Entanglement-assisted quantum MDS codes from cyclic codes

Abstract

Entanglement-assisted quantum error-correcting codes are a generalization of standard stabilizer quantum error-correcting codes, which can be possibly constructed from any classical codes by relaxing the duality condition and utilizing pre-shared entanglement between the sender and the receiver. In this paper, we construct seven new families of entanglement-assisted quantum maximum-distance-separable codes from cyclic codes by exploiting less pre-shared entangled states. Most of these codes are new in the sense that their parameters are not covered by the codes available in the literature.

1 Introduction

Quantum error-correcting (QEC) codes were introduced to reduce decoherence during quantum communications and quantum computations. The stabilizer formalism makes QEC codes that can be constructed from classical codes with certain self-orthogonality (dual-containing) properties. However, the need for such dual-containing forms an obstacle in the development of quantum coding theory. In Brun et al. [1, 12], a more general framework named entanglement-assisted stabilizer formalism was introduced and it increases the communication capacity. The related codes are called entanglement-assisted quantum error-correcting (EAQEC) codes which can be possibly constructed from any classical codes by relaxing the duality condition and utilizing pre-shared entanglement between the sender and the receiver. After that, many scholars have constructed lots of EAQEC codes with good parameters. (see, for example, [6, 10, 11, 16, 17, 35, 36] and the relevant references therein).

Let q be a prime power. A q-ary EAQEC code, denoted by $[[n,k,d;c]]_q$, encodes k information qudits into n channel qudits with the help of c pairs of maximally entangled states and can correct up to $\lfloor \frac{d-1}{2}\rfloor$ errors, where d is the minimum distance of the code. Actually, if $c=0$, it is the standard $[[n,k,d]]_q$ quantum codes. Moreover, if $c=n-k$, it is called a maximal-entanglement EAQEC code.

Similar to quantum Singleton bound, there is also a so-called entanglement-assisted (EA) quantum Singleton bound for EAQEC codes.

Theorem 1

(EA quantum Singleton bound) [1, 7, 15] For any $[[n,k,d;c]]_q$ EAQEC code with $d\leqslant \frac{n+2}{2}$, its parameters satisfy

$$n+c-k\geqslant 2(d-1),$$

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

An EAQEC code achieving this bound is called an EAQMDS code. If $c=0$, it is the quantum Singleton bound. As we said before, EAQEC codes can be constructed from classical codes without dual-containing condition, but it is still hard to do so, since it is not an easy task to determine the number of shared pairs in the construction of EAQEC codes. Scholars have proposed several methods to solve this problem, and many EAQEC codes with good parameters have been constructed.

Maximal-entanglement EAQEC codes with small lengths constructed from quaternary zero radical codes were presented in [25]. Qian and Zhang [31] constructed maximal-entanglement EAQEC codes from arbitrary binary linear codes and proved that asymptotically good EAQEC codes exist in binary case. Very recently, Liu et al. [22] generalized [31] to linear codes with k-Galois product and they also constructed some EAQEC codes from matrix-product codes in [21]. Recently, a relationship between the number of maximally shared qudits required to construct an EAQEC code from a classical code and hull of the classical code was obtained in [8], in which EAQEC codes with flexible parameters were also constructed. Meanwhile, codes based on linear codes with complementary duals were also used to construct EAQEC codes in [9] and [32], respectively. In addition, EAQMDS codes were constructed from Reed–Solomon and generalized Reed–Solomon codes in [18, 28, 29].

Lu et al. [24] utilized the decomposition of the defining set of codes which was introduced in [20], to construct EAQEC codes from BCH codes. Recently, Lu et al. [26] and Chen et al. [2] proposed the concept of decomposition of the defining set of constacyclic codes, which makes the shared qudits c that can be easily determined, and they also constructed some new EAQMDS codes. Since then, many EAQMDS codes have been constructed from constacyclic codes (including cyclic codes and negacyclic codes). Among the obtained results, the lengths of these EAQMDS codes divide $q^2+1$ (see, for example, $q^2+1$ in [2, 27, 33, 34]; $\frac{q^2+1}{2}$ in [2]; $\frac{q^2+1}{5}$ in [3, 13, 26]; $\frac{q^2+1}{10}$ in [13, 27], etc.), and $q^2-1$ (see, for example, [2, 19, 23, 26, 27, 34]). Very recently, Chen et al. [4] also used negacyclic BCH codes to construct EAQEC codes.

In this paper, through the analysis of the intersection of the defining set D of cyclic codes and $-qD$, we obtain several new families of EAQMDS codes of lengths that divide $q^2+1$ with q being an odd prime power as follows:

• (1) $[[\frac{q^2+1}{2},\frac{q^2+1}{2}-2d+6,d;4]]_q$, where $q+2\le d\le 2q-1$ is odd.

• (2) $[[\frac{q^2+1}{5},\frac{q^2+1}{5}-2d+3,d;1]]_q$, where $q\equiv 3~({{\,\mathrm{mod~}\,}}10)$, $q>3$, $2\le d\le \frac{4q-2}{5}$ is even.

• (3) $[[\frac{q^2+1}{5},\frac{q^2+1}{5}-2d+7,d;5]]_q$, where $q\equiv 3~({{\,\mathrm{mod~}\,}}10)$, $q>3$, $\frac{4q+8}{5}\le d\le \frac{6q+2}{5}$ is even.

• (4) $[[\frac{q^2+1}{5},\frac{q^2+1}{5}-2d+3,d;1]]_q$, where $q\equiv 7~({{\,\mathrm{mod~}\,}}10)$, $2\le d\le \frac{4q+2}{5}$ is even.

• (5) $[[\frac{q^2+1}{5},\frac{q^2+1}{5}-2d+7,d;5]]_q$, where $q\equiv 7~({{\,\mathrm{mod~}\,}}10)$, $\frac{4q+12}{5}\le d\le \frac{6q-2}{5}$ is even.

• (6) $[[\frac{q^2+1}{10},\frac{q^2+1}{10}-2d+6,d;4]]_q$, where $q\equiv 3~({{\,\mathrm{mod~}\,}}10),~q>3$, $\frac{2q+9}{5}\le d\le \frac{4q+3}{5}$ is odd.

• (7) $[[\frac{q^2+1}{10},\frac{q^2+1}{10}-2d+6,d;4]]_q$, where $q\equiv 7~({{\,\mathrm{mod~}\,}}10)$, $\frac{2q+11}{5}\le d\le \frac{4q-3}{5}$ is odd.

The paper is organized as follows. In Sect. 2, some notations and basic results of cyclic codes and EAQEC codes are presented. In Sect. 3, some new families of EAQMDS codes with small pre-shared entangled states are constructed from cyclic codes. The conclusion is given in Sect. 4.

2 Preliminaries

Let q be a prime power and $\mathbb {F}_{q^2}$ be the Galois field with $q^2$ elements. A $q^2$-ary linear code $\mathscr {C}$ of length n with dimension k, denoted by $[n,k]_{q^2}$, is a linear subspace of $\mathbb {F}_{q^2}^n$ with dimension k. The number of nonzero components of $\mathbf {c}\in \mathscr {C}$ is said to be the weight $\mathrm{wt}(\mathbf {c})$ of the codeword $\mathbf {c}$. The minimum nonzero weight of all codewords in $\mathscr {C}$ is said to be the minimum distance of $\mathscr {C}$, denoted by $d(\mathscr {C})$. An $[n,k]_{q^2}$ linear code with minimum distance d is denoted by $[n,k,d]_{q^2}$.

Given two vectors $\mathbf {x}=(x_0, x_1,\ldots ,x_{n-1})$, and $\mathbf {y}=(y_0, y_1,\ldots ,y_{n-1})\in \mathbb {F}_{q^2}^n$, their Hermitian inner product is defined as

$$\begin{aligned} \langle \mathbf {x}, \mathbf {y}\rangle :=x_0y_0^q+x_1y_1^q+\cdots +x_{n-1}y_{n-1}^q. \end{aligned}$$

The vectors $\mathbf {x}$ and $\mathbf {y}$ are called orthogonal with respect to the Hermitian inner product if $\langle \mathbf x , \mathbf y \rangle =0$. For a $q^2$-ary linear code $\mathscr {C}$ of length n, the Hermitian dual code of $\mathscr {C}$ is defined as

$$\begin{aligned} \mathscr {C}^{\bot _H}:=\{\mathbf {x}\in \mathbb {F}_{q^2}^n:\langle \mathbf x , \mathbf {y}\rangle =0 ~\text {for all} ~ \mathbf {y}\in \mathscr {C}\}. \end{aligned}$$

It is clear that $\mathscr {C}^{\bot _H}$ is a $q^2$-ary linear code with dimension $n-\dim (\mathscr {C})$.

A $q^2$-ary linear code $\mathscr {C}$ of length n is called cyclic if it is invariant under the cyclic shift of $\mathbb {F}_{q^2}^n$: $(c_0,c_1,\ldots ,c_{n-1})\rightarrow ( c_{n-1},c_0,\ldots ,c_{n-2})$. Each codeword $\mathbf c =(c_0,c_1,\ldots ,c_{n-1})$ is customarily identified with its polynomial representation $c(x):=c_0+c_1x+\cdots +c_{n-1}x^{n-1}$, and the code $\mathscr {C}$ is in turn identified with the set of all polynomial representations of its codewords. Then in the ring $\mathscr {R}:=\mathbb {F}_{q^2}[x]/(x^n-1)$, xc(x) corresponds to a cyclic shift of c(x). It is well known that a $q^2$-ary linear code $\mathscr {C}$ of length n is cyclic if and only if $\mathscr {C}$ is an ideal of the quotient ring $\mathscr {R}$. Moreover, $\mathscr {R}$ is a principal ideal ring, whose ideals are generated by monic factors of $x^n-1$, i.e., $\mathscr {C}=(f(x))$ and $f(x)|(x^n-1)$.

Suppose that $\gcd (n,q)=1$. Let i be an integer with $0\le i\le n-1$. The $q^2$-cyclotomic coset of i modulo n is defined by $C_i:=\{ i q^{2l} ~({{\,\mathrm{mod~}\,}}n): 0\le l\le l_i-1 \}$, where $l_i$ is the smallest positive integer such that $iq^{2l_i}\equiv i~({{\,\mathrm{mod~}\,}}n)$. The smallest number in $C_i$ is called the coset leader of $C_i$.

Let $\mathscr {C}$ be a $q^2$-ary cyclic code of length n with generator polynomial g(x), then the set $D=\{0\leqslant i\leqslant n-1: g(\alpha ^i)=0\}$ is called the defining set of $\mathscr {C}$, where $\alpha$ is a primitive n-th root of unity in some extension field of $\mathbb {F}_{q^2}$. Obviously, the defining set D is a union of some $q^2$-cyclotomic cosets and $\text {dim}(\mathscr {C})=n-|D|$, where |D| denotes the cardinality of the set D. The minimum distance of $\mathscr {C}$ has the following well-known bound.

Theorem 2

(BCH bound) [30] Let $\delta$ be an integer in the range $2\leqslant \delta \leqslant n$. Assume that C is a cyclic code of length n with defining set D. If D consists of $\delta -1$ consecutive elements, then $d(\mathscr {C})\ge \delta$.

As we said before, scholars had proposed several methods to construct EAQMDS codes. Among these methods, the most frequently used one is to decompose the defining set of the codes based on [2, 26] et al. Similar to this method, we have the following result.

Theorem 3

Let $\mathscr {C}$ be a $q^2$-ary cyclic code of length n with defining set D. Suppose $\mathscr {D}=D\cap (-qD)$, where $-qD=\{-qz~({{\,\mathrm{mod~}\,}}n):z\in D\}$. If $\mathscr {C}$ has parameters $[n,n-|D|,d]_{q^2}$, then there exists an EAQEC code with parameters $[[n,n-2|D|+|\mathscr {D}|,d;|\mathscr {D}|]]_q$.

3 Constructions of entanglement-assisted quantum MDS codes

In this section, we will construct some new EAQMDS codes with lengths that divide $q^2+1$. We first give a useful lemma in the following which will play an important role in our construction.

Lemma 1

[14] Let $n\mid (q^2+1)$ and $s=\lfloor \frac{n}{2}\rfloor$. If n is odd, then the $q^2$-cyclotomic cosets modulo n containing integers from 0 to n are: $C_0=\{0\}, C_i=\{i, -i\}=\{i, n-i\}$, where $1\leqslant i\leqslant s$. If n is even, then the $q^2$-cyclotomic cosets modulo n containing integers from 0 to n are: $C_0=\{0\}, C_s=\{s\}$ and $C_i=\{i, -i\}=\{i, n-i\}$, where $1\leqslant i\leqslant s-1$.

3.1 Entanglement-assisted quantum MDS codes of length $\frac{q^2+1}{2}$

Throughout this subsection, let q be an odd prime power, $n=\frac{q^2+1}{2}$ and $s=\frac{n-1}{2}$. From Lemma 1, the $q^2$-cyclotomic cosets modulo n are: $C_0=\{0 \}$ and for every i with $0\le i \le s-1$,

$$\begin{aligned} C_{s-i}=\{s-i, s+1+i \}. \end{aligned}$$

(1)

For every t with $0\le t\le s-1$, let $\mathscr {C}_{I, t}$ be the $q^2$-ary cyclic code of length n with defining set

$$\begin{aligned} D_{I, t}=\bigcup _{i=0}^{t}C_{s-i}. \end{aligned}$$

(2)

We have the following basic property for the defining set $D_{I, t}$.

Lemma 2

Let $D_{I, t}$ be defined as above. If $\frac{q-1}{2}\le t\le q-2$, then $|D_{I, t}\cap (-qD_{I, t})|=4$.

Proof

Clearly, for every $q^2$-cyclotomic coset C, $-qC$ is also a $q^2$-cyclotomic coset modulo n. Consequently, $D_{I, t}\cap (-qD_{I, t})$ is either an empty set or a union of some $q^2$-cyclotomic cosets. Next, by analyzing the $q^2$-cyclotomic coset modulo n represented by $-qC_{s-i}$, for $0 \le i \le t$, we will determine $D_{I, t}\cap (-q D_{I, t})$. Thereby, the desired result follows.

For every i with $0\le i\le s-1$, from (1), there is a unique $a_i$ with $0\le a_i\le s-1$ such that $-qC_{s-i}=C_{s-a_i}$. Now, we determine $a_i$ for $0\le i\le t$. Since $s=\frac{n-1}{2}$ and q is odd,

$$\begin{aligned} -q (s-i)\equiv s+\frac{q+1}{2}+qi~({{\,\mathrm{mod~}\,}}n), \end{aligned}$$

(3)

for each i in the range $0\le i\le s-1$. It follows from (1) and (3) that

$$\begin{aligned} a_i\equiv \frac{q-1}{2}+q i~({{\,\mathrm{mod~}\,}}n), \end{aligned}$$

(4)

or

$$\begin{aligned} a_i\equiv -\left( \frac{q+1}{2}+q i\right) ~({{\,\mathrm{mod~}\,}}n). \end{aligned}$$

(5)

Notice that $\frac{q+1}{2}\le \frac{q+1}{2}+q i\le 2n-\frac{3q+1}{2}$ for $0\le i\le q-2$, so we have the following four cases.

• Case 1: $i\in \Gamma _1:=\{i: \frac{q+1}{2}\le \frac{q+1}{2}+qi\le s\}$. From (4), $a_i=\frac{q-1}{2}+qi$. Hence,

  $$\begin{aligned} -qC_{s-i}=C_{s-\frac{q-1}{2}-qi}. \end{aligned}$$

  From (2), $(-q C_{s-i})\cap D_{I, t}\ne \emptyset$ if and only if $0\le \frac{q-1}{2}+q i\le t\le q-2$. Since i is an integer, there is only an $i=0$ such that $(-q C_{s-i})\cap D_{I, t}\ne \emptyset$ for $i\in \Gamma _1$. Hence,

  $$\begin{aligned} (\cup _{i\in \Gamma _1} (-qC_{s-i}))\cap D_{I, t}=C_{s-\frac{q-1}{2}}. \end{aligned}$$

• Case 2: $i\in \Gamma _2:=\{i: s+1\le \frac{q+1}{2}+q i\le n \}$. We claim $\frac{q+1}{2}+q i\ne s+1$. Otherwise,

  $$\begin{aligned} \frac{n-q}{2}=s+1-\frac{q+1}{2}\equiv 0~({{\,\mathrm{mod~}\,}}q). \end{aligned}$$

  It implies that $n \equiv 0~({{\,\mathrm{mod~}\,}}q)$ since q is odd, which contradicts to the fact that $\gcd (n, q)=1$. It follows from (5) that

  $$\begin{aligned} a_i =n-\frac{q+1}{2}-qi=q\left( \frac{q-1}{2}-i\right) . \end{aligned}$$

  From (2), $(-q C_{s-i})\cap D_{I, t}\ne \emptyset$ if and only if $0\le a_i \le t\le q-2$. Since i is an integer, there is only an $i=\frac{q-1}{2}$ such that $(-q C_{s-i})\cap D_{I, t}\ne \emptyset$ for $i\in \Gamma _2$. In this case, $a_i=0$. Therefore,

  $$\begin{aligned} (\cup _{i\in \Gamma _2} (-qC_{s-i}))\cap D_{I, t}=C_{s}. \end{aligned}$$

• Case 3: $i\in \Gamma _3:=\{i: n+1\le \frac{q+1}{2}+qi\le n+s\}$. It follows from (4) that

  $$\begin{aligned} a_i =\frac{q+1}{2}+qi-n-1=q\left( i-\frac{q-1}{2}\right) -1. \end{aligned}$$

  From (2), $(-q C_{s-i})\cap D_{I, t}\ne \emptyset$ if and only if $0\le a_i\le t\le q-2$. Since i is an integer, there is no $i\in \Gamma _3$ such that $0\le a_i\le q-2$. That is to say, $\cup _{i\in \Gamma _3} (-qC_{s-i})\cap D_{I, t}=\emptyset$.

• Case 4: $i\in \Gamma _4:=\{i: n+s+1\le \frac{q+1}{2}+q i\le 2n-\frac{3q+1}{2} \}$. Clearly, if $\Gamma _4\ne \emptyset$, we have $q>3$. Now, assume that $q>3$. We claim $\frac{q+1}{2}+q i\ne n+s+1$ for $i\in \Gamma _4$. Otherwise,

  $$\begin{aligned} \frac{q^2+3}{4}=\frac{n+1}{2}=s+1=qi+\frac{q+1}{2}-n =q(i-\frac{q-1}{2}), \end{aligned}$$

  which implies that $3\equiv 0~({{\,\mathrm{mod~}\,}}q)$ since q is odd. This contradicts to the fact that $q>3$. It follows from (5) that

  $$\begin{aligned} \frac{3q+1}{2}\le a_i=2n-\frac{q+1}{2}-qi=q(q-i)-\frac{q-1}{2}. \end{aligned}$$

  From (2), $(-q C_{s-i})\cap D_{I, t}\ne \emptyset$ if and only if $\frac{3q+1}{2}\le a_i \le t\le q-2$. Since i is an integer, there is no $i\in \Gamma _4$ such that $\frac{3q+1}{2}\le a_i\le q-2$. Then,

  $$\begin{aligned} (\cup _{i\in \Gamma _4} (-qC_{s-i}))\cap D_{I, t}=\emptyset . \end{aligned}$$

In conclusion, we have $D_{I, t}\cap (-qD_{I, t})=C_{s}\cup C_{s-\frac{q-1}{2}}$. The result follows $\square$

Theorem 4

Let q be an odd prime power. For each odd integer d with $q+2\le d\le 2q-1$, there exists an EAQMDS code with parameters $[[\frac{q^2+1}{2},\frac{q^2+1}{2}-2d+6,d;4]] _q$.

Proof

For each odd integer d with $q+2\le d\le 2q-1$, let $t=\frac{d-3}{2}$, then $\frac{q-1}{2}\le t\le q-2$. Consider the $q^2$-ary cyclic code $\mathscr {C}_{I, t}$ of length $n=\frac{q^2+1}{2}$ with defining set $D_{I, t}$. It follows from (1) and (2) that $|D_{I, t}|=2t+2=d-1$. Hence, $\dim (\mathscr {C}_{I, t})=n-|D_{I, t}|=n-d+1$. By the definition of $D_{I, t}$ (see (2)), the defining set $D_{I, t}$ consists of $d-1$ consecutive integers

$$\begin{aligned} \left\{ s-\frac{d-3}{2}, s-\frac{d-5}{2},\cdots , s-1, s, s+1,\cdots , s+\frac{d-3}{2}, s+\frac{d-1}{2} \right\} . \end{aligned}$$

Then by Theorem 2, the minimum distance of $\mathscr {C}_{I, t}$ is at least d. Thus, $\mathscr {C}_{I, t}$ is a cyclic code with parameters $[n,n-d+1,\geqslant d]_{q^2}.$ From Lemma 2, $|D_{I, t}\cap (-qD_{I, t})|=4$. Combining Theorem 3 with EA quantum Singleton bound, there is an EAQMDS code with parameters

$$\begin{aligned}{}[[n,n-2d+6,d;4]]_q. \end{aligned}$$

The result follows. $\square$

Table 1 New entanglement-assisted quantum MDS codes of length $\frac{q^2+1}{2}$

Example 1

We list some new EAQMDS codes of length $\frac{q^2+1}{2}$ obtained from Theorem 4 in Table 1.

Remark 1

EAQMDS codes of length $\frac{q^2+1}{2}$ with $c=1$ and $c=5$ had been constructed in [2] and [5] using cyclic codes and negacyclic codes, respectively. We list them in Table 2.

Table 2 Entanglement-assisted quantum MDS codes of length $\frac{q^2+1}{2}$

3.2 Entanglement-assisted quantum MDS codes of length $\frac{q^2+1}{5}$

Throughout this subsection, let q be an odd prime power with $q\equiv \pm 3 ~({{\,\mathrm{mod~}\,}}10)$ and $q>3$. Let $n=\frac{q^2+1}{5}$, then n is even. From Lemma 1, the $q^2$-cyclotomic cosets modulo n are: $C_0=\{0 \}$, $C_{\frac{n}{2}}=\{\frac{n}{2} \}$, and for $1\le i\le \frac{n}{2}-1$,

$$\begin{aligned} C_{i}=\{ i, n-i \}. \end{aligned}$$

(6)

For each t with $0\le t \le \frac{n}{2}-1$, let $\mathscr {C}_{II,t}$ be the $q^2$-ary cyclic code of length n with defining set

$$\begin{aligned} D_{II,t}=\bigcup _{i=0}^{t}C_{i}. \end{aligned}$$

(7)

We have the following basic property for the defining set $D_{II,t}$.

Lemma 3

Let $D_{II,t}$ be defined as above. If $0\le t\le \lfloor \frac{3q-4}{5} \rfloor$, then

$$\begin{aligned} |D_{II,t}\cap (-q D_{II,t})|={\left\{ \begin{array}{ll}1 &{}\mathrm{if}~0\le t\le \lfloor \frac{2q-4}{5} \rfloor ,\\ 5 &{} \mathrm{if}~\lfloor \frac{2q+1}{5} \rfloor \le t\le \lfloor \frac{3q-4}{5} \rfloor . \end{array}\right. } \end{aligned}$$

Proof

Let $t_0=\lfloor \frac{3q-4}{5} \rfloor$. We now prove

$$\begin{aligned} D_{II,t_0}\cap (-q D_{II,t_0})=C_0\cup C_{\lceil \frac{q-2}{5} \rceil }\cup C_{\lfloor \frac{2q+1}{5} \rfloor } \end{aligned}$$

and $-q C_{\lceil \frac{q-2}{5} \rceil }=C_{\lfloor \frac{2q+1}{5} \rfloor }$. The main idea is to analyze the $q^2$-cyclotomic coset modulo n represented by $-qC_{i}$ for $0\le i\le t_0$. Clearly, $-qC_0=C_0$. For every i in the range $1\le i\le t_0$,

$$\begin{aligned} -q(n-i)\equiv qi ~({{\,\mathrm{mod~}\,}}n). \end{aligned}$$

(8)

Let $a_i$ be an integer with $1\le a_i\le \frac{n}{2}$ such that $-qC_{i}=C_{a_i}$. From (6) and (8), we have

$$\begin{aligned} a_i\equiv qi~({{\,\mathrm{mod~}\,}}n), \end{aligned}$$

(9)

or

$$\begin{aligned} a_i\equiv -qi ~({{\,\mathrm{mod~}\,}}n). \end{aligned}$$

(10)

Notice that $0<q i\le qt_0<3n$ for $1\le i\le t_0$, we now analyze $a_i$ in the following six cases.

• Case 1: $i\in \Gamma _1:=\{i: 1\le qi\le \frac{n}{2}\}$. From (9), $a_i=qi$, i.e., $-qC_{i}=C_{qi}$. From (7), $(-q C_{i})\cap D_{II,t_0}\ne \emptyset$ if and only if $1\le qi\le t_0$. Since i is an integer and $t_0<q$, then

  $$\begin{aligned} (\cup _{i\in \Gamma _1} (-qC_{i}))\cap D_{II,t_0}=\emptyset . \end{aligned}$$

• Case 2: $i\in \Gamma _2:=\{i: \frac{n}{2}+1\le qi\le n \}$. Since $\gcd (n, q)=1$, $qi\ne n$. It follows from (10) that $a_i=n- q i$, i.e., $-qC_{i}=C_{n-q i}$. From (7), $(-q C_{i})\cap D_{II, t_0}\ne \emptyset$ if and only if $1\le n-qi \le t_0$. Since i is an integer, $1\le n-qi \le t_0$ is equivalent to

  $$\begin{aligned} \left\lceil \frac{q-2}{5} \right\rceil =\left\lceil \frac{n-t_0}{q} \right\rceil \le i\le \left\lfloor \frac{n-1}{q} \right\rfloor = \left\lfloor \frac{q-2}{5} \right\rfloor . \end{aligned}$$

  Hence, if $q\equiv -3~({{\,\mathrm{mod~}\,}}10)$, there is only an $i=\frac{q-2}{5}$ such that $(-q C_{i})\cap D_{II, t_0}\ne \emptyset$ for $i\in \Gamma _2$. In this case, $a_i=n-qi=\frac{2q+1}{5}$, and

  $$\begin{aligned} (\cup _{i\in \Gamma _2} (-qC_{i}))\cap D_{II,t_0}=C_{\frac{2q+1}{5}}. \end{aligned}$$

  If $q\equiv 3~({{\,\mathrm{mod~}\,}}10)$, $(\cup _{i\in \Gamma _2} (-qC_{i}))\cap D_{II,t_0}=\emptyset$.

• Case 3: $i\in \Gamma _3:=\{i: n+1\le qi\le \frac{3n}{2}\}$. From (9), $a_i=qi-n$, i.e., $-qC_{i}=C_{qi-n}$. It follows from (7) that $(-q C_{i})\cap D_{II,t_0}\ne \emptyset$ if and only if $1\le qi-n\le t_0$. Note that i is an integer, $1\le qi-n\le t_0$ is equivalent to

  $$\begin{aligned} \left\lceil \frac{q+2}{5} \right\rceil = \left\lceil \frac{n+1}{q} \right\rceil \le i\le \left\lfloor \frac{n+t_0}{q} \right\rfloor = \left\lfloor \frac{q+2}{5} \right\rfloor . \end{aligned}$$

  Hence, if $q\equiv 3~({{\,\mathrm{mod~}\,}}10)$, there is only an $i=\frac{q+2}{5}$ such that $(-q C_{i})\cap D_{II, t_0}\ne \emptyset$ for $i\in \Gamma _3$. In this case, $a_i=qi-n=\frac{2q-1}{5}$, and

  $$\begin{aligned} (\cup _{i\in \Gamma _3} (-qC_{i}))\cap D_{II,t_0}=C_{\frac{2q-1}{5}}. \end{aligned}$$

  If $q\equiv -3 ~({{\,\mathrm{mod~}\,}}10)$, we have $(\cup _{i\in \Gamma _3} (-qC_{i}))\cap D_{II,t_0}=\emptyset$.

• Case 4: $i\in \Gamma _4:=\{i: \frac{3n}{2}+1\le qi\le 2n \}$. Since q is an odd prime power and $\gcd (n, q)=1$, one can get $qi\ne 2n$. Hence, $1\le 2n-qi\le \frac{n}{2}-1$. From (10), $a_i=2n-qi$, i.e., $-qC_{i}=C_{2n-qi}$. It follows from (7) that $(-q C_{i})\cap D_{II,t_0}\ne \emptyset$ if and only if $1\le 2n-qi \le t_0$. Since i is an integer, $1\le 2n-qi\le t_0$ is equivalent to

  $$\begin{aligned} \left\lceil \frac{2q-1}{5}\right\rceil =\left\lceil \frac{2n-t_0}{q}\right\rceil \le i\le \left\lfloor \frac{2n-1}{q} \right\rfloor =\left\lfloor \frac{2q-1}{5} \right\rfloor . \end{aligned}$$

  Hence, if $q\equiv 3~({{\,\mathrm{mod~}\,}}10)$, there is only an $i=\frac{2q-1}{5}$ such that $(-q C_{i})\cap D_{II,t_0}\ne \emptyset$, for $i\in \Gamma _4$. In this case, $a_i=2n-qi=\frac{q+2}{5}$, and

  $$\begin{aligned} (\cup _{i\in \Gamma _4} (-qC_{i}))\cap D_{II,t_0}=C_{\frac{q+2}{5}}. \end{aligned}$$

  If $q\equiv -3~({{\,\mathrm{mod~}\,}}10)$, we have $(\cup _{i\in \Gamma _4} (-qC_{i}))\cap D_{II,t_0}=\emptyset$.

• Case 5: $i\in \Gamma _5:=\{i: 2n+1\le qi\le \frac{5n}{2} \}$. Similar to Case 4, one can get $qi\ne \frac{5n}{2}$. It implies that $1\le qi-2n\le \frac{n}{2}-1$. From (9), $a_i=qi-2n$, i.e., $-qC_{i}=C_{qi-2n}$. From (7), $(-q C_{i})\cap D_{II,t_0}\ne \emptyset$ if and only if $1\le qi-2n \le t_0$. Since i is an integer, $1\le qi-2n \le t_0$ is equivalent to

  $$\begin{aligned} \left\lceil \frac{2q+1}{5}\right\rceil = \left\lceil \frac{2n+1}{q}\right\rceil \le i\le \left\lfloor \frac{2n+t_0}{q} \right\rfloor =\left\lfloor \frac{2q+1}{5} \right\rfloor . \end{aligned}$$

  Therefore, if $q\equiv -3~({{\,\mathrm{mod~}\,}}10)$, there is only an $i=\frac{2q+1}{5}$ such that $(-q C_{i})\cap D_{II,t_0}\ne \emptyset$, for $i\in \Gamma _5$. In this case, $a_i=qi-2n=\frac{q-2}{5}$, and

  $$\begin{aligned} (\cup _{i\in \Gamma _5} (-qC_{i}))\cap D_{II,t_0}=C_{\frac{q-2}{5}}. \end{aligned}$$

  If $q\equiv 3~({{\,\mathrm{mod~}\,}}10)$, we have $(\cup _{i\in \Gamma _5} (-qC_{i}))\cap D_{II,t_0}=\emptyset$.

• Case 6: $i\in \Gamma _6:=\{i: \frac{5n}{2}+1\le qi\le qt_0 \}$. Obviously, $1<3n-qt_0\le 3n-qi\le \frac{n}{2}-1$. From (10), $a_i=3n-qi$, i.e., $-qC_{i}=C_{3n-qi}$. From (7), $(-q C_{i})\cap D_{II,t_0}\ne \emptyset$ if and only if $3n-qt_0\le 3n-qi \le t_0$. Since i is an integer, $3n-qt_0\le 3n-qi\le t_0$ is equivalent to

  $$\begin{aligned} \left\lceil \frac{3q-1}{5}\right\rceil =\left\lceil \frac{3n-t_0}{q}\right\rceil \le i\le t_0 =\left\lfloor \frac{3q-4}{5} \right\rfloor \end{aligned}$$

  Therefore, $(\cup _{i\in \Gamma _6} (-qC_{i}))\cap D_{II,t_0}=\emptyset$.

In conclusion, we have $D_{II,t_0}\cap (-q D_{II,t_0})=C_0\cup C_{\lceil \frac{q-2}{5} \rceil }\cup C_{\lfloor \frac{2q+1}{5} \rfloor }$. The result follows. $\square$

Theorem 5

Let q be an odd prime power with $q\equiv \pm 3~({{\,\mathrm{mod~}\,}}10)$ and $q>3$. For each even integer d with $2\le d\le 2 \lfloor \frac{2q+1}{5} \rfloor$, there exists an EAQMDS code with parameters

$$\begin{aligned} \left[ \left[ \frac{q^2+1}{5},\frac{q^2+1}{5}-2d+3,d;1\right] \right] _q. \end{aligned}$$

For each even integer d with $2 \lfloor \frac{2q+6}{5} \rfloor \le d\le 2 \lfloor \frac{3q+1}{5} \rfloor$, there exists an EAQMDS code with parameters

$$\begin{aligned} \left[ \left[ \frac{q^2+1}{5},\frac{q^2+1}{5}-2d+7,d;5\right] \right] _q. \end{aligned}$$

Proof

For each even integer d with $2\le d\le 2 \lfloor \frac{3q+1}{5} \rfloor$, let $t=\frac{d-2}{2}$, then $0\le t\le \lfloor \frac{3q-4}{5} \rfloor$. Now consider the $q^2$-ary cyclic code $\mathscr {C}_{II,t}$ of length $n=\frac{q^2+1}{5}$ with defining set $D_{II,t}$. From (7), $|D_{I, t}|=2t+1=d-1$. Hence, $\dim (\mathscr {C}_{II,t})=n-|D_{II,t}|=n-d+1$. According to the definition of $D_{II,t}$, it consists of $d-1$ consecutive integers

$$\begin{aligned} \left\{ -\frac{d-2}{2}, \cdots ,-1, 0,1, \cdots , \frac{d-2}{2} \right\} . \end{aligned}$$

Then by Theorem 2, $d(\mathscr {C}_{II,t})\geqslant d$. Therefore, $\mathscr {C}_{II,t}$ is a cyclic code with parameters $[n,n-d+1,\geqslant d]_{q^2}$. From Lemma 3,

$$\begin{aligned} |D_{II,t}\cap (-q D_{II,t})|={\left\{ \begin{array}{ll}1 &{}\mathrm{if}~0\le t\le \lfloor \frac{2q-4}{5} \rfloor ,\\ 5 &{} \mathrm{if}~\lfloor \frac{2q+1}{5} \rfloor \le t\le \lfloor \frac{3q-4}{5} \rfloor . \end{array}\right. } \end{aligned}$$

Combining Theorem 3 with the EA quantum Singleton bound, there is an EAQMDS code with parameters

$$\begin{aligned} \left[ \left[ \frac{q^2+1}{5},\frac{q^2+1}{5}-2d+3,d;1\right] \right] _q, \end{aligned}$$

for even d with $2\le d\le 2 \lfloor \frac{2q+1}{5} \rfloor$; and

$$\left[ \left[ \frac{q^2+1}{5},\frac{q^2+1}{5}-2d+7,d;5\right] \right] _q,$$

for even d with $2 \lfloor \frac{2q+6}{5} \rfloor \le d\le 2 \lfloor \frac{3q+1}{5} \rfloor$. The result follows. $\square$

Example 2

We list some new EAQMDS codes of length $\frac{q^2+1}{5}$ obtained from Theorem 5 in Table 3.

Remark 2

EAQMDS codes of length $\frac{q^2+1}{5}$ with $c=1$ and $c=5$ had been constructed in [26] from negacyclic codes, where $q=10m+3$, $q=10m+7$ and m is an even integer. However, in this paper, we construct EAQMDS codes of length $\frac{q^2+1}{5}$ with $c=1$ and $c=5$ under the case $q=10m+3$, $q=10m+7$ and m is any positive integer. Hence, our results are more general. It is easy to see that our results coincide with theirs under the case m is even. But when m is odd, our results are new. EAQMDS codes of such length with other cases also had been studied (see Table 4).

Table 3 New entanglement-assisted quantum MDS codes of length $\frac{q^2+1}{5}$

Table 4 Entanglement-assisted quantum MDS codes of length $\frac{q^2+1}{5}$

3.3 Entanglement-assisted quantum MDS codes of length $\frac{q^2+1}{10}$

Throughout this subsection, let q be an odd prime power with $q\equiv \pm 3 ~({{\,\mathrm{mod~}\,}}10)$ and $q>3$. Let $n=\frac{q^2+1}{10}$ and $s=\frac{n-1}{2}$. From Lemma 1, the $q^2$-cyclotomic cosets modulo n are: $C_0=\{0 \}$,

$$\begin{aligned} C_{s-i}=\{ s-i, s+1+i \}, \end{aligned}$$

(11)

for $1\le i\le s-1$. For every t with $0\le t\le s-1$, let $\mathscr {C}_{III,t}$ be the $q^2$-ary cyclic code of length n with defining set

$$\begin{aligned} D_{III,t}=\bigcup _{i=0}^{t}C_{s-i}. \end{aligned}$$

(12)

We have the following basic property for the defining set $D_{III,t}$.

Lemma 4

Let $D_{III,t}$ be defined as above. If $\lceil \frac{q-3}{5} \rceil \le t\le \lfloor \frac{2q-6}{5} \rfloor$, then

$$\begin{aligned} |D_{III,t}\cap (-q D_{III,t})|=4. \end{aligned}$$

Proof

Let $t_0= \lfloor \frac{2q-6}{5} \rfloor$. It is clear that we only have to prove that

$$\begin{aligned} D_{III,t_0}\cap (-qD_{III,t_0})=C_{s- \lfloor \frac{q-3}{10}\rfloor }\cup C_{s- \lceil \frac{q-3}{5} \rceil }. \end{aligned}$$

The main idea is similar to Lemma 3, that is, to analyze the $q^2$-cyclotomic coset modulo n represented by $-qC_{s-i}$ for $0\le i\le t_0$. From (11), there is a unique integer $a_i$ with $0\le a_i\le s-1$ such that $-qC_{s-i}=C_{s-a_i}$. For every i with $0\le i\le t_0$, it is easy to check that

$$\begin{aligned} -q (s-i)\equiv s+\frac{q+1}{2}+qi~({{\,\mathrm{mod~}\,}}n). \end{aligned}$$

(13)

From (11), we have

$$\begin{aligned} a_i\equiv \frac{q-1}{2}+qi ~({{\,\mathrm{mod~}\,}}n) , \end{aligned}$$

(14)

or

$$\begin{aligned} a_i\equiv -(\frac{q+1}{2}+qi)~({{\,\mathrm{mod~}\,}}n). \end{aligned}$$

(15)

Note that $\frac{q+1}{2}\le \frac{q+1}{2}+q i\le \frac{q+1}{2}+qt_0<4n$, for $0\le i\le t_0$. We now analyze $D_{III,t_0}\cap (-qD_{III,t_0})$ in the following eight cases.

• Case 1: $i\in \Gamma _1:=\{i: \frac{q+1}{2}\le \frac{q+1}{2}+qi\le s\}$. From (14), $a_i=\frac{q-1}{2}+qi$, i.e., $-qC_{s-i}=C_{s-\frac{q-1}{2}-qi}$. From (12), $(-q C_{s-i})\cap D_{III,t_0}\ne \emptyset$ if and only if $0\le \frac{q-1}{2}+qi\le t_0$. Since i is an integer and $0<t_0<\frac{q-1}{2}$, we have $(\cup _{i\in \Gamma _1} (-qC_{s-i}))\cap D_{III,t_0}=\emptyset$.

• Case 2: $i\in \Gamma _2:=\{i: s+1\le \frac{q+1}{2}+qi\le n \}$. Similar to Case 2 in the proof of Lemma 2, $\frac{q+1}{2}+qi\ne s+1$. From (15), $a_i =n-\frac{q+1}{2}-qi$. Thereby, from (12), $(-q C_{s-i})\cap D_{III,t_0}\ne \emptyset$ if and only if $0\le a_i\le t_0$. Since i is an integer, $0\le a_i\le t_0$ is equivalent to

  $$\begin{aligned} \left\lceil \frac{q-7}{10} \right\rceil =\left\lceil \frac{2n-q-1-2t_0}{2q} \right\rceil \le i\le \left\lfloor \frac{q^2-5q-4}{10q}\right\rfloor =\left\lfloor \frac{q-7}{10} \right\rfloor . \end{aligned}$$

  Therefore, if $q\equiv -3~({{\,\mathrm{mod~}\,}}10)$, there is only an $i=\frac{q-7}{10}$ such that $0\le a_i\le t_0$ for $i\in \Gamma _2$. In this case, $a_i=n-\frac{q+1}{2}-qi=\frac{q-2}{5}$. Hence,

  $$\begin{aligned} (\cup _{i\in \Gamma _2} (-qC_{s-i}))\cap D_{III,t_0}=C_{s-\frac{q-2}{5}}. \end{aligned}$$

  If $q\equiv 3~({{\,\mathrm{mod~}\,}}10)$, there is no $i\in \Gamma _2$ such that $0\le a_i\le t_0$. Hence,

  $$\begin{aligned} \cup _{i\in \Gamma _2} (-qC_{s-i})\cap D_{III,t_0}=\emptyset . \end{aligned}$$

• Case 3: $i\in \Gamma _3:=\{i: n+1\le \frac{q+1}{2}+qi\le n+s\}$. From (14), $a_i =\frac{q-1}{2}+qi-n$. It follows from (12) that $(-q C_{s-i})\cap D_{III,t_0}\ne \emptyset$ if and only if $0\le a_i \le t_0$. Since i is an integer, $0\le a_i\le t_0$ is equivalent to

  $$\left\lceil \frac{q-3}{10} \right\rceil = \left\lceil \frac{2n-q+1}{2q} \right\rceil \le i\le \left\lfloor \frac{2n-q+1+2t_0}{2q} \right\rfloor = \left\lfloor \frac{q-3}{10} \right\rfloor .$$

  Hence, if $q\equiv 3~({{\,\mathrm{mod~}\,}}10)$, there is only an $i=\frac{q-3}{10}$ such that $(-q C_{s-i})\cap D_{III,t_0}\ne \emptyset$ for $i\in \Gamma _3$. Therefore,

  $$\begin{aligned} (\cup _{i\in \Gamma _3} (-qC_{s-i}))\cap D_{III,t_0}=C_{s-\frac{q-3}{5}}. \end{aligned}$$

  If $q\equiv -3~({{\,\mathrm{mod~}\,}}10)$, $\cup _{i\in \Gamma _3} (-qC_{s-i})\cap D_{III,t_0}=\emptyset$.

• Case 4: $i\in \Gamma _4:=\{i: n+s+1\le \frac{q+1}{2}+qi\le 2n \}$. Similar to Case 4 in the proof of Lemma 2, we have $\frac{q+1}{2}+qi\ne n+s+1$. It follows from (15) that $a_i=2n-\frac{q+1}{2}-qi$. From (12), $(-q C_{s-i})\cap D_{III,t_0}\ne \emptyset$ if and only if $0\le a_i\le t_0$. Since i is an integer, $0\le a_i\le t_0$ is equivalent to $\lceil \frac{q-3}{5} \rceil \le i\le \lfloor \frac{q-3}{5} \rfloor$. Hence, if $q\equiv 3~({{\,\mathrm{mod~}\,}}10)$, there is only an $i=\frac{q-3}{5}$ such that $(-q C_{s-i})\cap D_{III,t_0}\ne \emptyset$ for $i\in \Gamma _4$, i.e.,

  $$\begin{aligned} (\cup _{i\in \Gamma _4} (-qC_{s-i}))\cap D_{III,t_0}=C_{s-\frac{q-3}{10}}. \end{aligned}$$

  Otherwise, if $q\equiv -3~({{\,\mathrm{mod~}\,}}10)$, we have $\cup _{i\in \Gamma _4} (-qC_{s-i})\cap D_{III,t_0}=\emptyset$.

• Case 5: $i\in \Gamma _5:=\{i: 2n+1\le \frac{q+1}{2}+qi\le 2n+s \}$. From (14), $a_i=\frac{q-1}{2}+qi-2n$. From (12), $(-q C_{s-i})\cap D_{III,t_0}\ne \emptyset$ if and only if $0\le a_i \le t_0$. Since i is an integer, $0\le a_i\le t_0$ is equivalent to $\lceil \frac{q-2}{5} \rceil \le i\le \lfloor \frac{q-2}{5} \rfloor$. Hence, if $q\equiv -3~({{\,\mathrm{mod~}\,}}10)$, there is only an $i=\frac{q-2}{5}$ such that $(-q C_{s-i})\cap D_{III,t_0}\ne \emptyset$ for $i\in \Gamma _5$. Thereby,

  $$\begin{aligned} (\cup _{i\in \Gamma _5} (-qC_{s-i}))\cap D_{III,t_0}=C_{s-\frac{q-7}{10}}. \end{aligned}$$

  Otherwise, if $q\equiv 3~({{\,\mathrm{mod~}\,}}10)$, we have $\cup _{i\in \Gamma _5} (-qC_{s-i})\cap D_{III,t_0}=\emptyset$.

• Case 6: $i\in \Gamma _6:=\{i: 2n+s+1\le \frac{q+1}{2}+qi\le 3n \}$. We claim $\frac{q+1}{2}+qi\ne 2n+s+1$. Otherwise, $q+1+2qi=5n+1$. It implies that $5n\equiv 0~({{\,\mathrm{mod~}\,}}q)$, which contradicts to the fact that $\gcd (5n, q)=1$. From (15), $a_i=3n-\frac{q+1}{2}- qi$. It follows from (12) that $(-q C_{s-i})\cap D_{III,t_0}\ne \emptyset$ if and only if $0\le a_i \le t_0$. Since i is an integer, $0\le a_i\le t_0$ is equivalent to $\lceil \frac{3q-1}{10} \rceil \le i\le \lfloor \frac{3q-9}{10} \rfloor$. Therefore,

  $$\begin{aligned} (\cup _{i\in \Gamma _6} (-qC_{s-i}))\cap D_{III,t_0}=\emptyset . \end{aligned}$$

• Case 7: $i\in \Gamma _7:=\{i: 3n+1\le \frac{q+1}{2}+qi\le 3n+s \}$. From (14), $a_i=\frac{q-1}{2}+qi-3n$. It follows from (12) that $(-q C_{s-i})\cap D_{III,t_0}\ne \emptyset$ if and only if $0\le a_i \le t_0$. Since i is an integer, $0\le a_i\le t_0$ is equivalent to $\lceil \frac{3q-1}{10} \rceil \le i\le \lfloor \frac{3q-9}{10} \rfloor$. Hence,

  $$\begin{aligned} (\cup _{i\in \Gamma _7} (-qC_{s-i}))\cap D_{III,t}=\emptyset . \end{aligned}$$

• Case 8: $i\in \Gamma _8:=\{i: 3n+s+1\le \frac{q+1}{2}+qi\le \frac{q+1}{2}+qt_0 \}$. We claim $\frac{q+1}{2}+qi\ne 3n+s+1$. Otherwise, $q+1+2qi=7n+1$. It implies that $7n\equiv 0~({{\,\mathrm{mod~}\,}}q)$. Note that $\gcd (n,q)=1$, we have $q=7$. However, if $q=7$, we have $\Gamma _8=\emptyset$. It follows from (15) that $a_i=4n-\frac{q+1}{2}-qi$. From (12), $(-q C_{s-i})\cap D_{III,t_0}\ne \emptyset$ if and only if $0\le a_i\le t_0$, that is, $\lceil \frac{2q-4}{5} \rceil \le i\le \lfloor \frac{2q-4}{5} \rfloor$. Notice that $i\le \lfloor \frac{2q-6}{5} \rfloor$, we have $(\cup _{i\in \Gamma _8} (-qC_{s-i}))\cap D_{III,t_0}=\emptyset$.

According to all the cases above, we have $D_{III,t_0}\cap (-qD_{III,t_0})=C_{s- \lfloor \frac{q-3}{10}\rfloor }\cup C_{s- \lceil \frac{q-3}{5} \rceil }$. The result follows. $\square$

Theorem 6

Let q be an odd prime power with $q\equiv \pm 3~({{\,\mathrm{mod~}\,}}10)$ and $q>3$. For each odd integer d with $2\lceil \frac{q+2}{5} \rceil +1\le d\le 2\lfloor \frac{2q-1}{5} \rfloor +1$, there is an EAQMDS code with parameters

$$\begin{aligned} \left[ \left[ \frac{q^2+1}{10},\frac{q^2+1}{10}-2d+6,d;4\right] \right] _q. \end{aligned}$$

Proof

For each odd integer d with $2\lceil \frac{q+2}{5} \rceil +1\le d\le 2\lfloor \frac{2q-1}{5} \rfloor +1$, let $t=\frac{d-3}{2}$, then $\lceil \frac{q-3}{5} \rceil \le t\le \lfloor \frac{2q-6}{5} \rfloor$. Consider the $q^2$-ary cyclic code $\mathscr {C}_{III,t}$ of length $n=\frac{q^2+1}{10}$ with defining set $D_{III,t}$. From (12), $|D_{III,t}|=2(t+1)=d-1$. Hence, $\dim (\mathscr {C}_{III,t})=n-|D_{III,t}|=n-d+1$. Clearly, the defining set $D_{III,t}$ consists of $d-1$ consecutive integers $\{s-\frac{d-3}{2},\cdots , s-1, s, s+1,\cdots , s+\frac{d-1}{2} \}$. Then by Theorem 2, $d(\mathscr {C}_{III,t})\geqslant d$. Therefore, $\mathscr {C}_{III,t}$ is a cyclic code with parameters $[n,n-d+1,\geqslant d]_{q^2}$. From Lemma 4, $c=|D_{III,t}\cap (-q D_{III,t})|=4$. Combining Theorem 3 with the EA quantum Singleton bound, $\mathscr {C}_{III,t}$ is an EAQMDS code with parameters

$$\begin{aligned} \left[ \left[ \frac{q^2+1}{10},\frac{q^2+1}{10}-2d+6,d;4\right] \right] _q. \end{aligned}$$

The result follows. $\square$

Example 3

We list some new EAQMDS codes of length $\frac{q^2+1}{10}$ obtained from Theorem 6 in Table 5.

Table 5 New entanglement-assisted quantum MDS codes of length $\frac{q^2+1}{10}$

Remark 3

EAQMDS codes of length $\frac{q^2+1}{10}$ with $c=1$ had been constructed in [2] from negacyclic codes. EAQMDS codes of the same length with $c=5$ and $c=9$ had been constructed in [13] utilizing constacyclic codes with order $q+1$. We list all the known results of EAQMDS codes of length $\frac{q^2+1}{10}$ in Table 6.

Table 6 Entanglement-assisted quantum MDS codes of length $\frac{q^2+1}{10}$

4 Conclusion

In this paper, EAQMDS codes of three different lengths, i.e., $\frac{q^2+1}{2}$, $\frac{q^2+1}{5}$,$\frac{q^2+1}{10}$, have been constructed by exploiting less pre-shared maximally entangled states. Comparing the parameters of the obtained EAQMDS codes with all known EAQMDS codes of such lengths, one can find that these EAQMDS codes are new in the sense that their parameters are not covered by the codes available in the literature, except the length $\frac{q^2+1}{5}$, where $q=10m+3$ and $q=10m+7$ with m even, which is the same as the results in [27].