Optimal p-ary cyclic codes with two zeros

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 p-ary cyclic codes which leads to three explicit constructions. In addition, another class of p-ary optimal cyclic codes are presented.

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 $0 \le t, s, h \le m-1$ and $\gcd (m, t)=\gcd (m, s-h)=1$ such that the congruence

$$\begin{aligned} (p^{t}+1)v\equiv p^{s}+p^{h}\pmod {p^{m}-1} \end{aligned}$$

(5)

has solutions. Let v be a solution of it, then $\mathcal {C}_{p}(1,v)$ is optimal with parameters $[n, n-2m, 4]$ if $\gcd (p^{s}-v, n)=\gcd (v-1, n)=\gcd (p^{h}-v, n)=1$.

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 $n=\frac{2(p^{m}-1)}{p-1}$. 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 $0\le \lambda \le m-1$. 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

$$\begin{aligned} p^{\lambda }(p^{t}+1)v \equiv (p^{t}+1)(p^{k}+1) \pmod {n}. \end{aligned}$$

Since $(p^{t}+1)v\equiv p^{s}+p^{h}\pmod {p^{m}-1}$, then

$$\begin{aligned} p^{\lambda }(p^{s}+p^{h}) \equiv (p^{t}+1)(p^{k}+1) \pmod {n}. \end{aligned}$$

(6)

So we have

$$\begin{aligned} \frac{2(p^{m}-1)}{p-1}| (p^{t+k}+ p^{t}+ p^{k}+1 -p^{\lambda +s}-p^{\lambda +h}) \end{aligned}$$

which implies

$$\begin{aligned} 2(p^{m}-1) | (p^{t+k}+ p^{t}+ p^{k}+1 -p^{\lambda +s}-p^{\lambda +h})(p-1). \end{aligned}$$

In the following, for any integer $\kappa$, let $\kappa _{m}$ be the residue of $\kappa$ modulo m, $0 \le \kappa _{m}\le m-1$. Then by the expression above, we have

$$\begin{aligned} (p^{m}-1) | (p^{(t+k)_{m}}+ p^{t}+ p^{k}+1 -p^{(\lambda +s)_{m}}-p^{(\lambda +h)_{m}})(p-1). \end{aligned}$$

(7)

It is easy to check that $|(p^{(t+k)_{m}}+ p^{t}+ p^{k}+1 -p^{(\lambda +s)_{m}}-p^{(\lambda +h)_{m}})(p-1)|< 2(p^{m}-1)$ and $|(p^{(t+k)_{m}}+ p^{t}+ p^{k}+1 -p^{(\lambda +s)_{m}}-p^{(\lambda +h)_{m}})(p-1)|\ne p^{m}-1$. So if (7) holds, $(p^{(t+k)_{m}}+ p^{t}+ p^{k}+1 -p^{(\lambda +s)_{m}}-p^{(\lambda +h)_{m}})(p-1)=0$, i.e.,

$$\begin{aligned} p^{(t+k)_{m}}+ p^{t}+ p^{k}+1 -p^{(\lambda +s)_{m}}-p^{(\lambda +h)_{m}}=0. \end{aligned}$$

(8)

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.