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!

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 $(a,c)=(2,-1)$
$$x^4+x^3+x^2+x+1=(x+2)(x-3)(x-4)(x+6)$$
