Irreducibilty of $x^2+1$ over $\mathbb{F}_p$ with $p \equiv 3 \mod 4$

Question

$p$ is prime number and $\mathbb{F}_p$ is the $\mathbb{Z}/p$ field.

The polynomial given is $x^2+1$, and I don't understand how you can conclude that when $x^2+1$ is irreducible over $\mathbb{F}_p[x]$, then $p \equiv 3 \mod 4$?

Thanks for the help!

Answer 1

Recall that the unit group $(\mathbb{F}_p)^\times$ is cyclic of order $p-1$. If $p \equiv 1 \pmod{4}$, then $4 \mid p-1$, hence $(\mathbb{F}_p)^\times$ has an element $i$ of order $4$. Then $i^4 = 1$, so $i^2 = -1$ since the order of $i$ is not $2$. Then $x^2 + 1 = (x-i)(x+i)$ is reducible.

(In the remaining case $p=2$, we have $x^2+1 = (x+1)^2$ by the Freshman's Dream.)

Answer 2

I wish to help you, where this solution in number theory.

$x^2+1\equiv0 \pmod p \iff x^2\equiv-1\pmod p \iff (\frac{-1}{p})=1 \iff p\equiv 1\pmod 4 $

$x^2+1\not\equiv 0 \pmod p \iff x^2≢-1\pmod p \iff (\frac{-1}{p})=-1\iff p\equiv 3\pmod 4 $

where $(\frac{-1}{p})$ is Legendre symbol.