Solving quadratic equation in GF(2)

Question

I have the following equation: $$x^2+x+1 = 0$$

And I should try to solve it in GF(2), GF(3) and GF(5) by trial and error. GF(3) was pretty easy, but is there even an answer for GF(2) and GF(5)? As far as I see it, you can't solve the equation in these two fields. Am I right?

Answer 1

Note that $GF(p)$ is isomorphic to $\mathbb{Z}_p$ for any prime $p$. In your case you can go through all elements and see whether it's a root of $f(x)=x^2+x+1$.

In $\mathbb{Z}_2$ we have $f(0)=f(1)=1$ so $f$ is irreducible over $\mathbb{Z}_2$.

In $\mathbb{Z}_3$ we have $f(1)=0$ and $x^2+x+1=(x-1)^2$.

In $\mathbb{Z}_5$ we have $f(0)=f(4)=1$, $f(1)=f(3)=3$, $f(2)=2$, so $f$ is irreducible over $\mathbb{Z}_5$.

Answer 2

The equation $x^2 + x + 1 = 0$ admits $1$ as a root iff $3 =0$, i.e. the characteristic is 3, i.e. your finite field is of the form $\mathbf F_{3^m}$. Otherwise, finding the solutions $x\neq 1$ is the same as finding a primitive cubic root of $1$ in your finite field $\mathbf F_q$ , where $q = p^n$ is a power of a prime $p$. But it is known that $\mathbf F_q^{*}$ is a cyclic group, so a necessary and sufficient condition is that $3$ divides $p^n - 1$.

NB: I take this opportunity to dispute the use, instead of $\mathbf Z /p\mathbf Z$, of the notation $\mathbf Z_p$, which is reserved for the $p$-adic integers (Bourbaki's custom).