It is well known that $f(x+a) \in \mathbb{Z}[x]$ is irreducible when $f(x)$ is irreducible. I was wondering whether $f(2x)$ and in general whether $f(nx)$ is irreducible as well.
Though it is relatively easy to find counterexamples for $f(x^2)$ reducible with $f(x)$ irreducible (consider $f(x) = x^2+4$), I haven't been able to find counterexamples for $f(2x)$.
I tried to write $f(2x) = g(x)h(x)$ so that $f(x) = g(\frac{x}{2})h(\frac{x}{2})$ which are polynomials in $\mathbb{Q}[x]$. Then if $f(x) \in \mathbb{Z}[x]$ then maybe somehow $g(\frac{x}{2})$ and $h(\frac{x}{2})$ are both in $\mathbb{Z}[x]$ too?
Edit: by "irreducible" I meant that $f$ cannot be written as a product of two nonconstant polynomials.
Thanks in advance!
It makes a difference if you are considering irreducibility in $\mathbb{Z}[x]$ or in $\mathbb{Q}[x]$.
A counterexample for the first case: $f(x) = x$ is irreducible in $\mathbb{Z}[x]$, but $f(2x) = 2\cdot x$ is not.
In the second case your statement is true for all $n\neq 0$, since then $n$ is a unit in $\mathbb{Q}[X]$, so $f(x)\mapsto f(nx)$ is an automorphism of $\mathbb{Q}[x]$.
UPDATE
After giving this answer,the following was added to the question
First of all, this is a bit unfortunate, since it differs from the established definition of irreducibility in ring theory.
However by Gauss' Lemma, it turns out that your notion of "irreducibility" of a polynomial $f\in \mathbb Z[x]$ is exactly the same as the irreducibility of $f$ in $\mathbb Q[x]$, which is covered by the above discussion.