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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 preliminaries. 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 nondegenerate 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