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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:

  1. 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?

  2. What is the ring $\mathbb{F}_p[X]$? What is the relationship between this ring and $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$?

  3. 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?

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    $\begingroup$ There are not too many ways in which one can detect irreducibility of polynomials by hand. If I see a (and only one) prime $p$ then I'm trying to reduce mod $p$ for sure, and see if $f$ is irreducible modulo $p$. If not, then I'm trying to contradict the definition of reducibility by finding structure in the factors of $f$. In this case, reducing modulo $p$ provides structure to possible factors of $f$ which aids in proving irreducibility. $\endgroup$ Jan 27 at 17:59

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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 $\mathbb F_p$.

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  • $\begingroup$ Thank you for the great solution! $\endgroup$ Jan 27 at 17:59

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