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$p$ 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!

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    Have you known the famous theorem that a prime is a sum of two squares if and only if $p\equiv 1\pmod 4$? – 2017-02-04
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    @HanulJeon An odd prime. $p=2$ is not equivalent to $1$ mod $4$ and $2=1^2+1^2$. – 2017-02-04

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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.)

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    @learnmore I've added a link to the Wikipedia page on the Freshman's Dream. Of course, it's easily seen in this one case: $(x+1)^2 = x^2 + 2x + 1 = x^2 + 1$. – 2017-02-04
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I wish to help you, where this solution in number theory.

$x^2+1=0 (mod p) \Leftrightarrow x^2\equiv-1(mod p) \Leftrightarrow (\frac{-1}{p})=1 \Leftrightarrow p\equiv 1(mod 4) $

$x^2+1\ne 0 (mod p) \Leftrightarrow x^2≢-1(mod p) \Leftrightarrow (\frac{-1}{p})=-1 \Leftrightarrow p\equiv 3(mod 4) $

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