[求助]二次剩余的求解

awei

fungarwai

3x^3+2x^2+2x+5≡0(mod11)
3(3)^3+2(3)^2+2(3)+5=110
3x^3+2x^2+2x-105≡0(mod11)
(x-3)(3x^2+11x+35)≡0(mod11)
3x^2+2≡0(mod11)
15x^2-1≡0(mod11)
(2x+1)(2x-1)≡0(mod11)
x=5,6

fungarwai

ax^2+bx+c≡0(modk)
直接变换成(2ax+b)^2≡b^2-4ac(modk)
x^2+5x+8≡0(mod11)
(2x+5)^2≡4(mod11)
(2x+3)(2x+7)≡0(mod11)
x=4,2