We determined the linear complexity of a family of p2-periodic binary threshold sequences and a family of p2-periodic binary sequences constructed using the Legendre symbol,both of which are derived from Fermat quotie...We determined the linear complexity of a family of p2-periodic binary threshold sequences and a family of p2-periodic binary sequences constructed using the Legendre symbol,both of which are derived from Fermat quotients modulo an odd prime p.If 2 is a primitive element modulo p2,the linear complexity equals to p2-p or p2-1,which is very close to the period and it is large enough for cryptographic purpose.展开更多
设p,q,r为奇素数,p≡13 mod 24,q≡19 mod 24,(p/q)=-1.利用同余式、平方剩余、递归序列、Legendre符号的性质、Pell方程解的性质等证明了:(A)若r≡5 mod 12,则方程G:x3-1=2pqry2仅有平凡解(x,y)=(1,0);若r≡11 mod 12,则方程G最多有2...设p,q,r为奇素数,p≡13 mod 24,q≡19 mod 24,(p/q)=-1.利用同余式、平方剩余、递归序列、Legendre符号的性质、Pell方程解的性质等证明了:(A)若r≡5 mod 12,则方程G:x3-1=2pqry2仅有平凡解(x,y)=(1,0);若r≡11 mod 12,则方程G最多有2组正整数解.(B)若r≡11 mod 12,则方程H:x3+1=2pqry2仅有平凡解(x,y)=(-1,0);若r≡5 mod 12且(pq/r)=-1,则方程H最多有2组正整数解.展开更多
基金supported by NSFC(No.10901002)supported by NSFC(No.11126173)+2 种基金the NSF of Anhui Province Education Committee(No.KJ2011Z151)the Research Culture Funds of Anhui Normal University(No.2012xmpy009)Anhui Province Natural Science Foundation(No.1208085QA02)
基金the National Natural Science Foundation of China，the Open Funds of State Key Laboratory of Information Security (Chinese Academy of Sciences)，the Program for New Century Excellent Talents in Fujian Province University
文摘We determined the linear complexity of a family of p2-periodic binary threshold sequences and a family of p2-periodic binary sequences constructed using the Legendre symbol,both of which are derived from Fermat quotients modulo an odd prime p.If 2 is a primitive element modulo p2,the linear complexity equals to p2-p or p2-1,which is very close to the period and it is large enough for cryptographic purpose.
文摘设p,q,r为奇素数,p≡13 mod 24,q≡19 mod 24,(p/q)=-1.利用同余式、平方剩余、递归序列、Legendre符号的性质、Pell方程解的性质等证明了:(A)若r≡5 mod 12,则方程G:x3-1=2pqry2仅有平凡解(x,y)=(1,0);若r≡11 mod 12,则方程G最多有2组正整数解.(B)若r≡11 mod 12,则方程H:x3+1=2pqry2仅有平凡解(x,y)=(-1,0);若r≡5 mod 12且(pq/r)=-1,则方程H最多有2组正整数解.
