Root of $x^2+1$ in field of positive characteristic

Let $F$ be a finite field of characteristic $p$ of order $p^s$ with $s > 1$. Show that the polynomial $x^2 + 1$ has a root in $F$ if and only if $p^s$ is congruent to $1$ mod $4$.

Answer

I guess you're assuming $p$ odd, because if $p=2$ then $x^2+1 = (x+1)^2$ and so it has a root in $\mathbb{F}_2$ (and so in $F$).

By Euler's Criterion https://en.wikipedia.org/wiki/Euler%27s_criterion we have that $-1$ is a quadratic residue modulo $p$ (i.e. there exists $a \in \mathbb{Z}/p \mathbb{Z}$ such that $a^2 \equiv -1 \pmod p$) if and only if $p \equiv 1 \pmod 4$.

Now, if $p^s \equiv 1 \pmod 4$, there are two possibilities: either $p \equiv 1 \pmod 4$ or $p \equiv 3 \pmod 4$ and $s$ is even. In the first case, $x^2 + 1$ factors as $(x-a)(x+a)$ for $a$ such that $a^2 \equiv -1 \pmod p$, so $x^2 + 1$ has a root in $F$. In the second case, $s$ is even so $F$ contains $\mathbb{F}_{p^2}$, and all polynomials of degree two with coefficients in $\mathbb{F}_p$ completely factor in $\mathbb{F}_{p^2}$ and so in $F$.

If $p^s \equiv 3 \pmod 4$ instead, then $p \equiv 3 \pmod 4$ and $s$ is odd. It follows that $x^2+1$ has no roots in $F$ (otherwise $-1$ would be a quadratic residue modulo $p$) and since $s$ is odd $F$ does not contain any quadratic extension of $\mathbb{F}_p$ and so $x^2+1$ is irreducible in $F$, so in particular it does not have a root in $F$.

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The answer is accepted.