Irreducibility of $x^2 + 1$ in $\Bbb F_p$

Question

I can show that in field formed by roots of $x^{4 \cdot 2^n + 1} - x$, the polynomial $x^2 + 1$ factors into linear factors. This gives $p = 4 \cdot 2^n+1\equiv 1 \pmod 4$. But how if $p \equiv 3 \pmod 4$, $x^2 + 1$ is irreducible in $\Bbb F_p$?

So far, I think if $x^2 + 1$ is irreducible in $\Bbb F_p$ then $x(x^p-1)$ should not contain a factor $x^2 + 1$. Any value other than $p=4n+1$ does not give this factor. But why $p \equiv 3 \pmod 4$ in particular? Why not $p \equiv 0 \pmod 4$ or $p \equiv 2 \pmod 4$?

Here it is done in example 1.0.3.

Workout: I was too stupid to realize that $p$ was prime. No primes have the form $p = 4n$ or $p = 4n+2 $

Answer 1

The polynomial $x^2+1$ factors into linear polynomials if the field has a fourth root of unity.