TITLE: Why is $X^4+1$ reducible over $\mathbb F_p$ with $p \geq 3,$ prime
QUESTION [9 upvotes]: I have proven that in $\mathbb F_{p^2}^*$ exists an element $\alpha$ with $\alpha^8 = 1$. 

Let $f(X) := X^4+1 \in \mathbb F_p[X]$. How can I prove that $f$ is reducible over $\mathbb F_p$? 

Has $f$ a zero in $\mathbb F_p$ ?

REPLY [17 votes]: I think I know, finally!, what you meant with this exercise. Be sure you can follow and prove all the following:
Claim: $\,x^4+1\in\Bbb F_p[x]\,$ is reducible for any prime $\,p\,$
Proof : For $\,p=2\,$ the claim follows from $\,x^4+1=(x+1)^4\bmod 2\,$ , so let us suppose $\,p\,$ is odd.
Note that for any such prime, $\,p^2-1=0\bmod 8\,$ (why?) , so that the cyclic multiplicative group $\,\Bbb F_{p^2}^*\,$ has a cyclic subgroup of order $\;8\;$ , and from here that you have an element of order $\,8\,$ in $\,\Bbb F_{p^2}^*\;$ (and not merely an element s.t. $\,\alpha^8=1\,$ , which is completely trivial). Since
$$x^8-1=(x^4+1)(x^4-1)$$
over any field, all the roots of the right hand side are contained in $\;\Bbb F_{p^2}^*\;$ , and since this is an extension of degree two over the prime field $\,\Bbb F_p\,$ (and observe that $\,x^4+1\,$ is always a polynomial over this prime field!), it cannot be this polynomial is irreducible over the prime field since then any of its roots would form an extension of $\,\Bbb F_p\,$ of degree $\;4\;$ , which is absurd. $\;\;\;\;\;\;\;\;\;\;\;\;\;\square\;$