✔ 最佳答案
1+x+x^2+…+x^r=0 的解是 1^[1/(r+1)], r 個複數根, x =/= 1.-----(1) , 因(1-x)(1+x+x^2+…+x^r )=1-x^(r+1). so
1+x+x^2+…+x^m=0 的解是 1^[1/(m+1)], m 個複數根, x =/= 1-----(2) ; and
1+x^n+x^(2n)+…+x^(mn)=0 的解是 1^[1/(mn+n)], mn+ n 個複數根, 但x^n =/= 1 -----(3)
Therefore,” 1+x^n+x^(2n)+....+x^(mn)能被1+x+x^2+....+x^m整除 ” iff “ (2)之根能被(3)之根完全覆蓋”
解集合就是 {(m,n), with (m+1, n)=1 , (m+1 and n 互質)}, including (m,n)=(m,1) the trivial one.