Abstract
As a subclass of linear codes, cyclic codes have efficient encoding and decoding algorithms, so they are widely used in many areas such as consumer electronics, data storage systems and communication systems. In this paper, we give a general construction of optimal pary cyclic codes which leads to three explicit constructions. In addition, another class of pary optimal cyclic codes are presented.
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Acknowledgements
The authors are very grateful to the anonymous reviewer for the comments which improved the presentation and quality of this paper.
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Y. Liu: is supported by the National Natural Science Foundation of China (No. 12001475 and 11701498), Natural Science Foundation of Jiangsu Province (No. BK20201059) and the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (No. 19KJB120014). X. Cao is supported by the National Natural Science Foundation of China (No. 11771007)
Appendix
Appendix
According to the definition of the cyclic code \(\mathcal {C}_{p}(u,v)\), \(\mathcal {C}_{p}(1,i)\) and \(\mathcal {C}_{p}(1,j)\) are the same code if and only if \(m_{i}(x)=m_{j}(x)\), i.e., \(C_{i}=C_{j}\). Based on this fact, in the following, we will prove the codes in Sect. 3 are new. To this end, we need to show that the codes in this paper and thoes in [16] are different. To give a clear explanation, we list the main result of [16] in the following.
Theorem 4.1
([16]) Let \(m> 2\) be an integer. Let t, s, h be integers with \(0 \le t, s, h \le m1\) and \(\gcd (m, t)=\gcd (m, sh)=1\) such that the congruence
has solutions. Let v be a solution of it, then \(\mathcal {C}_{p}(1,v)\) is optimal with parameters \([n, n2m, 4]\) if \(\gcd (p^{s}v, n)=\gcd (v1, n)=\gcd (p^{h}v, n)=1\).
First, we will prove the codes in Sect. 3.1 are new. By the discussion above, we need to prove \(p^{k}+1\) and v which is an even solution of (5) are not in different cyclotomic cosets modulo n where \(n=\frac{2(p^{m}1)}{p1}\). Therefore, we need to prove \(p^{\lambda }v \not \equiv p^{k}+1 \pmod {n}\) for any integer \(\lambda \). Since m is the multiplicative order of p modulo n, we can assume that \(0\le \lambda \le m1\). Suppose on the contrary that there is an integer \(\lambda \) such that \(p^{\lambda } v \equiv p^{k}+1 \pmod {n}\). Notice that \(\gcd (p^{t}+1, p^{m}1)=2\), then \(\gcd (p^{t}+1, n)=2\). Therefore, when m is odd, \(p^{\lambda }v \equiv p^{k}+1 \pmod {n}\) is equivalent to
Since \((p^{t}+1)v\equiv p^{s}+p^{h}\pmod {p^{m}1}\), then
So we have
which implies
In the following, for any integer \(\kappa \), let \(\kappa _{m}\) be the residue of \(\kappa \) modulo m, \(0 \le \kappa _{m}\le m1\). Then by the expression above, we have
It is easy to check that \((p^{(t+k)_{m}}+ p^{t}+ p^{k}+1 p^{(\lambda +s)_{m}}p^{(\lambda +h)_{m}})(p1)< 2(p^{m}1)\) and \((p^{(t+k)_{m}}+ p^{t}+ p^{k}+1 p^{(\lambda +s)_{m}}p^{(\lambda +h)_{m}})(p1)\ne p^{m}1 \). So if (7) holds, \((p^{(t+k)_{m}}+ p^{t}+ p^{k}+1 p^{(\lambda +s)_{m}}p^{(\lambda +h)_{m}})(p1)=0\), i.e.,
So at least one of \((t+k)_{m}\), t, k, \((\lambda +s)_{m}\), \((\lambda +h)_{m}\) is zero. Otherwise, from (8), \(1\equiv 0 \pmod {p}\) which is a contradiction. So we discuss these cases one by one.

(1)
If \((t+k)_{m}=0\), then (8) becomes
$$\begin{aligned} p^{t}+ p^{k}p^{(\lambda +s)_{m}}p^{(\lambda +h)_{m}}+2=0. \end{aligned}$$Similarly as above, at least one of t, k, \((\lambda +s)_{m}\), \((\lambda +h)_{m}\) is zero. If \(t=0\), then \(k=0\) and then \(p^{(\lambda +s)_{m}}+p^{(\lambda +h)_{m}}=4\) which is impossible when \(p>3\). So \(t\ne 0\). Similarly, we can obtain \(k\ne 0\). If \((\lambda +s)_{m}=0\), \(t\ne 0\), \(k\ne 0\), then (8) becomes \(p^{t}+ p^{k}p^{(\lambda +h)_{m}}+1=0\). Therefore, \((\lambda +h)_{m}=0\) and then \(p^{t}+ p^{k}=0\) which is impossible.

(2)
If one of t or k is zero but not both, (8) becomes
$$\begin{aligned} 2p^{k}p^{(\lambda +s)_{m}}p^{(\lambda +h)_{m}}+2=0 \end{aligned}$$or
$$\begin{aligned} 2p^{t}p^{(\lambda +s)_{m}}p^{(\lambda +h)_{m}}+2=0. \end{aligned}$$According to similar discussion as above, this case can not happen.

(3)
If \((t+k)_{m}\ne 0\), \(t\ne 0\), \(k\ne 0\) and \((\lambda +s)_{m}=0\), then (8) becomes
$$\begin{aligned} p^{(t+k)_{m}}+ p^{t}+ p^{k} p^{(\lambda +h)_{m}}=0 \end{aligned}$$(9)which implies \((\lambda +h)_{m}>0\). Otherwise, \(1\equiv 0 \pmod {p}\) which is impossible. By simplification, (9) becomes \(p^{\alpha _{1}}+p^{\alpha _{2}}+p^{\alpha _{3}}1=0\), \(p^{\alpha _{1}}+p^{\alpha _{2}}p^{\alpha _{3}}+1=0\), \(p^{\alpha _{1}}p^{\alpha _{2}}+2=0\), \(p^{\alpha _{1}}+p^{\alpha _{2}}=0\) or \(p^{\alpha _{1}}+1=0\) which are all impossible when \(p>3\) where \(\alpha _{1}, \alpha _{2}, \alpha _{3}\) are all positive integers.

(4)
Similarly as above, we can have \((\lambda +h)_{m}\ne 0\).
Summarizing the discussion above, there is a contradiction which shows that \(\mathcal {C}_{p}(1,p^{k}+1)\) is different from the codes in [16].
Furthermore, we can prove the codes in Sect. 3.2 are new by the same method. Since the proof is similar and lengthy, we omit the details.
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Liu, Y., Cao, X. Optimal pary cyclic codes with two zeros. AAECC (2021). https://doi.org/10.1007/s00200021004895
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Keywords
 Cyclic code
 optimal code
 Sphere packing bound
Mathematics Subject Classification
 94B15
 11T71