Weight Distribution of A p-ary Cyclic Code

0.18 MB·English

by  Xiangyong Zeng

Weight Distribution of A p-ary Cyclic Code
Xiangyong Zeng1, Lei Hu2, Wenfeng Jiang2, Qin Yue3, Xiwang Cao3
1. Faculty of Mathematics and Computer Science, HubeiUniversity,Wuhan,430062, China
E-mail: [email protected]
2. StateKey Laboratory of Information Security,Graduate School of
Chinese Academy of Sciences, Beijing, 100049, China
E-mail:{hu,wfjiang}@is.ac.cn
3. Department of Math, School of Sciences
Nanjing Universityof Aeronautics and Astronautics, Nanjing 210016, China
9
0 E-mail:{yueqin,xwcao}@nuaa.edu.cn
0
2 Abstract: For an odd prime p and two positive integers n 3 and k with n being
≥ gcd(n,k)
n odd, the paper determines the weight distribution of a p-ary cyclic code over F with
p
a C
J nonzeros α−1, α−(pk+1) and α−(p3k+1), where α is a primitive element of Fpn.

6
Keywords: Cyclic code, exponentialsum, quadraticform, weightdistribution, linearized
1
polynomial

]
T
I 1 Introduction
.
s
c
[ Nonlinearfunctionsoverfinitefieldshaveusefulapplicationsincodingtheoryandcryptography[2,15].

1 Some linear codes having good properties [3, 5, 7, 13, 15, 17] were constructed from highly nonlinear
v
functions [4, 6, 8, 16].
1
9 Let q be a power of a prime p, and Fqn be a finite field with qn elements. A p-ary [m,l] linear
3 code is a linear subspace of Fm with dimension l. The Hamming weight of a codeword c c c is
2 p 1 2··· m
. the number of nonzero ci for 1 i m. In this paper, we study a [pn 1,3n] cyclic code given by
1 ≤ ≤ − C
0
09 C =nc(ǫ,γ,δ)=(cid:16)Trn1(ǫx+γxpk+1+δxp3k+1)(cid:17)x∈F∗pn |ǫ, γ, δ ∈Fpno,
:
v where k is a positive integer and Trn is the trace function from F to F . This code is constructed
1 pn p
Xi from the function Trn(ǫx+γxpk+1 +δxp3k+1), which can have high nonlinearity if either γ or δ is
1
r nonzero. It is easy to know that α−1, α−(pk+1), α−(p3k+1) and their Fp-conjugates are all nonzeros of
a
the cyclic code , where α is a primitive element of F [15].
pn
C
Inthispaperweassumethatpand n arebothodd,andwedeterminetheweightdistribution
gcd(k,n)
of the code . To this goal, we will focus on determining the ranks of a class of quadratic forms and
C
calculating two classes of exponential sums. The ranks of quadratic forms are determined through
finding the number of solutions to a class of linearized polynomials
L (z)=γzpk +γp−kzp−k +δzp3k +δp−3kzp−3k
γ,δ
over the field F . By applying the theory of quadratic forms, two classes of exponential sums are
pn
evaluated and the weight distribution of the cyclic code is determined. Throughout the paper, we
C
set d=gcd(k,n), s= n and n 3.
gcd(k,n) ≥
The remainder of this paper is organized as follows. Section 2 gives some definitions and prelimi-
naries. Section 3 studies the rank distribution of a class of quadratic forms. Section 4 determines the
weight distribution of .
C
1
2 Preliminaries
Identifying F with the n-dimensional F -vector space Fn, a univariable polynomial f(x) defined on
qn q q
F can be regarded as an n-variable polynomial on F . The former is called a quadratic form if the
qn q
latter is a homogeneous polynomial of degree two:
f(x , ,x )= a x x ,
1 n jk j k
···
1 j k n
≤X≤ ≤
here we use a basis of Fnq over Fq and identify x ∈ Fqn with a vector (x1,···,xn) ∈ Fnq. The rank of
the quadratic form f(x) is defined as the codimension of the F -vector space
q
W = w F f(x+w)=f(x) for all x F , (1)
qn qn
{ ∈ | ∈ }
denoted by rank(f). Then W =qn rank(f).
−
| |
Foraquadraticformf(x),thereexistsasymmetricmatrixAsuchthatf(x)=XTAX,whereX is
written as a column vector and its transpose is XT =(x ,x , ,x ) Fn. The determinant det(f)
1 2 ··· n ∈ q
of f(x) is defined to be the determinant of A, and f(x) is nondegenerate if det(f)=0. By Theorem
6
6.21 of [14], there exists a nonsingular matrix B such that BTAB is a diagonal matrix. Making a
nonsingular linear substitution X =BY with YT =(y ,y , ,y ), one has
1 2 n
···
r
f(x)=YTBTABY = a y2 (2)
i i
i=1
X
where r n is the rank of f(x) and a ,a , ,a F .
≤ 1 2 ··· r ∈ ∗q
Let m be a positive factor of the integer n. The trace function Trn from F to F is defined by
m pn pm
n/m 1
Trn(x)= − xpmi, x F .
m ∈ pn
i=0
X
Let e(y)=e2π√−1Trn1(y)/p and ζp =e2π√−1/p.
The following lemmas will be used throughout this paper.
Lemma 1 (Theorems 5.33 and 5.15 of [14]): Let F be a finite field with q =pl, where p is
q
an odd prime. Let η be the quadratic character of F . Then for a=0,
q
6
Trl(ax2) η(a)( 1)l−1p2l, if p 1(mod4),
ζ 1 = − ≡
xX∈Fq p ( η(a)(−1)l−1(√−1)lp2l, if p≡3(mod4).
Lemma 2 (Theorems 6.26 and 6.27 of [14]): Let q be an odd prime power, f be a nonde-
generate quadratic form over F . Define a function υ(ρ) over F as υ(0) = q 1 and υ(ρ) = 1 for
ql q − −
ρ F . Then for ρ F the number of solutions to the equation f(x , ,x )=ρ is
∈ ∗q ∈ q 1 ··· l
ql−1+ql−21η ( 1)l−21ρ det(f)
− ·
(cid:16) (cid:17)
for odd l, and
ql−1+υ(ρ)ql−22η ( 1)2ldet(f)
−
(cid:16) (cid:17)
for even l.
Lemma 3: (i) (Theorems 5.15 of [14]) Let ζ be a complex primitive p-th root of unity and η
p
be the quadratic character of F . Then
p
p 1
− η(ρ)ζpρ = (−1)p−21p.
ρ=1 q
X
2
(ii) Let the function υ(ρ) over F be defined by υ(0)=p 1 and υ(ρ)= 1 for ρ F . Then
p − − ∈ ∗p
p 1
−
υ(ρ)ζρ =p.
p
ρ=1
X
3 Rank distribution of a class of quadratic forms
This section studies the rank distribution of the quadratic forms Trn(γxpk+1+δxp3k+1) for nonzero
d
γ or δ.
To determine the distribution, we define a related exponential sum
S(ǫ,γ,δ)= e ǫx+γxpk+1+δxp3k+1 , ǫ, γ, δ Fpn. (3)
x∈Fpn (cid:16) (cid:17) ∈
P
Then the possible values of the ranks are measured by evaluating the exponential sum S(ǫ,γ,δ). For
discussion of the exponential sum of a generalquadratic form, please refer to references [10] and [12].
Proposition 1: For odd s and δ F , the exponential sum S(ǫ,γ,δ) satisfies
∈ ∗pn
S(ǫ,γ,δ) =0, pn2, pn+2d, or pn2+d.
| |
Proof: Notice that
S(ǫ,γ,δ)2
| |
= S(ǫ,γ,δ)S(ǫ,γ,δ)
= e ǫx γxpk+1 δxp3k+1 e ǫy+γypk+1+δyp3k+1
x∈Fpn (cid:16)− − − (cid:17)y∈Fpn (cid:16) (cid:17) (4)
= P e ǫz+γzpk+1+δzp3k+1+γPzpkx+γzxpk +δzp3kx+δzxp3k by y =x+z
x,z∈Fpn (cid:16) (cid:17)
= P e ǫz+γzpk+1+δzp3k+1 e(xL (z))
γ,δ
z∈Fpn (cid:16) (cid:17)x∈Fpn
P P
where
L (z)=γzpk +γp−kzp−k +δzp3k +δp−3kzp−3k
γ,δ
is a linearized polynomial in z. Let V be the set of all roots of L (z) = 0. (By abuse of notation,
γ,δ
we use V to denote the set in despite of it depending on γ and δ.) Thus, V is an F -vector space.
pd
By (4), we have
S(ǫ,γ,δ)2 =pn e ǫz+γzpk+1+δzp3k+1 . (5)
| |
(cid:12)z V (cid:12)
(cid:12)P∈ (cid:16) (cid:17)(cid:12)
Let (cid:12) (cid:12)
(cid:12) (cid:12)
Φ (x)=γxpk+1+δxp3k+1 δp−kxp2k+p−k +δp−2kxpk+p−2k. (6)
γ,δ
−
By (6), we have
Trn(Φ (z))=Trn γzpk+1+δzp3k+1 (7)
1 γ,δ 1
(cid:16) (cid:17)
and
Φ (z)+Φ (z)p−k =zL (z). (8)
γ,δ γ,δ γ,δ
If z V, then by (8),
∈
Φ (z)pk = Φ (z). (9)
γ,δ γ,δ
−
3
′
Bygcd(k,n)=d,thereisanintegerk suchthatkk d(modn)andhence,Φ (z)pd =Φ (z)pkk =
′ ′ γ,δ γ,δ
≡
( 1)k′Φ (z), where the last equality is derived from (9). If k is even, Φ (z)pd =Φ (z) and then
γ,δ ′ γ,δ γ,δ
−
Φ (z)pk =Φ (z), which together with (9) again implies Φ (z)=0. If k is odd, then
γ,δ γ,δ γ,δ ′
Φ (z)pd = Φ (z). (10)
γ,δ γ,δ
−
By the property Trn(Φ (z))=Trn(Φ (z)p−d) of trace function and (8), we have
d γ,δ d γ,δ
0 = Trn(zL (z))
d γ,δ
= Trn(Φ (z))+Trn(Φ (z)p−k)
d γ,δ d γ,δ
= 2Trn(Φ (z))
d γ,δ
= 2(Φ (z)+Φ (z)pd + +Φ (z)p(s−1)d)
γ,δ γ,δ γ,δ
···
= 2Φ (z),
γ,δ
wherethelastequalsignholdsdueto(10)andsbeingodd. ThisimpliesΦ (z)=0andTrn(γzpk+1+
γ,δ 1
δzp3k+1) = Trn(Φ (z)) = 0 by (7). Conversely, if Φ (z) = 0, then by (8), L (z) = 0 and
1 γ,δ γ,δ γ,δ
Trn(Φ (z)) = 0. Therefore, z V if and only if Φ (z) = 0. Further, in this case Trn(γzpk+1 +
1 γ,δ ∈ γ,δ 1
δzp3k+1)=0. Thus, by (5),
S(ǫ,γ,δ) = pn ζTrn1(ǫz) . (11)
p
| | vuut (cid:12)(cid:12)(cid:12)zX∈V (cid:12)(cid:12)(cid:12)
Since V is an F -vector space, we can assume V =(cid:12)pdm for an i(cid:12)nteger m 0.
pd | | (cid:12) (cid:12) ≥
If m 3, then Φ (z)= 0 has at least p3d solutions. For a fixed z V 0 and for any z V,
γ,δ 0
≥ ∈ \{ } ∈
we have Φ (z)=Φ (z )=0 and Φ (z+z )=0 since z+z is also in the vector space V. Thus,
γ,δ γ,δ 0 γ,δ 0 0
the equation
(z+z )(z Φ (z)+zΦ (z )) zz Φ (z+z )=0 (12)
0 0 γ,δ γ,δ 0 0 γ,δ 0
−
has at least p3d solutions.
By (6), Equation (12) becomes
δp−2k(zpkz zzpk)(zp−2kz zzp−2k) δp−k(zp2kz zzp2k)(zp−kz zzp−k)=0, (13)
0− 0 0− 0 − 0− 0 0− 0
which has at least p3d roots on variable z. Let z =wz , then
0
δp−2kzpk+p−2k+2(wpk w)(wp−2k w) δp−kzp2k+p−k+2(wp2k w)(wp−k w)=0.
0 − − − 0 − −
Let u=wp−k w, the above equation can be rewritten as
−
δp−2kzpk+p−2kupk(up−k +u)+δp−kzp2k+p−k(up2k +upk)u=0,
− 0 0
which has at least p2d roots on u since wp−k w=u has at most pd roots on w for each u. Define
−
Ψ (x)=δp−2kzpk+p−2kxpk(xp−k +x) δp−kzp2k+p−k(xp2k +xpk)x.
δ,z0 0 − 0
Similarly, for each nonzero root u of Ψ (u)=0, the equation
0 δ,z0
(u+u )(u Ψ (u)+uΨ (u )) uu Ψ (u+u )=0
0 0 δ,z0 δ,z0 0 − 0 z,z0 0
has at least p2d solutions on u. By the definition of Ψ (x), the above equation is equivalent to
δ,z0
δp−2kzpk+p−2k(upku uupk)(up−ku uup−k)=0.
0 0− 0 0− 0
This shows that u=vu where v F . Consequently, for eachgiven u =0, the above equation has
0 ∈ pd 0 6
at most pd roots. This gives a contradiction and then m 2.
≤
4
This finishes the proof. (cid:4)
Remark 1: Thepossibleranksofsomequadraticformscanbedeterminedbydirectlycalculating
the number of the solutions to their related linearized polynomials [18, 11]. The number of the roots
to the linearized polynomial L (z) in Proposition 1 is discussed by studying that of an associated
γ,δ
nonlinear polynomial. The method was first presented to study a linear mapping over a finite field
of characteristic2 [9] and further used to discuss some triple error correcting binary codes with BCH
parameters [1]. In Proposition 1, we used this method to the cases of odd characteristic.
From Proposition 1, the value of the dimension m determines the rank of the following quadratic
form.
Corollary 1: For odd s and δ F , the quadratic form
∈ ∗pn
Ω (x)=Trn γxpk+1+δxp3k+1
γ,δ d
(cid:16) (cid:17)
has rank s, s 1, or s 2.
− −
When there is exactly one nonzero element in γ,δ , the rank of Ω (x) can be determined by
γ,δ
{ }
directly calculating the number of solutions to L (z)=0.
γ,δ
Proposition 2: For odd s and γ,δ F , the quadratic forms Ω (x) = Trn γxpk+1 and
∈ ∗pn γ,0 d
Ω (x)=Trn δxp3k+1 have rank s. (cid:16) (cid:17)
0,δ d
Proof: We(cid:16)onlygiv(cid:17)etheproofofrank(Ω )=s,andanothercasecanbeproveninasimilarway.
0,δ
It is sufficient to determine the number of solutions to δzp3k +δp−3kzp−3k =0. This equation has
nonzero solutions if and only if (δzp3k+1)p3k 1 = 1. If the latter holds, then gcd(p3k 1,pn 1)=
−
− − −
(pgcd(3k,n)−1)|pn2−1. Let s1 = gcd(n3k,n) and then
pn 1= pgcd(3k,n) 1 p(s1−1)gcd(3k,n)+p(s1−2)gcd(3k,n)+ +pgcd(3k,n)+1 .
− − ···
(cid:16) (cid:17)(cid:16) (cid:17)
Noticethats1isafactoroftheoddintegers. Asaconsequence,p(s1−1)gcd(3k,n)+p(s1−2)gcd(3k,n)+ +
···
pgcd(3k,n)+1 is odd and pn2−1 can not be divided by pgcd(3k,n)−1. Thus, −1 is not (pgcd(3k,n)−1)-th
powerof any elementin F andthen δzp3k+δp−3kzp−3k =0 has only the zerosolution. This finishes
∗pn
the proof. (cid:4)
Remark 2: For γ,δ F , Trd(Ω (x)) and Trd(Ω (x)) are p-ary bent functions.
∈ ∗pn 1 γ,0 1 0,δ
To study the rank distribution of the quadratic form Ω , for i 0,1,2 ,we define
γ,δ
∈{ }
R = (γ,δ) rank(Ω )=s i, (γ,δ) F F (0,0) . (14)
i γ,δ pn pn
| − ∈ × \{ }
n o
Lemma 4: |R2|= (pn−dp−2d1)(1pn−1).
−
Proof: If(γ,δ) R ,thenγδ =0byPropositions1and2,andV isatwo-dimensionalvectorspace
2
∈ 6
overFpd. Let{v1,v0}beabasisofV overFpd. Then,v1v0−1 ∈/ Fpd and(v1p4kv0p2k−v1p2kv0p4k)(v1pkv0p2k−
vp2kvpk)=0. By (13),
1 0 6
δpk−1 = (v(pv41pk3kvpv20pk2k−vpv21pk2vkpv40pk3k)()v(pvk1vv0pp22kk−vv1pp22kkvv0p)k)
1 vp02kv−pk1 vpk0vp2k 1 p0k−−1 1 0
= 1 0 − 1 0 .
(cid:18)(v1p2kv0−v1v0p2k)pk+1(cid:19)
Thus,
vp2kvpk vpkvp2k
δ =λ 1 0 − 1 0 (15)
(vp2kv v vp2k)pk+1
1 0− 1 0
5
for an element λ F . Since Φ (v ) = γvpk+1+δvp3k+1 δp−kvp2k+p−k +δp−2kvpk+p−2k = 0, we
∈ ∗pd γ,δ 1 1 1 − 1 1
have
γ =−δv1p3k−pk +δp−kv1p2k+p−k−pk−1−δp−2kv1p−2k−1. (16)
From (15) and (16), γ and δ are uniquely determined by v ,v and λ. Further, there are exactly
1 0
pd 1 pairs (γ,δ) corresponding to a given pair (v ,v ).
1 0
−
Onthe other hand, for anyv F andβ / F , let v =βv . If δ and γ are defined by (15) and
0 ∈ ∗pn ∈ pd 1 0
(16), respectively, then Φ (v )=0. In the sequel, we will prove v L (v )=0.
γ,δ 1 0 γ,δ 0
From (15), we have
δvp3k+1 = λ(βp2k −βpk). (17)
0 (β2k β)pk+1
−
Then
(δvp3k+1)(β βp2k)= λ(βp2k −βpk) and (δvp3k+1)(βp2k β)pk = λ(βp2k −βpk).
0 − (β β2k)pk 0 − βp2k β
− −
Thus, by (16) and (17),
v L (v ) = γvpk+1+(γvpk+1)p−k +δvp3k+1+(δvp3k+1)p−3k
0 γ,δ 0 0 0 0 0
= (δvp3k+1)βp3k pk +(δvp3k+1)p−kβp2k+p−k pk 1 (δvp3k+1)p−2kβp−2k 1 +
− 0 − 0 − − − 0 −
(cid:16) (δvp3k+1)p−kβp2k 1+(δvp3k+1)p−2kβpk+p−2k 1 p−k (δvp3k+1)p−3kβp−(cid:17)3k p−k
− 0 − 0 − − − 0 −
(cid:16)+δvp3k+1+(δvp3k+1)p−3k (cid:17)
0 0
= (δvp3k+1)(1 βp3k pk)+(δvp3k+1)p−k(βp2k+p−k pk 1 βp2k 1)
0 − − 0 − − − −
+(δvp3k+1)p−2k(βpk+p−2k 1 p−k βp−2k 1)+(δvp3k+1)p−3k(1 βp−3k p−k)
0 − − − − 0 − −
= β pk(δvp3k+1)(β βp2k)pk +βp2k pk 1(δvp3k+1)p−k(β βp2k)p−k
− 0 − − − 0 −
+βp−2k 1 p−k(δvp3k+1)p−2k(βp2k β)p−k +β p−k(δvp3k+1)p−3k(βp2k β)p−3k
− − 0 − − 0 −
= λ βpβ2k−βpp2kk−1 + βp2k−β1−ββp2pk2k−pk + βp−2kβ−p−βkp−−2βkp−2k−1 + 1−ββp−β2pk−−2pk−k
(cid:18) − − − − (cid:19)
= λ −β−β1(ββ−p2βkp2k) + β−β1(ββ−pβ−p2−k2k)
(cid:18) − − (cid:19)
= λ β 1+β 1
− −
−
= 0.
(cid:0) (cid:1)
This shows L (v ) = 0, and hence Φ (v ) = 0. Thus v ,v is a basis of the F -vector space
γ,δ 0 γ,δ 0 { 1 0} pd
consisting of all solutions to Φ (x)=0.
γ,δ
There are totally ((pp2nd−11))((ppn2d−ppdd)) two-dimensional vector subspaces of Fpn over Fpd, thus,
− −
(pn 1)(pn pd) (pn 1)(pn d 1)
R =(pd 1) − − = − − − .
| 2| − × (p2d 1)(p2d pd) p2d 1
− − −
(cid:4)
The values of S(0,γ,δ) can be discussed in terms of rank(Ω ) as below.
γ,δ
For (γ,δ) R , rank(Ω ) = s and by a nonsingular linear substitution as in (2), Ω (x) =
0 γ,δ γ,δ
∈
6
s
h y2, where h F and (y ,y , ,y ) Fn . Then
i i i ∈ ∗pd 1 2 ··· s ∈ pd
i=1
P
S(0,γ,δ) = ζTrd1(Ωγ,δ(x))
p
x∈Fpn
= P ζpTrd1(h1y12+h2y22+···+hsys2)
y1,y2,···,ys∈Fpd
= s P ζTrd1(hiyi2)
p
i=1yi∈Fpd
Q P s
η(hi)( 1)d−1pd2 , p 1(mod4) (18)
− ≡
=  i=1 by Lemma1
s (cid:16) (cid:17)
 Qη(hi)(−1)d−1(√−1)dpd2 , p≡3(mod4)
i=1
 Q (cid:16)( 1)d−1η( s hi)pn2, (cid:17) p 1(mod4)
− ≡
=  i=1
s
 (−1)d−1η( hQi)(√−1)npn2, p≡3(mod4).
i=1
Similarly, we have  Q
S(0,γ,δ) = ζTrd1(h1y12+h2y22+···+hs−1ys2−1)
p
y1,y2,···,ys∈Fpd
= pds−1P ζTrd1(hiyi2)
p
i=1yi∈Fpd (19)
Q Ps 1
η( − hi)pn+2d, p 1(mod4)
≡
=  i=1
 η(s−1hi)Q(√ 1)n−dpn+2d, p 3(mod4)
− ≡
i=1
for (γ,δ) R , and  Q
1
∈
s 2
( 1)d−1η( − hi)pn2+d, p 1(mod4)
− ≡
S(0,γ,δ)= i=1 (20)
 ( 1)d−1η(s−2hi)Q(√ 1)n−2dpn2+d, p 3(mod4)
− − ≡
i=1
for (γ,δ) R .  Q
2
∈
From (18), (19) and (20), for (γ,δ) R with i 0,2 , we have
i
∈ ∈{ }
S(0,γ,δ)= ( 1)pd2−1θipn+2id, θi 1 , (21)
− ∈{± }
q
and for (γ,δ) R ,
1
∈
n+d
S(0,γ,δ)=θ1p 2 , θ1 1 . (22)
∈{± }
Two subsets R of R for i 0,1,2 are defined as
i,j i
∈{ }
R = (γ,δ) R θ =j (23)
i,j i i
∈ |
n o
where j = 1.
±
The following result can be obtained based on Equalities (18), (20) and the fact that s is odd.
Lemma 5: For i 0,2 , R = R .
i,1 i, 1
∈{ } | | | − |
Proof: For i 0,2 , let (γ,δ) R and u F such that η(u)= 1. Then
∈{ } ∈ i ∈ ∗pd −
Ω (x)=Trn(uγxpk+1+uδxp3k+1)=uTrn(γxpk+1+δxp3k+1)=uΩ (x).
uγ,uδ d d γ,δ
7
By (18) and (20),
S(0,uγ,uδ)=η(u)s iS(0,γ,δ)=( 1)s iS(0,γ,δ)= S(0,γ,δ).
− −
− −
The above equality shows that for j 1, 1 , if (γ,δ) R , then (uγ,uδ) R . This finishes
i,j i, j
∈ { − } ∈ ∈ −
the proof. (cid:4)
Proposition 3: (i) S(0,γ,δ)=p2n.
γ,δ∈Fpn
P
p2n(2pn 1), pd 1(mod4),
(ii) S(0,γ,δ)2 = − ≡
γ,δ∈Fpn ( p2n, pd ≡3(mod4).
P
Proof: The result in (i) can be directly verified, and we only give the proof of (ii).
Notice that
S(0,γ,δ)2
γ,δ∈Fpn
= P ζTrn1“γ(xpk+1+ypk+1)” ζTrn1“δ(xp3k+1+yp3k+1)”
p p
x,y∈Fpnγ∈Fpn δ∈Fpn
= p2nPT , P P
1
| |
where T consists of all solutions (x,y) F F to the equation xpk+1+ypk+1 =0 since xpk+1+
1 pn pn
∈ ×
ypk+1 =0 implies xp3k+1+yp3k+1 =0.
If xy =0, (x,y)=(0,0) is the only solution of xpk+1+ypk+1 =0.
Ifxy =0,wehave(x)pk+1 = 1. If this equationhassolution,say x =αj fora primitiveelement
6 y − y
α of Fpn and 1 ≤ j < pn−1, then j(pk +1)≡ pn2−1(modpn−1). This equality holds if and only if
gcd(pk+1,pn 1) pn−1. Noticethatgcd(pk+1,pn 1)=2andsisodd. Consequently,(x)pk+1 = 1
− | 2 − y −
has solutions if andonly if pn 1(mod4). Further,in this case the number ofsolutions is equalto 2.
≡
Thus, xpk+1+ypk+1 =0 has 2(pn 1) solutions if pn 1(mod4), and no solution if pn 3(mod4).
− ≡ ≡
The above analysis and the equality pn pd(mod4) finish the proof. (cid:4)
≡
With the above preparations, the rank distribution of Ω (x) can be determined as below.
γ,δ
Proposition 4: (i) For i 0,1,2 and j 1, 1 ,R satisfies
i,j
∈{ } ∈{ − }
 ||||RRRR1012,,,,−111|||1===|=||(RRp(n02p−,,n−−d−+11d||p−==np2−2n2d−(2(p)pdn(np)−+(2npd(2−pn−d2−1−1d))p1−,(n)p12+,n)(d−p−21dp)−n.1+)p2d)(pn−1),
(ii)Forodds,when(γ,δ)runsthroughF F (0,0) ,the rankdistributionofthe quadratic
pn pn
× \{ }
form Ω (x) is given as follows:
γ,δ
s, (pn+2d−pn+pd2−dpn1+p2d)(pn−1) times,
 s 1, pn d(pn 1)−times,
−
 s−−2, (pn−dp−2d1)(−1pn−1) times.
−
Proof: By Propositions1, 2, 3, Lemmas 4, and 5, we have the following identities of parameters
8
R with i 0,1,2 and j 1 :
i,j
| | ∈{ } ∈{± }
R + R + R =p2n 1,
0 1 2
| | | | | | −
 pn+2d(R1,1 R1, 1 )+pn = S(0,γ,δ),
 (−1)p|d2−1pn||−R0||+−pn|+d|R1|+(γ−,δP1∈)Fppdn2−1pn+2d|R2|+p2n =γ,δ∈FpnS(0,γ,δ)2,
R = R , P
0,1 0, 1
This finishes t||Rhe2,1p||ro=of||.R2,−−1||= (pn−2d(p−21d)−(p1n)−1). (cid:4)
By (14), (18)-(23) and Proposition 4, an immediate result is given as below.
Corollary 2: For odd s, when (γ,δ) runs through F F (0,0) , the exponential sum
pn pn
× \ { }
S(0,γ,δ) defined in (3) has the following distribution:
(−1)pd2−1pn2, (pn+2d−pn2+(dp−2dpn1+)p2d)(pn−1) times,
 −qq(−−pp1n)n+2+p2ddd,2−,1pn2, (((pppnnn−−+dd2+−d−pppnnn22−−222+dd(dp))−2((ppdpnn−−n−−1+)11p))2dtt)ii(mmpnee−ss,1,) times,
 −qq(−(−11))pdp2d−2−11ppnn+2+22d2d,, ((ppnn−−22dd((pp−−2211dd))−−((pp11nn))−−11)) ttiimmeess,.
4 Weight distribution of the p-ary code
C
This section studies the distribution of the exponential sum S(ǫ,γ,δ) and the weight distribution of
the code .
C
If either γ or δ is nonzero, then Trd(Ω (x)) is also a quadratic form. By (1), Propositions 1, 2
1 γ,δ
and Corollary 1, rank(Trd(Ω ))=d rank(Ω )=n, n d, or n 2d.
1 γ,δ · γ,δ − −
For ρ F , let N (ρ) denote the number of solutions to Trd(Ω (x))+Trn(ǫx)=ρ. Then, (3)
∈ p ǫ,γ,δ 1 γ,δ 1
can be written as
p 1
S(ǫ,γ,δ)= − N (ρ)ζρ. (24)
ǫ,γ,δ p
ρ=0
P
n
Let α1,α2, ,αn be a basis of Fpn over Fp, and ǫ = ǫiαi with ǫi Fp. Then the matrix
{ ··· } ∈
i=1
C =(Trn(α α )) is nonsingular. Let DT =(ǫ ,ǫ , ,Pǫ ) Fn and X =BY be defined as in
1 i j 1 i,j n 1 2 ··· n ∈ p
Section 2, then Tr≤n(ǫ≤x)=DTCX. Denote DTCB =(b ,b , ,b ), and we have
1 1 2 ··· n
n n
Trd(Ω (x))+Trn(ǫx)=YTBTABY +DTCBY = a y2+ b y . (25)
1 γ,δ 1 i i i i
i=1 i=1
P P
By application of the quadratic form theory, the distribution of S(ǫ,γ,δ) is discussed and the
weight distribution of is determined.
C
Theorem 1: For two positive integers n and k with d=gcd(n,k), if s is odd, then when (ǫ,γ,δ)
9
runsthroughF F F ,theexponentialsumS(ǫ,γ,δ)definedin(3)hasthefollowingdistribution
pn pn pn
× ×
pn, 1 time,
 0, (pn 1)(p2n−d p2n−2d+p2n−3d pn−2d+1) times,
− − −
for odd d,an−dq−qq(q−((−−(−1−p1)1p)n1p)n+−p22)p+−22d1−p2d1−ζp21ζppρ1pnpρ,pn+n22,+2n22ζd2ζpdρζpρ,ζpρ,pρ,, ((((((ppppppnnnnnn−−−−−−21d21ddd−−+−−−11ηη11−+((+−−−υυηηρ(ρ((ρρ())−−p)p)ppρnρnnn))−−22pp−−2211n2n(2(dd))pp−−−−((22p2p22222dd2222(n(ndd))−−pp−−++((22pp111212ddnn))dd))−−−−−−((pp11ddppn))n+−nn−−++ppddddnn−−−−−−2211pdpd))nn(())pp((++ppnnppnn−−22−−dd11)1))1)(())ppttnnttiimm−−iimm11ee))eessss,tt,,iimmeess,,
pn, 1 time,
 0, (pn 1)(p2n−d p2n−2d+p2n−3d pn−2d+1) times,
− − −
for even d, where ρ−−pp−=pppnnpn+n+n2220+2n+222dζd,2dζpζdρ1ζpζpρρ,ζp,ρpρ,,pρ,,, ,p((((((ppppppnnnnnn−−−−−−11d221ddd−−−+,−−11υυη11−+((−+ρρυυi))sυυ((ppρρ((tρnρn))hpp))−−22ppenn2222nn−−))((q−−((22ppddppu22−−222222nndddda2((2−−++−2−2ppd))222222((11rddddpp))))a−−(−(−nnppt−−pp11nninn)d)dc−−+++−ddcddpp−−h−−nn11app−−22))rnn((ddapp++))nnc((pp−t−pp22enndd11r−)−)))((o1p1ptt))nnfiimm−−Ftt1i1ieepmm))ss,aeettnisismm,,deeυss,,(0)= p 1, υ(ρ) = 1
··· − − −
for ρ F .
∈ ∗p
Proof: Since s is odd, the integer n d is always even. If d is odd, then n and n 2d are both
− −
odd. The proof in this case is divided into the following subcases.
(i) For (γ,δ)=(0,0), S(ǫ,0,0)=0 for ǫ=0, and pn for ǫ=0.
6
(ii) For (γ,δ)=(0,0), the discussion is divided into three subcases.
6
n
In the case of (γ,δ) R , for 1 i n, let y =z bi . Then (a y2+b y )=ρ is equivalent
∈ 0 ≤ ≤ i i− 2ai i i i i
i=1
to n a z2 =λ +ρ, where λ = n b2i . Let ∆ = n a , thenPLemma 2 implies
i i ǫ,γ,δ ǫ,γ,δ 4ai 0 i
i=1 i=1 i=1
P P Q
Nǫ,γ,δ(ρ)=pn−1+pn−21η ( 1)n−21(λǫ,γ,δ+ρ)∆0 . (26)
−
(cid:16) (cid:17)
Notice that the matrix CB in (25) is nonsingular. As a consequence, (b ,b , ,b ) runs through
1 2 n
···
Fnp as ǫ runs through Fpn. λǫ,γ,δ is also a quadratic form with n variables bi for 1≤i≤n. Again by
Lemma 2, as ǫ runs through F ,
pn
λǫ,γ,δ = n 4ba2ii =ρ′ occurring pn−1+pn−21η (−1)n−21ρ′∆0 times (27)
i=1
(cid:16) (cid:17)
P
n n
for each ρ F since η (4n a ) 1 =η( a ).
′ p i − i
∈
(cid:18) i=1 (cid:19) i=1
By (24), (26) and LemmaQ3 (i), we have Q
S(ǫ,γ,δ)=η ( 1)n−21∆0 pn2 ( 1)p−21ζp−λǫ,γ,δ. (28)
− −
(cid:16) (cid:17) q
10