New Kloosterman sum identities from the Helleseth-Zinoviev result on $ Z_{4}$-linear Goethals codes

TL;DR

This paper corrects and extends a key theorem on solutions of equations related to $Z_4$-linear Goethals codes, deriving new Kloosterman sum identities and simplifying existing formulas through these insights.

Contribution

It corrects a previous theorem for even $m$, completes its statement, and introduces new Kloosterman sum identities with simpler proofs.

Findings

Corrected Theorem 4 for even $m$

Derived new Kloosterman sum identities

Simplified proofs of existing formulas

Abstract

In the paper of Tor Helleseth and Victor Zinoviev (Designs, Codes and Cryptography, \textbf{17}, 269-288(1999)), the number of solutions of the system of equations from $ Z_{4} $-linear Goethals codes $ G_{4} $ was determined and stated in Theorem 4. We found that Theorem 4 is wrong for $ m $ even. In this note, we complete Theorem 4, and present a series of new Kloosterman sum identities deduced from Theorem 4. Moreover, we show that several previously established formulas on the Kloosterman sum identities can be rediscovered from Theorem 4 with much simpler proofs.

Equations

$$K(a)=\sum _{x\in F^{\ast }}(-1{)}^{\mathrm{Tr}(ax+1/x)}$$

$$\{\begin{array}{ll}x+y+z+u& =1\\ u^2+xy+xz+xu+yz+yu+zu& =b^2\\ x^3+y^3+z^3+u^3& =c\end{array}$$

$$\mu_{2}(b,c)=\frac{1}{6}\bigl{(}q-8+(-1)^{\mathrm{Tr}(b)}(K(k_{1}k_{2})-3)\bigr{)}.$$

$$\mu_{2}(b,c)=\frac{1}{6}\bigl{(}q-2-(-1)^{\mathrm{Tr}(b)}(K(k_{1}k_{2})+3)\bigr{)}.$$

$$K(k_1{k}_2)=K(k_1{k}_2+k_2).$$

$$\{\begin{array}{ll}{k}_1& \in \{c{b}^n,b^2{c}^n,c{b}^n+b^k,b^2{c}^n+c^k\},\\ k_2& \in \{c{b}^n+\sqrt{c},\sqrt{c}{b}^n+c,b^2{c}^n+b,b{c}^n+b^2\}.\end{array}$$

$${\mu }_2(b,c)={\mu }_2(b+1,c)$$

$${\mu }_2(b,c)={\mu }_2(b,c+1).$$

$${\mu }_2(b,c)+{\mu }_2(b+1,c)=\frac{1}{3}\times \{\begin{array}{ll}(2^m-2)& \text{if}\mathrm{Tr}(c)=0,\\ (2^m-8)& \text{if}\mathrm{Tr}(c)=1,\end{array}$$

$${\mu }_2(b,c)+{\mu }_2(b,c+1)=\frac{1}{3}\times \{\begin{array}{ll}(2^m-2)& \text{if}\mathrm{Tr}(c)=1,\\ (2^m-8)& \text{if}\mathrm{Tr}(c)=0.\end{array}$$

$$\begin{split}\mu_{2}(b+1,c)&=\frac{1}{6}\bigl{(}q-8+(-1)^{\mathrm{Tr}(b+1)}(K(k_{1}^{{}^{\prime}}k_{2}^{{}^{\prime}})-3)\bigr{)}\\ &=\frac{1}{6}\bigl{(}q-8+(-1)^{\mathrm{Tr}(b)}(K(k_{1}k_{2}+k_{2})-3)\bigr{)}\ \text{and},\\ \mu_{2}(b,c)&=\frac{1}{6}\bigl{(}q-8+(-1)^{\mathrm{Tr}(b)}(K(k_{1}k_{2})-3)\bigr{)}.\end{split}$$

$$\begin{split}\mu_{2}(b,c+1)&=\frac{1}{6}\bigl{(}q-8+(-1)^{\mathrm{Tr}(b)}(K(k_{1}^{{}^{\prime\prime}}k_{2}^{{}^{\prime\prime}})-3)\bigr{)}\\ &=\frac{1}{6}\bigl{(}q-8+(-1)^{\mathrm{Tr}(b)}(K(k_{1}k_{2}+k_{2})-3)\bigr{)}\ \text{and},\\ \mu_{2}(b,c)&=\frac{1}{6}\bigl{(}q-8+(-1)^{\mathrm{Tr}(b)}(K(k_{1}k_{2})-3)\bigr{)}.\end{split}$$

$$K\biggl{(}\dfrac{(c^{n+1}+c^{n})^{3}(c^{n+1}+1)}{c^{4n}+1}\biggr{)}=K\biggl{(}\dfrac{(c^{n+1}+c^{n})(c^{n+1}+1)^{3}}{c^{4n}+1}\biggr{)}.$$

$$\begin{array}{rl}{k}_2& =b^2+b+c+\sqrt{c}=\frac{c+1}{c^n+1}+\sqrt{\frac{c+1}{c^n+1}}+c+\sqrt{c};\\ k_2^2& =\frac{c^2+1}{c^{2n}+1}+\frac{c+1}{c^n+1}+c^2+c=\frac{(c^{n+1}+c^n)(c^{n+1}+1)}{c^{2n}+1};\\ (k_1{k}_2{)}^2& =\frac{(c^{n+1}+c^n{)}^3(c^{n+1}+1)}{c^{4n}+1};(k_1{k}_2{)}^2+k_2^2=\frac{(c^{n+1}+c^n)(c^{n+1}+1{)}^3}{c^{4n}+1}.\end{array}$$

$$\begin{array}{rl}L(1)& =\frac{(c^2+c{)}^3(c^2+1)}{c^4+1}=\frac{c^3(c+1{)}^5}{(c+1{)}^4}=c^3(c+1),\\ R(1)& =\frac{(c^2+c)(c^2+1{)}^3}{c^4+1}=\frac{c(c+1{)}^7}{(c+1{)}^4}=c(c+1{)}^3.\end{array}$$

$$\begin{array}{rl}L(-1/4)& =\frac{(c^{3/4}+c^{-1/4}{)}^3(c^{3/4}+1)}{c^{-1}+1},R(-1/4)=\frac{(c^{3/4}+c^{-1/4})(c^{3/4}+1{)}^3}{c^{-1}+1}\\ L(-1/4{)}^4& =\frac{(c^3+c^{-1}{)}^3(c^3+1)}{c^{-4}+1}=\frac{(c^4+1{)}^3(c^4+c)}{c^4+1}=(c+1{)}^8(c^4+c),\\ R(-1/4{)}^4& =\frac{(c^3+c^{-1})(c^3+1{)}^3}{c^{-4}+1}=\frac{(c^4+1)(c^4+c{)}^3}{c^4+1}=c^3(c^3+1{)}^3.\end{array}$$

$$K\biggl{(}\dfrac{(b^{n+2}+b^{k}+1)(b^{n+2}+b^{n}+b^{k})^{3}}{b^{4n}+1}\biggr{)}=K\biggl{(}\dfrac{(b^{n+2}+b^{k}+1)^{3}(b^{n+2}+b^{n}+b^{k})}{b^{4n}+1}\biggr{)}.$$

$$\begin{array}{rl}{k}_1^2& =(b^{n+2}+b^n+b^k{)}^2/(b^{2n}+1),\\ k_2^2& =(b^{n+2}+b^k+1)(b^{n+2}+b^n+b^k)/(b^{2n}+1),\\ (k_1{k}_2{)}^2& =(b^{n+2}+b^k+1)(b^{n+2}+b^n+b^k{)}^3/(b^{4n}+1),\\ (k_1{k}_2{)}^2+k_2^2& =(b^{n+2}+b^k+1{)}^3(b^{n+2}+b^n+b^k)/(b^{4n}+1).\end{array}$$

$$K\biggl{(}\dfrac{b^{3}(b^{3}+b+1)^{3}}{(1+b)^{4}}\biggr{)}=K\biggl{(}\dfrac{b^{9}(b^{3}+b+1)}{(1+b)^{4}}\biggr{)}.$$

$$K\biggl{(}\dfrac{(b^{3}+b^{2}+1)(b^{3}+b^{2}+b)^{3}}{(1+b)^{4}}\biggr{)}=K\biggl{(}\dfrac{(b^{3}+b^{2}+1)^{3}(b^{3}+b^{2}+b)}{(1+b)^{4}}\biggr{)}.$$

$$K\biggl{(}\dfrac{(b^{k+1}+b^{2}+b)(b^{k+1}+b^{2}+1)^{3}}{(1+b)^{4}}\biggr{)}=K\biggl{(}\dfrac{(b^{k+1}+b^{2}+b)^{3}(b^{k+1}+b^{2}+1)}{(1+b)^{4}}\biggr{)}.$$

$$K\biggl{(}\dfrac{(b^{n+1}+b^{2})(b^{n+1}+b^{n}+b^{2}+1)^{3}}{b^{4n}+1}\biggr{)}=K\biggl{(}\dfrac{(b^{n+1}+b^{2})^{3}(b^{n+1}+b^{n}+b^{2}+1)}{b^{4n}+1}\biggr{)}.$$

$$K\biggl{(}\dfrac{(b^{3}+b^{2})(b^{3}+b+1)^{3}}{b^{8}+b^{4}+1}\biggr{)}=K\biggl{(}\dfrac{(b^{3}+b^{2})^{3}(b^{3}+b+1)}{b^{8}+b^{4}+1}\biggr{)}.$$

Full text

1 Introduction

Let $m$ be a positive integer, $q=2^{m},F:=\mathbb{F}_{q}$ be the finite field of $q$ elements, $F^{*}:=F\setminus\{0\}$, and $F^{**}:=F\setminus\{0,1\}$. The well known Kloosterman sums carlitz ; charpin are defined by

[TABLE]

where $a\in F^{*}$, and $\mathrm{Tr}(\cdot)$ is the trace function of $F$ over $\mathbb{F}_{2}$.

Let $b,c\in F$. The problem of finding the coset weight distribution of $Z_{4}$-linear Goethals codes $G_{4}$, is transformed into solving the following nonlinear system of equations over $F$ tor-victor :

[TABLE]

where $x,y,z$ and $u$ are pairwise distinct elements of $F$. The number of solutions of (2) (see p.284 of tor-victor ), denoted by $\mu_{2}(b,c)$, is given by Theorem 4 of tor-victor .

We found that Theorem 4 of tor-victor , which gives the explicit evaluation of $\mu_{2}(b,c)$, is wrong for $m$ even. It is obvious that the authors of tor-victor forgot to take account of the fact, that $\mathrm{Tr}(1)=1$ for $m$ odd and 0 for $m$ even, in the last step of the proof of Theorem 4. The following is the correct version of Theorem 4 of tor-victor :

Theorem 1

Let $\mu_{2}(b,c)$ be the number of different 4-tuples $(x,y,z,u)$, where $x,y,z,u$ are pairwise distinct elements of $F$, which are solutions to the system (2) over $F$, where $b,c$ are arbitrary elements of $F$.

1. (1)

If $m$ is odd and $\mathrm{Tr}(c)=1$ or $m$ is even and $\mathrm{Tr}(c)=0$, then

[TABLE] 2. (2)

If $m$ is odd and $\mathrm{Tr}(c)=0$ or $m$ is even and $\mathrm{Tr}(c)=1$, then

[TABLE]

Where $k_{1}=b^{2}+c+1$ and $k_{2}=b^{2}+b+c+\sqrt{c}$.

In this note, we will show that Theorem 4 of tor-victor implies not only a series of new Kloosterman sum identities, but almost all previously discovered formulas on the identities for Kloosterman sums, thanks to the following theorem which is a corollary of that theorem, too:

Theorem 2

Let $b,c\in F,k_{1}=b^{2}+c+1$ and $k_{2}=b^{2}+b+c+\sqrt{c}$. If $k_{1}k_{2}\neq 0$, then

[TABLE]

The idea is to affect a special value to $k_{1}$ or $k_{2}$ and then substitute them into (5) to arrive at a special Kloosterman sum identity. Let $b,c\in F$, and $n,k$ be rational numbers such that $b^{n},b^{k},c^{n},c^{k}$ are elements of $F$. In the sequel, $k_{1}$ and $k_{2}$ are affected to the following values:

[TABLE]

The previously discovered formulas on the identities for Kloosterman sums are stated in the following theorem:

Theorem 3

Let $a\in F^{**}$. Then

1. (1)

K\bigl{(}a^{3}(a+1)\bigr{)}=K\bigl{(}a(a+1)^{3}\bigr{)}* shin-kumar-helleseth ; shin-sung ; tor-victor-ksi1-dm ; tor-victor-ksi12-ff (Helleseth-Zinoviev Formula I).* 2. (2)

K\bigl{(}a^{5}(a+1)\bigr{)}=K\bigl{(}a(a+1)^{5}\bigr{)}* tor-victor-ksi12-ff (Helleseth-Zinoviev Formula II).* 3. (3)

K\bigl{(}a^{8}(a^{4}+a)\bigr{)}=K\bigl{(}(a+1)^{8}(a^{4}+a)\bigr{)}hollmann-xiang * (Hollmann-Xiang Formula).* 4. (4)

K\bigl{(}a/(a+1)^{4}\bigr{)}=K\bigl{(}a^{3}/(a+1)^{4}\bigr{)}* shin-kumar-helleseth (Shin-Kumar-Helleseth Formula).*

Note that Shin-Kumar-Helleseth Formula can be deduced from Helleseth-Zinoviev Formula I by the variable change $a=b/(b+1)$. In this note, we will show that except for Helleseth-Zinoviev Formula II, every formula of Theorem 3 can be obtained from Theorem 4 of tor-victor with help of Theorem 2.

In Section 2, we at first prove Theorem 2 with help of Theorem 1 (Theorem 4 of tor-victor ) and two lemmas of tor-victor-ksi1-dm , treat in details some cases of (6) for $k_{1}$ and $k_{2}$ which bring us new Kloosterman sum identities, prove Helleseth-Zinoviev Formula I and Hollmann-Xiang Formula, and finally generalize Shin-Kumar-Helleseth Formula.

2 New identities for Kloosterman sums

2.1 Proof of Theorem 2

Before proving Theorem 2, we need several preliminary lemmas:

Lemma 1** (carlitz )**

Let $a\in F^{*}$. Then, $K(a)=K(a^{2})$.

Lemma 2** (lidl-nieder )**

Let $a,b\in F$. Then,

1. (1)

$\mathrm{Tr}(a)=\mathrm{Tr}(a^{2}),\mathrm{Tr}(a+b)=\mathrm{Tr}(a)+\mathrm{Tr}(b)$. 2. (2)

$\mathrm{Tr}(1)=1$* if $m$ is odd, and $\mathrm{Tr}(1)=0$ if $m$ is even.*

The symmetry of the solutions of the nonlinear system (2) is characterized by the following two lemmas:

Lemma 3

(tor-victor-ksi1-dm*, *, Lemma 10)** Let $b,c$ be any elements of $F$, where $F$ has order $2^{m}$. Let $m$ be even. Let $\mu_{2}(b,c)$ denote the number of solutions to system (2). Then the following symmetry conditions are valid:

[TABLE]

and

[TABLE]

Lemma 4

(tor-victor-ksi1-dm*, *, Lemma 11)** Let $b,c$ be any elements of $F$, where $F$ has order $2^{m}$. Let $m$ be odd. Let $\mu_{2}(b,c)$ denote the number of solutions to system (2). Then the following symmetry conditions are valid:

[TABLE]

and

[TABLE]

Now, we are ready to prove Theorem 2. The idea of proof is to substitute the formulas for $\mu_{2}(b,c)$ from Theorem 1 into the formulas of Lemma 3 for $m$ even, and into the formulas of Lemma 4 for $m$ odd.

Proof (of Theorem 2)

1. (I)

Case for $m$ even.

1. (a)

Subcase that $\mathrm{Tr}(c)=0$.

Note that $\mathrm{Tr}(1)=0$ for $m$ even by Lemma 2, and $\mathrm{Tr}(b+1)=\mathrm{Tr}(b)+\mathrm{Tr}(1)=\mathrm{Tr}(b)$. Let $k_{1}^{{}^{\prime}}:=(b+1)^{2}+c+1,k_{2}^{{}^{\prime}}:=(b+1)^{2}+(b+1)+c+\sqrt{c}$. It is clear that $k_{1}^{{}^{\prime}}=k_{1}+1,k_{2}^{{}^{\prime}}=k_{2}$ and $k_{1}^{{}^{\prime}}k_{2}^{{}^{\prime}}=k_{1}k_{2}+k_{2}$, where $k_{1}=b^{2}+c+1$ and $k_{2}=b^{2}+b+c+\sqrt{c}$. From (3) of Theorem 1, we have

[TABLE]

Theorem 2 follows from (7) of Lemma 3 for this subcase.

Let $k_{1}^{{}^{\prime\prime}}:=b^{2}+(c+1)+1,k_{2}^{{}^{\prime\prime}}:=b^{2}+b+(c+1)+\sqrt{c+1}$. It is clear that $k_{1}^{{}^{\prime\prime}}=k_{1}+1,k_{2}^{{}^{\prime\prime}}=k_{2}$ and $k_{1}^{{}^{\prime\prime}}k_{2}^{{}^{\prime\prime}}=k_{1}k_{2}+k_{2}$. Since $\mathrm{Tr}(c+1)=\mathrm{Tr}(c)=0$, from (3) of Theorem 1, we obtain

[TABLE]

Again, Theorem 2 follows from (8) of Lemma 3 for this subcase. 2. (b)

Subcase that $\mathrm{Tr}(c)=1$.

For this subcase, we use (4) of Theorem 1 and (7) or (8) of Lemma 3 to prove Theorem 2, and omit the details due to limited space. 2. (II)

Case for $m$ odd.

For this case, we use (9) or (10) of Lemma 4 and Theorem 1 to prove Theorem 2, which is similar to the case for $m$ even and omitted.

∎

2.2 New Kloosterman sum identities and the proof of Theorem 3

Theorem 4

Let $c\in F$, and $n$ be a rational number such that $c^{n}$ is an element of $F$. If $c^{4n}+1\neq 0$, then

[TABLE]

Proof

Let $k_{1}=b^{2}+c+1=b^{2}c^{n}$. Then, $b^{2}=(c+1)/(c^{n}+1)$, and $k_{1}=(c^{n+1}+c^{n})/(c^{n}+1)$. Further, we can obtain

[TABLE]

The actual theorem follows from Lemma 1 and Theorem 2.

∎

Let $L(n):=(c^{n+1}+c^{n})^{3}(c^{n+1}+1)/(c^{4n}+1),R(n):=(c^{n+1}+c^{n})(c^{n+1}+1)^{3}/(c^{4n}+1)$, then, $K(L(n))=K(R(n))$. We now prove Helleseth-Zinoviev Formula I by Theorem 4.

Proof (of Helleseth-Zinoviev Formula I)

[TABLE]

Helleseth-Zinoviev Formula I follows from the case $K(L(1))=K(R(1))$ of Theorem 4.

∎

Corollary 1

Let $c\in F$ such that the rational functions in $c$ occurring in the following formulas are valid. Then,

1. (1)

K\bigl{(}c^{6}(c^{2}+c+1)/(c+1)^{4}\bigr{)}=K\bigl{(}c^{2}(c^{2}+c+1)^{3}/(c+1)^{4}\bigr{)}, 2. (2)

K\bigl{(}c^{9}(c+1)^{3}/(c^{8}+c^{4}+1)\bigr{)}=K\bigl{(}c^{3}(c+1)^{9}/(c^{8}+c^{4}+1)\bigr{)}, 3. (3)

K\bigl{(}(c+1)^{8}(c^{4}+c)\bigr{)}=K\bigl{(}c^{3}(c^{3}+1)^{3}\bigr{)}, 4. (4)

K\bigl{(}(c^{11}+c^{3})(c^{5}+1)\bigr{)}=K\bigl{(}c(c^{5}+1)^{3}\bigr{)}, 5. (5)

K\bigl{(}(c+1)^{20}(c^{8}+c)\bigr{)}=K\bigl{(}(c+1)^{4}(c^{8}+c)^{3}\bigr{)}.

Proof

The first formula of Corollary 1 corresponds to the case $K(L(2))=K(R(2))$ of Theorem 4, and the second formula to $K(L(3))=K(R(3))$, which the proofs are analogous to that of Helleseth-Zinoviev Formula I and omitted. We now prove the third formula:

[TABLE]

The third formula follows from $K(L(-1/4))=K(R(-1/4))$ of Theorem 4 and Lemma 1. The fourth formula arises from the case $K(L(1/4))=K(R(1/4))$ and the fifth formula from $K(L(-1/8))=K(R(-1/8))$, which the proofs are similar to that of the third formula and omitted.

∎

Proof (of Hollmann-Xiang Formula)

From Helleseth-Zinoviev Formula I, we obtain $K(c^{3}(c^{3}+1)^{3})=K(c^{9}(c^{3}+1))=K(c^{8}(c^{4}+c))$. From the third formula of Corollary 1, we have $K((c+1)^{8}(c^{4}+c))=K(c^{3}(c^{3}+1)^{3})=K(c^{8}(c^{4}+c))$.

∎

Remark that Shin-Kumar-Helleseth Formula is the specific case $K(L(-2))=K(R(-2))$ of Theorem 4, which we omit the proof.

2.3 New Kloosterman sum identities and the generalization of Shin-Kumar-Helleseth Formula

In this subsection, we establish several identities for Kloosterman sums which generalize Shin-Kumar-Helleseth Formula.

Theorem 5

Let $b\in F$, and $n,k$ be rational numbers such that $b^{n},b^{k}$ are elements of $F$. If $b^{4n}+1\neq 0$, then,

[TABLE]

Proof

Let $k_{1}=b^{2}+c+1=cb^{n}+b^{k}$, then $c=(b^{k}+b^{2}+1)/(b^{n}+1)$. After an elementary algebraic computation over $\mathbb{F}_{2}$, we get

[TABLE]

The actual theorem follows from Lemma 1 and Theorem 2.

∎

We obtain the following corollaries by affecting particular values to $n,k$ in Theorem 5, which the proofs are similar to Corollary 1 and omitted.

Corollary 2

Set $n=1,k=0$ in Theorem 5. Then

[TABLE]

Corollary 3

Set $n=1,k=2$ in Theorem 5. Then

[TABLE]

Corollary 4

Set $n=-1$ in Theorem 5. Then

[TABLE]

We can obtain analogous results as Theorem 4 and Theorem 5 if we affect one of remainder values from (5) to $k_{1}$ or $k_{2}$. For instance, set $k_{2}:=b^{2}+b+c+\sqrt{c}=b^{n}\sqrt{c}+c$, we obtain

Theorem 6

Let $b\in F$, and $n$ be rational number such that $b^{n}$ is an element of $F$. If $b^{4n}+1\neq 0$, then,

[TABLE]

Proof

The proof for the actual theorem is similar to that of Theorem 4 and Theorem 5 and omitted.

∎

Set $n=3$ in the formula of Theorem 6, we obtain an interesting corollary which the proof is similar to Corollary 1 and omitted:

Corollary 5

Let $b\in F$ such that $b^{8}+b^{4}+1\neq 0$. Then,

[TABLE]

3 Conclusion

In this note, we obtained a series of new Kloosterman sum identities and rediscovered several previously found formulas with much simpler proof from Theorem 4 of tor-victor . The exception is that the formula, $K(a^{5}(1+a))=K(a(1+a)^{5})$, we believe, could not be deduced from Theorem 4 of tor-victor and Theorem 2.