Let $p$ an odd prime number, congruent to $3$ mod $4$. Prove that the polynomial $f(x) = (X^2 + 1)^n + p$ is irreducible over the ring $\mathbb{Q}[X]$, regardless of the value of $n$ (natural number).
Doing some research, I have found a similar thread in AoPS community, that asks about a similar question (link: https://artofproblemsolving.com/community/c6h1455017p8368425) about the same polynomial being irreducible over $\mathbb{Z}[X]$. My specific questions are:
If I know that a polynomial is irreducible in $\mathbb{Z}[X]$, can we deduce something about being irreducible in $\mathbb{Q}[X]$. What conditions should we impose such that this transition may be made. What about the converse one?
What is the ring $\mathbb{F}_p[X]$? What is the relationship between this ring and $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$?
If none of the concepts stated have any relevance, how can we approach this problem? In particular, how to apply irreducibility criteria to this type of problem?
As you know (from the link) that $f$ is irreducible in $\mathbb Z[x]$, the irreducibility of $f$ in $\mathbb Q[x]$ is a direct consequence of Gauss's lemma for polynomials and the fact that the leading coefficient of $f$ is equal to $1$.
$\mathbb F_p[x]$ is the ring of polynomials over the finite field $\mathbb F_p$.