Primitive polynomials I am revising for a discrete mathematics exam and as quite stuck on this question.
Show that the polynomial $f = x^2 + 2 x + 3 \in \mathbb{Z}_5[x]$ is primitive. How many monic primitive quadratic polynomials are there in $\mathbb{Z}_5[x]$?
Any help would be greatly appreciated.

A polynomial is called primitive (in the context of finite fields), iff its zero is a generator of the multiplicative group of the field it generates. In this case the polynomial is quadratic, so a root $\alpha$ will generate the field $L=\mathbb{F}_{25}$.
The multiplicative group of $L$ is cyclic of order 24. By the well known facts about cyclic groups, the group $L^{\times}$ has $\phi(24)=\phi(2^3\cdot3)=2^2(3-1)=8$ generators. Another way of seeing this is to observe that if $g$ is a primitive root of unity of order $24$, all of them are in the list $g$, $g^5$, $g^7$, $g^{11}$, $g^{13}$, $g^{17}$, $g^{19}$ and $g^{23}$. Anyway, between them, these eight elements have four distinct minimal polynomials (the conjugate of a primitive element is also primitive, and the conjugates come in pairs in a quadratic extension).
The answer is thus that there are exactly $4$ quadratic primitive polynomials.

Looking at the specific polynomial $p(x)=x^2+2x+3$. Its discriminant is non-square, so it is irreducible (alternatively one may check that it has no zeros in $\mathbb{F}_5$). Let $g$ be one of its roots in the extension field
$\mathbb{F}_{25}$. We want to prove that it is of order $24$. We know that the order is a factor of $24$, so it suffices to rule out factors of $12$ and $8$
as orders.
The other zero of $p(x)$ is the Frobenius conjugate $g^5$ (basic Galois theory).
The product of the zeros of a monic quadratic is its constant term, so we know that
$$
g^6=g\cdot g^5=3.
$$
Therefore $g^{12}=3^2=4=-1$, and we can conclude that the order of $g$ is not a factor of $12$. We still have to rule out the possibility that $g$ could be of order $8$. If $g$ were of order $8$, then $g^2$ is of order four. But all the fourth roots of unity are in the prime field $\mathbb{F}_5$. OTOH from the minimal polynomial we read that $g^2=-(2g+3)=3g+2$ is not $\in \mathbb{F}_5$. 
The claim follows from this.
