<s>[INST] How to show $x^5-x+3$ is irreducible in $\Bbb Z_5[x]$? How to show $x^5-x+3$ is irreducible in $\Bbb Z_5[x]$?
I checked that there are no linear factors since there is no root. How to know that there is no quadratic factor? I checked from Mathematica that there are $32$ irreducible quadratic in $\Bbb Z_5[x]$.
 [/INST] By long division, if an $x^2 + a x + b$ divided $x^5 - x + 3$, you would need $a^4 - 3 a^2 b + b^2 - 1 \equiv 0$ and $a^3 b - 2 a b^2 + 3 \equiv  0 \mod 5$.
You can "complete the square" in $a^3 b - 2 a b^2 + 3 \mod 5$ to obtain
$3 a (b + a^2)^2 + 2 a^5 + 3 \equiv 3 a \left((b+a^2)^2 +4 a^4 + 1/a\right) $.  For this to be $0$ mod $5$, we need $ a^4 - 1/a$ to be a quadratic residue mod $5$ (thus $0$, $1$ or $4$); the possibilities are then 
$(a,b) = (1,4), (3,3)$ and $(3,4)$.  But none of those cases makes
$a^4 - 3 a^2 b + b^2 - 1 \equiv 0 \mod 5$.
</s>