# Thread: How do I solve this congruence?

1. ## How do I solve this congruence?

$x^3-9x^2+23x-15=0(mod 143)$

2. Originally Posted by Willie_Trombone
$x^3-9x^2+23x-15=0(mod 143)$
You need to solve $x^3 - 9x^2 + 23x - 15\equiv 0(\bmod 11)$ and $x^3 - 9x^2 + 23x - 15\equiv 0 (\bmod 13)$ simultaneously.

In the first congruence we can factor, $(x+6)(x+8)(x+10)\equiv 0(\bmod 11)$ this gives the solutions, $x\equiv 1,3,5(\bmod 11)$. In the second congruence we can factor, $(x+8)(x+10)(x+12)\equiv 0(\bmod 13)$ this gives the solutions, $x\equiv 1,3,5(\bmod 13)$. Now use the Chinese remainder theorem to get possibly 9 solutions up to congruence modulo $11\cdot 13 = 143$.