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 $\Bbb Z_5[x]$.
 A: 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
$a^4 - 3 a^2 b + b^2 - 1 \equiv 0 \mod 5$.
A: We will show that $P(x):=x^5 -x+r$ is irreducible in $\Bbb Z_5[x]$ for $r=1,2,3,4$.
Since $P(k)\equiv r\pmod{5}$ for $k\in \Bbb Z_5[x]$, $P$ has no linear factors and if it is reducible then $P(x)=Q(x)T(x)$ where $Q$ has degree $2$ and $T$ has degree $3$.
Consider the isomorphism $f: \Bbb Z_5[x]\to \Bbb Z_5[x]$, given by $f(P)(x)=P(x+1)$. Then the polynomial $P(x)=x^5-x+r$ is a fixed point of $f$:
$$f(P)=f(x^5-x+r)=(x+1)^5-(x+1)+r=x^5+1-x-1+r=x^5-x+r=P.$$
Therefore we should have $f(Q)=Q$ and $f(T)=T$ ($f$ is degree-invariant). Now we show that
$f$ has not any fixed polynomial of degree $2$:
$$f(ax^2+bx+c)=a(x+1)^2+b(x+1)+c=ax^2+(2a+b)x+(a+b+c)\\=ax^2+bx+c$$
iff $2a\equiv 0 \pmod{5}$ and $a+b\equiv 0 \pmod{5}$ which is impossible because $a\not\equiv 0 \pmod{5}$.
A: The only possibilities that your polynomial can factor is to degree 1 and 4 or two degree 2 and 3. You already know that there is no linear factor, so the first case is omitted. For the second case, do the polynomial division for the 32 irreducibles and if it does not factor the polynomial is irreducible :-)
