1) the algorithm of quadratic
平方剩余的计算
2) quadratic remainder
平方剩余
1.
With a recursive sequence,quadratic remainder and congruence,the diophantine equation x2-3y4=97 is proved that it has only positive integral solutions(x,y)=(10,1).
运用递归数列,同余式和平方剩余证明了不定方程x2-3y4=97仅有正整数解(x,y)=(10,1)。
2.
This paper proves that the Diophantine Equation has only positive integral solution with the methods of recursive sequence,congruence and quadratic remainder.
利用一种初等的证明方法,即递推序列、同余式和平方剩余的方法,对不定方程x2-11y4=38的正整数解进行了研究,证明了不定方程x2-11y4=38仅有正整数解(x,y)=(7,1)。
3.
Defined are the characters of quadratic remainder while the arithmetic is provided for choosing X coordinate of base point G .
结合椭圆曲线域参数属性 ,讨论了平方剩余的定义、性质 ,完整地设计出选取基点G的X坐标的算法 。
3) quadratic residue
平方剩余
1.
In this paper,the author has proved that the Diophantine equation x3+64=21y2 has only an integer solution(x,y)=(-4,0),(5,±3) and then gives all integer solution of x3+64=21y2 by using the elementary methods such as recursive sequence,congruent fomula and quadratic residue.
利用递归数列、同余式和平方剩余几种初等方法,证明了不定方程x3+64=21y2仅有整数解(x,y)=(-4,0),(5,±3);给出了x3+64=21y2的全部整数解。
2.
In this paper,the author has proved that the Diophantine equation x3+27=7y2 has only an integer solution(x,y)=(-3,0),(1,±2) and then gives all integer solution of x3+27=7y2 by using the elementary methods such as recursive sequence,congruent fomula and quadratic residu
利用递归数列、同余式和平方剩余几种初等方法,证明了不定方程x3+27=7y2仅有整数解(x,y)=(-3,0),(1,±2);给出了x3+27=7y2的全部整数解。
3.
In this paper the author has proved that the Diophantine equation x3+27=26y2 has only integer solutions(-3,0),(-1,±1),(719,±3781)with the methods of recurrent sequence,congruence and quadratic residue.
利用递归数列、同余式和平方剩余证明了不定方程x3+27=26y2仅有整数解(-3,0),(-1,±1),(719,±3781)。
4) congruence
[英]['kɔŋgruəns] [美]['kɑŋgrʊəns]
平方剩余
1.
In this paper,it has proved that the Diophantine equationh x2-3y4=222 has only positive integral solutions(x,y)=(15,1) with the methods of recursive sequence,quadratic remainder and congruence.
运用递归数列,同余式和平方剩余证明了不定方程x2-3y4=222仅有正整数解(x,y)=(15,1)。
2.
This paper proves that the diophantine equation x2-3y4=118 includes 3 positive integer solutions at least (x,y)=(11,1),(19,3),(650851,613) with the primary methods of recursive sequence,quadratic remainder and congruence.
本文利用一种初等的证明方法,即递归数列,同余式和平方剩余的方法,对一个不定方程x2-3y4=118的正整数解进行了研究。
6) square residue
平方剩余
1.
Let n is a positive integer, and b2(n) is the square residue of n .
设n为任一正整数,b2(n)为n的平方剩余数。
补充资料：平方剩余
假设p是素数，a是整数。 如果存在一个整数x使得x^2≡a(mod p) （即x^2-a可以被p整除）， 那么就称a在p的剩余类中是平方剩余的。
欧拉定理说：a平方剩余当且仅当 a^{(p-1)/2}≡1 (mod p).
在{1,2,...,p-1}中恰好有（p-1）/2 个数是平方剩余的。
拉格朗日符号： 【a/p】=1 (相应的，-1) 如果 a是平方剩余(相应的， 如果 a不是平方剩余）。
高斯著名的二次互反律告诉我们：假设p和q都是素数，则
【q/p】*【p/q】=(-1)^{(p-1)*(q-1)/4}.
说明：补充资料仅用于学习参考，请勿用于其它任何用途。