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Take a reducible polynomial $f(x)\in\mathbb{Q}[x]$. I am interested in the question: is $f(x)+1$ irreducible over $\mathbb{Q}$?

For $f (x) = (x −a_1) · · · (x −a_m)$ with distinct $a_1,\ldots, a_m\in \mathbb{Z}$ this is a question raised by Schur. For further examples in this direction see e.g. Györy, Hajdu, Tijdeman "Irreducibility criteria of Schur-type and Pólya-type".

Looking in a different direction, the question reminds me somewhat of Hilbert's irreducibility theorem. Take $F(x,t)=f(x)+t \in \mathbb{Q}[x,t]$ and then specialize the variable $t$ to $t_0=1$. But then, this needs an effective version of Hilbert's irreducibility theorem allowing one to show that $t_0=1$ belongs to the set of specializations where $F(x,t_0)$ remains irreducible.

My question: what about literature, known examples (single polynomials or whole families) in this direction?

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    $\begingroup$ How does the question, whether $f(x)+1$ is irreducible over the rationals, differ from the question of whether an arbitrary polynomial is irreducible over the rationals? I see that it differs when $f(x)$ is known to be reducible; is that the only case that interests you? $\endgroup$
    – Gerry Myerson
    Commented Apr 15, 2021 at 11:56
  • $\begingroup$ Yes, thanks, that's the question. $f$ is reducible. $\endgroup$
    – borntomath
    Commented Apr 15, 2021 at 12:20
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    $\begingroup$ There are surely many examples of polynomials $f(x)$ such that $f(x)$ and $f(x)+1$ are both reducible. For example $f(x) = -x^2$. $\endgroup$
    – Will Sawin
    Commented Apr 15, 2021 at 12:42
  • $\begingroup$ Or $−x^2+x(x−1)g$ for an arbitrary polynomial $g$ (except $g=1$). $\endgroup$
    – R.P.
    Commented Apr 15, 2021 at 19:46
  • $\begingroup$ As indicated in previous comments, this can only possibly become an interesting question when the shape of $f$ is very concretely restricted. Otherwise, every two-variable polynomial $g(x)-t$ can be turned into an example: choose two rational values $t_0$ and $t_1$ rendering $g(x)-t$ reducible (e.g., $t_0=g(0)$, $t_1=g(1)$) and then do a linear shift by defining $f(x) = (g(x)-t_0)/(t_0-t_1)$. $\endgroup$
    – Joachim König
    Commented Apr 19, 2021 at 13:30

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