Entanglement-assisted quantum MDS codes from negacyclic codes

Abstract

The entanglement-assisted formalism generalizes the standard stabilizer formalism, which can transform arbitrary classical linear codes into entanglement-assisted quantum error-correcting codes (EAQECCs) by using pre-shared entanglement between the sender and the receiver. In this work, we construct six classes of q-ary entanglement-assisted quantum MDS (EAQMDS) codes based on classical negacyclic MDS codes by exploiting two or more pre-shared maximally entangled states. We show that two of these six classes q-ary EAQMDS have minimum distance more larger than $q+1$. Most of these q-ary EAQMDS codes are new in the sense that their parameters are not covered by the codes available in the literature.

1 Introduction

Quantum error-correcting codes (QECCs) were introduced to protect quantum information from decoherence during quantum computations [1]. The stabilizer formalism allows standard quantum codes to be constructed from dual-containing (or self-orthogonal) classical codes [2]. However, the dual-containing condition forms a barrier in the development of quantum coding theory. Recently, a breakthrough is the entanglement-assisted (EA) stabilizer formalism proposed by Brun et al. in Ref. [3]. They prove that if shared entanglement is available between the sender and receiver, non-dual-containing classical quaternary codes can be used to construct EAQECCs, this leads to a more general framework for construction of quantum codes [4,5,6]. Currently, many works have focused on the construction of binary EAQECCs based on classical binary or quaternary linear codes, see [7,8,9,10,11,12,13,14]. Just as in the classical error-correcting codes and QECCs, EAQECCs over higher alphabets have many wide applications, such as constructing easily decodable binary EAQECCs. However, little attention has been paid to non-binary EAQECCs, let alone EA-quantum MDS codes which can achieve entanglement-assisted quantum singleton bound$^{3}$.

Let q be a prime power. A q-ary [[n, k, d; c]] EAQECC that encodes k information qubits into n channel qubits with the help of c pairs of maximally entangled Bell states (ebits) can correct up to $\lfloor \frac{d-1}{2}\rfloor$ errors, where d is the minimum distance of the code. If $c=0$, then it is called a q-ary standard [[n, k, d]] quantum code $\mathcal {Q}$. We denote a q-ary [[n, k, d; c]] EAQECC by $[[n,k,d;c]]_{q}$, and q-ary [[n, k, d]] QECC by $[[n,k,d]]_{q}$.

As in classical coding theory, one of the central tasks in quantum coding theory is to construct good quantum codes and EA-quantum codes.

Theory 1.1

[3](EA-Quantum Singleton Bound) An $[[n,k,d;c]]_{q}$ EAQECC satisfies

$$\begin{aligned} n+c-k\ge 2(d-1), \end{aligned}$$

where $0\le c \le n-1$.

A EAQECC achieving this bound is called a EA-quantum maximum-distance-separable (EAQMDS) code. If $c=0$, then this bound is quantum singleton bound, and a code achieving the bound is called quantum maximum-distance-separable (QMDS) code. Just as in the classical linear codes, QMDS codes and EAQMDS codes form an important family of quantum codes. Constructing QMDS codes and EAQMDS codes had become a central topic for quantum error correction codes in rent years. Many classes of QMDS codes have been constructed by different methods, in particular the constructions obtained from constacyclic codes or negacyclic codes containing their Hermitian dual over $F_{q^{2}}$ [15,16,17,18,19,20,21,22,23]. According to the MDS conjecture in [24], the maximum-distance-separable (MDS) code cannot exceed $q^{2}+1$. Many QMDS codes with lengths between $q+1$ and $q^{2}+1$ have been constructed [16,17,18,19,20,21,22,23, 25]. However, the problem of constructing QMDS codes with length n larger than $q+1$ is much more difficult.

It seems that there is a barrier for constructing more QMDS codes with distance larger than $q+1$. For larger distance than $q+1$ of code length $n\le q^2+1$, one need to construct a EAQMSD code.

The following Proposition is one of the most frequently used construction methods.

Proposition 1.2

[3, 5] If $\mathcal {C}=[n,k,d]_{q^{2}}$ is a classical code over $F_{q^{2}}$ and H is its parity check matrix , then $\mathcal {C}^{\perp _{h}}$ EA stabilizes an $[[n,2k-n+c,d;c]]_{q}$ EAQECC, where $c=$rank$(HH^{\dagger })$ is the number of maximally entangled states required and $H^{\dagger }$ is the conjugate matrix of H over $F_{q^{2}}$.

Until now, little attention has been paid to q-ary EA-quantum MDS codes. In [26], Fan et al. proposed several constructions of q-ary EAQMDS codes with minimum distance greater than $q+1$ based on classical MDS codes.

In this paper, we propose a concept of decomposition of the defining set of negacyclic codes. Recently, Chen et al. [30] proposed a same concept at the very same moment. Based on the concept, they construct some EA-quantum MDS codes which are all different with the codes in this paper. More precisely, based on concept of decomposition of the defining set of negacyclic codes, we construct several classes of EA-quantum MDS codes as follows:

• (1) Let q be an odd prime power of the form $q=atm+1$, a be even, or a be odd and t be even, then there exists a q-ary $[[\frac{q^{2}-1}{at},\frac{q^{2}-1}{at}-2d+2,d;0]]$ EA-quantum MDS codes, where $2 \le d \le (\frac{at}{2}+1)m+1$; there exists a q-ary $[[\frac{q^{2}-1}{at},\frac{q^{2}-1}{at}-2d+4,d;2]]$- EA-quantum MDS codes, where $(\frac{at}{2}+1)m+2 \le d \le (\frac{at}{2}+2)m+1$; and there exists a q-ary $[[\frac{q^{2}-1}{at},\frac{q^{2}-1}{at}-2d+6,d;4]]$- EA-quantum MDS codes, where $(\frac{at}{2}+2)m+2 \le d \le (\frac{at}{2}+3)m+1$.

• (2) Let q be an odd prime power of the form $q=30m+11$, then there exists a q-ary $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+4,d;2]]$- EA-quantum MDS codes, where $8m+4 \le d \le 11m+5$; and there exists a q-ary $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+6,d;4]]$- EA-quantum MDS codes, where $11m+6\le d \le 14m+7$.

• (3) Let q be an odd prime power of the form $q=30m+19$, then there exists a q-ary $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+4,d;2]]$- EA-quantum MDS codes, where $8m+6 \le d \le 11m+7$; there exists a q-ary $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+6,d;4]]$- EA-quantum MDS codes, where $11m+8\le d \le 13m+8$; and there exists a q-ary $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+8,d;6]]$- EA-quantum MDS codes, where $13m+9\le d \le 16m+10$.

• (4) Let q be an odd prime power of the form $q=12m+5$, then there exists a q-ary $[[\frac{q^{2}-1}{12},\frac{q^{2}-1}{12}-2d+4,d;2]]$- EA-quantum MDS codes, where $5m+3 \le d \le 7m+3$; and there exists a q-ary $[[\frac{q^{2}-1}{12},\frac{q^{2}-1}{12}-2d+6,d;4]]$- EA-quantum MDS codes, where $7m+4\le d \le 8m+3$.

• (5) Let q be an odd prime power of the form $q=10m+3$. (a) If $m=2t+1$ is odd, then there exists a q-$[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]$- EA-quantum MDS codes, where $4m+3\le d \le 6m+1$ is odd and $6m+4\le d\le 10m+4$ is even. (b) If $m=2t$ is even, then there exists a q-ary $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+3,d;1]]$-EA-quantum MDS codes, where $2\le d \le 8m+1$ is even; there exists a q-ary $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]$-EA-quantum MDS codes, where $4m+3\le d \le 6m+1$ is odd; and there exists a q-ary $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+7,d;5]]$-EA-quantum MDS codes, where $8m+4\le d \le 12m+4$ is even.

• (6) Let q be an odd prime power of the form $q=10m+7$. (a)If $m=2t+1$ is odd, then there exists a q-$[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]$- EA-quantum MDS codes, where $8m+7\le d \le 14m+11$ is odd; and $6m+6\le d\le 10m+8$ is even.(b) If $m=2t$ is even, then there exists a q-$[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+3,d;1]]$- EA-quantum MDS codes, where $2\le d \le 8m+6$ is even; there exists a q-$[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]$- EA-quantum MDS codes, where $8m+7\le d \le 14m+11$ is odd; and there exists a q-$[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+7,d;5]]$- EA-quantum MDS codes, where $8m+8\le d \le 12m+8$ is even.

The first class of EAQMDS codes has minimum distance upper limit greater than $\frac{q}{2}+1$ by consuming a few ebits. EAQMDS codes in (2)–(4) have minimum distance upper limit closed to $\frac{q}{2}+1$. EAQMDS codes in (5) and (6) have minimum distance upper limit more greater than $q+1$. Briefly, most of these EAQMDS codes are new in the sense that their parameters are not covered by the codes available in the literature.

This paper is organized as follows. In Sect. 2, we introduce some basic notations and definitions of classical negacyclic codes and EAQECCs. In Sect. 3, we give some new classes of EA-quantum MDS codes. The conclusion is given in Sect. 4.

2 Preliminaries

In this section, we review some basic results on negacyclic codes, BCH codes, decomposition of defining sets of codes and EAQECCs for the purpose of this paper. Details on BCH codes and negacyclic codes can be found in standard textbook on coding theory [27], and for EAQECCs please see Refs. [3,4,5,6,7,8,9].

Let p be a prime number and q a power of p, i.e., $q=p^{l}$ for some $l>0$. $F_{q^{2}}$ denotes the finite field with $q^{2}$ elements. For any $\alpha \in F_{q^{2}}$, the conjugation of $\alpha$ is denoted by $\overline{\alpha }=\alpha ^{q}$. Given two vectors $\mathbf {x}=(x_{1},x_{2},\ldots ,x_{n})$ and $\mathbf {y}=(y_{1},y_{2},\ldots ,y_{n})\in F_{q^{2}}^{n}$, their Hermitian inner product is defined as

$$\begin{aligned} (\mathbf {x},\mathbf {y})_{h}=\sum \overline{x_{i}}y_{i}=\overline{x_{1}}y_{1}+\overline{x_{2}}y_{2}+\cdots +\overline{x_{n}}y_{n}. \end{aligned}$$

For a linear code $\mathcal {C}$ over $F_{q^{2}}$ of length n, the Hermitian dual code $\mathcal {C}^{\bot _{h}}$ is defined as

$$\begin{aligned} \mathcal {C}^{\bot _{h}}=\{x\in F_{q^{2}}^{n} | (x,y)_{h}=0, \forall y \in \mathcal {C}\} \end{aligned}$$

If $\mathcal {C} \subseteq \mathcal {C}^{\bot _{h}}$, then $\mathcal {C}$ is called a Hermitian dual-containing code, and $\mathcal {C}^{\bot _{h}}$ is called a Hermitian self-orthogonal code.

We now recall some results about classical negacyclic codes. For any vector $(c_{0},c_{1},\ldots ,c_{n-1})\in F_{q^{2}}^{n}$, if a $q^{2}$-ary linear code $\mathcal {C}$ of length n is invariant under the permeation of $F_{q^{2}}$, i.e.,

$$\begin{aligned} \lambda (c_{0}, c_{1}, \ldots , c_{n-1})=(\lambda c_{n-1}, c_{0},\ldots , c_{n-2}), \end{aligned}$$

where $\lambda$ is a nonzero element of $F_{q^{2}}$, then $\mathcal {C}$ is a constacyclic code. If $\lambda =1$, then $\mathcal {C}$ is called a cyclic code , and if $\lambda =-1$, then $\mathcal {C}$ is called a negacyclic code.

For a negacyclic code $\mathcal {C}$, each codeword $c = (c_{0}, c_{1}, \ldots , c_{n-1})$ is customarily represented in its polynomial form: $c(x) = c_{0} + c_{1}x + \cdots + c_{n-1}x_{n-1},$ and the code $\mathcal {C}$ is in turn identified with the set of all polynomial representations of its codewords. The proper context for studying negacyclic codes is the residue class ring $\mathcal {R}_{n}=\mathbb {F}_{q}[x]/(x^{n}+1)$. xc(x) corresponds to a negacyclic shift of c(x) in the ring $\mathcal {R}_{n}$. As we all know, a linear code $\mathcal {C}$ of length n over $F_{q^{2}}$ is negacyclic if and only if C is an ideal of the quotient ring $\mathcal {R}_{n}=\mathbb {F}_{q}[x]/(x^{n}+1)$. It follows that $\mathcal {C}$ is generated by monic factors of $(x^{n}+1)$, i.e., $\mathcal {C}=\langle f(x) \rangle$ and $f(x)|(x^{n}+1)$. The f(x) is called the generator polynomial of $\mathcal {C}_{n}$.

Let $gcd(n,q)=1$ and m be the multiplicative order of $q^{2}$ modulo 2n. Let $\beta \in F_{q^{2m}}$ be a primitive 2n-th root of unity. Then, $\xi$ be a primitive 2n-th root of unity such that $\xi =\beta ^{2} \in F_{q^{2m}}$. Hence, $x^{n} + 1 =\prod ^{n-1} _{i=0} (x-\beta ^{2i+1}).$ Let $Z_{2n}$ denote the set of odd integers from 1 to 2n, i.e., $Z_{2n}=\{1,3,\ldots ,2n-1\}$. For each $i\in Z_{2n}$, let s be an integer with $0\le s < 2n$, the $q^{2}$-cyclotomic coset modulo 2n that contains s is defined by the set $C_{s}=\{s, sq^{2}, sq^{2\cdot 2}, \ldots , sq^{2(k-1)} \}$ (mod 2n), where k is the smallest positive integer such that $xq^{2k}$ $\equiv x$ (mod 2n).

The defining set of a negacyclic code $\mathcal {C}= <g(x)>$ of length n is the set $T= \{i \in Z_{2n} | \beta ^{i}$ is a root of $g(x) \}$. We can see that the defining set T is a union of some $q^{2}$-cyclotomic cosets module 2n and $dim(\mathcal {C}) = n-|T|$.

Lemma 2.1

[18] Let $\mathcal {C}$ be a $q^{2}$-ary negacyclic code of length n with defining set T. Then, $\mathcal {C}$ contains its Hermitian dual code if and only if $T \bigcap T^{-q}=\emptyset$, where $T^{-q}$ denotes the set $Z^{-q}=\{-qz($mod $2n)| z\in T\}$.

Let $\mathcal {C}$ be a negacyclic code with a defining set $T = \bigcup \limits _{s \in S} C_{s}$. Denoting $T^{-q}=\{2n-qs | s\in T \}$, then we can deduce that the defining set of $\mathcal {C}^{\bot _{h}}$ is $T^{\perp _{h}} =\mathbb {Z}_{n} \backslash T^{-q}$, see Ref. [16].

Since there is a striking similarity between cyclic codes and negacyclic code, we give a correspondence defining of skew-symmetric and skew-asymmetric as follows.

A cyclotomic coset $C_{s}$ is skew-symmetric if $2n-qs$ mod $2n\in C_{s}$, and otherwise is skew-asymmetric. Skew-asymmetric cosets $C_{s}$ and $C_{2n-qs}$ come in pair, and we use $(C_{s},C_{2n-qs})$ to denote such a pair.

Thus, one has the following lemma.

Lemma 2.2

[28] If $\mathcal {C}$ is a negacyclic code of length n over $F_{q^{2}}$ with defining set T, then $\mathcal {C}^{\perp _{h}}\subseteq \mathcal {C}$ if and only if one of the following holds:

• (1) $T \cap T^{-q}=\emptyset$, where $T^{-q}=\{2n-qs \mid s\in T\}$.

• (2) If $i,j,k\in T$, then $C_{i}$ is a skew-asymmetric coset and ($C_{j}$, $C_{k}$) is not a skew-asymmetric cosets pair.

Using above-mentioned Lemma 2.2, one can get that $\mathcal {C}^{\perp _{h}}\subseteq$ $\mathcal {C}$ can be described by the relationship of its cyclotomic coset $C_{s}$. Firstly, we introduce a fundamental definition.

Definition 2.3

[12] Let $\mathcal {C}$ be a negacyclic code of length n with defining set T. Denote $T_{ss}=T \cap T^{-q}$ and $T_{sas}=T \setminus T_{ss}$, where $T^{-q}=\{2n-qx | x\in T \}$. $T=T_{ss} \cup T_{sas}$ is called decomposition of the defining set of $\mathcal {C}$.

To determine $T_{ss}$ and $T_{sas}$, we give the following lemma to characterize them.

Lemma 2.4

[12] Let $\mathcal {C}$ be a negacyclic code of length n over $F_{q^{2}}$ with defining set T, $T=T_{ss} \cup T_{sas}$ be decomposition of T.

• (1) If $i,j\in T_{sas}$, then $C_{i}$ is skew-asymmetric coset, and $C_{i}$ and $C_{j}$ cannot form a skew-asymmetric cosets pair.

• (2) If $l\in T_{ss}$, then either $C_{l}$ is a skew-symmetric coset, or $C_{l}$ is a skew-asymmetric coset and there is a $p\in T$ such that $C_{l}$ and $C_{p}$ form a skew-asymmetric cosets pair.

To determine $T_{ss}$ and $T_{sas}$, we give the following lemma to characterize them.

Lemma 2.5

[12, 18, 29] Let $gcd(q, n) = 1$, $ord_{2n}(q^{2})=m$, $0 \le x, y$, $z \le n-1$.

• (1) $C_{x}$ is skew-symmetric if and only if there is a $t\le \lfloor \frac{m}{2}\rfloor$ such that $x \equiv xq^{2t+1}$(mod 2n).

• (2) If $C_{y}\ne C_{z}$, $(C_{y}, C_{z})$ form a skew-asymmetric pair if and only if there is a $t\le \lfloor \frac{m}{2}\rfloor$ such that $y \equiv zq^{2t+1}$ (mod n) or $z \equiv yq^{2t+1}$(mod 2n).

Using decomposition of the defining set T of a negacyclic code $\mathcal {C}$, one can give a decomposition of $\mathcal {C}^{\perp _{h}}$ as follow.

Lemma 2.6

[12] Let $\mathcal {C}$ be a negacyclic code with defining set T, $T=T_{ss} \cup T_{sas}$ be decomposition of T. Denote the negacyclic codes with defining set $T_{sas}$ and $T_{ss}$ be $\mathcal {C}_{R}$ and $\mathcal {C}_{E}$, respectively. Then, $\mathcal {C}^{\perp _{h}}_{R}\subseteq \mathcal {C}_{R}$, $\mathcal {C}_{E}\cap$ $\mathcal {C}^{\perp _{h}}_{E}$ $=\{0\}$, $\mathcal {C}^{\perp _{h}}_{R}$ $\subset \mathcal {C}_{E}$, $\mathcal {C}_{R}\cap$ $\mathcal {C}_{E}$ $=\mathcal {C}$ and $\mathcal {C}^{\perp _{h}}_{R}+\mathcal {C}^{\perp _{h}}_{E}$ $=\mathcal {C}^{\perp _{h}}$.

Lemma 2.7

[12] Let T be a defining set of a negacyclic code $\mathcal {C}$, $T=T_{ss}\cup T_{sas}$ be decomposition of T. Using $\mathcal {C}^{\perp _{h}}$ as EA stabilizer, the optimal number of needed ebits is $c=\mid T_{ss} \mid$.

Proof

According to Definition 2.3, we denote the defining sets of negacyclic codes $\mathcal {C}_{1}$ and $\mathcal {C}_{2}$ into $T_{ss}$ and $T_{sas}$, respectively. The parity check matrix of $\mathcal {C}_{1}$ and $\mathcal {C}_{2}$ is $H_{1}$ and $H_{2}$, respectively. Let $H=\left( \begin{array}{lllllll} H_{1}\\ H_{2}\\ \end{array} \right)$ be the parity check matrix of $\mathcal {C}$. Then,

$$\begin{aligned} HH^{\dag }=\left( \begin{array}{ll} H_{1}H_{1}^{\dag } &{}\quad H_{1}H_{2}^{\dag }\\ H_{2}H_{1}^{\dag } &{}\quad H_{2}H_{2}^{\dag }\\ \end{array} \right) . \end{aligned}$$

Since $H_{2}$ is the parity check matrix of $\mathcal {C}_{2}$ with defining set of $T_{sas}$, $H_{2}H_{2}^{\dag }=0$. Because of $\mathcal {C}_{1}^{\bot h} \subseteq \mathcal {C}_{2}$, $H_{1}H_{2}^{\dag }=0$ and $H_{2}H_{1}^{\dag }=0$. Therefore,

$$\begin{aligned} HH^{\dag }=\left( \begin{array}{ll} H_{1}H_{1}^{\dag } &{}\quad 0\\ 0 &{}\quad 0\\ \end{array} \right) . \end{aligned}$$

According to Refs. [3,4,5], one obtains that $c=rank(HH^{\dag })=rank(H_{1}H_{1}^{\dag })$. Since $H_{1}$ is the parity check matrix of $\mathcal {C}_{2}$ with defining set of $T_{ss}$, $H_{1}$ is a full-rank matrix. Hence, $c=rank(H_{1}H_{1}^{\dag })=|T_{ss}|$. $\square$

Lemma 2.8

Let $\mathcal {C}$ be an $[n,k,d]_{q^{2}}$ negacyclic code with defining set T, and the decomposition of T be $T=T_{ss}\cup T_{sas}$. Then, $\mathcal {C}^{\perp _{h}}$ EA stabilizes an q-ary $[[n,n-2|T|+|T_{ss}|,d \ge \delta ; |T_{ss}|]]$ EAQECC.

Proof

The dimension of $\mathcal {C}$ is $k=n-|T|$. From Proposition 1 and Lemma 2.6, we know $\mathcal {C}^{\perp _{h}}$ EA stabilizes an EAQECC with parameters $[[n,2k-n+c,d ;c]]=[[n,n-2|T|+|T_{ss}|,d; |T_{ss}|]]$.

If $\mathcal {C}$ is a negacyclic BCH code $BCH(n,\delta )$, denote its defining set T as $T=T(\delta )$, the decomposition of T as $T(\delta )=T_{ss}(\delta )\cup T_{sas}(\delta )$. According to Lemma 2.7, $\mathcal {C}^{\perp _{h}}$ EA stabilizes an EAQECC with parameters $[[n,k^{ea},d ;c]]$ $=[[n,n-2|T(\delta )|+|T_{ss}(\delta )|,d\ge \delta ; |T_{ss}|]]$. In the following two sections, we will discuss how to determine $|T_{ss}(\delta )|$, $|T(\delta )|$. $\square$

3 New EA-quantum MDS codes

3.1 New EA-quantum MDS codes of length $n=\frac{q^{2}-1}{at}$

In this subsection, we construct some classes of q-ary EA-quantum MDS codes of length $n=\frac{q^{2}-1}{at}$, where q be an odd prime power of the form $q=atm+1$, a be a even number, or a be an odd number and t be a even number. Since $2n|q^{2}-1$, then for each odd x in the range $1\le x \le 2n$, the $q^{2}$-cyclotomic coset $C_{x}$ modulo 2n is $C_{x}=x$.

Lemma 3.1

Let q be an odd prime power of the form $q=atm+1$, a be a even number, or a be an odd number and t be a even number, $n=\frac{q^{2}-1}{at}$. If $\mathcal {C}$ is a $q^{2}$-ary negacyclic code of length n with defining set $T=\bigcup _{j=0}^{s} C_{1+2j}$, where $0\le s\le (\frac{at}{2}+1)m-1$, then $\mathcal {C}^{\bot h} \subseteq \mathcal {C}$.

Proof

For $0 \le s \le (\frac{at}{2}+1)m-1$, it is sufficient to prove $T \cap (-qT)=\emptyset$. According to Lemma 2.2 and Definition 2.3, one obtains that $\mathcal {C}^{\perp _{h}}$ $\subseteq \mathcal {C}$ if and only if there is no skew-symmetric cyclotomic coset and any two cyclotomic cosets do not form a skew-asymmetric pair in the defining set T. Suppose there exist integers $0 \le x \le y \le (at+2)m-1$ such that $C_{x}=-qC_{y}$, that is $x\equiv -qy$ mod 2n. In other words $x+qy\equiv 0$ mod 2n. Since $q=atm+1$, $2n=2m(q+1)$.

If $1\le x\le y\le 2m-1$, then $0\le x+qy\le (2m-1)(q+1)\le 2m(q+1)-(q+1)<2n$, a contradiction.

Similarly, we have for $1\le i\le \frac{at}{2}$, if $[at+2-2(i-1)]m+1\le x< y\le (at+2-2i)m-1$, then $(i-1)2n<[(at-2i)(m+1)](q+1)\le x+qy \le [(at+2-2i)m-1](q+1)\le 2im(q+1)-(q+1)<i2n$, a contradiction.

Hence, for $0\le s \le (\frac{at}{2}+1)m-1$, there are no skew-symmetric cyclotomic cosets and skew-asymmetric cosets pairs in defining set $T=\bigcup _{j=0}^{s} C_{1+2j}$. It means that $T_{ss}=\emptyset$. $\mathcal {C}^{\bot h} \subseteq \mathcal {C}$ holds. $\square$

Theory 3.2

Let q be an odd prime power of the form $q=atm+1$, a be even, or a be odd and t be even, $n=\frac{q^{2}-1}{at}$. Then, there exists a q-ary $[[\frac{q^{2}-1}{at},\frac{q^{2}-1}{at}-2d+2,d]]$ quantum MDS codes, where $2 \le d \le (\frac{at}{2}+1)m+1$.

Proof

Consider the negacyclic codes over $F_{q^{2}}$ of length $n=\frac{q^{2}-1}{at}$ with defining set $T=\bigcup _{i=0}^{s}C_{1+2i}$, where $0\le s \le (\frac{at}{2}+1)m-1$ for q be an odd prime power of the form $q=atm+1$, a be even, or a be odd and t be even. By Lemma 3.1, there is $\mathcal {C}^{\perp _{h}}$ $\subseteq \mathcal {C}$. Since every $q^{2}$-cyclotomic coset $C_{x}$ has exactly one element and x must be odd number, we can obtain that T consists of $s+1$ integers $\{1,3,\ldots ,1+2s\}$. It implies that $\mathcal {C}$ has minimum distance at least $s+2$. Hence, $\mathcal {C}$ is a $q^{2}$-ary negacyclic code with parameters $[n,n-(s+1),\ge s+2]$. Combining the Hermitian construction with quantum singleton bound, we can obtain a quantum MDS code with parameters $[[\frac{q^{2}-1}{at},\frac{q^{2}-1}{at}-2d+2,d]]$, where $2 \le d \le (\frac{at}{2}+1)m+1$, q be an odd prime power of the form $q=atm+1$, a be even, or a be odd and t be even. $\square$

Example 1

Let $a=3,t=2$. Then, $q=19$, $n=60$ applying Theory 3.2 produces quantum MDS codes with parameters $[[60,58,2]]_{19}$, $[[60,56,3]]_{19}$, $[[60,54,4]]_{19}$, $[[60,52,5]]_{19}$,$[[60,50,6]]_{19}$, $[[60,48,7]]_{19}$, $[[60,46,8]]_{19}$, $[[60,44,9]]_{19}$, $[[60,42,10]]_{19}$, $[[60,40,11]]_{19}$, $[[60,38,12]]_{19}$, $[[60,36,13]]_{19}$.

Lemma 3.3

Let q be an odd prime power of the form $q=atm+1$, a be a even number, or a be an odd number and t be a even number, $n=\frac{q^{2}-1}{at}$.

• (i) For $1\le i \le 3$, $(C_{1+(at+2i)m},C_{1+(2im-2)})$ forms a skew-asymmetric pair.

• (ii)
  $$\begin{aligned} |T_{ss}(\delta )| = \left\{ \begin{array}{lll} 0, &{}\quad \hbox {if }2 \le \delta \le (\frac{at}{2}+1)m+1;\\ 2, &{}\quad \hbox {if }(\frac{at}{2}+1)m+2 \le \delta \le (\frac{at}{2}+2)m+1;\\ 4, &{}\quad \hbox {if }(\frac{at}{2}+2)m+2 \le \delta \le (\frac{at}{2}+3)m+1.\\ \end{array} \right. \end{aligned}$$

Proof

• (i) For $q=atm+1$, then $n=\frac{q^{2}-1}{at}=m(q+1)$. $2n-q(1+(2im-2))= 2m(q+1)-[2mi(q+1)-2mi-q] \equiv 1+(at+2i)m$ mod $2m(q+1)$ . Hence, for $1\le i \le 3$, $(C_{1+(at+2i)m},C_{1+(2im-2)})$ forms a skew-asymmetric pair.

• (ii) According to Lemma 3.1, for $2 \le \delta \le (\frac{at}{2}+1)m+1$ in the defining set $T(\delta )=\bigcup _{j=0}^{\delta -1} C_{1+2j}$, we have $T_{ss}(\delta )=\emptyset$.

$\square$

For $(\frac{at}{2}+1)m+2 \le \delta \le (\frac{at}{2}+2)m+1$, according to Lemma 2.3 and Lemma 2.4, one determine the set $T_{ss}$ and $T_{sas}$ by decomposition of the defining set T. In $T_{sas}$, there is no skew-symmetric cyclotomic coset and any two cyclotomic cosets do not form a skew-asymmetric pair; and in $T_{ss}$, there are skew-symmetric cyclotomic cosets or there exist skew-asymmetric pairs. Suppose there exist integers $y \in [(at+2)m+3, (at+4)m-1]$ and $x\in [0,(at+2)m-1]\cup [(at+2)m+3,(at+4)m-1]$ such that $x\equiv -qy$ mod 2n. We find a contradiction as follows.

If $(at+2)m+1 \le x < y \le (at+4)m-1$, then $(\frac{at}{2}+1)2n< (\frac{at}{2}+1)2m(q+1)+3(q+1)=[(at+2)m+3](q+1)\le x+yq \le [(at+4)m-1](q+1)=(\frac{at}{2}+2)2m(q+1)-(q+1)<(\frac{at}{2}+2)2n$, that means that$(\frac{at}{2}+1)2n< x+yq <(\frac{at}{2}+2)2n$, a contradiction. And if $(at+2)m+3 \le y \le (at+4)m-1$, $0 \le y \le (at+2)m-1$, then $(\frac{at}{2}+1)2n< (\frac{at}{2}+1)2m(q+1)+2atm+2m+3=[(at+2)m+3]q\le x+yq \le (at+2)m-1+[(at+4)m-1]q=(\frac{at}{2}+2)2n-q<(\frac{at}{2}+2)2n$, a contradiction.

Similarly, for $(\frac{at}{2}+2)m+2 \le \delta \le (\frac{at}{2}+3)m+1$, suppose there exist integers $y \in [(at+4)m+3, (at+6)m-1]$ and $x \in [0,(at+2)m-1]\cup [(at+2)m+3,(at+4)m-1]\cup [(at+4)m+3, (at+6)m-1]$ such that $x\equiv -qy$ mod 2n. Using the same methods, we can deduce that $(\frac{at}{2}+2)2n<x+qy<(\frac{at}{2}+3)2n$, a contradiction.

Theory 3.4

Let q be an odd prime power of the form $q=atm+1$, a be even, or a be odd and t be even, $n=\frac{q^{2}-1}{at}$. Then, there exists a q-ary $[[\frac{q^{2}-1}{at},\frac{q^{2}-1}{at}-2d+4,d;2]]$- EA-quantum MDS codes, where $(\frac{at}{2}+1)m+2 \le d \le (\frac{at}{2}+2)m+1$; and there exists a q-ary $[[\frac{q^{2}-1}{at},\frac{q^{2}-1}{at}-2d+6,d;4]]$- EA-quantum MDS codes, where $(\frac{at}{2}+2)m+2 \le d \le (\frac{at}{2}+3)m+1$.

Proof

Consider the negacyclic codes over $F_{q^{2}}$ of length $n=\frac{q^{2}-1}{at}$ with defining set $T=\bigcup _{i=0}^{s}C_{1+2i}$, where $0 \le s \le (\frac{at}{2}+3)m-1$ for q be an odd prime power of the form $q=atm+1$, a be even, or a be odd and t be even. By Lemma 3.3, there is $c=|T_{ss}(\delta )|=2$ if $(\frac{at}{2}+1)m \le s\le (\frac{at}{2}+2)m-1$ and $c=|T_{ss}(\delta )|=4$ if $(\frac{at}{2}+2)m \le s \le (\frac{at}{2}+3)m-1$. Since every $q^{2}$-cyclotomic coset $C_{x}$ has exactly one element and x must be odd number, we can obtain that T consists of $s+1$ integers $\{1,3,\ldots ,1+2s\}$. It implies that $\mathcal {C}$ has minimum distance at least $s+2$. Hence, $\mathcal {C}$ is a $q^{2}$-ary negacyclic code with parameters $[n,n-(s+1),\ge s+2]$. Combining Lemma 2.8 with EA-quantum singleton bound, we can obtain a quantum MDS code with parameters $[[\frac{q^{2}-1}{at},\frac{q^{2}-1}{at}-2d+4,d;2]]$, where $(\frac{at}{2}+1)m+2 \le d \le (\frac{at}{2}+2)m+1$; $[[\frac{q^{2}-1}{at},\frac{q^{2}-1}{at}-2d+6,d;4]]$, where $(\frac{at}{2}+2)m+2 \le d \le (\frac{at}{2}+3)m+1$, for q be an odd prime power of the form $q=atm+1$, a be even, or a be odd and t be even. $\square$

Example 2

Let $a=3,t=2$. Then, $q=19$, $n=60$ applying Theory 3.4 produces:

• (1) new $2-ebits$ EA-quantum MDS codes with parameters $[[60,36,14;2]]_{19}$, $[[60,34,15;2]]_{19}$, $[[60,32,16;2]]_{19}$.

• (2) new 4-ebits EA-quantum MDS codes with parameters $[[60,32,17;4]]_{19}$, $[[60,30,18;4]]_{19}$, $[[60,28,19;4]]_{19}$.

3.2 New EA-quantum MDS codes of length $n=\frac{q^{2}-1}{2s_{1}s_{2}}$

In this subsection, we construct some classes of q-ary EA-quantum MDS codes of length $n=\frac{q^{2}-1}{2s_{1}s_{2}}$, where q be an odd prime power, $2s_{1}|(q-1)$, $s_{2}|(q+1)$ and $s_{2}$ is an odd integer. Let $n=\frac{q^{2}-1}{2s_{1}s_{2}}$, $r=2$. Since $2n|q^{2}-1$, then for each odd x in the range $1\le x \le 2n$, the $q^{2}$-cyclotomic coset $C_{x}$ modulo 2n is $C_{x}=x$.

Lemma 3.5

Let q be an odd prime power of the form $q=30m+11$, $n=\frac{q^{2}-1}{30}$.

• (i) $(C_{1+2(10m+3)},C_{1+2(5m+1)})$, $(C_{1+2(13m+4)},C_{1+2(8m+4)})$ and $(C_{1+2(16m+5)},C_{1+2(11m+3)})$ form skew-asymmetric pairs.

• (ii)
  $$\begin{aligned} |T_{ss}(\delta )| = \left\{ \begin{array}{lll} 0, &{}\quad \hbox {if }2 \le \delta \le 8m+3;\\ 2, &{}\quad \hbox {if }8m+4 \le \delta \le 11m+5;\\ 4, &{}\quad \hbox {if }11m+6\le \delta \le 14m+7.\\ \end{array} \right. \end{aligned}$$

Proof

• (i) Let $q=30m+11$ and $2n=2\frac{q^{2}-1}{30}=2(3m+1)(10m+4)$. Since $[1+2(5m+1)]q=(10m+4-1)(30m+10+1)=5\cdot 2(10m+4)(3m+1)-[1+2(10m+3)]$, $-[1+2(5m+1)]q \equiv 1+2(10m+3)$ mod 2n.

  Since $[1+2(8m+4)]q=(10m+4+6m+1)(30m+10+1)=8\cdot 2(10m+4)(3m+1)-[1+2(13m+4)]$, $-(16m+5)q \equiv 1+2(13m+4)$ mod 2n.

  Since $[1+2(11m+3)]q=(20m+8+2m-7)(30m+10+1)=11\cdot 2(10m+4)(3m+1)-[1+2(16m+5)]$, $-(22m+7)q \equiv 1+2(16m+5)$ mod 2n.

• (ii) According to Lemma 17 in Ref. 22, if the defining set $T=\bigcup _{j=2m+1}^{l} C_{1+2j}$, where $2m+1\le l \le 10m+2$, then $T_{ss}(\delta )=\emptyset$ for $2 \le \delta \le 8m+3$.

  Let $T_{1}=\bigcup _{j=2m+1}^{10m+2} C_{1+2j}$. If the defining set $T=\bigcup _{j=10m+4}^{l} C_{1+2j} \bigcup T_{1}$, where $10m+4\le l \le 13m+3$, then $8m+4 \le \delta \le 11m+5$ holds.

  Let $I_{0}=[1+2(2m+1),1+2(10m+2)]$, $I_{1}=[1+2(10m+4),1+2(13m+3)]$ and $I_{2}=[1+2(13m+5),1+2(16m+4)]$.

$\square$

According to Lemmas 2.4 and 2.1 in [12], one needs to testify that for $x,y \in I_{0}\cup I_{1}\cup I_{2}$, there $x+yq \not \equiv 0$ mod 2n holds.

For $8m+4 \le \delta \le 11m+5$, suppose there exist integers $y \in I_{1}$, $x\in I_{0}\cup I_{1}$, such that $x\equiv -qy$ mod 2n. We find a contradiction in the following.

We divided $I_{1}$ into three parts such as $[1+2(10m+4),1+2(10m+4)+2(m-1)]\cup$ $[1+2(10m+4)+2m,1+2(10m+4)+4(m-1)]\cup [1+2(10m+4)+4m-2,1+2(13m+3)]$.

If $x,y\in [1+2(10m+4),1+2(10m+4)+2(m-1)]$, then $10(2n)<10(2n)+70m+28=(20m+19)(30m+12)\le y(q+1)\le (22m+7)(30m+12)<11(2n)$; if $x,y\in [1+2(10m+4)+2m,1+2(10m+4)+4(m-1)]$, then $11(2n)<(22m+9)(30m+12)\le y(q+1)\le (24m+5)(30m+12)<12(2n)$; and if $x,y\in [1+2(10m+4)+4m-2,1+2(13m+3)]$, then $12(2n)<(24m+7)(30m+12)\le y(q+1)\le (26m+7)(30m+12)=13(2n)-50m-20<13(2n)$, a contradiction.

Similarly, for $11m+6\le \delta \le 14m+7$, suppose there exist integers $y \in I_{2}$, $x\in I_{0}\cup I_{1} \cup I_{2}$, such that $x\equiv -qy$ mod 2n. Using the same method, one also finds a contradiction.

Theory 3.6

Let q be an odd prime power of the form $q=30m+11$, $n=\frac{q^{2}-1}{30}$. There exists a q-ary $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+4,d;2]]$- EA-quantum MDS codes, where $8m+4 \le d \le 11m+5$; and there exists a q-ary $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+6,d;4]]$- EA-quantum MDS codes, where $11m+6\le d \le 14m+7$.

Proof

Consider the negacyclic codes over $F_{q^{2}}$ of length $n=\frac{q^{2}-1}{30}$ with defining set $T=\bigcup _{i=2m+1}^{s}C_{1+2i}$, where $2m+1 \le s \le 16m+4$ for q be an odd prime power of the form $q=30m+11$, m is integer number. By Lemma 3.5, there is $c=|T_{ss}(\delta )|=2$ if $10m+3 \le s \le 13m+3$ and $c=|T_{ss}(\delta )|=4$ if $13m+4\le s \le 16m+4$. Since every $q^{2}$-cyclotomic coset $C_{x}$ has exactly one element and x must be odd number, we can obtain that T consists of $s+1$ integers $\{1+2(2m+1),1+2(2m+2),\ldots ,1+2s\}$. It implies that $\mathcal {C}$ has minimum distance at least $s-(2m+1)+1$. Hence, $\mathcal {C}$ is a $q^{2}$-ary negacyclic code with parameters $[n,n-(s-(2m+1)+1),\ge s-(2m+1)+2]$. Combining Lemma 2.8 with EA-quantum singleton bound, we can obtain a quantum MDS code with parameters $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+4,d;2]]$, where $8m+4 \le d \le 11m+5$; $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+6,d;4]]$, where $11m+6\le d \le 14m+7$, for q be an odd prime power of the form $q=30m+11$. $\square$

Example 3

Let $q=43$, applying Theory 3.6 produces:

• (1) new 2-ebits EA-quantum MDS codes with parameters $[[56,36,12;2]]_{43}$, $[[56,34,13;2]]_{43}$, $[[56,32,14;2]]_{43}$, $[[56,30,15;2]]_{43}$, $[[56,28,16;2]]_{43}$.

• (2) new 4-ebits EA-quantum MDS codes with parameters $[[56,28,17;4]]_{43}$, $[[56,26,18;4]]_{43}$, $[[56,24,19;4]]_{43}$, $[[56,22,20;4]]_{43}$, $[[56,20,21;4]]_{43}$.

Lemma 3.7

Let q be an odd prime power of the form $q=30m+19$, $n=\frac{q^{2}-1}{30}$.

• (i) $(C_{1+2(9m+5)},C_{1+2(6m+3)})$, $(C_{1+2(12m+7)},C_{1+2(3m+1)})$, $(C_{1+2(14m+8)},C_{1+2 (11m+6)})$ and $(C_{1+2(16m+10)},C_{1+2(8m+4)})$ form skew-asymmetric pairs.

• (ii)
  $$\begin{aligned} |T_{ss}| = \left\{ \begin{array}{lll} 0, &{}\quad \hbox {if }2 \le \delta \le 8m+5;\\ 2, &{}\quad \hbox {if }8m+6 \le \delta \le 11m+7;\\ 4, &{}\quad \hbox {if }11m+8\le \delta \le 13m+8;\\ 6, &{}\quad \hbox {if }13m+9\le \delta \le 16m+10.\\ \end{array} \right. \end{aligned}$$

Proof

• (i) Let $q=30m+19$ and $2n=2\frac{q^{2}-1}{30}=2(3m+1)(10m+6)$. Since $[1+2(6m+3)]q=(12m+8-1)(30m+18+1)=6\cdot 2(10m+6)(3m+2)-[1+2(9m+5)]$, $-[1+2(6m+3)]q \equiv 1+2(9m+5)$ mod 2n.

  Since $[1+2(3m+1)]q=(6m+4)(30m+18+1)=3\cdot 2(3m+2)(10m+6)-[1+2(12m+7)]$, $-[1+2(3m+1)]q \equiv [1+2(12m+7)]$ mod 2n.

  Since $[1+2(11m+6)]q=(21m+14+m-1)(30m+18+1)=12\cdot 2(3m+2)(10m+6)-[1+2(14m+8)]$, $-[1+2(11m+6)]q \equiv [1+2(14m+8)]$ mod 2n.

  Since $[1+2(8m+4)]q=(15m+14+m-1)(30m+18+1)=12\cdot 2(3m+2)(10m+6)-[1+2(17m+10)]$, $-[1+2(17m+10)]q \equiv [1+2(8m+4)]$ mod 2n.

• (ii) According to Lemma 17 in Ref. [22], if the defining set $T=\bigcup _{j=m+1}^{l} C_{1+2j}$, where $m+1\le l \le 9m+4$, then $T_{ss}(\delta )=\emptyset$ for $2 \le \delta \le 8m+5$.

  Let $I_{0}=[1+2(m+1),1+2(9m+4)]$, $I_{1}=[1+2(9m+6),1+2(12m+6)]$, $I_{2}=[1+2(12m+8),1+2(14m+7)]$, and $I_{3}=[1+2(14m+9),1+2(17m+9)]$.

$\square$

According to Lemmas 2.4 and 2.1 in [12], one needs to testify that for $x,y \in I_{0}\cup I_{1}\cup I_{2}\cup I_{3}$, there $x+yq \not \equiv 0$ mod 2n holds.

Let $T_{1}=\bigcup _{j=m+1}^{10m+2} C_{1+2j}$. If the defining set $T=\bigcup _{j=9m+5}^{l} C_{1+2j} \bigcup T_{1}$, where $9m+5\le l \le 12m+6$, then $8m+6 \le \delta \le 11m+7$ holds.

For $8m+6 \le \delta \le 11m+7$, suppose there exist integers $y \in I_{1}$, $x\in I_{0}\cup I_{1}$, such that $x\equiv -qy$ mod 2n. We find a contradiction in the following.

We divided $I_{1}$ into three parts such as $[1+2(9m+6),1+2(10m+4)+2(m-1)]\cup$ $[1+2(10m+4)+2m,1+2(10m+4)+4(m-1)]\cup [1+2(10m+4)+4m-2,1+2(13m+3)]$.

If $x,y\in [1+2(10m+4),1+2(10m+4)+2(m-1)]$, then $10(2n)<10(2n)+70m+28=(20m+19)(30m+12)\le y(q+1)\le (22m+7)(30m+12)<11(2n)$; if $x,y\in [1+2(10m+4)+2m,1+2(10m+4)+4(m-1)]$, then $11(2n)<(22m+9)(30m+12)\le y(q+1)\le (24m+5)(30m+12)<12(2n)$; and if $x,y\in [1+2(10m+4)+4m-2,1+2(13m+3)]$, then $12(2n)<(24m+7)(30m+12)\le y(q+1)\le (26m+7)(30m+12)=13(2n)-50m-20<13(2n)$, a contradiction.

Similarly, for $\delta \in [8m+6,11m+7]\cup [11m+8,13m+8]\cup [13m+9,16m+10]$, using the same method, the lemma holds.

Theory 3.8

Let q be an odd prime power of the form $q=30m+19$, $n=\frac{q^{2}-1}{30}$. There exists a q-ary $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+4,d;2]]$- EA-quantum MDS codes, where $8m+6 \le d \le 11m+7$; there exists a q-ary $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+6,d;4]]$- EA-quantum MDS codes, where $11m+8\le d \le 13m+8$; and there exists a q-ary $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+8,d;6]]$- EA-quantum MDS codes, where $13m+9\le d \le 16m+10$.

Proof

Consider the negacyclic codes over $F_{q^{2}}$ of length $n=\frac{q^{2}-1}{30}$ with defining set $T=\bigcup _{i=m+1}^{s}C_{1+2i}$, where $m+1 \le s \le 16m+9$ for q be an odd prime power of the form $q=30m+19$, m is integer number. By Lemma 3.7, there is $c=|T_{ss}(\delta )|=2$ if $9m+5\le s \le 12m+6$, $c=|T_{ss}(\delta )|=4$ if $12m+7\le s \le 14m+7$, and $c=|T_{ss}(\delta )|=6$ if $14m+8\le s \le 17m+9$. Since every $q^{2}$-cyclotomic coset $C_{x}$ has exactly one element and x must be odd number, we can obtain that T consists of $s+1$ integers $\{1+2(m+1),1+2(m+2),\ldots ,1+2s\}$. It implies that $\mathcal {C}$ has minimum distance at least $s-(m+1)+1$. Hence, $\mathcal {C}$ is a $q^{2}$-ary negacyclic code with parameters $[n,n-(s-(m+1)+1),\ge s-(m+1)+2]$. Combining Lemma 2.8 with EA-quantum singleton bound, we can obtain a EA-quantum MDS code with parameters $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+4,d;2]]_{q}$ where $8m+6 \le d \le 11m+7$; $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+6,d;4]]_{q}$s, where $11m+8\le d \le 13m+8$; and $[[\frac{q^{2}-1}{30},\frac{q^{2}-1}{30}-2d+8,d;6]]_{q}$, where $13m+9\le d \le 16m+10$, for q be an odd prime power of the form $q=30m+19$. $\square$

Example 4

Let $q=49$, applying Theory 3.8 produces:

• (1) 2-ebits EA-quantum MDS codes with parameters $[[80,56,14;2]]_{49}$, $[[80,54,15;2]]_{49}$, $[[80,52,16;2]]_{49}$, $[[80,50,17;2]]_{49}$.

• (2) 4-ebits EA-quantum MDS codes with parameters $[[80,48,19;4]]_{49}$, $[[80,46,20;4]]_{49}$, $[[80,44,21;4]]_{49}$.

• (3) 6-ebits EA-quantum MDS codes with parameters $[[80,44,22;6]]_{49}$, $[[80,42,23;6]]_{49}$, $[[80,40,24;6]]_{49}$, $[[80,38,25;6]]_{49}$, $[[80,36,26;6]]_{49}$.

Lemma 3.9

Let q be an odd prime power of the form $q=12m+5$, $n=\frac{q^{2}-1}{12}$.

• (i) $(C_{1+2(7m+2)},C_{1+2(5m+1)})$, $(C_{1+2(9m+3)},C_{1+6m})$ and $(C_{1+2(10m+3)},C_{1+2(8m+2)})$ form skew-asymmetric pairs.

• (ii)
  $$\begin{aligned} |T_{ss}| = \left\{ \begin{array}{lll} 0, &{}\quad \hbox {if }2 \le \delta \le 5m+2;\\ 2, &{}\quad \hbox {if }5m+3 \le \delta \le 7m+3;\\ 4, &{}\quad \hbox {if }7m+4\le \delta \le 8m+3.\\ \end{array} \right. \end{aligned}$$

Proof

• (i) Let $q=12m+5$ and $2n=2\frac{q^{2}-1}{30}=4(2m+1)(3m+1)$. Since $[1+2(5m+1)]q=(10m+5-2)(12m+4+1)=5\cdot 4(2m+1)(3m+1)-[1+2(7m+2)]$, $-[1+2(7m+2)]q \equiv 1+2(5m+1)$ mod 2n.

  Since $(6m+1)q=(6m+3-2)(12m+4+1)=3\cdot 4(2m+1)(3m+1)-[1+2(9m+3)]$, $-[1+2(9m+3)]q \equiv 6m+1$ mod 2n.

  Since $(6m+1)q=(6m+3-2)(12m+4+1)=3\cdot 4(2m+1)(3m+1)-[1+2(9m+3)]$, $-[1+2(9m+3)]q \equiv 6m+1$ mod 2n.

  Since $[1+2(8m+2)]q=(16m+8-3)(12m+4+1)=8\cdot 4(2m+1)(3m+1)-[1+2(10m+3)]$, $-[1+2(10m+3)]q \equiv 1+2(8m+2)$ mod 2n.

• (ii) According to Lemma 17 in Ref. 22, if the defining set $T=\bigcup _{j=2m+1}^{l} C_{1+2j}$, where $2m+1\le l \le 7m+2$, then $T_{ss}(\delta )=\emptyset$ for $2 \le \delta \le 5m+2$.

  Let $I_{0}=[1+2(m+1),1+2(7m+1)]$, $I_{1}=[1+2(7m+3),1+2(9m+2)]$ and $I_{2}=[1+2(9m+4),1+2(10m+2)]$.

$\square$

According to Lemmas 2.4 and 2.1 in [12], one needs to testify that for $x,y \in I_{0}\cup I_{1}\cup I_{2}$, there $x+yq \not \equiv 0$ mod 2n holds.

Let $T_{0}=\bigcup _{j=m+1}^{7m+2} C_{1+2j}$. If the defining set $T=\bigcup _{j=7m+3}^{l} C_{1+2j} \bigcup T_{0}$, where $7m+3\le l \le 9m+3$, then $5m+3 \le \delta \le 7m+3$ holds.

For $5m+3 \le \delta \le 7m+3$, suppose there exist integers $y \in I_{1}$, $x\in I_{0}\cup I_{1}$, such that $x\equiv -qy$ mod 2n. We find a contradiction in the following.

We divided $I_{1}$ into two parts such as $[1+2(7m+3),1+2(8m+2)]\cup [1+2(8m+3),1+2(9m+2)]$.

If $x,y\in [1+2(7m+3),1+2(8m+2)]$, then $7(2n)<7(2n)+28m+14=(14m+7)(12m+6)\le y(q+1)\le (16m+5)(12m+6)=8(3m+1)(8m+4)-(4m+2)<8(2n)$; and if $x,y\in [1+2(8m+3),1+2(9m+2)]$, then $8(2n)<(16m+7)(12m+6)\le y(q+1)\le (18m+5)(12m+6)<9(2n)$, a contradiction.

If $x\in I_{0}$, $y\in [1+2(7m+3),1+2(8m+2)]$, since $7(2n)+14m+7=(14m+7)(12m+5)\le yq\le (16m+5)(12m+5)=7(2n)+24m^{2}-3$, then $2n-yq > x$ mod 2n;

and if $x\in I_{0}$, $y\in [1+2(8m+3),1+2(9m+2)]$, since $8(2n)+4m+3=(16m+7)(12m+5)\le yq\le (18m+5)(12m+5)=9(2n)-30m^{2}-11$, then $2n-yq > x$ mod 2n, a contradiction.

Similarly, for $\delta \in [7m+4, 8m+3]$, using the same method, the lemma holds.

Theory 3.10

Let q be an odd prime power of the form $q=12m+5$, $n=\frac{q^{2}-1}{12}$. There exists a q-ary $[[\frac{q^{2}-1}{12},\frac{q^{2}-1}{12}-2d+4,d;2]]$- EA-quantum MDS codes, where $5m+3 \le d \le 7m+3$; and there exists a q-ary $[[\frac{q^{2}-1}{12},\frac{q^{2}-1}{12}-2d+6,d;4]]$- EA-quantum MDS codes, where $7m+4\le d \le 8m+3$.

Proof

Consider the negacyclic codes over $F_{q^{2}}$ of length $n=\frac{q^{2}-1}{12}$ with defining set $T=\bigcup _{i=2m+1}^{s}C_{1+2i}$, where $2m+1 \le s \le 10m+2$ for q be an odd prime power of the form $q=12m+5$, m is integer number. By Lemma 3.9, there is $c=|T_{ss}(\delta )|=2$ if $7m+2 \le s \le 9m+2$ and $c=|T_{ss}(\delta )|=4$ if $9m+3\le s \le 10m+2$. Since every $q^{2}$-cyclotomic coset $C_{x}$ has exactly one element and x must be odd number, we can obtain that T consists of $s+1$ integers $\{1+2(2m+1),1+2(2m+2),\cdots ,1+2s\}$. It implies that $\mathcal {C}$ has minimum distance at least $s-(2m+1)+1$. Hence, $\mathcal {C}$ is a $q^{2}$-ary negacyclic code with parameters $[n,n-(s-(2m+1)+1),\ge s-(2m+1)+2]$. Combining Lemma 2.8 with EA-quantum singleton bound, we can obtain a quantum MDS code with parameters $[[\frac{q^{2}-1}{12},\frac{q^{2}-1}{12}-2d+4,d;2]]_{q}$, where $5m+3 \le d \le 7m+3$; and $[[\frac{q^{2}-1}{12},\frac{q^{2}-1}{12}-2d+6,d;4]]_{q}$, where $7m+4\le d \le 8m+3$; for q be an odd prime power of the form $q=12m+5$. $\square$

Example 5

Let $q=29$, applying Theory 3.10 produces:

• (1) 2-ebits EA-quantum MDS codes with parameters $[[70,48,13;2]]_{29}$, $[[70,46,14;2]]_{29}$, $[[70,44,15;2]]_{29}$, $[[70,42,16;2]]_{29}$, $[[70,40,17;2]]_{29}$.

• (2) 4-ebits EA-quantum MDS codes with parameters $[[70,40,18;4]]_{29}$, $[[70,38,19;4]]_{29}$.

3.3 New EA-quantum MDS codes of length $n=\frac{q^{2}+1}{5}$

In this section, let q be an odd prime power of the $q=10m+3$ or $q=10m+7$, where m is a positive integer. Let $n=\frac{q^{2}+1}{5}$, $r=2$ and $\eta \in F_{q^{2}}$ be a primitive rth root of unity. Since 5 and 2 are two factors of $q^2+1$ and $2n|q^{4}-1$, then for each odd x in the range $1\le x \le n$, the $q^{2}$-cyclotomic coset $C_{x}$ modulo 2n is $C_{x}=\{x,n-x\}$. Then, we discuss negacyclic codes of length n over $F_{q^{2}}$ to construct EA-quantum MDS codes.

Lemma 3.11

Let q be an odd prime power of the form $q=10m+3$, $n=\frac{q^{2}+1}{5}$, $s=\frac{n}{2}$.

• (1) If $1\le x\le 12m+3$, then $(C_{1+4m},C_{1+2m})$ and $(C_{1+2(6m+1)},C_{1+2(3m+1)})$ form skew-asymmetric pairs, respectively;

  $$\begin{aligned} |T_{ss}(\delta )| = \left\{ \begin{array}{lll} 0, &{}\quad \hbox {if } 3 \le \delta \le 4m+1, \quad \hbox { for }\delta \hbox { is odd};\\ 4, &{}\quad \hbox {if }4m+3 \le \delta \le 12m+3,\quad \hbox { for }\delta \hbox { is odd}.\\ \end{array} \right. \end{aligned}$$

• (2) If $m=2t+1$ is odd, and $s\le x\le s+10m+4$, then $(C_{s+6m+2},C_{s+2m})$, $(C_{s+10m+4}, C_{s+10m+2})$ form skew-asymmetric pairs, respectively;

  $$\begin{aligned} |T_{ss}(\delta )| = \left\{ \begin{array}{lll} 0, &{}\quad \hbox {if }2 \le \delta \le 6m+2, \quad \hbox { for }\delta \hbox { is even};\\ 4, &{}\quad \hbox {if }6m+4 \le \delta \le 10m+4, \quad \hbox { for }\delta \hbox { is even}.\\ \end{array} \right. \end{aligned}$$

  If $m=2t$ is even, and $s\le x\le s+12m+4$, then $C_{s}$ is skew-symmetric, and $(C_{s+8m+2},C_{s+4m+2}), (C_{s+12m+4},C_{s+4m})$ form skew-asymmetric pairs, respectively;

  $$\begin{aligned} |T_{ss}(\delta )| = \left\{ \begin{array}{lll} 1, &{}\quad \hbox {if }2 \le \delta \le 8m+2, \quad \hbox { for } \delta \hbox { is even};\\ 5, &{}\quad \hbox {if }8m+4 \le \delta \le 12m+4, \quad \hbox { for } \delta \hbox { is even}.\\ \end{array} \right. \end{aligned}$$

Proof

Let $q=10m+3$. Since $2n=40m^2+24m+4$.

(1) (i) Let $1\le x\le 12m+2$. Since $(4m+1)q=40m^2+24m+3-(2m+1)$, $(C_{1+4m},C_{1+2m})$ form skew-asymmetric pairs. Since $(12m+3)q=120m^2+66m+9=4n+2n-(6m+3)\equiv 2n-(6m+3)$ mod 2n, $(C_{1+2(6m+1)},C_{1+2(3m+1)})$ form skew-asymmetric pairs.

(ii) (a) If the defining set $T=\bigcup _{j=1}^{l} C_{1+2j}$, where $1\le l \le 2m-1$, we testify that $T_{ss}(\delta )=\emptyset$ for $2 \le \delta \le 4m+1$. According to Lemmas 2.4 and 2.1 in [12], one needs to testify that for $x,y \in [1,4m-1]$, there $x+yq \not \equiv 0$ mod 2n holds.

If $x,y\in [1,4m-1]$, $1<y(q+1)<(4m-1)(q+1)=40m^2+6m-4<2n$. Hence, if the defining set $T=\bigcup _{j=1}^{l}C_{1+2j}$, where $1\le l\le 2m-1$, the $T_{ss}(\delta )=\emptyset$ for $2 \le \delta \le 6m+2$.

(b) Let $I_{0}=[1,4m-1]$, $I_{1}=[4m+3,12m+1]$.

According to Lemmas 2.4 and 2.1 in [12], one needs to testify that for $x,y \in I_{0}\cup I_{1}$, there $x+yq \not \equiv 0$ mod 2n holds.

Let $T_{0}=\bigcup _{1}^{j=4m-1}C_{1+2j}$, and the defining set $T=\bigcup _{j=2m}^{l}C_{1+2j} \bigcup T_{0}$, where $2m\le l\le 6m$.

For $4m+3 \le \delta \le 12m+3$, suppose there exist integers $y\in I_{1}, x\in I_{0}\cup I_{1}$, such that $x\equiv -qy$ mod 2n. We find a contradiction in the following.

We divided $I_{1}$ into four parts such as $[4m+3,6m+1]\cup [6m+3,8m+1]\cup [8m+3,10m+1]\cup [10m+3,12m+1]$.

If $x,y\in [4m+3,6m+1]$, then $(2n)<40m^{2}+24m+4+22m+8=(4m+3)(q+1)\le y(q+1)\le (6m+1)(q+1)=60m^2+34m+4<2(2n)$; if $x,y\in [6m+3,8m+1]$, $2(2n)<(6m+3)(q+1)\le y(q+1)\le (8m+1)(q+1)<4(2n)$; if $x,y\in [8m+3,10m+1]$, $4(2n)<(8m+3)(q+1)\le y(q+1)\le (8m+1)(q+1)<6(2n)$; if $x,y\in [10m+3,12m+1]$, $6(2n)<(10m+3)(q+1)\le y(q+1)\le (12m+1)(q+1)<8(2n)$, a contradiction.

If $x\in I_{0}$, $y\in [4m+3,6m+1]$, since $2n<2n+18m+5=(4m+3)q\le yq\le (6m+1)q=2(2n)-20m^2-20m-3$, then $2n-yq>x$ mod 2n, a contradiction. Similarly, if $x\in I_{0}$, $y\in [8m+3,10m+1]$ and $x\in I_{0}$, $y\in [10m+3,12m+1]$, using the same method, one can deduce a contradiction.

(2) (i) If $m=2t+1$, $s\le x\le s+10m+4$, $(s+6m+2)q=(\frac{n}{2}+6m+2)q=5mn+4n+\frac{n}{2}+2m$. Since $m=2t+1$, $(s+6m+2)q\equiv \frac{3}{2}n+2m$ mod 2n. $-(s+6m+2)q\equiv \frac{n}{2}+2m$ mod 2n.

Since $(s+10m+4)q=(\frac{n}{2}+10m+4)q=5mn+6n+\frac{1}{2}n+10m+2$, $-(s+10m+4)q\equiv \frac{1}{2}n+10m+2$ mod 2n. Hence, $(C_{s+6m+2},C_{s+2m})$, $(C_{s+10m+4}, C_{s+10m+2})$ form skew-asymmetric pairs, respectively;

If the defining set $T=\bigcup _{j=1}^{l} C_{j}$, where $s\le l \le s+6m$, we testify that $T_{ss}(\delta )=\emptyset$ for $2 \le \delta \le 6m+2$. According to Lemma 2.4 and Lemma 2.1 in [12], one needs to testify that for $x,y \in [s,s+6m]$, there $x+yq \not \equiv 0$ mod 2n holds.

If $x,y\in [s,s+6m]$, $1<y(q+1)<(4m-1)(q+1)=40m^2+6m-4<2n$. Hence, if the defining set $T=\bigcup _{j=1}^{l}C_{1+2j}$, where $1\le l\le 2m-1$, the $T_{ss}(\delta )=\emptyset$ for $2 \le \delta \le 6m+2$.

Let $I_{0}=[s,s+6m]$, $I_{1}=[s+6m+4,s+10m+4]$.

According to Lemmas 2.4 and 2.1 in [12], one needs to testify that for $x,y \in I_{0}\cup I_{1}$, there $x+yq \not \equiv 0$ mod 2n holds.

Let $T_{0}=\bigcup _{j=0}^{3m}C_{s+2j}$, and the defining set $T=\bigcup _{j=3m+2}^{l}C_{s+2j} \bigcup T_{0}$, where $3m+4\le l\le 5m+2$.

For $6m+4 \le \delta \le 10m+4$, suppose there exist integers $y\in I_{1}, x\in I_{0}\cup I_{1}$, such that $x\equiv -qy$ mod 2n. We find a contradiction in the following.

We divided $I_{1}$ into four parts such as $[4m+3,6m+1]\cup [6m+3,8m+1]\cup [8m+3,10m+1]\cup [10m+3,12m+1]$.

If $x,y\in [4m+3,6m+1]$, then $(2n)<40m^{2}+24m+4+22m+8=(4m+3)(q+1)\le y(q+1)\le (6m+1)(q+1)=60m^2+34m+4<2(2n)$; if $x,y\in [6m+3,8m+1]$, $2(2n)<(6m+3)(q+1)\le y(q+1)\le (8m+1)(q+1)<4(2n)$; if $x,y\in [8m+3,10m+1]$, $4(2n)<(8m+3)(q+1)\le y(q+1)\le (8m+1)(q+1)<6(2n)$; if $x,y\in [10m+3,12m+1]$, $6(2n)<(10m+3)(q+1)\le y(q+1)\le (12m+1)(q+1)<8(2n)$, a contradiction.

(ii) For $m=2t$ is even, since $2n-sq=2n-(5mn+n+\frac{n}{2})\equiv s$ mod 2n, $C_{s}$ is skew-symmetric. $2n-(s+8m+2)q\equiv s+4m+2$ mod 2n and $2n-(s+12m+4)q\equiv s+4m$ mod 2n.

If the defining set $T=\bigcup _{1}^{j=l} C_{s+2j}$, where $1\le l \le 4m$, we testify that $T_{ss}(\delta )=\{C_{s}\}$ for $2 \le \delta \le 8m+2$. Since $C_{s}$ is skew-symmetric, according to Lemma 2.4 and Lemma 2.1 in [12], one needs to testify that for $x,y \in [s+2,s+8m]$, there $x+yq \not \equiv 0$ mod 2n holds.

If $x,y\in [s+2,s+8m]$, $5mn+2n+20m+8=(s+2)(q+1)<y(q+1)<(4m-1)(q+1)=5mn+2n+4n-16m-8$. For $x,y\in [s+2,s+8m]$, there is no skew-symmetric cyclotomic coset and any two cyclotomic cosets are not skew-asymmetric cosets. Hence, if the defining set $T=\bigcup _{1}^{j=l}C_{s+2j}$, where $1\le l\le 4m$, the $T_{ss}(\delta )=\{C_{s}\}$ for $2 \le \delta \le 8m+2$.

Let $T_{0}=\bigcup _{2}^{j=4m}C_{s+2j}$, and the defining set $T=\bigcup _{4m+2}^{j=l}C_{s+2j} \bigcup T_{0}$, where $4m+2\le l\le 6m+1$. We testify that for $x,y\in T$, there is no skew-symmetric cyclotomic coset and any two cyclotomic cosets are not skew-asymmetric cosets.

Let $I_{0}=[s+2,s+8m]$, $I_{1}=[s+8m+4,s+12m+2]$. According to Lemmas 2.4 and 2.1 in [12], one needs to testify that for $x,y \in I_{0}\cup I_{1}$, there $x+yq \not \equiv 0$ mod 2n holds. Using the same above-mentioned method, one can easily testify that the lemma holds. $\square$

Theory 3.12

Let q be an odd prime power of the form $q=10m+3$.

• (1) If $m=2t+1$ is odd, then there exists a q-$[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]$- EA-quantum MDS codes, where $4m+3\le d \le 6m+1$ be odd and $6m+4\le d\le 10m+4$ be even.

• (2) If $m=2t$ is even, then there exists a q-ary $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+3,d;1]]$-EA-quantum MDS codes, where $2\le d \le 8m+1$ be even; there exists a q-ary $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]$-EA-quantum MDS codes, where $4m+3\le d \le 6m+1$ be odd; and there exists a q-ary $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+7,d;5]]$-EA-quantum MDS codes, where $8m+4\le d \le 12m+4$ be even.

Proof

• (a) Consider the negacyclic codes over $F_{q^{2}}$ of length $n=\frac{q^{2}+1}{5}$ with defining set $T=\bigcup _{i=0}^{s}C_{1+2i}$, where $0\le s \le 6m$ for q be an odd prime power of the form $q=10m+3$. If m is odd, by Lemma 3.11 (i), there is $c=|T_{ss}(\delta )|=4$ if $2m\le s \le 6m$. Since every $q^{2}$-cyclotomic coset $C_{x}=\{x,n-x\}$ and x must be odd number, we can obtain that T consists of $2s+1$ integers $\{n-(1+2s),\ldots ,n-1,1,3,\ldots ,1+2s\}$. It implies that $\mathcal {C}$ has minimum distance at least $2s+2$. Hence, $\mathcal {C}$ is a $q^{2}$-ary negacyclic code with parameters $[n,n-2(s+1)+1,\ge 2s+2]$. Combining Lemma 2.8 with EA-quantum singleton bound, we can obtain a EA-quantum MDS code with parameters $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]_{q}$, where $4m+3\le d \le 6m+1$ be odd. If m is even, using the same method, one can obtain that $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]_{q}$, where $4m+3\le d \le 6m+1$ be odd.

• (b) Consider the negacyclic codes over $F_{q^{2}}$ of length $n=\frac{q^{2}+1}{5}$ with defining set $T=\bigcup _{i=0}^{k}C_{s+2i}$, where $0\le k \le 5m+1$, $s=\frac{n}{2}$ for q be an odd prime power of the form $q=10m+3$. If m is odd, by Lemma 3.11 (ii), there is $c=|T_{ss}(\delta )|=4$ if $3m+1\le s \le 5m+1$. Since every $q^{2}$-cyclotomic coset $C_{x}=\{x,n-x\}$ and x must be odd number, we can obtain that T consists of $2k+1$ integers $\{n-(s+2k),\ldots ,n-s,s,s+2,\ldots ,s+2k\}$. It implies that $\mathcal {C}$ has minimum distance at least $2k+2$. Hence, $\mathcal {C}$ is a $q^{2}$-ary negacyclic code with parameters $[n,n-2(s+1)+1,\ge 2s+2]$. Combining Lemma 2.8 with EA-quantum singleton bound, we can obtain a EA-quantum MDS code with parameters $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]_{q}$, $6m+4\le d\le 10m+4$ be even.

  If m is even, by Lemma 3.11 (ii), there is $c=|T_{ss}(\delta )|=1$ if $1\le k \le 4m$ and $c=|T_{ss}(\delta )|=5$ if $4m+1\le k \le 6m+1$. Since every $q^{2}$-cyclotomic coset $C_{x}=\{x,n-x\}$ and x must be odd number, we can obtain that T consists of $2k+1$ integers $\{n-(s+2k),\ldots ,n-s,s,s+2,\ldots ,s+2k\}$. It implies that $\mathcal {C}$ has minimum distance at least $2k+2$. Hence, $\mathcal {C}$ is a $q^{2}$-ary negacyclic code with parameters $[n,n-2(s+1)+1,\ge 2s+2]$. Combining Lemma 2.8 with EA-quantum singleton bound, we can obtain a EA-quantum MDS code with parameters $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;1]]_{q}$, where $2\le d \le 8m+1$ be even; and $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+7,d;5]]_{q}$ codes, where $8m+4\le d \le 12m+4$ be even.

$\square$

Example 6

Let $m=1$, $q=13$, applying Theory 3.12 (1) produces 4-ebits EA-quantum MDS codes with parameters $[[34,26,7;4]]_{13}$ for d odd; and $[[34,20,10;4]]_{13}$, $[[34,16,12;4]]_{13}$, $[[34,12,14;4]]_{13}$ for d even.

Example 7

Let $m=2$, $q=23$, applying Theory 3.12 (2) produces 1-ebits EA-quantum MDS codes with parameters $[[106,105,2;1]]_{23}$, $[[106,101,4;1]]_{23}$, $[[106,97,6;1]]_{23}$, $[[106,93,8;1]]_{23}$, $[[106,89,10;1]]_{23}$, $[[106,85,12;1]]_{23}$, $[[106,81,14;1]]_{23}$, $[[106,77,16;1]]_{23}$ for d even; 4-ebits EA-quantum MDS codes with parameters $[[106,90,11;4]]_{23}$, $[[106,86,13;4]]_{23}$ for d odd; and 5-ebits EA-quantum MDS codes with parameters $[[106,73,20;5]]_{23}$, $[[106,69,22;5]]_{23}$, $[[106,65,24;5]]_{23}$, $[[106,61,26;5]]_{23}$, $[[106,57,28;5]]_{23}$ for d even.

Lemma 3.13

Let $q=10m+7$, $n=\frac{q^{2}+1}{5}$, $s=\frac{n}{2}$.

• (1) If $1\le x\le 14m+9$, then $(C_{8m+5},C_{6m+5}), (C_{14m+11},C_{12m+7})$ form skew-asymmetric pairs, respectively.

  $$\begin{aligned} |T_{ss}(\delta )| = \left\{ \begin{array}{lll} 0, &{}\quad \hbox {if }3 \le \delta \le 8m+5,\quad \hbox { for }\delta \hbox { is odd};\\ 4, &{}\quad \hbox {if }8m+7 \le \delta \le 14m+11,\quad \hbox { for }\delta \hbox { is odd}.\\ \end{array} \right. \end{aligned}$$

• (2) If $m=2t+1$ is odd, and $s\le x\le s+10m+8$, then $(C_{s+6m+4},C_{s+2m+2})$, $(C_{s+10m+8},C_{s+10m+6})$ form skew-asymmetric pairs;

  $$\begin{aligned} |T_{ss}(\delta )| = \left\{ \begin{array}{lll} 0, &{}\quad \hbox {if }2 \le \delta \le 6m+4, \quad \hbox { for } \delta \hbox { is even};\\ 4, &{}\quad \hbox {if }6m+6 \le \delta \le 10m+8, \quad \hbox { for } \delta \hbox { is even}.\\ \end{array} \right. \end{aligned}$$

  If $m=2t$ is even,and $s\le x\le s+12m+8$, then $C_{s}$ is skew-symmetric, and $(C_{s+8m+6},C_{s+4m+2})$, $(C_{s+12m+8},C_{s+4m+4})$ form skew-asymmetric pairs;

  $$\begin{aligned} |T_{ss}(\delta )| = \left\{ \begin{array}{lll} 1, &{}\hbox {if }2 \le \delta \le 8m+6, \quad \quad \quad \quad \hbox { for } \delta \hbox { is even};\\ 5, &{}\hbox {if }8m+8 \le \delta \le 12m+8,\quad \hbox { for } \delta \hbox { is even}.\\ \end{array} \right. \end{aligned}$$

Proof

Let $q=10m+7$. Since $2n=40m^2+56m+20$.

• (1) Let $1\le x\le 14m+10$. Since $(8m+5)q=40m^2+56m+20-(6m+5)$, $(C_{8m+5},C_{6m+5})$ form skew-asymmetric pairs.

  Since $(8m+5)q=40m^2+56m+20-(6m+5)$, $(C_{8m+5},C_{6m+5})$ form skew-asymmetric pairs.

  • (a) If the defining set $T=\bigcup _{j=1}^{l} C_{1+2j}$, where $1\le l \le 4m+1$, we testify that $T_{ss}(\delta )=\emptyset$ for $3 \le \delta \le 8m+5$. According to Lemmas 2.4 and 2.1 in [12], one needs to testify that for $x,y \in [1,8m+3]$, there $x+yq \not \equiv 0$ mod 2n holds.

    We divided $[1,8m+3]$ into $[1,4m+1]\cup [4m+3,8m+3]$. If $x,y\in [1,4m+1]$, $1<y(q+1)<(4m+1)(q+1)=40m^2+42m+8<2n$; if $x,y\in [4m+3,8m+3]$, $2n<4m^2+62m+24=(4m+1)(q+1)<y(q+1)<(8m+3)(q+1) =80m^2+94m+24<4n$. Hence, if the defining set $T=\bigcup _{j=1}^{l}C_{1+2j}$, where $1\le l\le 4m+1$, the $T_{ss}(\delta )=\emptyset$ for $3 \le \delta \le 8m+5$.

  • (b) Let $I_{0}=[1,8m+3]$, $I_{1}=[8m+7,14m+9]$.

    According to Lemmas 2.4 and 2.1 in [12], one needs to testify that for $x,y \in I_{0}\cup I_{1}$, there $x+yq \not \equiv 0$ mod 2n holds.

    Let $T_{0}=\bigcup _{j=1}^{4m+1}C_{1+2j}$, and the defining set $T=\bigcup _{j=4m+3}^{l}C_{1+2j} \bigcup T_{0}$, where $4m+3\le l\le 7m+4$.

    For $8m+7 \le \delta \le 14m+11$, suppose there exist integers $y\in I_{1}, x\in I_{0}\cup I_{1}$, such that $x\equiv -qy$ mod 2n. We find a contradiction in the following.

    We divided $I_{1}$ into four parts such as $[8m+7,12m+7]\cup [12m+9,14m+9]$.

    If $x,y\in [8m+7,12m+7]$, $2\cdot 2n<80m^2+134m+56=(8m+7)(10m+8)\le y(q+1)\le (12m+7)(10m+8)=120m^2+166m+56<3\cdot 2n$; if $x,y\in [12m+9,14m+9]$, $3\cdot 2n<120m^2+186m+72=(12m+9)(q+1)\le y(q+1)\le (14m+9)(q+1) =140m^2+202m+72<4\cdot 2n$. Hence, if the defining set $T=\bigcup _{j=1}^{l}C_{1+2j}$, where $1\le l\le 7m+4$, besides $(C_{s+8m+6},C_{s+4m+2})$, there is no skew-symmetric cyclotomic coset and any two cyclotomic cosets are not skew-asymmetric cosets in T.

• (2) For $m=2t$ is even, since $2n-sq=2n-(5mn+3n+\frac{n}{2})\equiv s$ mod 2n, $C_{s}$ is skew-symmetric. $2n-(s+8m+6)q\equiv s+4m+2$ mod 2n and $2n-(s+12m+8)q\equiv s+4m+4$ mod 2n.

  • (a) Let $T_{0}=\bigcup _{j=s}^{4m+3}C_{1+2j}$, and the defining set $T=\bigcup _{j=4m+5}^{l}C_{1+2j} \bigcup T_{0}$, where $4m+3\le l\le 7m+4$.

    If the defining set $T=T_{0}$, we testify that $T_{ss}(\delta )=\{C_{s}\}$ for $2 \le \delta \le 8m+6$ and $\delta$ is even. Since $C_{s}$ is skew-symmetric cyclotomic coset, and according to Lemma 2.4 and Lemma 2.1 in [12], one needs to testify that for $x,y \in [s+2,s+8m+4]$, there $x+yq \not \equiv 0$ mod 2n holds.

    We divided $[s+2,s+8m+4]$ into three parts such as $[s+2,s+3m+2]\cup [s+3m+4,s+7m+4]\cup [s+7m+6,s+8m+4]$.

    If $x,y\in [s+2,s+3m+2]$, $(5t+2)\cdot 2n<(5m+4)n+20m+16=(s+2)(10m+8)\le y(q+1)\le (s+3m+2)(10m+8)<(5t+3)\cdot 2n$; if $x,y\in [s+3m+4,s+7m+4]$, $(5t+3)\cdot 2n<(s+3m+4)(q+1)\le y(q+1)\le (s+7m+4)(q+1) <(5t+4)\cdot 2n$. If $x,y\in [s+7m+6,s+8m+4]$, $(5t+5)\cdot 2n<(s+7m+6)(q+1)\le y(q+1)\le (s+8m+4)(q+1) <(5t+5)\cdot 2n$. Hence, if the defining set $T=\bigcup _{j=s}^{l}C_{s+2j}$, where $1\le l\le 4m+2$, besides $C_{s}$, there is no skew-symmetric cyclotomic coset and any two cyclotomic cosets are not skew-asymmetric cosets in the defining set T.

  • (b) We divided $[s+8m+8,s+12m+6]$ into two parts such as $[s+8m+8,s+11m+6]\cup [s+11m+8,s+12m+6]$.

    If $x,y\in [s+8m+8,s+11m+6]$, $(5t+4)\cdot 2n<(5m+8)n+32m+24=(s+8m+8)(10m+8)\le y(q+1)\le (s+11m+6)(10m+8)=(5m+8)n+30m^2+36m+8<(5t+8)\cdot 2n+2n$; if $x,y\in [s+11m+8,s+12m+6]$, $(5t+9)\cdot 2n<(s+11m+8)(q+1)\le y(q+1)\le (s+12m+6)(q+1) <(5t+9)\cdot 2n+2n$.

  Hence, using the same method, one can obtain that if the defining set $T=\bigcup _{j=s}^{l}C_{s+2j}$, where $1\le l\le 6m+3$, besides $C_{s}$ and $(C_{s+8m+6},C_{s+4m+2})$, there is no skew-symmetric cyclotomic coset and any two cyclotomic cosets are not skew-asymmetric cosets in the defining set T.

$\square$

Theory 3.14

Let q be an odd prime power of the form $q=10m+7$.

If $m=2t+1$ is odd, then there exists a q-$[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]$- EA-quantum MDS codes, where $8m+7\le d \le 14m+11$ be odd; and $6m+6\le d\le 10m+8$ be even.

If $m=2t$ is even, then there exists a q-$[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+3,d;1]]$- EA-quantum MDS codes, where $2\le d \le 8m+6$ be even; there exists a q-$[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]$- EA-quantum MDS codes, where $8m+7\le d \le 14m+11$ be odd; and there exists a q-$[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+7,d;5]]$- EA-quantum MDS codes, where $8m+8\le d \le 12m+8$ be even.

Proof

• (a) Consider the negacyclic codes over $F_{q^{2}}$ of length $n=\frac{q^{2}+1}{5}$ with defining set $T=\bigcup _{i=0}^{s}C_{1+2i}$, where $0\le s \le 7m+4$ for q be an odd prime power of the form $q=10m+7$. By Lemma 3.13 (i), there is $c=|T_{ss}(\delta )|=4$ if $4m+2\le s \le 7m+4$. Since every $q^{2}$-cyclotomic coset $C_{x}=\{x,n-x\}$ and x must be odd number, we can obtain that T consists of $2s+1$ integers $\{n-(1+2s),\ldots ,n-1,1,3,\ldots ,1+2s\}$. It implies that $\mathcal {C}$ has minimum distance at least $2s+2$. Hence, $\mathcal {C}$ is a $q^{2}$-ary negacyclic code with parameters $[n,n-2(s+1)+1,\ge 2s+2]$. Combining Lemma 2.8 with EA-quantum singleton bound, we can obtain a EA-quantum MDS code with parameters $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]_{q}$, $8m+7\le d \le 14m+11$ be odd.

• (b) If m is odd, consider the negacyclic codes over $F_{q^{2}}$ of length $n=\frac{q^{2}+1}{5}$ with defining set $T=\bigcup _{i=0}^{k}C_{s+2i}$, where $0\le k \le 5m+4$, $s=\frac{n}{2}$ for q be an odd prime power of the form $q=10m+7$. By Lemma 3.11 (ii), there is $c=|T_{ss}(\delta )|=4$ if $3m+2\le s \le 5m+4$. Since every $q^{2}$-cyclotomic coset $C_{x}=\{x,n-x\}$ and x must be odd number, we can obtain that T consists of $2k+1$ integers $\{n-(s+2k),\ldots ,n-s,s,s+2,\ldots ,s+2k\}$. It implies that $\mathcal {C}$ has minimum distance at least $2k+2$. Hence, $\mathcal {C}$ is a $q^{2}$-ary negacyclic code with parameters $[n,n-2(s+1)+1,\ge 2s+2]$. Combining Lemma 2.8 with EA-quantum singleton bound, we can obtain a EA-quantum MDS code with parameters $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;4]]_{q}$, $6m+6\le d\le 10m+8$ be even.

If m is even, consider the negacyclic codes over $F_{q^{2}}$ of length $n=\frac{q^{2}+1}{5}$ with defining set $T=\bigcup _{i=0}^{k}C_{s+2i}$, where $0\le k \le 6m+3$, $s=\frac{n}{2}$ for q be an odd prime power of the form $q=10m+7$. By Lemma 3.11 (ii), there is $c=|T_{ss}(\delta )|=1$ if $1\le k \le 4m+2$ and $c=|T_{ss}(\delta )|=5$ if $4m+3\le k \le 6m+3$. Since every $q^{2}$-cyclotomic coset $C_{x}=\{x,n-x\}$ and x must be odd number, we can obtain that T consists of $2k+1$ integers $\{n-(s+2k),\ldots ,n-s,s,s+2,\ldots ,s+2k\}$. It implies that $\mathcal {C}$ has minimum distance at least $2k+2$. Hence, $\mathcal {C}$ is a $q^{2}$-ary negacyclic code with parameters $[n,n-2(s+1)+1,\ge 2s+2]$. Combining Lemma 2.8 with EA-quantum singleton bound, we can obtain a EA-quantum MDS code with parameters $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+6,d;1]]_{q}$, $2\le d \le 8m+6$ be even; and $[[\frac{q^{2}+1}{5},\frac{q^{2}+1}{5}-2d+7,d;5]]_{q}$ codes, where $8m+8\le d \le 12m+8$ be even. $\square$

Example 8

Let $m=1$, $q=17$, applying Theory 3.14 (1) produces 4-ebits EA-quantum MDS codes with parameters $[[58,34,15;4]]_{17}$, $[[58,30,17;4]]_{17}$, $[[58,26,19;4]]_{17}$, $[[58,22,21;4]]_{17}$, $[[58,18,23;4]]_{17}$, $[[58,14,25;4]]_{17}$ for d odd; and $[[58,40,12;4]]_{17}$, $[[58,36,14;4]]_{17}$, $[[58,32,16;4]]_{17}$, $[[58,28,18;4]]_{17}$ for d even.

Example 9

Let $m=2$, $q=27$, applying Theory 3.14 (2) produces 1-ebits EA-quantum MDS codes with parameters $[[146,145,2;1]]_{27}$, $[[146,141,4;1]]_{27}$, $[[146,137,6;1]]_{27}$, $[[146,133,8;1]]_{27}$, $[[146,129,10;1]]_{27}$, $[[146,125,12;1]]_{27}$, $[[146,121,14;1]]_{27}$, $[[146,117,16;1]]_{27}$, $[[146,113,18;1]]_{27}$, $[[146,109,20;1]]_{27}$, $[[146,105,22;1]]_{27}$ for d even; 4-ebits EA-quantum MDS codes with parameters $[[146,106,23;4]]_{27}$, $[[146,102,25;4]]_{27}$, $[[146,98,27;4]]_{27}$, $[[146,94,29;4]]_{27}$, $[[146,90,31;4]]_{27}$, $[[146,86,33;4]]_{27}$, $[[146,82,35;4]]_{27}$, $[[146,78,37;4]]_{27}$, $[[146,74,39;4]]_{27}$ for d odd; and 5-ebits EA-quantum MDS codes with parameters $[[146,105,24;5]]_{27}$, $[[146,101,26;5]]_{27}$, $[[146,97,28;5]]_{27}$, $[[146,93,30;5]]_{27}$, $[[146,89,32;5]]_{27}$ for d even.

Table 1 New EA-Quantum MDS codes

4 Discussion and Conclusion

In this paper, we have constructed six families of entanglement-assisted quantum MDS (EAQMDS) codes based on classical negacyclic MDS codes. Two of these six classes q-ary EAQMDS have minimum distance more larger than $q+1$. Most of these q-ary EAQMDS codes are new in the sense that their parameters are not covered by the codes available in the literature. In Table 1, we list the EA-quantum MDS codes constructed in this paper. The results show that using entanglement, EAQMDS codes have a larger minimum distance than QMDS codes. We look forward to seeing that some special types of [[n, k, d; c]] EAQMDS codes that better perform than $[[n+c,k,d]]$ QMDS codes even if these [[n, k, d; c]] EAQMDS codes are equivalent to those $[[n+c,k,d]]$ QMDS codes.