What are the solutions of $x^2+1=0$ over the finite field $F$ with ${\rm char}~F\neq 2$?

Question

What are the solutions of $x^2+1=0$ over the finite field $F$ with ${\rm char}~F\neq 2$? As we know that the solutions of $x^2+1=0$ are $\pm i$ over the complex field, then how about when over the finite field $F$ with ${\rm char}~F\neq 2$? I have no idea, if we should consider field extension? Can someone give me some prompts? Thanks in advance!

Answer 1

Let $\Bbb{F}$ be a finite field with odd characteristic $\operatorname{char}(\Bbb{F}) = p > 2$. $\Bbb{F}$ contains the field with $p$ elements $\Bbb{F}_p$ and is equal to $\Bbb{F}_{p^k}$ for some $k \geq 1$. There are two relevant cases:

• $p \equiv 1 \pmod 4$: In this case, $x^2 + 1$ factors over $\Bbb{F}_p$ (this is a more or less classical result) and therefore also factors over $\Bbb{F}$. So, $\Bbb{F}$ already contains solutions to $x^2 + 1 = 0$.

• $p \equiv 3 \pmod 4$: In this case, $x^2 + 1$ is irreducible over $\Bbb{F}_p$ (for the same reason as above). The splitting field of $x^2 + 1$ over $\Bbb{F}_p$ is isomorphic to $\Bbb{F}_{p^2}$, as it has dimension $2$ as a vector space over $\Bbb{F}_p$ (we can consider it as the quotient $\Bbb{F}_p[x]/(x^2 + 1)$). Since $\Bbb{F}_{p^m} \subseteq \Bbb{F}_{p^n}$ iff $m$ divides $n$, this means our original field $\Bbb{F} = \Bbb{F}_{p^k}$ either contains a solution of $x^2 + 1 = 0$ (when $k$ is even) or has a degree $2$ extension $\Bbb{F}(i) \cong \Bbb{F}[x]/(x^2 + 1)$ (when $k$ is odd).

Notice that telling you when $\Bbb{F} = \Bbb{F}_{p^k}$ contains a root of $x^2 + 1 = 0$ is different than telling you what that root is in terms of $p^k$. Indeed, if I were able to do this for every $p$ and every $k$, I would in particular be able to give a positive or negative answer to Landau's fourth problem: is $n^2 + 1$ prime for infinitely many $n$? (I would just check if the solution to $x^2 + 1 = 0$ mod $p$, $p \equiv 1 \pmod 4$, was equal to $\sqrt{p-1}$ infinitely often.)