Why is $X^4+1$ reducible over $\mathbb F_p$ with $p \geq 3,$ prime 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 $\mathbb F_p$ ?

We can use three different factorizations:
If $2$ is a square mod $p$. (which occurs for $p=\pm1$ mod $8$)
$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2$ Assume $q^2=2$ mod $p$. Then $x^4+1=(x^2+1+qx)(x^2+1-qx)$. 
If $-2$ is a square mod $8$. (which occurs for $p=1$ or $3$ mod $8$)
$x^4+1=x^4-2x^2+1+2x^2=(x^2-1)^2-(-2)x^2=(x^2-1+rx)(x^2-1-rx)$, where $r^2=-2$ mod $8$.
Finally, if $-1$ is a square mod $8$. (which happens for $p=1$ mod $4$)
$x^4+1=x^4-(-1)=(x^2+r)(x^2-r)$, where $r^2=-1$ mod $p$.
Checking if all the cases are covered. 
All (odd) primes are congruent to either $1,3,5,7$ mod $8$. The cases $1$ and $7$ are covered by the first factorization. The case $3$ by the second. The case $5$ by the third, since $p=5$ mod $8$ implies $p=1$ mod $4$.
