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]

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.