1. ## Need Little Help

Hello Everybody,

I need help in solving this question. The equation is:

$\displaystyle 5x^2 + 4x + 1 \equiv 0$

The question is to:

determine all the solutions in Z(17) and in Z(20).
[In Z(17): 9 and 14, in Z(20): none ]

NOTE::
The Z(17) is suppose to be the whole numbers symbol with subscript 17 but i don't know how to do it in Latex. The same applies to Z(20).

2. Originally Posted by kap
Hello Everybody,

I need help in solving this question. The equation is:

$\displaystyle 5x^2 + 4x + 1 \equiv 0$

The question is to:

determine all the solutions in Z(17) and in Z(20).
[In Z(17): 9 and 14, in Z(20): none ]

NOTE::
The Z(17) is suppose to be the whole numbers symbol with subscript 17 but i don't know how to do it in Latex. The same applies to Z(20).

$\displaystyle 5x^2 + 4x + 1 \not \equiv 0$

4. Originally Posted by kap
You have no real solution but complex.

5. Hello, kap!

I never studied modulo arithmetic formally,
. . but I just "invented" a method . . .

Solve: .$\displaystyle 5x^2 + 4x + 1 \:\equiv\: 0 \text{ (mod 17)}$

We have: .$\displaystyle 5x^2 + 4x + 1 \:\equiv\:0\text{ (mod 17)}$

Multiply by 7: .$\displaystyle 35x^2 + 28x + 7 \:\equiv\:0\text{ (mod 17)}$

. . . . . . . . . . . $\displaystyle x^2 + 11x + 24 \:\equiv\:0\text{ (mod 17)}$

. . . . . . . . . . . $\displaystyle (x + 8)(x+3) \:\equiv\:0\text{ (mod 17)}$

Therefore: .$\displaystyle \begin{Bmatrix}\;x &\equiv& -8 & \equiv & 9 & \text{(mod 17)}\;\\ x &\equiv& -3 & \equiv & 14 & \text{(mod 17)}\end{Bmatrix}$