假设x1,x2是2个不同的根
x1^3 = a(mod b)
x2^3 = a(mod b)
->
x1^3 - x2^3 = 0(mod b)
(x1-x2)(x1^2 + x1x2+x2^2)= 0 (mod b)
而x1,x2<b,b是素数,所以
x1^2 + x1x2+x2^2= 0 (mod b)

现在题目给了你2个根x1,x2
x1^2 + x1x3+x3^2= 0 (mod b)
x2^2 + x2x3+x3^2= 0 (mod b)
->
x1^2-x2^2+x3(x1-x2)=0(mod b)
x3(x1-x2)=(x2^2 - x1^2)(mod b)
x3 = -x1-x2(mod b)
而x1,x2<b
所以ans = (2*b -x1 -x2) % b

• 很详细的证明，懂了。十分感谢！ (0) KOWA 2009-10-24 17:32:25