This question is based on this other question from @Malkoun, and more specifically the comments.
Context.
Let's recall the following proof which shows that $-1$ is a square if $\mathbb F_p$ for $p\ne 2$ if, and only if, $p\equiv 1\pmod 4$.
Consider the morphism $x\mapsto x^2\in \mathbb F_p$ for $p\ne 2$, considering its kernel, you can show that there is
$$\frac{p-1}2$$
squares in $\mathbb F_p^*$.
Then let's define
$$X:=\{x\in \mathbb F_p^*,\ x^{(p-1)/2}=1\}.$$
You can show that all squares are in $X$, and since $\mathbb F_p$ is a field, $\vert X\vert \leqslant \frac {p-1}2$.
So $X$ contain all the non-null squares.
And you have:
$$-1\in X\iff \frac{p-1}2\equiv 0\pmod 4\iff p\equiv 1\pmod 4.$$
The question.
Let's take $n$ such that $n\mid p-1$.
We want to extend the previous proof to get information about when $-1$ is a $n$'th power in $\mathbb F_p$.
I think that if $\mathbb F_p$ contain all $n$'th roots of $1$, then this argument shows that
$$-1\text{ is a $n$-th power } \iff p\equiv 1\pmod {2n}.$$
How can we characterize the primes $p$ such that $\mathbb F_p$ contain all $n$'th roots of unity?
