I'm doing a question, which asks to prove $x^4+1$ is not irreducible in any field of polynomials over finite fields.
According to the hint, there're three cases. First, there exists $a$ such that $a^2=-1$; secondly, there exists $b$, $b^2=2$; finally, $c$ such that $c^2=-2$. I managed to find 3 proper factorizations under these three cases, and the final step should be that one of these cases always hold for any odd prime(p=2 is trivial).
I found some answers to this question in the site, but none of them explained why these 3 cases cover all odd prime numbers. So this is not a duplicate question
If $p=3$, this is trivial.
If $p\geq 5$, this is because the quotient group $\mathbb{F}_p^\times/\mathbb{F}_p^{\times 2}$ has order $2$.
IfHence amongst the three distinct classes of $-1,2,-2$ modulo $p$ (they are distinct since $p\geq 5$) , you have at least one which becomes trivial in the quotient group by the pigeonhole principle, meanning that at least one of your three integers is a square modulo $p$.
To prove the above claim, you can look at the morphism $\bar{x}\in\mathbb{F}_p^\times\mapsto \bar{x}^2\in \mathbb{F}_p^{\times 2}$.
This morphism is surjective, and the kernel is $\{\pm\bar{1}\}$ since $\mathbb{F}_p$ is a field. Hence the factorisation theorem tells you that $\mathbb{F}_p^\times/\{\pm \bar{1}\}\simeq \mathbb{F}_p^{\times 2},$ so playing with cardinalities shows that $\vert \mathbb{F}_p^\times/\mathbb{F}_p^{\times 2}\vert =2$.
Alternaltively, you could use the nontrivial fact that $\mathbb{F}_p^\times$ is cyclic of order $p-1$. If $\bar{u}$ is a generator, then $\mathbb{F}_p^{\times 2}$ is generated bu $u^2$, which has order $\dfrac{p-1}{2}$ (because $2\mid p-1$).