Get the solution to "Quadratic over Z7 no solutions?3x2+x+1=0"

High School
Algebra
Algebra I
Quadratic function and equation

Oxinailelpels3t14

Answered question

2022-04-02

Quadratic over Z7 no solutions?

3 x^{2} + x + 1 = 0

Answer & Explanation

Ruben Gibson

Beginner2022-04-03Added 9 answers

Step 1
Mod 7
$5 (3 x^{2} + x + 1) \equiv x^{2} + 5 x + 5 \equiv {(x - 1)}^{2} - 3 \pm \mod 7$
and $(3 ∣ 7) = - 1; 3$ is not a square mod 7. The squares are $0, 1, 2, 4 \pm \mod 7$
Mod 5
$2 (3 x^{2} + x + 1) \equiv x^{2} + 2 x + 2 \equiv {(x + 1)}^{2} + 1 \equiv {(x + 1)}^{2} - 4 \pm \mod 5$
and 4 has square roots $2, - 2 \equiv 2, 3 \pm \mod 5,$ these are $x \equiv 1, 2 \pm \mod 5$

shvatismop1rj

Beginner2022-04-04Added 10 answers

You are correct that there is no solution, but not for the reason you're giving here. The quadratic equation only holds if we're working in

R

C

or something similar (like a subring of one of those). In the case of

\frac{Z}{7 Z}

there are only seven elements to consider, so you can simply try plugging each of them into

3 x^{2} + x + 1

and verify that none of them is a zero.

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