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1.Let p be an odd prime, integer e ≥ 2, and Z/(p~e) be the integer residue ring modulo p~e.
设p是奇素数,整数e≥2,Z/(p~e)是整数模p~e的剩余类环。 环Z/(p~e)上序列(?)收藏指正
2.A method of C-implementing the perations on a finite field whose radix is a power of prime integer was given by means of shifting division for polynomials, and some main C-functions requred by implementing the operations were listed.
利用多项式的移位相除法,给出了素数琴元有限域运算的一种C实现方法; 并列出了实现素数幂元有限域运算所需要的若干主要的C函数。收藏指正
3.The Relationship between Prime and Composite Number as well Positive Integer being a Condition of Prime Number
素数与复合数的关系、正整数素数的条件收藏指正
4.Suppose G contains no sections isonaorphic to the extension qPn: (Zm,Zp) of an elementary abelian q-group of order qpn by the group Zm:Zp for any prime number q and any integer n with m= (qpn-1)/(qn-1).
其中q~(pn):(Z_m:Z_n))是初等交换q-群q~(pn)被Z_m:Z_p的扩张,而m=(q~(pn)-1)/(q~n-1)。收藏指正
5.In this paper, the relation among the polynomial in several elements and the periodand linear complexity of the clock controlled sequences over the finite field GF(q)(q=p~a, p≥2 is a prime number, a≥1 is a positive integer number)is discussed.
本文讨论了有限域GF(q)(q=p~α,p≥2为素数,α≥1为正整数)上多元多项式与钟控序列的周期和线性复杂度的关系。收藏指正
6.Let r be an odd integer with r > 1, m be an even integer. Let Ur, Vr beintegers satisfying and c was a power of prime, then the equation ax + by = cz had only the positive integer solution (x,y,z) = (2, 2,r).
设r是大干1 的奇数,m是偶数,Ur和Vr是适合 的整数,a=|Vr1,|b=|Ur|,c=m2+1.证明了:当r≡3(mod 4),m≡2(mod 4),m>r/π,且c是素数方幂时,方程az+by=cz仅有正整数解(x,y,z)=(2,2,r).收藏指正
7.5(mod8), c is a power of a prime and the equationax+by=cz has a positive integer solution (x, y, z) = (2,2, r), where r is an odd integer with r>1, then the exceptional solutions (x, y, z) of the equation satisfy x=2 and y=z=1(mod 2).
5(mod 8)且c是素数方幂时,如果ax+by=cz有正整数解(x,y,z)=(2,2,r),其中r是大于1的奇数,则该方程的例外解(x,y,z)都满足x=2以及y(?) z(?)收藏指正
8.The integer 1 is divisible only by itself; it is not a prime.
整数1只能被它本身整除,所以不是素数收藏指正
9.On the integer represented as the product of k prime numbers in arithmetic progression
关于表整数为算术数列中k个素数的乘积收藏指正
10.We prove that if a+b2l-1=c2,b ≡ 5(mod 12) and c is an odd prime with c≡-1(mod b2l),where l is a positive integer,then the equation ax+by=cz has only the positive integer solution(x,y,z)=(1,2l-1,2).
本文证明了:当a+b2 l-1=c2,b≡5(m od 12),c是适合c≡-1(m od b2 l)的奇素数,其中l是正整数时,方程ax+by=cz仅有正整数解(x,y,z)=(1,2 l-1,2).收藏指正
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