This has been discussed a couple times, and in general I understand the logic to showing that something is a primitive polynomial. However, I repeatedly become lost at a step that seems to be taken as trivial. This question illustrates the process for this polynomial quite nicely, but I can't quite understand a component of the calculation.
If I understand correctly, the roots of $x^2 + x + 2$ lie in $\mathbb{F}_{25}$, i.e. if $\alpha$ is a root of $x^2 + x + 2 \in \mathbb{F}_5$ and $\alpha$ is a primitive element in $\mathbb{F}_{25}$, it has order $24$. The divisors of $24$ are $1, 2, 3, 4, 6, 8, 12$.
Trivially, we have $\alpha^2 + \alpha = -2 \Rightarrow \alpha^2 = -\alpha - 2$. Further, I can see why we have $\alpha^3 = -\alpha ^2 - 2 \alpha$.
This is the point where I get lost in this process. Our reasoning continues:
$\alpha^4 = \alpha ^2 + 4 \alpha + 4 = -2 \alpha + 2 = -2(\alpha - 1)$
$\alpha^6 = \alpha^2 - 4\alpha + 4 = \alpha^2 + \alpha - 1 = 2$
$\alpha ^8 = -\alpha^2 + 2\alpha - 1 = -2\alpha + 1$.
From these conclusions, I can understand the remaining logic required to justify that $x^2 + x + 2$ is a primitive polynomial in $\mathbb{F}_5$. However, I can't really understand how these calculations are performed. For instance, in the case of $\alpha^4$, how do we have that $\alpha ^2 + 4 \alpha + 4 = -2\alpha + 2$? Similarly, how do we get that $\alpha^6 = 2$? There looks to be some sort of modular arithmetic or division occurring here that I'm failing to understand.
What calculation am I missing that makes these equalities hold?
Remember that $\alpha^2=-\alpha-2$. So $$\alpha^4=(\alpha^2)^2=(-\alpha-2)^2\\ =\alpha^2+4\alpha+4=-\alpha-2+4\alpha+4=3\alpha+2 $$ Coupled with the fact that $3=-2$, we do indeed get $-2(\alpha-1)$. All the other powers are calculated similarly.