Given the polynomial $f(x)=x^4+1$ in the field $\mathbb{F}_p$, prove that $[F: \mathbb{F}_p]\leq 2$, where $F$ is the splitting field of $f$.
Suppose $a$ is one of the roots of $f$. Then (if $p>2$, case $p=2$ is trivial) all different roots of $f$ are $a,a^3,a^5,a^7$ and therefore $x^4+1=(x-a)(x-a^3)(x-a^5)(x-a^7)$. Clearly $F=\mathbb{F}_p(a)$.
Now I want to prove that $a$ has minimal polynomial of degree not greater than 2. Because minimal polynomial divides other polynomials that have $a$ as a root, I need to prove that one of the following polynomials: $x-a$, $x^2-(a+a^3)x+a^4$, $x^2-(a+a^5)x+a^6$, $x^2-(a+a^7)x+a^8$ is in $\mathbb{F}_p [x]$. Or, equivalently (after simplifications), that at least one of the elements from $\{a,a+a^3,a^2,a-a^3\}$ is in $\mathbb{F}_p$. And here I am stuck. Not even sure this was the right way to start.
So, how can I prove the aforementioned statement?
My favorite way to prove it is via the following lemma: $f$ splits in $\mathbb F_q$, of characteristic $\neq 2$, if and only if $8 \mid q-1$. Then simply use the fact that $p^2 \equiv 1 \mod 8$ for $p$ an odd prime.