Why is $X^4-X^2+1$ reducible over $\Bbb F_5$? I have checked $X^4-X^2+1=0$ and got the solution that the polynom is never equal 0 so it should be irreducibel. 
However it is. 
My prof gave me that hint: 
$X^4-X^2+1=(X^2+aX+b)(X^2+cX+d)$
Unfortunately that did not really help me. How can I solve that equation and where does it come from ?
 A: What you've mentioned only holds if the degree of the polynomial $f$ is less or equal to $3$. Consider for example
$$
(x^2+1)(x^2+1)=x^4+2x^2+1
$$
over $\Bbb R$. The polynomial $x^4+2x^2+1$ obviously has no roots over $\Bbb R$ but is nevertheless reducible. 
Your prof gave you the hint because a reducible polynomial of degree $4$ can be the procuct of two polynomials of degree $2$ or a product of one with degree $3$ and one with degree $1$. You've only checked that it can't be the second case, because it has no roots. That didn't rule out the first case.
A: When we say a polynomial $f(x)$ is irreducible, that means we cannot write it as $f(x) = g(x)h(x)$ unless one of $g$ or $h$ is constant. You are confusing this with meaning "the polynomial has no roots," which is different. Now if the degree of $f$ happens to be $1$, $2$, or $3$, these will be the same (because then at least one of $g$ and $h$ would have degree $1$). However, it is not the same in general. 
Now let's follow your professor's hint. Suppose $f(x) = h(x)g(x)$. Since you've shown that $f(x)$ has no roots, you can skip to the case where $h$ and $g$ are quadratic. Either we will get a contradiction, and conclude $f$ is irreducible, or we will not, in which case we will have found a factorization of $f$.
Write $h(x) = x^2 + ax + b$ and $g(x) = x^2 + cx + d$. Now multiplying out gives $f(x) = x^4 + (a+c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd$.
Compare this with the actual coefficients of $f$, which are known. Then you are solving a system of equations in $\mathbb{F}_5$:


*

*$a + c = 0$

*$ac + b + d = -1$

*$ad + bc = 0$

*$bd = 1$


Substituting $a = -c$, this is equivalent to the three equations


*

*$-a^2 + b + d = -1$

*$a(d-b) = 0$

*$bd = 1$


Now by inspection, $a = 2$ and $b = d = -1$ is a solution. So $f(x) = (x^2 + 2x  - 1)(x^2 - 2x - 1)$.
A: $$x^{4} - x^{2} + 1= x^{4} - 2 x^{2} + 1 + x^{2} = x^{4} - 2 x^{2} + 1 - 4 x^{2} =\\= (x^{2} - 1)^{2} - (2 x)^{2} = (x^{2} - 1 + 2 x) (x^{2} - 1 - 2 x)$$
