$\alpha \in \overline{\mathbb{F}}_q$ satisfying $\alpha^{q+1}+\alpha=-1$ Let $\overline{\mathbb{F}}_q$ be the algebraic closure of $\mathbb{F}_q$. Assume that $\alpha \in \overline{\mathbb{F}}_q$ satisfies at
$$\alpha^{q+1}+\alpha=-1.$$
Show that $\alpha \in \mathbb{F}_{q^3}$.
 A: As I just did the calculation, I will answer, albeit the hints in the comments nearly told it all.
So start with
$$(1)\quad\alpha^{q+1} + \alpha + 1 = 0$$
Raising to the $q$-th power
$$(2)\quad \alpha^{q^2+q} + \alpha^q + 1 = 0$$
and again
$$(3)\quad\alpha^{q^3+q^2} + \alpha^{q^2} + 1 = 0$$
Multiplying the latter with $\alpha^q$
$$(4)\quad\alpha^{q^3+q^2+q} + \alpha^{q^2 + q} + \alpha^q = 0$$
Substituting for $\alpha^{q^2+q}$ from (2)
$$(5)\quad \alpha^{q^3+q^2+q} - 1 - \alpha^q + \alpha^q = 0$$
Again substituting from (2)
$$(6)\quad \alpha^{q^3} (-\alpha^q - 1) - 1 = 0$$
Multiplying with $\alpha$
$$(7)\quad \alpha^{q^3} (-\alpha^{q+1} - \alpha) - \alpha = 0$$
Substituting from (1) $\alpha^{q+1} + \alpha = -1$ we are done.
A: For an easier typing let us use $a$ instead of $\alpha$. You are looking for an $s$ such that
$$a^{q^s} = a .$$
Rewrite your equation as $$a^{q+1} = -a -1$$
and raise to the power $q$:
$$a^{q^2+q} = (-1)^q (a^q + 1) = (-1)^q (-\frac 1 a) ,$$
so
$$a^{q^2+q+1} = (-1)^{q+1} = 1$$
(if $q$ is odd this is clear; if $q$ is even, $-1 =1$). But
$$1 = a^{q^s - 1}$$
too, so
$$a ^{q^s - q^2 - q -2} = 1$$
and therefore
$$\text{ord} a | q^s - q^2 - q -2$$
in $\Bbb F _{q^s}$.
But the order of an element divides the order of the group ($q^s - 1$), so you are left with
$$\text{ord} a| q^2 + q +1$$
or equivalently
$$(q-1) \text{ord} a| (q-1) (q^2 + q +1) = q^3 - 1 ,$$
so
$$\text{ord} a | q^3 - 1$$
or equivalently
$$a ^{q^3-1} = 1 ,$$
so
$$a ^{q^3} = a$$
as desired, so $s=3$.
Two final notes: first, the proof is not long, it just seems so because of the typesetting and because I have written down even the most humble details, assuming that the OP is a beginner. Second, @Jürgen Böhm's proof begins by raising to the $q$th power too, but it is different in spirit since it obtains an additive relationship, while from the beginning mine tried to produce a purely multiplicative one in order to be able to use basic concepts from group theory.
A: The above answer is certainly correct.
To further understand what's going on, observe that the original equation implies
$$
\alpha^q = \frac {-\alpha-1}{\alpha},
$$
so the Galois conjugate of $\alpha$ is a Mobius transformation of order $3$
in $\alpha$. If you repeat it three times, you get $\alpha^{q^3}=\alpha$.
