多项式f(x)=multiply from i=1 to m(x-a_i)multiply from j=1 to n[(x-b_j)~2+c_j^2]的超级数根  

Super Number Roots for A Kind of Polynomial f(x)=multiply from i=1 to m(x-a_i)multiply from j=1 to n[(x-b_j)~2+c_j^2]

作　　者：邱学绍[1] 王靳辉[2] 王霞[1] 龙洪波[3] 魏萌[2] 

机构地区：[1]郑州轻工业学院信息与计算科学系,河南郑州450002 [2]解放军信息工程大学理学院,河南郑州450001 [3]广东药学院基础部,广东广州510224

出　　处：《重庆大学学报（自然科学版）》2004年第10期64-66,78,共4页Journal of Chongqing University

基　　金：河南省教育厅基金资助项目(2000110005)

摘　　要：用数论中同余和整除的方法证明了:如果a1,a2,Λ,am为互不相同的整数,而(bj,cj)(j=1,2,Λ,n)为互不相同的整数对,且cj≠0(j=1,2,Λ,n),则多项式f(x)在超级数环内有且仅有m2+2mn个根。从而将日格列维奇A·Б·,彼德罗夫Н·Н·的结论推广到了一个一般的情形。The conclusion is drawn that, if integer numbers a_1,a_2,Λ,a_m are different from one anther, even integer numbers (b_j,c_j)(j=1,2,Λ,n) are different from one anther and c_j≠0(j=1,2,Λ,n),then polynomial f(x) has and only has m^2+2mn different super number roots, by the method of exactly divisible and congruence. This conclusion be extended to a widespread case than the conclusion of Jiglevich.

关 键 词：多项式 根 整除 超级数 

分 类 号：O151.1[理学—数学] O156[理学—基础数学]