As the title says: I want t prove or disprove the irreducibility of $X^4+1$ in $\frac{\mathbb{Z}}{p\mathbb{Z}}[X]$, $ p$ prime.
I have already proven that $X^4+1$ can't have linear factors in $\frac{\mathbb{Z}}{p\mathbb{Z}}[X]$, since it doesn't have roots in $\mathbb{Z}.$ By contradiction, such a root $k$ with $k^4=-1$ would have order $8$ which can't be since $8 \not| p$.
But that still leaves the possibility of a nontrivial decomposition with quadratic factors. So say $X^4+1 = (a_1x^2+q_1)(a_2X^2+q_2) \implies a_1a_2 = 1, a_1q_2+a_2q_1 = 0, q_1q_2 =0$. My hunch is that there is a contradiction somewhere in this equations, but I can't find it. Maybe there is none?
For $p=2$ we have "freshman's dream" $$ x^4+1=(x+1)^4, $$ so it is certainly not irreducible. For $p=3$ we have $$ x^4+1=(x^2 + 2x + 2)(x^2 + x + 2). $$ In general see this duplicate:
Why is $X^4+1$ reducible over $\mathbb F_p$ with $p \geq 3,$ prime