Modular equations admitting a solution

Question

In a finite Field $\mathbb{F}_q$ with $q = p^n$ and $p$ a prime numbers I am interested in the modular equation $$ x^2+3 \equiv 0 \pmod p. $$ More precisely I would like to ask whether there is a closed formula for the pairs $(p,n)$ such that the above equation has a solution in $\mathbb{F}_{p^n}$.

Answer 1

If there is a root in $\mathbb F_p,$ there is a root in $\mathbb F_{p^n}.$ There is a root in $\mathbb F_p$ iff $-3$ is a square modulo $p,$ which, via a quadratic reciprocity calculation, would mean if $p=2,3$ or $p\equiv 1\pmod 6.$

If $n$ is even, then $\mathbb F_{p^n}$ contains $\mathbb F_{p^2}.$ But $\mathbb F_{p^2}\cong\mathbb F_p[x]/\langle x^2+3\rangle$ if $-3$ is not a square modulo $p,$ so $\mathbb F_{p^2}$ has a root of the polynomial.

You need to finally deal with the case $p\equiv 5\pmod 6$ and $n$ odd to show that in that case, $-3$ is not a square in $\mathbb F_{p^n}.$

If $p^n\equiv 5\pmod 6,$ and $u\in \mathbb F_q$ is a root of $x^2+3,$ then we must have $$1 =u^{q-1}=(-3)^{(p^n-1)/2}=\left((-3)^{(p-1)/2}\right)^{1+p+\cdots +p^{n-1}}.$$

But since $-3$ is not a square modulo $p,$ $(-3)^{(p-1)/2}=-1$ and since $p$ is odd and $n$ is odd, then $1+p+\cdots p^{n-1}$ is odd, so:

$$u^{p^n-1}=-1$$

So no such $u$ can exist.