= Solution
Let
$$
P_n(X)=1+X+\cdots+X^{n-1}.
$$
If $n=ab$ with $a,b>1$, then
$$
P_n(X)=\left(1+X+\cdots+X^{a-1}\right)
\left(1+X^a+\cdots+X^{a(b-1)}\right),
$$
so $P_n$ is reducible.
Conversely, let $n=p$ be prime. Translation by one is an automorphism of $\mathbb Z[X]$, and
$$
P_p(X+1)=\frac{(X+1)^p-1}{X}
=\sum_{k=1}^p\binom pkX^{k-1}.
$$
Its leading coefficient is one, every lower coefficient is divisible by $p$, and its constant coefficient is $p$, which is not divisible by $p^2$. The <Eisenstein criterion> proves that $P_p(X+1)$ is irreducible, hence so is $P_p(X)$. Therefore the <geometric-sum irreducibility criterion> is
$$
\boxed{1+X+\cdots+X^{n-1}\text{ is irreducible in }\mathbb Z[X]
\Longleftrightarrow n\text{ is prime}}.
$$
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