Solving quadratic equations over finite fields

Question

How does one solve the equations $x^2-3=0$ and $x^2+x-1=0$ in $\Bbb{F}_7[x] / (x^2-5)$?

Not sure how do do it, I know there is a solution for both, but I don't see how to get there.

Thanks.

Answer 1

Let $\alpha\in\Bbb{F}_7[x]/(x^2-5)$ such that $\alpha^2=5$. Then for any $c\in\Bbb{F}_7$ we have $$(c\alpha)^2=c^2\alpha^2=5c^2,$$ so for which $c\in\Bbb{F}_7$ is $c\alpha$ a root of $x^2-3$?

More generally every element of $k$ can be written as $a\alpha+b$ for some $a,b\in\Bbb{F}_7$. Plugging this expression into your polynomials yields two equations for $a$ and $b$, which aren't hard to solve in the quadratic case.

Answer 2

Note that $x^2-3=(x^2-5)+2$ and in this field is equal to 2. Idem with the other polynomial

Answer 3

Let $\omega$ a root of $x^-5=0$. A root $\;\xi$ of $x^2-3=0$ must satisfy $N(\xi)=3$. If we try $\xi=a\omega$, we have $\;N(a\omega)=a^2N(\omega)=5a^2$, whence the equation $5a^2=3$. As $5^{-1}=3$, this implies $a^2=3^2$, whence $a=\pm 3$. So the solutions are $$\xi=\pm 3\omega.$$