解同余式6x3+27x2+17x+20=0(mod30)

Question

解同余式6x3+27x2+17x+20=0(mod30)

Answer 1

x满足6x³+27x²+17x+20 ≡ 0 (mod 30),当且仅当其同时满足:
6x³+27x²+17x+20 ≡ 0 (mod 2),6x³+27x²+17x+20 ≡ 0 (mod 3),6x³+27x²+17x+20 ≡ 0 (mod 5).
也即x²+x ≡ 0 (mod 2),2x+2 ≡ 0 (mod 3),x³+2x²+2x ≡ 0 (mod 5).
x²+x = x(x+1) ≡ 0 (mod 2)恒成立.
2x+2 ≡ 0 (mod 3)等价于x ≡ 2 (mod 3).
x³+2x²+2x ≡ x³-3x²+2x = x(x-1)(x-2) ≡ 0 (mod 5),等价于x ≡ 0,1,2 (mod 5).
于是6x³+27x²+17x+20 ≡ 0 (mod 30)等价于x ≡ 2,5,11 (mod 15).
如果要求x mod 30,则有x ≡ 2,5,11,17,20,26 (mod 30).
WskTuuYtyh 指正.