For which $q$ does $x^2 = -3$ have a solution in $\mathbb{F}_q$, the finite field of $q$ elements? Let $q = p^h$ where $p$ is prime not equal to $3$. I would like to know
for which $h$ does $x^2 = -3$ have a solution in $\mathbb{F}_q$, the finite field of $q$ elements? I know that if $2|h$, then $\mathbb{F}_{p^2} \subseteq\mathbb{F}_q$ and the answer is yes. 
Is it possible that it has a solution for some $h$ with $2 \nmid h$? Thank you very much!
 A: You correctly argued that if $2\mid h$, then it always works, because the splitting field of $x^2+3$ over $\Bbb{F}_p$ is either the prime field itself or its unique quadratic extension. Any other extension field has a solution if and only if the said extension field contains the splitting field. Therefore it suffices to find the smallest $h$ that works for a given prime - the other possibilities are then its integer multiples.
The splitting field has extension degree $1$ or $2$, so the remaining question is whether $h=1$ works. This happens if and only if $-3$ is a quadratic residue modulo $p$.
The law of quadratic reciprocity would quickly settle this, but we can also argue as follows. The quadratic equation
$$
p(x)=x^2+x+1=0
$$
has discriminant $-3$, so it has zeros in $\Bbb{F}_p$, $p\neq3$, if and only if $-3$ is a quadratic residue modulo $p$. But
$$
p(x)(x-1)=x^3-1,
$$
so an element $a\in\Bbb{F}_p, a\neq1,$ such that $p(a)=0$ is of order three. Observe that $a=1$ is a zero only, if $p=3$ - a case we excluded. Converserly,
an element of order three is necessarily a zero of $p(x)$. The order of the multiplicative group $\Bbb{F}_p^*$ is $p-1$, so by the theorems of Lagrange and Cauchy from elementary group theory, there exists an element of order three if and only if $3\mid p-1$.
This gives the conclusion:

The equation $x^2=-3$ has a solution in $\Bbb{F}_q,q=p^h, p>3,$ if and only if either $p\equiv1\pmod3$ or $2\mid h$.

