$f(x)$ is irreducible but $f(x^n)$ is reducible

Question

Does there exist an irreducible polynomial $f(x)\in \mathbb{Z}[x]$ with degree greater than one such that for each $n>1$, $f(x^n)$ is reducible (over $\mathbb{Z}[x]$)?

Answer 1

There is no such polynomial.

It is clear that $f$ cannot be a cyclotomic polynomial (your condition $\deg{f} > 1$ excludes $x-1$). So suppose $f$ is non-cyclotomic and irreducible, of degree $d$, and consider a prime $p$ for which $f(x^p)$ is reducible. For $\alpha$ a root of $f(x^p)$, reducibility means that $[\mathbb{Q}(\alpha):\mathbb{Q}] < pd$. On the other hand $[\mathbb{Q}(\alpha^p):\mathbb{Q}] = d$ by the irreducibility of $f$, and we conclude that $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^p)] < p$; this means that the polynomial $x^p - \alpha^p$ is reducible over $\mathbb{Q}(\alpha^p)$. By the Vahlen-Capelli theorem, this is only possible if $\alpha^p = \eta^p$ with $\eta \in \mathbb{Q}(\alpha^p)$. We conclude that for every prime $p$, either $f(x^p)$ is irreducible, or else each root $\xi$ of $f$ has the property that $\xi$ is a $p$-th power from $K := \mathbb{Q}(\xi)$.

It remains to note that for any number field $K$, and any $\xi \in K^{\times} \setminus \mu_{\infty}$ not a root of unity, $\xi \in (K^{\times})^p$ for only finitely many primes $p$. This follows for instance from Dirichlet's unit theorem.