Factorization of $x^5-1$ over $\mathbb F_{19}$ I've come across this question Factorization of $x^5-1$ over $F_{11}$ and $F_{19}$. The answer was good but I don't understand how to actually solve it. 
It says that i can split $x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)$. If I expand this I get $x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$. So to my understanding this gives:
$a+c=1 \mod 19$ 
$ac+d+b=1 \mod 19$
$ad+bc=1 \mod 19$
$bd=1 \mod 19$
What I dont understand is how to solve this system of equations. Any pointers? Thank you!
 A: The equations in the field $\mathbb{Z}/p$ for $p$ prime are given by
$$
a+c-1=0,\; ac + b + d - 1=0,\; ad + bc - 1=0,\; bd=1.
$$
Setting $c=1-a$ and $d=b^{-1}$ we obtain from the third equation that
$$
(ab + a - b)(b - 1)=0.
$$
Since $\mathbb{Z}/p$ has no zero divisors, we obtain tow cases, either $b=1$, or
$ab+a-b=0$.
Now you will be able to finish.
A: Paying some time on this problem in order to give an elementary proof for the O.P. one has the cyclotomic $x^4+x^3+x^2+x+1$ is irreducible over $\Bbb F_{19}$.
We have
$$a+c=1\Rightarrow (a,c)\in S_1\subset \Bbb F_{19}\text { x }\Bbb F_{19}\\bd=1\Rightarrow (b,d)\in S_2\subset \Bbb F_{19}\text { x }\Bbb F_{19}$$
$$S_2=\{(1,1),(2,-9),(3,-6),(4,5),(5,4),(6,-3),(7,-8),(8,-7),(9,-2),(-9,2),(-8,7),(-7,8),(-6,3),(-5,-4),(-3,6),(-2,9),(-1,-1)\}$$
$$ad+bc=1\Rightarrow\begin{cases}a+b^2c=b\\ad^2+c=d\end{cases}\Rightarrow\begin{cases}c(b+1)=1\\a(d+1)=1\end{cases}\qquad (*)$$ Then we can evaluate directly all the cases  $0\le a\le 18$ using the explicit set $S_2$ and $(*)$ the following way:
Let $E=ac+b+d$; we need to have $E=1$
►$a=0,1$ is trivially discarded.
►$(a,c)=(2,-1)\Rightarrow d+1=10,b+1=-1\Rightarrow E=-2+9-2=5\ne 1$
►$(a,c)=(3,-2)\Rightarrow d+1=13,b+1=9\Rightarrow E=-6+12+8=14\ne 1$
►$(a,c)=(4,-3)\Rightarrow d+1=5,b+1=6\Rightarrow E=-12+4+5\ne 1$ 
An so on till
►$(a,c)=(18,2)\Rightarrow d+1=-1,b+1=10\Rightarrow E=36+9-2\ne 1$
This shows that $x^4+x^3+x^2+x+1$ does not factorize on $\Bbb F_{19}[x]$
$$***$$
For $\Bbb F_{11}$ the same method applies and we find at once for $(a,c)=(2,-1)$
$$x^4+x^3+x^2+x+1=(x+2)(x-3)(x-4)(x+6)$$
