+3=0.\\ \end{array} \right. \end{aligned}$$

It follows that $5(i-j)=0$, a contradiction.

Case III  $(w_H(c(x)), \,w_p(c(x)))=(5,\,7)$. In this case, we can assume that c(x) is of the form

$$\begin{aligned} \left( \mathbf {a},\,\mathbf {0},\,\mathbf {b},\,\mathbf {0}\right) \end{aligned}$$

where $\mathbf {a}$, $\mathbf {b}$ are row vectors with all entries of $\mathbf {a}$, $\mathbf {b}$ being nonzero. Then its certain cyclic shift must have the form

$$\begin{aligned} \left( \star ,\,\star ,\,\star ,\,\star ,\,\mathbf {0},\,\star ,\,\mathbf {0}\right) \end{aligned}$$

or

$$\begin{aligned} \left( \star ,\,\star ,\,\star ,\,\mathbf {0},\,\star ,\,\star ,\,\mathbf {0}\right) . \end{aligned}$$

• For $\left( \star ,\,\star ,\,\star ,\,\star ,\,\mathbf {0},\,\star ,\,\mathbf {0}\right)$, there are five subcases to be considered:

  (1) Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^{2}+a_{3}\,x^3+a_{4}\,x^{5i}$ with $1\le i\le p-1$ and $a_{1},\,a_{2},\,a_{3},\,a_{4}\in {\mathbb {F}}_{p}^{*}$. It can be verified that

  $$\begin{aligned} \left\{ \begin{array}{l} 1+a_{1}+a_{2}+a_{3}+a_{4}=0,\\ 1+a_{1}\,\beta +a_{2}\,\beta ^2+a_{3}\,\beta ^3+a_{4}=0,\\ 1+a_{1}\,\beta ^2+a_{2}\,\beta ^4+a_{3}\,\beta +a_{4}=0\\ \end{array} \right. \end{aligned}$$

  since $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$. Then one can derive that $p\not \mid \left( a_{4}+1\right)$. By Lemma 6, one can obtain

  $$\begin{aligned} a_{1}=-\frac{\beta ^2+\beta +1}{\beta ^2}(a_{4}+1), a_{2}=\frac{\beta ^2+\beta +1}{\beta ^3}(a_{4}+1), a_{3}=-\frac{1}{\beta ^3}(a_{4}+1). \end{aligned}$$

  (26)

  It follows from $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ that

  $$\begin{aligned} \left\{ \begin{array}{l} a_{1}+2\,a_{2}+3\,a_{3}+5i\,a_{4}=0,\\ a_{1}+2\,a_{2}\,\beta ^2+3\,a_{3}\,\beta ^4+5i\,a_{4}\,\beta ^3=0\\ \end{array} \right. \end{aligned}$$

  which indicates

  $$\begin{aligned} \left( \beta ^3-1\right) a_{1}+2\left( \beta ^3-\beta ^2\right) a_{2} +3\left( \beta ^3-\beta ^4\right) a_{3}=0. \end{aligned}$$

  Combining with (26), one can derive that

  $$\begin{aligned} -\left( \beta ^3-1\right) \beta \left( \beta ^2+\beta +1\right) +2\beta ^2\left( \beta -1\right) \left( \beta ^2+\beta +1\right) +3\beta ^3\left( \beta -1\right) =0. \end{aligned}$$

  Since $\beta$ is a primitive 5-th root of unity, by expanding the above equality, one can get $\beta ^2+3\beta +1=0$. This is contradictory with the inequality (3) in Lemma 6.

  (2) Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^2+a_{3}\,x^3+a_{4}\,x^{5i+1}$ with $1\le i\le p-1$ and $a_{1},\,a_{2},\,a_{3},\,a_{4}\in {\mathbb {F}}_{p}^{*}$. It follows from $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$ and Lemma 6 that

  $$\begin{aligned} a_{1}+a_{4}=-\frac{\beta ^2+\beta +1}{\beta ^2},\quad a_{2}=\frac{\beta ^2+\beta +1}{\beta ^3},\quad a_{3}=-\frac{1}{\beta ^{3}}. \end{aligned}$$

  (27)

  Then $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ induces that

  $$\begin{aligned} \left\{ \begin{array}{l} a_{1}+2\,a_{2}+3\,a_{3}+\left( 5i+1\right) a_{4}=0,\\ a_{1}+2\,a_{2}\,\beta ^2+3\,a_{3}\,\beta ^4+\left( 5i+1\right) a_{4}=0.\\ \end{array} \right. \end{aligned}$$

  This leads to

  $$\begin{aligned} 2\left( \beta ^2-1\right) a_{2}+3\left( \beta ^4-1\right) a_{3}=0. \end{aligned}$$

  Together with (27), one can immediately get $\left( \beta -1\right) ^2=0$, which is impossible.

  (3) Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^2+a_{3}\,x^3+a_{4}\,x^{5i+2}$ with $1\le i\le p-1$ and $a_{1},\,a_{2},\,a_{3},\,a_{4}\in {\mathbb {F}}_{p}^{*}$. The fact $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$ and Lemma 6 induces

  $$\begin{aligned} a_{1}=-\frac{\beta ^2+\beta +1}{\beta ^2},\quad a_{2}+a_{4}=\frac{\beta ^2+\beta +1}{\beta ^3},\quad a_{3}=-\frac{1}{\beta ^3}. \end{aligned}$$

  (28)

  It follows from $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ that

  $$\begin{aligned} \left\{ \begin{array}{l} a_{1}+2\,a_{2}+3\,a_{3}+\left( 5i+2\right) a_{4}=0,\\ a_{1}+2\,a_{2}\,\beta ^2+3\,a_{3}\,\beta ^4+\left( 5i+2\right) a_{4}\,\beta ^2=0\\ \end{array} \right. \end{aligned}$$

  which implies

  $$\begin{aligned} \left( \beta ^2-1\right) a_{1}+3\left( \beta ^2-\beta ^4\right) a_{3}=0. \end{aligned}$$

  By substituting (28) into the above equality, we have $\left( \beta -1\right) ^2=0$, a contradiction.

  (4) Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^2+a_{3}\,x^3+a_{4}\,x^{5i+3}$ with $1\le i\le p-1$ and $a_{1},\,a_{2},\,a_{3},\,a_{4}\in {\mathbb {F}}_{p}^{*}$. By $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$ and Lemma 6, one has

  $$\begin{aligned} a_{1}=-\frac{\beta ^2+\beta +1}{\beta ^2},\quad a_{2}=\frac{\beta ^2+\beta +1}{\beta ^3},\quad a_{3}+a_{4}=-\frac{1}{\beta ^3}. \end{aligned}$$

  (29)

  It follows from $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ that

  $$\begin{aligned} \left\{ \begin{array}{l} a_{1}+2\,a_{2}+3\,a_{3}+\left( 5i+3\right) a_{4}=0,\\[2mm] a_{1}+2\,a_{2}\,\beta ^2+3\,a_{3}\,\beta ^4+\left( 5i+3\right) a_{4}\,\beta ^4=0.\\ \end{array} \right. \end{aligned}$$

  This yields

  $$\begin{aligned} \left( \beta ^4-1\right) a_{1}+2\left( \beta ^4-\beta ^2\right) a_{2}=0. \end{aligned}$$

  Combining with (29), one can derive that $\left( \beta -1\right) ^2=0$, which is impossible.

  (5) Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^2+a_{3}\,x^3+a_{4}\,x^{5i+4}$ with $1\le i\le p-2$ and $a_{1},\,a_{2},\,a_{3},\,a_{4}\in {\mathbb {F}}_{p}^{*}$. It can be verified that

  $$\begin{aligned} \left\{ \begin{array}{l} 1+a_{1}+a_{2}+a_{3}+a_{4}=0,\\ 1+a_{1}\,\beta +a_{2}\,\beta ^2+a_{3}\,\beta ^3+a_{4}\,\beta ^4=0,\\ 1+a_{1}\,\beta ^2+a_{2}\,\beta ^4+a_{3}\,\beta +a_{4}\,\beta ^3=0\\ \end{array} \right. \end{aligned}$$

  since $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$. Then one can obtain that

  $$\begin{aligned} \left\{ \begin{array}{l} a_{1}=-\beta ^3\,a_{4}+\beta ^2+\beta ,\\ a_{2}=-\left( \beta ^4+1\right) a_{4}-\beta -1,\\ a_{3}=-\left( \beta ^2+\beta +1\right) a_{4}-\beta ^2.\\ \end{array} \right. \end{aligned}$$

  (30)

  It follows from $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ that

  $$\begin{aligned} \left\{ \begin{array}{l} a_{1}+2\,a_{2}+3\,a_{3}+\left( 5i+4\right) a_{4}=0,\\[2mm] a_{1}+2\,a_{2}\,\beta ^2+3\,a_{3}\,\beta ^4+\left( 5i+4\right) a_{4}\,\beta =0\\ \end{array} \right. \end{aligned}$$

  which implies

  $$\begin{aligned} \left( \beta -1\right) a_{1}+2\left( \beta -\beta ^2\right) a_{2} +3\left( \beta -\beta ^4\right) a_{3}=0. \end{aligned}$$

  This is equivalent to

  $$\begin{aligned} a_1-2\,\beta \,a_2-3\,\beta \left( \beta ^2+\beta +1\right) a_3=0. \end{aligned}$$

  Together with (30), one can immediately have

  $$\begin{aligned} (-\beta ^3+2\beta (\beta ^4+1)+3\,\beta (\beta ^2+\beta +1)^2)a_4+\beta ^2+\beta +2\,\beta (\beta +1) +3\,\beta ^3(\beta ^2+\beta +1)=0. \end{aligned}$$

  Then we get that

  $$\begin{aligned} -\beta ^3+2\beta \left( \beta ^4+1\right) +3\,\beta \left( \beta ^2+\beta +1\right) ^2=0 \end{aligned}$$

  due to $\beta ^4+\beta ^3+\beta ^2+\beta +1=0$ and $a_4\in {\mathbb {F}}_{p}^{*}$. By a straightforward computation, one has $\beta ^2+3\beta +1=0$. This contradicts with the inequality (3) in Lemma 6.

• For $\left( \star ,\,\star ,\,\star ,\,\mathbf {0},\,\star ,\,\star ,\,\mathbf {0}\right)$, there are also five subcases to be considered:

  (1) Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^2+a_{3}\,x^{5i}+a_{4}\,x^{5i+1}$ with $1\le i\le p-1$ and $a_{1},\,a_{2},\,a_{3},\,a_{4}\in {\mathbb {F}}_{p}^{*}$. It follows from $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$ that

  $$\begin{aligned} \left\{ \begin{array}{l} 1+a_{1}+a_{2}+a_{3}+a_{4}=0,\\ 1+a_{1}\,\beta +a_{2}\,\beta ^2+a_{3}+a_{4}\,\beta =0,\\ 1+a_{1}\,\beta ^2+a_{2}\,\beta ^4+a_{3}+a_{4}\,\beta ^2=0\\ \end{array} \right. \end{aligned}$$

  which implies

  $$\begin{aligned} \left\{ \begin{array}{l} \left( a_{1}+a_{4}\right) \left( \beta -1\right) +a_{2}\left( \beta ^2-1\right) =0,\\ \left( a_{1}+a_{4}\right) \left( \beta ^2-\beta \right) +a_{2}\left( \beta ^4-\beta ^2\right) =0.\\ \end{array} \right. \end{aligned}$$

  This indicates that $\beta \left( \beta ^2-1\right) a_{2}=\left( \beta ^4-\beta ^2\right) a_{2}$. Hence $\beta =1$, a contradiction.

  (2) Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^2+a_{3}\,x^{5i+1}+a_{4}\,x^{5i+2}$ with $1\le i\le p-1$ and $a_{1},\,a_{2},\,a_{3},\,a_{4}\in {\mathbb {F}}_{p}^{*}$. It can be verified that

  $$\begin{aligned} \left\{ \begin{array}{l} 1+a_{1}+a_{2}+a_{3}+a_{4}=0,\\ 1+a_{1}\,\beta +a_{2}\,\beta ^2+a_{3}\,\beta +a_{4}\,\beta ^2=0,\\ 1+a_{1}\,\beta ^2+a_{2}\,\beta ^4+a_{3}\,\beta ^2+a_{4}\,\beta ^4=0\\ \end{array} \right. \end{aligned}$$

  since $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$. Then one can derive that

  $$\begin{aligned} \left\{ \begin{array}{l} \left( a_{2}+a_{4}\right) \left( \beta ^2-\beta \right) =\beta -1,\\ \left( a_{2}+a_{4}\right) \left( \beta ^4-\beta ^3\right) =\beta -1.\\ \end{array} \right. \end{aligned}$$

  It follows that $\beta ^3=\beta$, which is impossible.

  (3) Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^2+a_{3}\,x^{5i+2}+a_{4}\,x^{5i+3}$ with $1\le i\le p-1$ and $a_{1},\,a_{2},\,a_{3},\,a_{4}\in {\mathbb {F}}_{p}^{*}$. The fact $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$ and Lemma 6 induces that

  $$\begin{aligned} a_{1}=-\frac{\beta ^2+\beta +1}{\beta ^2},\quad a_{2}+a_{3}=\frac{\beta ^2+\beta +1}{\beta ^3},\quad a_{4}=-\frac{1}{\beta ^3}. \end{aligned}$$

  (31)

  It follows from $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ that

  $$\begin{aligned} \left\{ \begin{array}{l} a_{1}+2\,a_{2}+\left( 5i+2\right) a_{3}+\left( 5i+3\right) a_{4}=0,\\ a_{1}+2\,a_{2}\,\beta ^2+\left( 5i+2\right) a_{3}\,\beta ^2 +\left( 5i+3\right) a_{4}\,\beta ^4=0.\\ \end{array} \right. \end{aligned}$$

  This yields

  $$\begin{aligned} \left( \beta ^2-1\right) a_{1}+\left( 5i+3\right) \left( \beta ^2-\beta ^4\right) a_{4}=0. \end{aligned}$$

  By substituting (31), one can deduce that

  $$\begin{aligned} \beta ^2-\left( 5i+2\right) \beta +1=0. \end{aligned}$$

  Let $t=5i+2$. Then $\beta ^2=t\beta -1$ and

  $$\begin{aligned} \beta ^4+\beta ^3+\beta ^2+\beta +1=(t\beta -1)(t^2+t-1)=0. \end{aligned}$$

  It follows that $t^2+t=1$. By $c^{\left( 2\right) }\left( 1\right) =0$ and (31), we get

  $$\begin{aligned} 5i\left( t+1\right) a_{3}=\left( t+2\right) \beta +1. \end{aligned}$$

  The fact $c^{\left( 1\right) }\left( 1\right) =0$ indicates $5i\,a_{3}=\left( 2-t\right) \left( \beta +1\right)$. Hence

  $$\begin{aligned} \left( t+2\right) \beta +1=\left( t+1\right) \left( 2-t\right) \left( \beta +1\right) . \end{aligned}$$

  This leads to $t^2\,\beta -2\,t=0$ due to $t^2+t=1$. It follows from $t\ne 0$ that $t\beta =2$ and $\beta ^2=t\beta -1=1$, a contradiction.

  (4) Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^2+a_{3}\,x^{5i+3}+a_{4}\,x^{5i+4}$ with $1\le i\le p-2$ and $a_{1},\,a_{2},\,a_{3},\,a_{4}\in {\mathbb {F}}_{p}^{*}$. It can be checked that

  $$\begin{aligned} \left\{ \begin{array}{l} 1+a_{1}+a_{2}+a_{3}+a_{4}=0,\\ 1+a_{1}\,\beta +a_{2}\,\beta ^2+a_{3}\,\beta ^3+a_{4}\,\beta ^4=0,\\ 1+a_{1}\,\beta ^2+a_{2}\,\beta ^4+a_{3}\,\beta +a_{4}\,\beta ^3=0\\ \end{array} \right. \end{aligned}$$

  since $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$. Then one can derive that

  $$\begin{aligned} \left\{ \begin{array}{l} a_{1}=-\beta ^3\,a_{4}+\beta ^2+\beta ,\\ a_{2}=-\left( \beta ^4+1\right) a_{4}-\beta -1,\\ a_{3}=-\left( \beta ^2+\beta +1\right) a_{4}-\beta ^2.\\ \end{array} \right. \end{aligned}$$

  (32)

  Let $t=5i+2$. By $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ and (32), we have

  $$\begin{aligned} \left\{ \begin{array}{l} t\beta ^2+\beta +2=\left( (t-1)\beta ^4+t\beta ^3+t\right) a_{4},\\ 2\beta ^2+\beta +t=\left( t\beta ^2+(t-1)\beta +t\right) a_{4}.\\ \end{array} \right. \end{aligned}$$

  (33)

  Then

  $$\begin{aligned} (t\beta ^2+\beta +2)\left( t\beta ^2+(t-1)\beta +t\right) =(2\beta ^2+\beta +t)\left( (t-1)\beta ^4+t\beta ^3+t\right) \end{aligned}$$

  which implies

  $$\begin{aligned} (t^2+t-1)(\beta ^2-1)=0. \end{aligned}$$

  Thus $t^2+t=1$. It follows from $c^{\left( 2\right) }\left( 1\right) =0$ that $2\,a_{2}+a_{3}+\left( 2t+3\right) a_{4}=0$. Together with (32), one can immediately get

  $$\begin{aligned} \left( -\beta ^4+\beta ^3+2t+1\right) a_{4}=\beta ^2+2\beta +2. \end{aligned}$$

  Combining with the second equality in (33), we can obtain

  $$\begin{aligned} \left( -\beta ^4+\beta ^3+2t+1\right) \left( 2\beta ^2+\beta +t\right) =\left( \beta ^2+2\beta +2\right) \left( t\beta ^2+(t-1)\beta +t\right) . \end{aligned}$$

  By expanding the above equality, one can deduce

  $$\begin{aligned} \left( \beta ^2-1\right) t+3\beta ^2+2=0 \end{aligned}$$

  which yields $t=\frac{3\beta ^2+2}{1-\beta ^2}$. The fact $t^2+t-1=0$ induces

  $$\begin{aligned} \left( \frac{3\beta ^2+2}{1-\beta ^2}\right) ^2+\frac{3\beta ^2+2}{1-\beta ^2}-1=0 \end{aligned}$$

  which is equivalent to

  $$\begin{aligned} \left( 3\beta ^2+2\right) ^2+\left( 3\beta ^2+2\right) \left( 1-\beta ^2\right) -\left( 1-\beta ^2\right) ^2=0. \end{aligned}$$

  It follows that

  $$\begin{aligned} \beta ^4+3\,\beta ^2+1=0 \end{aligned}$$

  which indicates

  $$\begin{aligned} 2\,\beta ^2-\beta ^3-\beta =0 \end{aligned}$$

  due to $\beta ^4+\beta ^3+\beta ^2+\beta +1=0$. Hence $\beta (\beta -1)^2=0$, which is impossible.

  (5) Consider the subcase $c(x)=1+a_{1}\,x+a_{2}\,x^2+a_{3}\,x^{5i+4}+a_{4}\,x^{5i+5}$ with $0\le i\le p-2$ and $a_{1},\,a_{2},\,a_{3},\,a_{4}\in {\mathbb {F}}_{p}^{*}$. It follows from $c\left( 1\right) =c\left( \beta \right) =c\left( \beta ^2\right) =0$ that $p\not \mid \left( a_{4}+1\right)$ and

  $$\begin{aligned} a_{1}=-\frac{1}{\beta }\left( a_{4}+1\right) ,\quad a_{2}=-\frac{\beta }{\beta +1}\left( a_{4}+1\right) ,\quad a_{3}=\frac{1}{\beta (\beta +1)}\left( a_{4}+1\right) \end{aligned}$$

  (34)

  due to Lemma 6. The fact $c^{\left( 1\right) }\left( 1\right) =c^{\left( 1\right) }\left( \beta ^2\right) =0$ leads to

  $$\begin{aligned} \left\{ \begin{array}{l} a_{1}+2\,a_{2}+\left( 5i+4\right) a_{3}+\left( 5i+5\right) a_{4}=0,\\[2mm] a_{1}+2\,a_{2}\,\beta ^2+\left( 5i+4\right) a_{3}\,\beta +\left( 5i+5\right) a_{4}\,\beta ^3=0\\ \end{array} \right. \end{aligned}$$

  which implies

  $$\begin{aligned} \left\{ \begin{array}{l} \left( \beta ^3-1\right) a_{1}+2\left( \beta ^3-\beta ^2\right) a_{2} +\left( 5i+4\right) \left( \beta ^3-\beta \right) a_{3}=0,\\[2mm] 2\left( \beta +1\right) a_{2}+\left( 5i+4\right) a_{3}+\left( 5i+5\right) \left( \beta ^2+\beta +1\right) a_{4}=0.\\ \end{array} \right. \end{aligned}$$

  By substituting (34), one can obtain that

  $$\begin{aligned} \beta ^4+\left( 5i+3\right) \beta ^3-\left( 5i+3\right) \beta -1=0 \end{aligned}$$

  (35)

  and

  $$\begin{aligned} \left( -2\beta ^2\left( \beta +1\right) +5i+4\right) \left( a_{4}+1\right) +\left( 5i+5\right) \beta \left( \beta +1\right) \left( \beta ^2+\beta +1\right) a_4=0. \end{aligned}$$

  (36)

  Let $t=5i+3$. It follows from (35) that

  $$\begin{aligned} \beta ^4-1+t\left( \beta ^3-\beta \right) =\left( \beta ^2-1\right) (\beta ^2+1+t\,\beta )=0 \end{aligned}$$

  which yields $\beta ^2=-t\beta -1$. Then we have

  $$\begin{aligned} 0=\beta ^4+\beta ^3+\beta ^2+\beta +1=-\left( t^2-t-1\right) \beta ^2 \end{aligned}$$

  which indicates $t^2=t+1$ due to $\beta ^2\ne 0$. It can be verified that

  $$\begin{aligned} \begin{aligned}&-2\beta ^2\left( \beta +1\right) +5i+4+\left( 5i+5\right) \beta \left( \beta +1\right) \left( \beta ^2+\beta +1\right) \\&\quad =-2\beta ^3-2\beta ^2+t+1-\left( t+2\right) \left( \beta ^4+\beta +2\right) \\&\quad =-2t\left( \beta +1\right) +2\left( t\beta +1\right) +\left( t+2\right) \left( \beta +t\right) -\left( t+2\right) \beta -t-3=0. \end{aligned} \end{aligned}$$

  Hence (36) and $a_4\in {\mathbb {F}}_{p}^{*}$ induces

  $$\begin{aligned} 0=-2\beta ^2\left( \beta +1\right) +5i+4=3-t \end{aligned}$$

  which means that $t=3$ and $\beta ^2=-3\beta -1$, a contradiction with the inequality (3) in Lemma 6.

As a consequence, ${\mathcal {C}}$ is an MDS $\left( 5p,\,8\right) _{p}$ symbol-pair code. The desired result follows.

$\square$