The root of $x^2+[1]=[0]$ in $\mathbb{Z}_p$

Question

In $\mathbb{Z}_p$, where $p$ is a prime, how many roots of $x^2+[1]=[0]$?

It is equivalent to show $[x^2]=[p-1]$,when p=3,there is non. When p=5, $x=2$,does there exist any rule of it

Answer 1

Modulo a prime $\;p\;$ we have the following:

$$p=2:\;\;x^2+1=(x+1)^2\implies\;\text{there's only one double root in}\;\;\Bbb F_2(=\Bbb Z/2\Bbb Z)\;$$

$$p=3\pmod 4\:\;\;\text{there are no roots, i.e. the polynomial is irreducible as its degree is}\;\le 3$$

$$p=1\pmod 4:\;\;\text{there are two different roots}$$