Showing $x^4 + 2$ is irreducible in $F_5[x]$ There is a similar question like this here
but I don't understand the solution. Since this is a fourth degree and since it has no root in $F_5$, it can only have a quadratic reduction. But how do we rule out the possibility that $x^4 + 2 = (x^2 + ax + b)(x^2 + cx + d)?$ Where the quadratic is irreducible in $F_5$?
There are a lot of similar questions like this on my practice exams.
Thanks

Assume $x^4+2$ factors in $\mathbb{Z}_5[x]$ as
$$x^4+2=(x^2+ax+b)(x^2+cx+d)$$
If the $\text{RHS}$ is expanded,

*

*The coefficient of the $x^3$ term is $a+c$, hence we must have $c=-a$.$\\[4pt]$

*The constant term is $bd$, hence we must have $d={\large{\frac{2}{b}}}$.  

hence the factorization can be rewritten as
$$x^4+2=(x^2+ax+b)(x^2-ax+\frac{2}{b})$$
Expanding, we get
$$x^4+2=x^4+\left(b+\frac{2}{b}-a^2\right)x^2+\left(\frac{2a}{b}-ab\right)x+2$$
hence we must have
$$
\left\lbrace
\begin{align*}
&\frac{2a}{b}-ab=0&&(\text{eq}1)\\[4pt]
&b+\frac{2}{b}-a^2=0&&(\text{eq}2)\\[4pt]
\end{align*}
\right.
$$
If $a\ne 0$, $(\text{eq}1)$ yields $b^2=2$, and if $a=0$, $(\text{eq}2)$ yields $b^2=-2$.

Either way we have a contradiction since $2$ and $-2$ are not squares in $\mathbb{Z}_5$.
