Factorisation of $X^n-a$ in $\mathbb{F}_p$ Do we know general formula  for factorisation of $X^n-a$ in $\mathbb{F}_p$ for any $a\in\mathbb{F}_p$? At least do we know the degree of the factors?
Some reflexions:

*

*if $a=1$ it is the "cyclotomic factorisation" with factorisation of the cyclomic polynomial $\phi_i$ in factor of degree the order of $p$ in $\mathbb{Z}/a\mathbb{Z}$.

*if $a=-1$ we can get the factorisation of $X^n+1$ from the factorisation of $X^{2n}-1$

*if $a$ is an $n$-power in $\mathbb{F}_p$ we can get a factorisation with
$$ X^n-a=X^n-b^n=b^n\left(\left(\frac{X}{b}\right)^n-1\right) $$

*if $-a$ is an $n$-power in $\mathbb{F}_p$ we can do the same thing with $X^n+1$

*if $n=p^k$ then $a$ is a $n$-power (and then Frobenius make the job)

*I have seen in Lang Algebra (Theo 9.1) that $X^n-a$ will be irreducible $a$ is never a $q$-power for all $q|n$.

Any Idea?
 A: I dont' consider it really as a complet answer but it's a begin. Let take an example: $X^6-3$ modulo 5.
We have $3^4=1$ so that if $\alpha$ is a root of $X^6-3$ ie $\alpha^6=6$ then $\alpha^{24}=1$ i.e. $\alpha$ is a root of $X^{24}-1$ and hence $X^6-3$ divides $X^{24}-1$. As a result $X^6-3$ is a product of cycltomic polynomial $\phi_{i,j}$ for $i|24$ (the index $j$ is because the $\phi_i$ are not irreducible in $\mathbb{F}_p$)
We can formalise that by denoting $E_{5,6,3}$ the subset of divisor of $24=6\times\text{ord}_5(3)$ (weighted with the number of subdivisor of $\phi_i$ for each $i$) with the property that
$$ X^6-3=\prod\limits_{i\in E_{24,3},j} \phi_{i,j}(X) $$
So the question is: how to find $E_{5,6,3}$?
One way to proceed, which do not give direct (I mean explicit) answer:
\begin{align*}
X^{24}-1 &= (X^6)^4-4^4\\
  &= 3^4\left[ \left( \frac{X^6}{3} \right)^4-1 \right]\\
  &= 3^4\prod\limits_{i|4}\phi_i\left( \frac{X^6}{3} \right)\\
  &= \prod\limits_{i|4}3^{\varphi(i)}\phi_i\left( \frac{X^6}{3} \right)\\
  &= (X^6-3)(X^6+3)(X^{12}+9)
\end{align*}
As a result,
$$ X^6-3=\frac{X^{24}-1}{(X^6+3)(X^{12}+9)} $$
And as we know the irreducible factorisation of $X^{24}-1$ it should be not to difficult (here is maybe a problem) to get this one of $X^6-3$:
Here $X^{12}+9=X^{12}-1$ so that
$$ X^6-3=\frac{X^{12}+1}{X^6+3} $$
But $X^{12}+1=\phi_{24}\phi_8$, these two factor decompose in $\mathbb{F}_5$ in factors of degree 2 because 2 is the order of 5 in $\mathbb{Z}/24\mathbb{Z}$ and $\mathbb{Z}/8\mathbb{Z}$ so that
$$ X^6-3=\frac{\phi_{24,1}\phi_{24,2}\phi_{24,3}\phi_{24,4}\phi_{8,1}\phi_{8,2}}{X^6+3} $$
so that for "symetric reasons" (hum...)
$$ X^6-3=\phi_{24,1}\phi_{24,2}\phi_{8,1} $$
So here $E_{5,6,3}=\{24_2,8_1\}$.
As I said (if the above is correct) it is a little "craft" and it doesn't give directly $E_{p,n,a}$.
A: I think I answered to my question in the two following personal paper: paper 1 and paper 2.
