How to find the roots in the field $\mathbb Z_p$?

Question

Considering the field $\mathbb Z_p$ in which $p$ is a prime; how many of roots satisfying $$x^2+1=0$$

I tried to look into $p=2,3,5,7$, but didn't find a rule for that. Is it that the number of roots depends on the value of $p$? Or there's a pattern that I didn't see. Thanks you.

Answer 1

Yes, the answer depends on $p$:

• If $p \equiv 1 \mod 4$, then there are two distinct roots of $x^2 + 1$,

• If $p \equiv 3 \mod 4$, then there are no roots,

• If $p = 2$, then $1$ is a double root of $x^2 + 1$.

To see the first case, remember that the multiplicative group of $\mathbb{Z}/(p)$ is cyclic of order $p-1$. Let $g$ be a generator of the cyclic group; then $(p-1)/4$ is an integer and $g^{(p-1)/4}$ will be a primitive fourth root of $1$, i.e., a square root of $-1$. In the second case, where $4$ does not divide $p-1$, there can be no primitive fourth root (i.e., an element "$i$" of order $4$).

Answer 2

Taking a cue directly from Wikipedia, we have the result "$x^2+1\equiv 0\mod p\text { is solveable }\iff p\equiv 1\mod 4.$" This has the additional constraint that $p\gt 2$ since that case has the double root $x=1$ while $2\equiv 2\ne 1\mod 4.$

Answer 3

Is $\mathbb{Z}_p$ the $p$-adic numbers, or the finite field with $p$ elements? $\mathbb{F}_p$ is an unambiguous notation for the latter. (both will have the same answer, though)

Let's solve:

$$ x^2 + 1 \equiv 0 $$ $$ x^2 \equiv -1 $$ $$ x \equiv \pm \sqrt{-1} $$

So, we need $-1$ to have a square root in $\mathbb{Z}_p$ for this to work.

There are simpleways to approach this:

• Use the standard tool to compute whether some number is a square modulo $p$
• Use what you know about the group of units to determine if $\mathbb{Z}_p$ has a fourth root of unity.