How to solve polynomial Congruence

Question

$21x^5 + 10x^3 + 14x \equiv 0 \pmod{3}$.

I want to use flt and crt to solve this but I am not sure where to start. Do you have any tips?

Mod 3 you have that 21 = 0, 10 = 1 and 14 = -1. And you only have to consider x = 0, 1 and -1. — Count Iblis, Mar 14 '15 at 15:36

score 2 · Answer 1 · answered Mar 14 '15 at 15:36

2

The given equation is equivalent mod $3$ to (using that $\Bbb Z_3$ is a field)

$$x^3+2x=0\iff x(x^2+2)=0\iff x=0\lor x=\pm 1$$

answered Mar 14 '15 at 15:36

score 2 · Answer 2 · answered Mar 14 '15 at 15:46

2

$$21x^5+10x^3+14x\equiv x^3-x\pmod3$$

But $x^3-x=(x-1)x(x+1)$ being product of three consecutive integer is always divisible by $3$

Hence $21x^5+10x^3+14x\equiv0\pmod3$ is an examaple of Identical Congruence

answered Mar 14 '15 at 15:46

lab bhattacharjee

correct me if I'm wrong but i can I say, x^3 - x = 0 (mod 3) is same as x-x = 0 (mod 3) by flt, and so 0 = 0 (mod 3), hence, identical congruence? – Joe Mar 14 '15 at 16:00
@Joe, Identical congruence can be defined as congruence equation that is satisfied by all possible values of $x$. Now you judge yourself – lab bhattacharjee Mar 14 '15 at 16:02

2 Answers2