# Need Little Help

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• March 30th 2010, 06:57 AM
kap
Need Little Help
Hello Everybody,

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

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

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).

Thanks for your help.
• March 30th 2010, 07:30 AM
novice
Quote:

Originally Posted by kap
Hello Everybody,

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

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

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).

Thanks for your help.

$5x^2 + 4x + 1 \not \equiv 0$
• March 30th 2010, 08:31 AM
kap
Could you please explain your answer to me a little bit.
• March 30th 2010, 09:48 AM
novice
Quote:

Originally Posted by kap
Could you please explain your answer to me a little bit.

You have no real solution but complex.
• March 30th 2010, 11:11 AM
Soroban
Hello, kap!

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

Quote:

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

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

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

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

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

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