\section{Introduction}\label{sec1}


Quantum information and quantum computing have become hot topics in recent years.
Quantum codes have an important effect on quantum communications and quantum computing.
Just as in classical coding theory,
a central theme of quantum error correction is to construct quantum codes with good parameters.
In \cite{RefJ (1998) introduct 1}, Calderbank et al. establish the fundamental to use additive codes over GF(4) to construct a class of quantum codes named stabilizer codes.
Then Rains \cite{RefJ (1999) introduct 2}, Ashikhmin and Knill \cite{RefJ (2001) introduct 3} generalized their results to the general finite fields.
Since then, many quantum codes with good parameters have been constructed by classical linear error-correcting codes
with certain self-orthogonality properties (see \cite{RefJ (2007) introduct 5,RefJ (2016) introduct 8,RefJ (2004) n<q+1(1),RefJ (2004) n<q+1(2),RefJ (2011) n=q^2+1(1),RefJ (2015) n=q^2+1(2)}).

Let $q$ be a prime power.
A $q$-ary quantum code $Q$ of length $n$ and size $K$ is a $K$-dimensional subspace of a $q^n$-dimensional Hilbert space
$\mathbb{H}=(\mathbb{C}^q)^{\otimes n}=\mathbb{C}^q\bigotimes \dots \bigotimes \mathbb{C}^q$.
Just as in classical coding theory, the third important parameter of a quantum code is its minimum distance besides its length and size.
If a quantum code has minimum distance $d$,
then it can detect any $d-1$ quantum errors and correct any $\lfloor \frac{d-1}{2}\rfloor$ quantum errors.
We use the notation $[[n,k,d]]_q$ to denote a quantum code of length $n$, dimension $K$ and minimum distance $d$, where $k=\log_qK$.
One of the main problems of quantum coding theory is to construct quantum codes with the minimum distance as large as possible.
However, just as in classical coding theory, there are many tradeoffs between $n$, $k$ and $d$ for a quantum code.
One of the most well-known tradeoffs is that an $[[n,k,d]]_q$ quantum code must satisfy the following $quantum\ Singleton\ bound$ (see \cite{RefJ (2001) introduct 3}):
\begin{equation}\label{eq singleton bound}
2d\leq n-k+2.
\end{equation}
A quantum code achieving the above bound (\ref{eq singleton bound}) is called a $quantum\ maximum-distance-separable$ (MDS) code.
In quantum error correction, quantum MDS codes are one of the particularly interesting and useful quantum codes.

\subsection{Related work}

In recent years, the construction of new quantum MDS codes has attracted many researchers and the relevant results are gradually growing.
More precisely,
All $q$-ary quantum MDS codes of length $n\leq q+1$ have been constructed in \cite{RefJ (2004) n<q+1(1),RefJ (2004) n<q+1(2)}.
Jin et al. in \cite{RefJ (2010) L.jin} proposed an efficient construction of which the code length $n$ satisfies $n\leq q^2-1$,
 however, the minimum distance of these codes are at most $q/2$.
So the main task of researchers is to construct quantum codes with length $n>q+1$ and minimum distance $d>q/2+1$.
Researchers have made a effort to construct quantum MDS codes via negacyclic codes (see \cite{RefJ (2013) n=q^2+1(3)}),
constacyclic codes (see \cite{RefJ (2014) kai,RefJ (2015) L.W,RefJ (2015) T.Zhang,RefJ (2016) L.Hu,RefJ (2015) B.chen}) and pseudo-cyclic codes (see \cite{RefJ (2016) S.Li}).
As one of the important MDS codes, GRS codes are also widely used to construct quantum MDS codes.
Li et.al \cite{RefJ (2008) n=q^2+1(4)} first proposed a unified framework for constructing quantum MDS codes through GRS codes.
Then the conclusions in \cite{RefJ (2008) n=q^2+1(4)} was developed by Jin and Xing \cite{RefJ (2014) L.jin}.
Later, GRS codes have been widely used to construct quantum MDS codes with distance greater than $q/2+1$ in recent years
(see \cite{RefJ (2018)W. two,RefJ (2019)W. some,RefJ (2017)Lem GRS,RefJ (2021) n=q^2+1(5),RefJ (2017) X.Shi,RefJ (2017) L.Jin,RefJ (2021) application,RefJ (2020) X.Fang,RefJ (2022) R.Jin}).
We summarize some known results of quantum MDS codes in Table \ref{tab:1} of Section \ref{sec5}.



\subsection{Our results}

In this paper, we focus on the constructions of quantum MDS codes via Hermitian self-orthogonal GRS codes.
The main idea of our constructions is to find suitable code locators $a_1,a_2,\dots,a_n\in \F_{q^2}$
and column multipliers $v_1,v_2,\dots,v_n\in \F_{q^2}^*$
such that a system of homogenous equations over $\F_{q^2}$ has a solution over $\F_q^*$ (see Lemma \ref{lem (a,v)=0}).
Firstly, we give all the lemmas required by the theorem.
(see Lemmas \ref{lem youjie 1}, \ref{lem youjie 2}, \ref{lem zhenchu}, \ref{lem zhenchu 2}, \ref{lem zhenchu 3}).
By employing this Lemmas,
we construct three new classes of Hermitian self-orthogonal GRS codes,
and further draw several classes of quantum MDS codes from these GRS codes (see Theorems \ref{th1}, \ref{th2}, \ref{th3}).
We find that the quantum MDS codes constructed in this paper have larger distance parameters than the previous results (see Remarks \ref{remrak 1}, \ref{remrak 2}, \ref{remrak 3}).
Further, combined with the code we constructed,
it is similar to the proof of Theorem 3 and 4 in \cite{RefJ (2020) X.Fang}.
We can directly obtain four classes of quantum MDS codes. These codes have new lengths (see Corollaries \ref{coro 2}, \ref{coro 3}).
Finally, the minimum distance of all the $q$-ary quantum MDS codes constructed in this paper can be larger than $q/2+1$.
We list our new constructions of quantum MDS codes in Table \ref{tab:3} of Section \ref{sec5}.



\subsection{Organization}

The rest of this paper is organized as follows.
In Section \ref{sec2}, we recall some basic results about Hermitian self-orthogonal GRS codes.
In Section \ref{sec3}, we introduce the main lemmas needed in this paper.
In Section \ref{sec4}, we obtain three new classes of quantum MDS codes with flexible parameters.
In Section \ref{sec5}, we give a comparison of our results with previous ones.
We use some conclusions to end this paper in Section \ref{sec6}.



\section{Preliminaries}\label{sec2}


In this section, we recall some definitions and basic theory of Hermitian self-orthogonal GRS codes over $\F_{q^2}$
and present some lemmas.

Let $q$ be a prime power and $\F_{q^2}$ be the finite field with $q^2$ element.
An $[n,k,d]_{q^2}$ linear code $\C$ over $\F_{q^2}$ is just a $k$-dimensional subspace of $\F_{q^2}^n$.
The minimum distance $d$ of $\C$
is equal to the the minimum nonzero Hamming weight of all codewords in $\C$.
For a vector $\bm{c}=(c_1,c_2,\dots,c_n)\in \F_{q^2}^n$, we always denote $\bm{c}^i$ is a vector $(c_1^i,c_2^i,\dots,c_n^i)\in \F_{q^2}^n$.
And $0^0$ is set to be $1$.
For any two vectors $\bm{x}=(x_1,x_2,\dots,x_n),\ \bm{y}=(y_1,y_2,\dots,y_n)\in \F_{q^2}^n$.
The Euclidean and Hermitian inner product of vectors $\bm{x}$, $\bm{y}$ are defined as
$$\langle \bm{x},\bm{y} \rangle_{E}=\sum_{i=1}^{n}x_iy_i,\quad \langle \bm{x},\bm{y} \rangle_{H}=\sum_{i=1}^{n}x_iy_i^q,$$
respectively.
And the Euclidean and Hermitian dual codes of $\C$ are defined as
$$\C^{\perp_{E}}=\{\bm{x}\in \F_{q^2}^n|\  \langle \bm{x}, \bm{y}\rangle_{E}=0, \ for \ all\ \bm{y}\in \C \}$$
and
$$\C^{\perp_{H}}=\{\bm{x}\in \F_{q^2}^n|\  \langle \bm{x}, \bm{y}\rangle_{H}=0, \ for \ all\ \bm{y}\in \C \},$$
respectively.
If $\C\subseteq \C^{\perp_H}$, the code $\C$ is called Hermitian self-orthogonal.
Particularly, if $\C=\C^{\perp_{H}}$,
we call $\C$ is a Hermitian self-dual code.
It is easy to check that $C^{\perp_H}=(C^{(q)})^{\perp_E}$, where $C^{(q)}=\{\bm{c}^q:\bm{c}\in C\}$.
For a matrix $A=(a_{ij})$ over $\F_{q^2}$,
we define matrix $(a_{ij}^q)$ as $A^{(q)}$.

Now, we recall some definitions of GRS codes.
Choose $\mathbf{a}=(a_{1},a_{2},\dots,a_{n})$
to be an $n$-tuple of distinct elements of $\F_{q^2}$.
And let $\bm{v}=(v_1,v_2,\dots,v_n)$ be a vector of $(\F_{q^2}^*)^n$, where $\F_{q^2}^*=\F_{q^2}\setminus \{0\}$.
For an integer $k$ satisfying $1\leq k\leq n$, the GRS code associated to $\bm{a}$ and $\bm{v}$ is defined by
$$GRS_{k}(\bm{a},\bm{v})=\{(v_{1}f(a_{1}),v_{2}f(a_{2}),\dots,v_{n}f(a_{n})):\ f(x)\in \F_{q^2}[x],\ \deg(f(x))\leq k-1\}.$$
It is well known that $GRS_{k}(\bm{a},\bm{v})$ is an $[n,k,n-k+1]_{q^2}$ MDS code.
And its dual code is also MDS.
The elements $a_1, a_2,\dots,a_n$ are called the $code\ locators$ of
$GRS_k(\bm{a},\bm{v})$,
and the elements $v_1,v_2,\dots,v_n$ are called the $column\ multipliers$ of $GRS_k(\bm{a},\bm{v})$.
Note that the vectors $(v_1a_1^j,v_2a_2^j,\dots,v_na_n^j),\ j=0,1,\dots,k-1$ form a basis of $GRS_k(\bm{a},\bm{v})$.
Thus we can directly obtain the following lemma, which also given in \cite{RefJ (2017)Lem GRS,RefJ (2014) L.jin,RefJ (2017) X.Shi}.

\begin{lemma}\label{lem (a,v)=0}
The two vectors $\bm{a}=(a_1,a_2,\dots,a_n)$ and $\bm{v}=(v_1,v_2,\dots,v_n)$
are defined above.
Then $GRS_k(\bm{a},\bm{v})$ is Hermitian self-orthogonal
if and only if $\langle \bm{a}^{qi+j}, \bm{v}^{q+1} \rangle_E=\sum_{l=1}^nv_la_l^j(v_la_l^i)^q=0$, for all $0\leq i,j \leq k-1$.
\end{lemma}

The following lemma is the most important one, which is used to construct quantum MDS code.

\begin{lemma}\label{Lem zhuyao}
(Hermitian construction of quantum MDS codes \cite{RefJ (2001) introduct 3})
If there exists a linear code $\C$ with parameters $[n,k,n-k+1]_{q^2}$ and satisfies $\C^{\bot_H}\subseteq \C$,
then there exists a quantum code $Q$ with parameters $[[n,n-2k,k+1]]_{q}$.
\end{lemma}

Through the next lemma, we can get a new quantum MDS code from the known code.

\begin{lemma}\label{lem propagation rule}
(Propagation Rule \cite{RefJ (2015) n=q^2+1(2)})
If there exists a quantum MDS code $Q_1$ with parameters $[[n,n-2k,k+1]]_q$,
then there exists a quantum MDS code $Q_2$ with parameters $[[n-1,n-2k+1,k]]_{q}$.
\end{lemma}


\section{Main Lemma}\label{sec3}

For our constructions, we still need some other basic properties.
In this section, we list them in the followings.

The following two lemmas give a sufficient condition for an equation to have a solution on $\F_{q}^*$.


\begin{lemma}(\cite{RefJ (2019)W. some})\label{lem youjie 1}
Suppose $A$ is an $(r-1)\times r$ matrix over $\F_{q^2}$, where $r>0$.
If $A$ satisfies the follows conditions:
\begin{itemize}
\item[(1)] any $r-1$ columns of $A$ are linearly independent;
\item[(2)]$A^{(q)}$ is row equivalent to $A$.
\end{itemize}
Then the following equation
\begin{equation}
A\bm{u}^T=\bm{0}^T
\end{equation}
has a nonzero solution $\bm{u}=(u_1,u_2,\dots,u_{r})\in (\F_q^*)^{r}$.
\end{lemma}


\begin{lemma}(\cite{RefJ (2019)W. some})\label{lem youjie 2}
Suppose $A$ is an $(r-2)\times r$ matrix over $\F_{q^2}$, where $3\leq r< q+1$.
If $A$ satisfies the follows conditions:
\begin{itemize}
\item[(1)] any $r-2$ columns of $A$ are linearly independent;
\item[(2)]$A^{(q)}$ is row equivalent to $A$.
\end{itemize}
Then the following equation
\begin{equation}
A\bm{u}^T=\bm{0}^T
\end{equation}
has a nonzero solution $\bm{u}=(u_1,u_2,\dots,u_{r})\in (\F_q^*)^{r}$.
\end{lemma}

In the following, we introduce two lemmas that have been proposed in \cite{RefJ (2019)W. some}, which gives the set of $qi+j$ under certain conditions.

\begin{lemma}\label{lem zhenchu}
\begin{itemize}
\item[(1)]
Denote by $m=\frac{q^2-1}{2s+1}$ where $(2s+1)\mid (q+1)$.
Suppose $1\leq k\leq (s+t+1)\frac{q+1}{2s+1}-1$, where $0\leq t\leq s-1$.
For any $0\leq i,j\leq k-1$, $m\mid (qi+j)$ if and only if
$qi+j\in \{0,(s-t+1)m,(s-t+2)m,\dots,(s+t)m\}$.
\item[(2)]
Denote by $m=\frac{q^2-1}{2s}$ where $2s\mid (q+1)$.
Suppose $1\leq k\leq (s+t+1)\frac{q+1}{2s}-1$, where $0\leq t\leq s-2$.
For any $0\leq i,j\leq k-1$, $m\mid (qi+j)$ if and only if
$qi+j\in \{0,(s-t)m,(s-t+1)m,\dots,(s+t)m\}$.
\end{itemize}
\end{lemma}

Similarly, we give the following lemmas.

\begin{lemma}\label{lem zhenchu 2}
Denote by $m=\frac{q^2-1}{2s}$ where $2s\mid (q+1)$.
Suppose $1\leq k\leq (s+t+2)\frac{q+1}{2s}-2$, where $0\leq t\leq s-2$.
Then for any $0\leq i,j\leq k-1$, $m\mid (qi+j+q+1-\frac{q+1}{2s})$ if and only if
$(qi+j+q+1-\frac{q+1}{2s})\in \{(s-t)m,(s-t+1)m,\dots,(s+t+1)m\}$.
\end{lemma}

\begin{proof}
We suppose that $qi+j+q+1-\frac{q+1}{2s}$ is divisible by $\frac{q^2-1}{2s}$ for some $0\leq i,j\leq k-1$.
Then there exists an integer $l$ such that
$$qi+j+q+1-\frac{q+1}{2s}=l\frac{q^2-1}{2s}.$$
Hence we have
$$qi+j=[\frac{l(q+1)}{2s}-2]q+q-1-\frac{l(q+1)}{2s}+\frac{q+1}{2s}.$$
For $1\leq l\leq 2s$,
it follows that
$$i=\frac{l(q+1)}{2s}-2,\ j=q-1-\frac{l(q+1)}{2s}+\frac{q+1}{2s}.$$

If $l\geq s+t+2$, then
$$i=\frac{l(q+1)}{2s}-2\geq (s+t+2)\frac{q+1}{2s}-2\geq k,$$
which contradicts to the fact that $i\leq k-1$;

If $l\leq s-t-1$, then
$$j=q-1-\frac{l(q+1)}{2s}+\frac{q+1}{2s}\geq (s+t+2)\frac{q+1}{2s}-2\geq k,$$
which contradicts to the fact that $j\leq k-1$.
Thus $s-t\leq l\leq s+t+1$.
This completes the proof.
\end{proof}


\begin{lemma}\label{lem zhenchu 3}
Denote by $m=\frac{q^2-1}{2s}$ where $2s\mid (q-1)$.
Suppose $1\leq k\leq (s+t)\frac{q-1}{2s}$, where $0\leq t\leq s$.
Then for any $0\leq i,j\leq k-1$, $m\mid (qi+j+\frac{q+1}{2})$ if and only if
$qi+j+\frac{q+1}{2}\in \{m,2m,\dots,(t-1)m,(s+1)m,(s+2)m,\dots,(s+t-1)m\}$.
\end{lemma}

\begin{proof}
We suppose that $qi+j+\frac{q+1}{2}$ is divisible by $\frac{q^2-1}{2s}$ for some $0\leq i,j\leq k-1$.
Then there exists an integer $l$ such that
$$qi+j+\frac{q+1}{2}=l\frac{q^2-1}{2s}.$$
Hence for $1\leq l\leq s$,
we have
\[\begin{split} qi+j&=l\frac{q^2-1}{2s}-\frac{q+1}{2}\\
                           &=(l\frac{q-1}{2s}-1)q+(s+l)\frac{q-1}{2s}.\\
	\end{split}\]
It follows that
$$i=l\frac{q-1}{2s}-1,\ j=(s+l)\frac{q-1}{2s}.$$

If $t\leq l\leq s$, then
$$j=(s+l)\frac{q-1}{2s}\geq (s+t)\frac{q-1}{2s}\geq k,$$
which contradicts to the fact that $j\leq k-1$;

On other hand, for $s+1\leq l\leq 2s$,
we have
\[\begin{split} qi+j&=l\frac{q^2-1}{2s}-\frac{q+1}{2}\\
                           &=(l\frac{q-1}{2s})q+(l-s)\frac{q-1}{2s}-1.\\
	\end{split}\]
It follows that
$$i=l\frac{q-1}{2s},\ j=(l-s)\frac{q-1}{2s}.$$

If $s+t\leq l\leq 2s$, then
$$i=l\frac{q-1}{2s}\geq (s+t)\frac{q-1}{2s}\geq k,$$
which contradicts to the fact that $i\leq k-1$.
Thus $l\in \{1,2,\dots,t-1,s+1,s+2,\dots,s+t-1\}$.
This completes the proof.
\end{proof}



\section{The construction of quantum MDS codes}\label{sec4}

In this section, we construct three new classes of quantum MDS codes by Hermitian self-orthogonal GRS codes.
Let $w$ be a primitive element of $\F_{q^2}$ and $q^2-1=mh$.
Suppose $\langle \theta\rangle$ be the cyclic subgroup of the multiplicative group $\F_{q^2}^*$ where $\theta=w^h\in \F_{q^2}$.
And let $\{\alpha_i\langle \theta\rangle \}_{i=1}^r$ be distinct cosets in $\F_{q^2}^*$.
It is easy to check that $w^{i_1}\theta^{j_1}\neq w^{i_2}\theta^{j_2}$ for any $1\leq i_1\neq i_2\leq r$ and $0\leq j_1\neq j_2\leq m-1$, where $1\leq r \leq h$.

\subsection{First construction of quantum MDS codes}

Let $h\mid q+1$. We set
$$\bm{a}=(0, w, w\theta,\dots,w\theta^{m-1},\dots,w^{r},w^{r}\theta,\dots,w^{r}\theta^{m-1})\in \F_{q^2}^n.$$
$$\bm{v}=(v_0,\underbrace{v_1,\dots,v_1}_{m\ times},\dots,\underbrace{v_r,\dots,v_r}_{m\ times}),$$
where $v_1,\dots,v_r\in \F_{q^2}^*$.
Then when $(i,j)=(0,0)$, we have
\begin{equation}\label{eq qiuhe 0}
\langle \bm{a}^{0},\bm{v}^{q+1}\rangle=v_0^{q+1}+m\sum_{l=1}^rv_l^{q+1}.
\end{equation}
And when $(i,j)\neq (0,0)$, we have
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=\sum_{l=1}^rw^{l(qi+j)}v_l^{q+1}\sum_{\nu=0}^{m-1}\theta^{\nu(qi+j)},$$
thus
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=0,\ if \ m\nmid (qi+j),$$
and
\begin{equation}\label{eq qiuhe 2}
\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=m\sum_{l=1}^rw^{l(qi+j)}v_l^{q+1},\ if \ m\mid (qi+j).
\end{equation}



\begin{theorem}\label{th1}
Let $n=r\frac{q^2-1}{h}+1$,
where $h\mid q+1$, $h\geq 3$ and $1\leq r\leq h$.
If $2\nmid (r+h)$,
then for any $1\leq k\leq (\frac{r+h-1}{2})\frac{q+1}{h}-1$,
there exists an $[[n,n-2k,k+1]]_q$-quantum MDS codes.
\end{theorem}


\begin{proof}
Keep the notations as above, we divide our proof into the following two parts according to whether $h$ is odd or even.

$\bullet$ $\textbf{Case 1:}$ $h=2s$.

Since $2\nmid r+h$, suppose $r=2t+3$, where $0\leq t\leq s-2$ and $1\leq k\leq (s+t+1)\frac{q+1}{2s}-1$.
By Lemma \ref{lem zhenchu}, $m\mid (qi+j)$ if and only if $qi+j\in \{0,(s-t)m,(s-t+2)m,\dots,(s+t)m\}$.
Thus by Eq. (\ref{eq qiuhe 2}), when $qi+j=\mu m\ (s-t\leq \mu\leq s+t)$,
we have
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=m\sum_{l=1}^{r}w^{l\mu m}v_l^{q+1}. $$
Let $\alpha=w^m\in \F_{q^2}^*$ be a primitive $2s$-th root of unity.
Put $a=s-t$,
it is easy to check that $\alpha^{a+v}\neq \alpha^{a+v'}\neq 1$
for any $0\leq v\neq v'\leq r-3$.
Let
$$A=\begin{pmatrix}
 1&   1    &  1  &  \dots  & 1  \\
 0& \alpha^a &  \alpha^{2a} & \dots & \alpha^{ra}\\
 0& \alpha^{a+1} &  \alpha^{2(a+1)}  &\dots & \alpha^{r(a+1)}\\
 \vdots& \vdots &  \vdots & \ddots & \vdots\\
 0& \alpha^{a+r-3} & \alpha^{2(a+r-3)} & \dots & \alpha^{r(a+r-3)}\\
\end{pmatrix}$$
be an $(r-1)\times (r+1)$ matrix over $\F_{q^2}$.
Note that $2s\mid (q+1)$,
then we have
$$\alpha^{i(a+j)q}=\alpha^{-i(s-t+j)}=\alpha^{i(s+t-j)}=\alpha^{i(a+r-3-j)},$$
for any $1\leq i\leq r$ and $0\leq j\leq r-3$.
Thus $A$ is row equivalent to $A^{(q)}$.
Let $A_{\mu\nu}\ (1\leq \mu\neq \nu\leq r+1)$ be the $(r-1)\times (r-1)$ matrix obtained from $A$
by deleting the $\mu$-th and $\nu$-th columns.
It is easy to check that $\det(A_{\mu\nu})\neq 0$.
By Lemma \ref{lem youjie 2},
the system of homogenous equations $A\bm{u}^T=\bm{0}^T$
has a solution $\bm{u}=(u_0,u_1,\dots,u_{r})\in (\F_q^*)^{r+1}$.

$\bullet$ $\textbf{Case 2:}$ $h=2s+1$.

Since $2\nmid r+h$, suppose $r=2t+2$, where $0\leq t\leq s-1$ and $1\leq k\leq (s+t+1)\frac{q+1}{2s}-1$.
By Lemma \ref{lem zhenchu}, $m\mid (qi+j)$ if and only if $qi+j\in \{0,(s-t+1)m,(s-t+2)m,\dots,(s+t)m\}$.
Thus by Eq. (\ref{eq qiuhe 2}), when $qi+j=\mu m\ (s-t+1\leq \mu\leq s+t)$,
we have
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=m\sum_{l=1}^{r}w^{l\mu m}v_l^{q+1}. $$
Let $\alpha=w^m\in \F_{q^2}^*$ be a primitive $(2s+1)$-th root of unity.
Put $a=s-t+1$
it is easy to check that $\alpha^{a+v}\neq \alpha^{a+v'}\neq 1$
for any $0\leq v\neq v'\leq r-3$.
Assume that the definition of $A$ is the same as that of Case 1.
Similar to Case 1,
the system of homogenous equations $A\bm{u}^T=\bm{0}^T$
has a solution $\bm{u}=(u_0,u_1,\dots,u_{r})\in (\F_q^*)^{r+1}$.

Combining $\textbf{Case 1}$ and $\textbf{Case 2}$,
let $v_l^{q+1}=u_l$ for $1\leq l\leq r$ and let $v_0\in \F_{q^2}^*$ such that $v_0^{q+1}=u_0m$.
Then by Eq. (\ref{eq qiuhe 0}), we have
$$\langle \bm{a}^{0},\bm{v}^{q+1}\rangle=v_0^{q+1}+m\sum_{l=1}^rv_l^{q+1}=m\sum_{l=0}^{r}u_l=0.$$
And by Eq. (\ref{eq qiuhe 2}), we have
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=m\sum_{l=1}^{r}w^{l\mu m}v_l^{q+1}=m\sum_{l=1}^{r}w^{l\mu m}u_l=0,$$
for $0\leq i,j\leq k-1$ and $(i,j)\neq (0,0)$.
Hence $\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=0$, for all $0\leq i,j\leq k-1$.
Therefore, by Lemma \ref{lem (a,v)=0},
the code $GRS_k(\bm{a},\bm{v})$ is Hermitian self-orthogonal.
Then the conclusion follows from Lemma \ref{Lem zhuyao}.
This completes the proof.
\end{proof}

Fang et al. [\cite{RefJ (2019)W. some}, Theorem 6.3] constructed a family of quantum MDS codes of length $n=(2t+1)\frac{q^2-1}{2s}$, where $2s\mid (q+1)$.
Applying the propagation rule (see Lemma \ref{lem propagation rule}) for Theorem \ref{th1} Class 1,
we can immediately obtain the following result.

\begin{corollary}
Let $n=r\frac{q^2-1}{2s}$,
where $2s\mid (q+1)$, $r=2t+1$ and $1\leq t\leq s-1$.
Then for any $1\leq k\leq (s+t)\frac{q+1}{2s}-2$,
there exists an $[[n,n-2k,k+1]]_q$-quantum MDS code.
\end{corollary}

\begin{example}
In this example,
we give some quantum MDS codes from Theorem \ref{th1}.
\begin{itemize}
\item[(1)]
When $8\mid (q+1)$, let $(r,h)=(7,8)$ in Theorem \ref{th1} Class 1.
Then for any $1\leq k\leq \frac{7}{8}(q+1)-1$,
there exists a $[[\frac{7}{8}(q^2-1)+1,\frac{7}{8}(q^2-1)+1-2k,k+1]]_q$-quantum MDS code;
\item[(2)]
When $9\mid (q+1)$, let $(r,h)=(8,9)$ in Theorem \ref{th1} Class 2.
Then for any $1\leq k\leq \frac{8}{9}(q+1)-1$, there exists
a $[[\frac{8}{9}(q^2-1)+1,\frac{8}{9}(q^2-1)+1-2k,k+1]]_q$-quantum MDS code.
\end{itemize}
\end{example}

\subsection{Second construction of quantum MDS codes}


Let $h\mid q+1$. We set
$$\bm{a}=(w, w\theta,\dots,w\theta^{m-1},\dots,w^{r},w^{r}\theta,\dots,w^{r}\theta^{m-1})\in \F_{q^2}^n.$$
$$\bm{v}=(v_1,v_1w^{h-1},\dots,v_1w^{(m-1)(h-1)},\dots,v_r,v_rw^{h-1}\dots,v_rw^{(m-1)(h-1)}),$$
where $v_1,\dots,v_r\in \F_{q^2}^*$.
Then for any $0\leq i,j\leq k-1$, we have
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=\sum_{l=1}^rw^{l(qi+j)}v_l^{q+1}\sum_{\nu=0}^{m-1}\theta^{\nu(qi+j+q+1-\frac{q+1}{h})},$$
thus
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=0,\ if \ m\nmid (qi+j+q+1-\frac{q+1}{h}),$$
and
\begin{equation}\label{eq qiuhe 3}
\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=m\sum_{l=1}^rw^{l(qi+j)}v_l^{q+1},\ if \ m\mid (qi+j+q+1-\frac{q+1}{h}).
\end{equation}


\begin{theorem}\label{th2}
Let $n=r\frac{q^2-1}{h}$,
where $h\mid q+1$, $h=2s$, $r=2t+3$ and $0\leq t\leq s-2$.
Then for any $1\leq k\leq (\frac{r+h+1}{2})\frac{q+1}{h}-2$,
there exists an $[[n,n-2k,k+1]]_q$-quantum MDS codes.
\end{theorem}

\begin{proof}
Keep the notations as above.
Since $1\leq k\leq (s+t+2)\frac{q+1}{2s}-2$,
by Lemma \ref{lem zhenchu 2}, $m\mid (qi+j+q+1-\frac{q+1}{h})$ if and only if $(qi+j+q+1-\frac{q+1}{h})\in \{(s-t)m,(s-t+2)m,\dots,(s+t+1)m\}$.
Thus by Eq. (\ref{eq qiuhe 3}), when $qi+j+q+1-\frac{q+1}{h}=\mu m\ (s-t\leq \mu\leq s+t+1)$,
we have
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=m\sum_{l=1}^{r}w^{l(\mu m+\frac{q+1}{h}-q-1)}v_l^{q+1}. $$
Let $\alpha=w^m\in \F_{q^2}^*$ be a primitive $2s$-th root of unity.
Put $a=s-t$, it is easy to check that $\alpha^{a+v}\neq \alpha^{a+v'}\neq 1$
for any $0\leq v\neq v'\leq r-2$.
Let
$$A=\begin{pmatrix}
 \alpha^aw^{\frac{q+1}{h}-q-1} &  \alpha^{2a}w^{2(\frac{q+1}{h}-q-1)} & \dots & \alpha^{ra}w^{r(\frac{q+1}{h}-q-1)}\\
 \alpha^{a+1}w^{\frac{q+1}{h}-q-1} &  \alpha^{2(a+1)}w^{2(\frac{q+1}{h}-q-1)}  &\dots & \alpha^{r(a+1)}w^{r(\frac{q+1}{h}-q-1)}\\
 \vdots &  \vdots & \ddots & \vdots\\
 \alpha^{a+r-2}w^{\frac{q+1}{h}-q-1} & \alpha^{2(a+r-2)}w^{2(\frac{q+1}{h}-q-1)} & \dots & \alpha^{r(a+r-2)}w^{r(\frac{q+1}{h}-q-1)}\\
\end{pmatrix}$$
be an $(r-1)\times r$ matrix over $\F_{q^2}$.
Note that $2s\mid (q+1)$,
then we have
\[\begin{split} \alpha^{i(a+j)q}w^{i(\frac{q+1}{h}-q-1)q}&=\alpha^{-i(s-t+j)}w^{i(\frac{q^2-1}{h}+\frac{q+1}{h}-q-1)}\\
                           &=\alpha^{i(s+t+1-j)}w^{i(\frac{q+1}{h}-q-1)}\\
                           &=\alpha^{i(a+r-2-j)}w^{i(\frac{q+1}{h}-q-1)},\\
	\end{split}\]
for any $1\leq i\leq r$ and $0\leq j\leq r-2$.
Thus $A$ is row equivalent to $A^{(q)}$.
Let $A_{\mu}\ (1\leq \mu\leq r)$ be the $(r-1)\times (r-1)$ matrix obtained from $A$
by deleting the $\mu$-th column.
It is easy to check that $\det(A_{\mu})\neq 0$.
By Lemma \ref{lem youjie 1},
the system of homogenous equations $A\bm{u}^T=\bm{0}^T$
has a solution $\bm{u}=(u_1,u_2,\dots,u_{r})\in (\F_q^*)^{r}$.

Let $v_l^{q+1}=u_l$ for $1\leq l\leq r$.
Hence
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=m\sum_{l=1}^{r}w^{l\mu m}v_l^{q+1}=m\sum_{l=1}^{r}w^{l\mu m}u_l=0,$$
for all $0\leq i,j\leq k-1$.
Therefore, by Lemma \ref{lem (a,v)=0},
the code $GRS_k(\bm{a},\bm{v})$ is Hermitian self-orthogonal.
Then the conclusion follows from Lemma \ref{Lem zhuyao}.
This completes the proof.
\end{proof}


Shi et al. [\cite{RefJ (2017) X.Shi}, Theorem 4.4] constructed a family of quantum MDS codes of length $n=(2t+2)\frac{q^2-1}{2s+1}$, where $(2s+1)\mid (q+1)$.
Combining theorem \ref{th2}, we can immediately obtain the following result.

\begin{corollary}\label{coro 1}
Let $n=r\frac{q^2-1}{h}$,
where $h\mid q+1$, $h\geq 3$ and $1\leq r <h$
If $2\nmid (r+h)$,
then for any $1\leq k\leq (\frac{r+h+1}{2})\frac{q+1}{h}-2$,
there exists an $[[n,n-2k,k+1]]_q$-quantum MDS codes.
\end{corollary}

In Corollary \ref{coro 1}, take $r=h-1$, we get the following result.

\begin{corollary}
Let $h\mid q+1$ and $h\geq 3$.
Then for any $1\leq k\leq q-1$,
there exists an $[[\frac{(h-1)(q^2-1)}{h},\frac{(h-1)(q^2-1)}{h}-2k,k+1]]_q$-quantum MDS codes.
\end{corollary}

\begin{example}
In this example,
we give some quantum MDS codes from Theorem \ref{th2}.
\begin{itemize}
\item[(1)]
When $8\mid (q+1)$, let $(r,h)=(5,8)$ in Theorem \ref{th2}.
Then for any $1\leq k\leq \frac{7}{8}(q+1)-2$, there exists
a $[[\frac{5}{8}(q^2-1),\frac{5}{8}(q^2-1)-2k,k+1]]_q$-quantum MDS code;
\item[(2)]
When $6\mid (q+1)$, let $(r,h)=(5,6)$ in Theorem \ref{th2}.
Then for any $1\leq k\leq q-1$, there exists
a $[[\frac{5}{6}(q^2-1),\frac{5}{6}(q^2-1)-2k,k+1]]_q$-quantum MDS code.
\end{itemize}
\end{example}


\subsection{Third construction of quantum MDS codes}

Let $q$ be an odd prime power,
suppose $h=2s$ and $h\mid q-1$, put
$$\bm{a}=(w^2, w^2\theta,\dots,w^2\theta^{m-1},\dots,w^{2r},w^{2r}\theta,\dots,w^{2r}\theta^{m-1})\in \F_{q^2}^n,$$
where $r\leq s$.
Set
$$\bm{v}=(v_1,v_1w^{s},\dots,v_1w^{(m-1)s},\dots,v_r,v_rw^{s}\dots,v_rw^{(m-1)s}),$$
where $v_1,\dots,v_r\in \F_{q^2}^*$.
Then for any $0\leq i,j\leq k-1$, we have
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=\sum_{l=1}^rw^{2l(qi+j)}v_l^{q+1}\sum_{\nu=0}^{m-1}\theta^{\nu(qi+j+\frac{q+1}{2})},$$
thus
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=0,\ if \ m\nmid (qi+j+\frac{q+1}{2}),$$
and
\begin{equation}\label{eq qiuhe 4}
\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=m\sum_{l=1}^rw^{2l(qi+j)}v_l^{q+1},\ if \ m\mid (qi+j+\frac{q+1}{2}).
\end{equation}


\begin{theorem}\label{th3}
Let $n=r\frac{q^2-1}{h}$, where
$h\mid q-1$, $h=2s$ and $1\leq r\leq s$.
Suppose $t=r-1$,
then for any $1\leq k\leq (s+t)\frac{q-1}{2s}$,
there exists an $[[n,n-2k,k+1]]_q$-quantum MDS codes.
\end{theorem}

\begin{proof}
Since $1\leq k\leq (s+t)\frac{q-1}{2s}$.
By Lemma \ref{lem zhenchu 3}, $m\mid (qi+j+\frac{q+1}{2})$ if and only if $(qi+j+\frac{q+1}{2})\in \{m,2m,\dots,(t-1)m,(s+1)m,(s+2)m,\dots,(s+t-1)m\}$.
Thus by Eq. (\ref{eq qiuhe 4}), when $qi+j+\frac{q+1}{2}=\mu m\ (\mu \in \{1,2,\dots,t-1,s+1,s+2,\dots,s+t-1\})$,
we have
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=m\sum_{l=1}^{r}w^{2l(\mu m-\frac{q+1}{2})}v_l^{q+1}=m\sum_{l=1}^{r}w^{l(2\mu m-q-1)}v_l^{q+1}. $$
Let $\alpha=w^{2m}\in \F_{q^2}^*$ be a primitive $s$-th root of unity.
It is easy to check that $\alpha^{v}\neq \alpha^{v'}\neq 1$
for any $1\leq v\neq v'\leq t-1$.
Let
$$A=\begin{pmatrix}
 \alpha w^{-(q+1)} &  \alpha^{2}w^{-2(q+1)} & \dots & \alpha^{r}w^{-r(q+1)}\\
 \vdots &  \vdots & \ddots & \vdots\\
 \alpha^{t-1}w^{-(q+1)} & \alpha^{2(t-1)}w^{-2(q+1)} & \dots & \alpha^{r(t-1)}w^{-r(q+1)}\\
  \alpha^{s+1} w^{-(q+1)} &  \alpha^{2(s+1)}w^{-2(q+1)} & \dots & \alpha^{r(s+1)}w^{-r(q+1)}\\
 \vdots &  \vdots & \ddots & \vdots\\
 \alpha^{s+t-1}w^{-(q+1)} & \alpha^{2(s+t-1)}w^{-2(q+1)} & \dots & \alpha^{r(s+t-1)}w^{-r(q+1)}\\
\end{pmatrix}$$
be an $2t\times r$ matrix over $\F_{q^2}$.
Let
$$B=\begin{pmatrix}
 \alpha w^{-(q+1)} &  \alpha^{2}w^{-2(q+1)} & \dots & \alpha^{r}w^{-r(q+1)}\\
 \alpha^2 w^{-(q+1)} &  \alpha^{4}w^{-2(q+1)} & \dots & \alpha^{2r}w^{-r(q+1)}\\
 \vdots &  \vdots & \ddots & \vdots\\
 \alpha^{t-1}w^{-(q+1)} & \alpha^{2(t-1)}w^{-2(q+1)} & \dots & \alpha^{r(t-1)}w^{-r(q+1)}\\
\end{pmatrix}$$
be an $t\times r$ matrix over $\F_{q^2}$.
Since $\alpha^s=1$,
then the matrix $A$ is row equivalent to the matrix $B$.
Hence equations set $A\bm{u}^T=\bm{0}^T$ has same solutions as equations set $B\bm{u}^T=\bm{0}^T$.
Note that $s\mid (q-1)$,
then we have $\alpha^q=\alpha$,
it follows that
$$\alpha^{ijq}w^{-i(q+1)q}=\alpha^{ij}w^{-i(q+1)}$$
for any $1\leq i\leq r$ and $j\in \{1,2,\dots,t-1\}$.
Thus $B$ is row equivalent to $B^{(q)}$.
Let $B_{\mu}\ (1\leq \mu\leq r)$ be the $t\times t$ matrix obtained from $B$
by deleting the $\mu$-th column.
It is easy to check that $\det(B_{\mu})\neq 0$.
By Lemma \ref{lem youjie 1},
the system of homogenous equations $B\bm{u}^T=\bm{0}^T$
has a solution $\bm{u}=(u_1,u_2,\dots,u_{r})\in (\F_q^*)^{r}$.
Then equations $A\bm{u}^T=\bm{0}^T$
has a solution $\bm{u}=(u_1,u_2,\dots,u_{r})\in (\F_q^*)^{r}$.
let $v_l^{q+1}=u_l$ for $1\leq l\leq r$.
Hence
$$\langle \bm{a}^{qi+j},\bm{v}^{q+1}\rangle=m\sum_{l=1}^{r}w^{l(2\mu m-q-1)}v_l^{q+1}=m\sum_{l=1}^{r}w^{l(2\mu m-q-1)}u_l=0,$$
for all $0\leq i,j\leq k-1$.
Therefore, by Lemma \ref{lem (a,v)=0},
the code $GRS_k(\bm{a},\bm{v})$ is Hermitian self-orthogonal.
Then the conclusion follows from Lemma \ref{Lem zhuyao}.
This completes the proof.
\end{proof}


\begin{example}
In this example,
we give some quantum MDS codes from Theorem \ref{th3}.
\begin{itemize}
\item[(1)]
When $12\mid (q-1)$, let $(r,h)=(5,12)$ in Theorem \ref{th3}.
Then for any $1\leq k\leq \frac{5}{6}(q-1)$, there exists
a $[[\frac{5}{12}(q^2-1),\frac{5}{12}(q^2-1)-2k,k+1]]_q$-quantum MDS code;
\item[(2)]
When $14\mid (q-1)$, let $(r,h)=(5,14)$ in Theorem \ref{th3}.
Then for any $1\leq k\leq \frac{11}{14}(q-1)$, there exists
a $[[\frac{5}{14}(q^2-1),\frac{6}{14}(q^2-1)-2k,k+1]]_q$-quantum MDS code.
\end{itemize}
\end{example}



\section{Comparison}\label{sec5}

In this section, we make some detailed comparisons of our results with the previous results.

In recent years, the construction of quantum MDS codes has received extensive attention.
In Table \ref{tab:1}, we summarized the parameters of most of quantum MDS codes previously known.
From Table \ref{tab:1}, we can see that the length form of the code in Theorem \ref{th1} is the same as that in Classes 10 and 11, the length form in Theorem \ref{th2} is the same as that in Classes 12, 13, 14 and 15 and the length form in Theorem \ref{th3} is the same as that in Classes 5, 6 and 7. So we only need to compare the same length form. Finally,
in Table \ref{tab:3},
we list our constructions of quantum MDS codes.
Now, we do some detailed comparisons in the following remark
and illustrate that our codes are new and better.

\begin{remark}(Comparison of Theorem \ref{th1})\label{remrak 1}
\begin{itemize}
\item[(1)] In [\cite{RefJ (2019)W. some} Theorem 4.3(i)] and [\cite{RefJ (2019)W. some} Theorem 5.3(i)], Fang et al. prove that there exists an $[[n=r\frac{q^2-1}{h}+1,n-2d,d]]$-quantum MDS code,
where $h\mid q+1$, $1\leq r\leq h$ and $2\leq d\leq (\lfloor\frac{h}{2}\rfloor+1)\frac{q+1}{h}$.
We can find that in Theorem \ref{th1}, the distance can reach $(\frac{r+h-1}{2})\frac{q+1}{h}$.
Therefore our result in Theorem \ref{th1} has larger minimum distance;
\item[(2)]In [\cite{RefJ (2019)W. some} Theorem 4.3(ii)] and [\cite{RefJ (2019)W. some} Theorem 5.3(ii)], Fang et al. prove that there exists an $[[n=r\frac{q^2-1}{h}+1,n-2d,d]]$-quantum MDS code,
where $h\mid q+1$, $1\leq r< h$, $2\mid (r+h)$ and $2\leq d\leq (\frac{r+h}{2})\frac{q+1}{h}$.
We can find that in Theorem \ref{th1}, our condition is changed to $2\nmid (r+h)$.
Therefore our result in Theorem \ref{th1} has different lengths.
\end{itemize}

So Theorem \ref{th1} generalizes the previous conclusion.
\end{remark}

\begin{remark}(Comparison of Theorem \ref{th2})\label{remrak 2}
\begin{itemize}
\item[(1)] In [\cite{RefJ (2019)W. some} Theorem 6.3], Fang et al. prove that there exists an $[[n=r\frac{q^2-1}{h},n-2d,d]]$-quantum MDS code,
where $2\mid h$, $h\mid q+1$, $1\leq r\leq h-1$, $2\nmid (r+h)$ and $2\leq d\leq (\frac{r+h-1}{2})\frac{q+1}{h}-1$.
We can find that in Theorem \ref{th2}, the distance can reach $(\frac{r+h+1}{2})\frac{q+1}{h}-1$.
Therefore our result in Theorem \ref{th2} has larger minimum distance;
\item[(2)]In [\cite{RefJ (2017) X.Shi} Theorem 4.4], Shi et al. prove that there exists an $[[n=r\frac{q^2-1}{h},n-2d,d]]$-quantum MDS code,
where $2\nmid h$, $h\mid q+1$, $1\leq r\leq h-1$, $2\nmid (r+h)$ and $2\leq d\leq (\frac{r+h+1}{2})\frac{q+1}{h}$.
We can find that in Theorem \ref{th2}, our condition is changed to $2\mid h$.
Therefore our result in Theorem \ref{th2} has different lengths.
\end{itemize}

So Theorem \ref{th2} generalizes the previous conclusion.
\end{remark}

\begin{remark}(Comparison of Theorems \ref{th3})\label{remrak 3}
\begin{itemize}
\item[(1)] In [\cite{RefJ (2017)Lem GRS} Theorem 3.2], Zhang et al. prove that there exists an $[[n=r\frac{q^2-1}{h},n-2d,d]]$-quantum MDS code,
where $2\mid h$,$h\mid q-1$, $1\leq r\leq h$, and $2\leq d\leq (\frac{h}{2}+1)\frac{q+1}{h}+1$.
We can find that in Theorem \ref{th3}, the distance can reach $(\frac{h}{2}+r-1)\frac{q-1}{h}+1$.
Therefore our result in Theorem \ref{th3} have larger minimum distance;
\item[(2)]In [\cite{RefJ (2019)W. some} Theorem 3.2] and [\cite{RefJ (2017) X.Shi} Theorem 4.12], Fang and Shi et al. prove that there exists an $[[n=r\frac{q^2-1}{h},n-2d,d]]$-quantum MDS code,
where $h\mid q-1$, $1\leq r\leq h$ and $2\leq d\leq r\frac{q-1}{h}$.
We can find that if $1\leq \frac{h}{2}$, then $r\frac{q-1}{h}< (\frac{h}{2}+r-1)\frac{q-1}{h}+1$.
Therefore our result in Theorems \ref{th3} have larger minimum distance.
\end{itemize}
So Theorems \ref{th3} generalize the previous conclusion.
\end{remark}


\newcommand{\tabincell}[2]{\begin{tabular}{@{}#1@{}}#2\end{tabular}}
\begin{table}
\caption{Some known results of $[[n,n-2d+2,d]]_q$-quantum MDS codes}
\label{tab:1}
\begin{center}
\resizebox{\textwidth}{75mm}{
	\begin{tabular}{cccc}
		\hline
		Class  & Length $n$ & Minimum Distance $d$ & References\\
		\hline
		1  &  $n\leq q+1$ & $2\leq d\leq \frac{n}{2}+1$ &  \cite{RefJ (2004) n<q+1(1),RefJ (2004) n<q+1(2)} \\
        \hline
        2 & $n=q^2+1$ & $2\leq d\leq q+1$, $d\neq q$ & \cite{RefJ (2011) n=q^2+1(1),RefJ (2015) n=q^2+1(2),RefJ (2013) n=q^2+1(3),RefJ (2008) n=q^2+1(4),RefJ (2021) n=q^2+1(5)} \\
        \hline
        3 & $n=t(q+1)+2$ & $2\leq d\leq t+2$, $(p,t,d)\neq(2,q-1,q)$& \cite{RefJ (2018)W. two} \\
        \hline
        4 &  $n=r\frac{q^2-1}{h}+1$, $h|q-1$, $1\leq r\leq h$ & $2\leq d\leq r\frac{q-1}{h}+1$ & \cite{RefJ (2019)W. some} \\
        \hline
        5 & $n=r\frac{q^2-1}{h}$, $h|q-1$, $1\leq r\leq h$ & $2\leq d\leq r\frac{q-1}{h}$ & \cite{RefJ (2019)W. some,RefJ (2017) X.Shi}  \\
        \hline
        6 &$n=r\frac{q^2-1}{h}$, $2\mid h$, $h|q-1$, $1\leq r\leq h$ & $2\leq d\leq (\frac{h}{2}+1)\frac{q-1}{h}+1$ & \cite{RefJ (2017)Lem GRS,RefJ (2014) kai} \\
        \hline
        7 & $n=\frac{q^2-1}{2}$ & $2\leq d\leq q$& \cite{RefJ (2014) kai} \\
        \hline
        8 & $n=q^2$ & $2\leq d\leq q$& \cite{RefJ (2008) n=q^2+1(4)} \\
        \hline
        9 & $n=tq$, $1\leq t\leq q$ & $2\leq d\leq \lfloor  \frac{tq+q-1}{q+1} \rfloor+1$& \cite{RefJ (2018)W. two,RefJ (2008) n=q^2+1(4),RefJ (2015) L.W} \\
        \hline
        10 & $n=r\frac{q^2-1}{h}+1$, $h|q+1$, $1\leq r\leq h$ & $2\leq d\leq (\lfloor \frac{h}{2}\rfloor+1)\frac{q+1}{h}$ & \cite{RefJ (2019)W. some,RefJ (2017) L.Jin} \\
        \hline
        11 &$n=r\frac{q^2-1}{h}+1$, $h|q+1$, $1\leq r\leq h-2$, $2|(h+r)$ & $2\leq d\leq (\frac{h+r}{2})\frac{q+1}{h}$ & \cite{RefJ (2019)W. some}  \\
        \hline
        12 & $n=r\frac{q^2-1}{h}$, $h|q+1$, $1\leq r< h$, $2\nmid (h+r)$, $2\nmid h$ &  $2\leq d\leq (\frac{h+r+1}{2})\frac{q+1}{h}-1$& \cite{RefJ (2017) X.Shi,RefJ (2017) L.Jin}\\
        \hline
        13 & $n=r\frac{q^2-1}{h}$, $h|q+1$, $1\leq r< h$, $2\nmid (h+r)$, $2\mid h$ &  $2\leq d\leq (\frac{h+r-1}{2})\frac{q+1}{h}-1$ & \cite{RefJ (2019)W. some}\\
        \hline
        14 &$n=r\frac{q^2-1}{h}$, $h|q+1$, $1\leq r\leq h-2$, $2|(h+r)$ & $2\leq d\leq (\frac{h+r}{2})\frac{q+1}{h}-1$ & \cite{RefJ (2019)W. some} \\
        \hline
        15 &$n=t(q-1)$, $1\leq t\leq q-1$ & $2\leq d\leq \lfloor \frac{tq-1}{q+1}\rfloor+1$ & \cite{RefJ (2021) application}  \\
        \hline
        16 & $n=\frac{q^2+1}{2}$ &  $2\leq d\leq q$, $d$ odd& \cite{RefJ (2013) n=q^2+1(3)} \\
        \hline
        17 &$n=\frac{q^2+1}{5}$, $q\equiv \pm 3({\rm mod}\ 10)$ & $2\leq d\leq \frac{3q\pm 1}{5}$, $d$ even & \cite{RefJ (2014) kai,RefJ (2015) T.Zhang,RefJ (2016) L.Hu} \\
        \hline
        18 &$n=\frac{q^2+1}{5}$,  $q\equiv \pm 2({\rm mod}\ 10)$ & $2\leq d\leq \frac{3q\mp 1}{5}$, $d$ odd & \cite{RefJ (2016) S.Li} \\
        \hline
        19 & \tabincell{c}{$n=r\frac{q^2-1}{s}+l\frac{q^2-1}{t}-rl\frac{q^2-1}{st}$, odd $s\mid q+1$,\\ even $t\mid q-1$, $r\leq s-1$, $l\leq t$, $rl\frac{q^2-1}{st}\leq q-1$ }
        & $2\leq d\leq \min\{\lfloor \frac{s+r}{2}\rfloor\frac{q+1}{s}-1, (\frac{t}{2}+1)\frac{q-1}{t}\}$ & \cite{RefJ (2020) X.Fang} \\
        \hline
        20 & \tabincell{c}{$n=r\frac{q^2-1}{s}+l\frac{q^2-1}{t}-rl\frac{q^2-1}{st}+1$, odd $s\mid q+1$,\\ even $t\mid q-1$, odd $r\leq s-1$, $l\leq t$, $rl\frac{q^2-1}{st}\leq q-1$ }
        & $2\leq d\leq \min\{ \frac{s+r}{2}\frac{q+1}{s}, (\frac{t}{2}+1)\frac{q-1}{t}\}$ & \cite{RefJ (2020) X.Fang} \\
        \hline
        21 & \tabincell{c}{$n=r\frac{q^2-1}{s}+l\frac{q^2-1}{t}-2rl\frac{q^2-1}{st}$, even $s\mid q+1$,\\ even $t\mid q-1$, $r\leq s-1$, $l\leq t$, $st> 2(q+1)$ }
        & $2\leq d\leq \min\{r\frac{q+1}{s}-1, (\frac{t}{2}+1)\frac{q-1}{t}\}$ & \cite{RefJ (2022) R.Jin} \\
        \hline
        22 & \tabincell{c}{$n=r\frac{q^2-1}{s}+l\frac{q^2-1}{t}-2rl\frac{q^2-1}{st}+1$, even $s\mid q+1$,\\ even $t\mid q-1$, $r\leq s-1$, $l\leq t$, $st> 2(q+1)$ }
        & $2\leq d\leq \min\{r\frac{q+1}{s}-1, (\frac{t}{2}+1)\frac{q-1}{t}\}$ & \cite{RefJ (2022) R.Jin} \\
        \hline

	\end{tabular}}
\end{center}
\end{table}



\begin{table}
\caption{Our new constructions of $[[n,n-2d+2,d]]_q$-quantum MDS codes}
\label{tab:3}
\begin{center}
\resizebox{\textwidth}{15mm}{
	\begin{tabular}{cccc}
		\hline
		Forms of $n$  & Length $n$ & Minimum Distance $d$ & References\\
        \hline
        $q-1\mid n-1$& $n=r\frac{q^2-1}{h}+1$, $h\mid q+1$, $1\leq r\leq h$, $2\nmid r+h$ & $2\leq d\leq (\frac{r+h-1}{2})\frac{q+1}{h}$ & Theorem \ref{th1} \\
        \hline
        $q-1\mid n$ & $n=r\frac{q^2-1}{h}$, $h\mid q+1$, $1\leq r\leq h$, $2\nmid r+h$, $2\mid h$ & $2\leq d\leq (\frac{r+h+1}{2})\frac{q+1}{h}-1$ & Theorem \ref{th2}\\
        \hline
        $q+1\mid n$ & $n=r\frac{q^2-1}{h}$, $2\mid h$, $h\mid q-1$, $1\leq r\leq \frac{h}{2}$ & $2\leq d\leq (\frac{h}{2}+r-1)\frac{q-1}{h}+1$ & Theorem \ref{th3}\\
        \hline
	\end{tabular}}
\end{center}
\end{table}

In \cite{RefJ (2020) X.Fang} and \cite{RefJ (2022) R.Jin}, we can find that many other quantum MDS codes of special length have been constructed.
Its main method is to construct it through two known Hermitian self-orthogonal GRS codes.
We find that they all use the similar construction in Theorem 3.2 in \cite{RefJ (2017)Lem GRS}, so there is one $(\frac{t}{2}+1)\frac{q-1}{t}$ in the distance of the obtained code. 
If we use Theorem \ref{th3}, we can get a larger distance $(\frac{t}{2}+l-1)\frac{q-1}{t}$.
We find that $\bm{a}$ and $\bm{v}$ in this paper are similar to those selected in \cite{RefJ (2020) X.Fang} and \cite{RefJ (2022) R.Jin}, so we use the known and Hermitian self-orthogonal GRS codes constructed in this paper.
Similar to the proof of Theorem 3 and Theorem 4 in \cite{RefJ (2020) X.Fang},
we can immediately draw the following conclusion.

Use Class 14, Corollary \ref{coro 1} and Theorem \ref{th3},
we can get the following quantum MDS code.


\begin{corollary}\label{coro 2}

\begin{itemize}
\item[(1)]
Set $n=r\frac{q^2-1}{s}+l\frac{q^2-1}{t}-rl\frac{q^2-1}{st}$,
where odd $s\mid q+1$, even $t\mid q-1$, $t\geq 2$,
$1\leq r\leq s-1$ and $1\leq l\leq \frac{t}{2}$.
If $q-1 > rl\frac{q^2-1}{st}$,
then for any $1\leq k\leq \min\{\lceil \frac{r+s}{2}\rceil\frac{q+1}{s}-2, (\frac{t}{2}+l-1)\frac{q-1}{t}\}$,
there exists an $[[n,n-2k,k+1]]_q$-quantum MDS codes.
\item[(2)]
Set $n=r\frac{q^2-1}{s}+l\frac{q^2-1}{t}-2rl\frac{q^2-1}{st}$,
where even $s\mid q+1$, even $t\mid q-1$, $t\geq 2$,
$1\leq r\leq s-1$ and $1\leq l\leq t$.
If $q-1 > 2rl\frac{q^2-1}{st}$,
then for any $1\leq k\leq \min\{\lceil \frac{r+s}{2}\rceil\frac{q+1}{s}-2, (\frac{t}{2}+l-1)\frac{q-1}{t}\}$,
there exists an $[[n,n-2k,k+1]]_q$-quantum MDS codes.
\end{itemize}
\end{corollary}


Use Class 11, Theorem \ref{th1} and Theorem \ref{th3},
we can get the following quantum MDS code.

\begin{corollary}\label{coro 3}
\begin{itemize}
\item[(1)]
Set $n=r\frac{q^2-1}{s}+l\frac{q^2-1}{t}-rl\frac{q^2-1}{st}+1$,
where odd $s\mid q+1$, even $t\mid q-1$, $t\geq 2$,
$1\leq r\leq s-1$ and $1\leq l\leq \frac{t}{2}$.
If $q-1 > rl\frac{q^2-1}{st}$,
then for any $1\leq k\leq \min\{\lfloor \frac{r+s}{2}\rfloor\frac{q+1}{s}-1, (\frac{t}{2}+l-1)\frac{q-1}{t}\}$,
there exists an $[[n,n-2k,k+1]]_q$-quantum MDS codes.
\item[(2)]
Set $n=r\frac{q^2-1}{s}+l\frac{q^2-1}{t}-2rl\frac{q^2-1}{st}+1$,
where even $s\mid q+1$, even $t\mid q-1$, $t\geq 2$,
$1\leq r\leq s-1$ and $1\leq l\leq \frac{t}{2}$.
If $q-1 > 2rl\frac{q^2-1}{st}$,
then for any $1\leq k\leq \min\{\lfloor \frac{r+s}{2}\rfloor\frac{q+1}{s}-1, (\frac{t}{2}+l-1)\frac{q-1}{t}\}$,
there exists an $[[n,n-2k,k+1]]_q$-quantum MDS codes.
\end{itemize}
\end{corollary}



\section{Conclusions}\label{sec6}

Quantum MDS codes are an important family of quantum codes.
In this paper,
we construct three new classes of $q$-ary quantum MDS codes by using Hermitian self-orthogonal GRS codes
(see Theorems \ref{th1}, \ref{th2}, \ref{th3}).
The quantum MDS codes constructed in this paper have larger distance parameters than the previous results (see Remarks \ref{remrak 1}, \ref{remrak 2}, \ref{remrak 3}).
The minimum distance of all the $q$-ary quantum MDS codes constructed in this paper can be larger that $q/2+1$.
Moreover, we generalize and improve some results in \cite{RefJ (2019)W. some,RefJ (2017)Lem GRS,RefJ (2017) X.Shi,RefJ (2017) L.Jin}.


\section*{Acknowledgments}

This research was supported by the National Natural Science Foundation of China (Nos.U21A20428 and 12171134).


\section*{References}

