= Solution
Let
$$
F(X)=a_nX^n+\cdots+a_1X+a_0\in\mathbb Z[X]
$$
be a <primitive polynomial>. The <Eisenstein criterion> states that if a <prime number> $p$ satisfies
$$
p\nmid a_n,\qquad p\mid a_j\ (0\leq j<n),\qquad p^2\nmid a_0,
$$
then $F$ is an <irreducible polynomial> in $\mathbb Z[X]$, equivalently in $\mathbb Q[X]$ by <Gauss lemma for polynomials>.
To prove it, suppose that $F=GH$ with $G,H\in\mathbb Z[X]$ of positive degree. Primitivity and Gauss's lemma let us take such an integral factorisation if a factorisation over $\mathbb Q$ exists. Reducing modulo $p$ gives
$$
\overline G\,\overline H=\overline{a_n}X^n.
$$
Because $p$ does not divide the leading coefficient of $F$, neither factor loses degree on reduction. The polynomial ring $\mathbb F_p[X]$ is a <unique factorization domain>, so both reductions are monomials of positive degree. In particular, $p$ divides both constant terms $G(0)$ and $H(0)$. It follows that $p^2$ divides
$$
F(0)=G(0)H(0)=a_0,
$$
a contradiction. This proves the criterion.
For a prime $p$, translate the geometric sum by one:
$$
f(X+1)=\frac{(X+1)^p-1}{X}
=\sum_{k=1}^{p}\binom pk X^{k-1}.
$$
Its leading coefficient is one, every other coefficient is divisible by $p$, and its constant coefficient is $p$, which is not divisible by $p^2$. It is therefore Eisenstein at $p$. Translation $X\mapsto X+1$ is an automorphism of $\mathbb Z[X]$, so
$$
\boxed{f(X)=1+X+\cdots+X^{p-1}\text{ is irreducible}.}
$$
This is the <geometric-sum irreducibility criterion>.
The <evaluation homomorphism>
$$
\operatorname{ev}_\zeta:\mathbb Z[X]\longrightarrow\mathbb C,
\qquad G\longmapsto G(\zeta),
$$
has image $\mathbb Z[\zeta]$ and contains $(f)$ in its kernel. Since $f$ is monic, division by $f$ in $\mathbb Z[X]$ writes every $G$ as $G=Qf+R$ with $\deg R<\deg f$. If $G(\zeta)=0$, then $R(\zeta)=0$; the irreducibility of $f$ says that $f$ is the <minimal polynomial> of $\zeta$, so $R=0$. Thus $\ker(\operatorname{ev}_\zeta)=(f)$, and the <first isomorphism theorem for rings> gives
$$
\boxed{\mathbb Z[\zeta]\cong\mathbb Z[X]/(f)}.
$$
Now take $p=3$. Then $\zeta^2+\zeta+1=0$, and the <Eisenstein integers> are
$$
\mathbb Z[\zeta]=\{a+b\zeta:a,b\in\mathbb Z\}.
$$
Complex conjugation sends $\zeta$ to $\zeta^2$, so the <field norm> is
$$
N(a+b\zeta)=(a+b\zeta)(a+b\zeta^2)=a^2-ab+b^2=|a+b\zeta|^2.
$$
Given $z=x+y\zeta\in\mathbb C$, choose integers $m,n$ with $|x-m|,|y-n|\leq\tfrac12$. For $q=m+n\zeta$,
$$
|z-q|^2=(x-m)^2-(x-m)(y-n)+(y-n)^2\leq\frac34<1.
$$
For $\alpha,\beta\in\mathbb Z[\zeta]$ with $\beta\ne0$, apply this to $z=\alpha/\beta$ and put $r=\alpha-q\beta$. Then
$$
N(r)=N(\beta)|z-q|^2<N(\beta).
$$
Hence the norm is a <Euclidean function>, proving that $\mathbb Z[\zeta]$ is a <Euclidean domain>. This is the <Euclidean norm on the Eisenstein integers>.
Finally suppose $A\in\mathrm{GL}_n(\mathbb Z)$ satisfies $A^2+A+I=0$. Make the free abelian group $M=\mathbb Z^n$ into a $\mathbb Z[\zeta]$-module by defining
$$
\zeta v=Av.
$$
This is well-defined precisely because $A$ obeys the same polynomial relation as $\zeta$. The module is finitely generated. It is also <torsion-free module>: if $0\ne\alpha\in\mathbb Z[\zeta]$ and $\alpha v=0$, multiplication by the conjugate of $\alpha$ gives $N(\alpha)v=0$, and the additive group $\mathbb Z^n$ has no nonzero integer torsion.
A Euclidean domain is a <principal ideal domain>, and the <structure theorem for finitely generated modules over a principal ideal domain> says that a finitely generated torsion-free module over one is free. Thus
$$
M\cong\mathbb Z[\zeta]^m
$$
for some $m$. Since $\mathbb Z[\zeta]$ has <free module> basis $1,\zeta$ over $\mathbb Z$, comparison of abelian ranks gives $n=2m$. There can consequently be no such matrix when $n$ is odd.
If $n=2m$, choose a $\mathbb Z[\zeta]$-basis $v_1,\ldots,v_m$. Then
$$
v_1,Av_1,\ldots,v_m,Av_m
$$
is a $\mathbb Z$-basis, and $A^2v_i=-v_i-Av_i$. In this basis the matrix of $A$ is a direct sum of $m$ copies of
$$
C=\begin{pmatrix}0&-1\\1&-1\end{pmatrix}.
$$
Every admissible matrix is therefore conjugate in $\mathrm{GL}_n(\mathbb Z)$ to $C^{\oplus m}$. Hence there is exactly one conjugacy class for even $n$, and none for odd $n$. This is the <classification of integral matrices satisfying the third cyclotomic polynomial>.
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