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I have a question in analytic number theory which is closely related to the open problem (Bunyakovsky conjecture and more generally, Schinzel's hypothesis H) that asks you if, any irreducible polynomial of degree at least 2 in $\mathbb{Z}[X]$ whose coefficients are mutually coprime, represents a prime infinitely often at integer arguments. I don't know if a similar question/answer already exists. Any information would be appreciated!

I phrase a weaker question in the following way:

For a positive integer $m>1$, let $\mathrm{LP}(m)$ denote the largest prime factor of $m$. For convenience, we make the following definitions \begin{gather*} \mathrm{LP}(m)= \begin{cases} 1\quad\text{if $m=1$}\\ 0\quad\text{if $m=0$}\\ \mathrm{LP}(|m|)\quad\text{if $m\leq-1$} \end{cases} \end{gather*} Does there exist a monic irreducible polynomial $P(X)\in\mathbb{Z}[X]$ of degree at least $2$ and a strictly increasing sequence of positive integers $\{a_n\}_{n\geq1}$ such that $$\limsup\limits_{n\rightarrow\infty}\frac{|P(a_n)|}{\mathrm{LP}(P(a_n))}<\infty?$$

I am not sure if this statement can be equivalent to the conjecture I mentioned. Still, a priori, it feels that it is only essentially asking if the values $P(n)$, as $n$ varies in $\mathbb{Z}$, have sufficiently large prime factors or not! I might be wrong in the sense that the answer to this question is actually 'simple'. Apologies in advance!

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    $\begingroup$ The weaker question os equivalent to "Is there integer $M>0$ such that there are infinitely many $n$ such that one of $P(n),2P(n),...,M\cdot P(n)$ is prime?", which is equivalent to "Is there integer $m>0$ such that $P(n)/m$ is prime for infinitely many $n$?". If true, then there is $P\in\mathbb Z[x]$ ($n\to mn+r$ for some $m\mid P(r)$) such that $\deg P>1$ and $P(n)$ is prime infinitely often. As far as I know, it's open $\endgroup$
    – te4
    Commented Nov 20 at 13:02
  • $\begingroup$ I believe this is open even if we replace $LP(m)$ with the radical of $m$, $\mathrm{rad}(m)=\prod_{p\mid m}p$. $\endgroup$
    – Ofir Gorodetsky
    Commented Nov 20 at 19:20

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te4 is quite right that this is open. A general comment is that asking whether there exists a subsequence whose lim sup is finite is equivalent to asking whether the lim inf of the original sequence is finite. So your question is whether there exists $P$ such that $$\liminf_{n\to\infty} \frac{|P(n)|}{LP( P(n))}<\infty$$ (which I think is a clearer statement) or equivalently $$\limsup_{n\to\infty} \frac{ LP(P(n))}{|P(n)|}>0$$ (which I find even a bit clearer).

If this is true for a given $P$ then we have $$\limsup_{n\to\infty} \frac{\log LP(P(n))}{\log |P(n)|}=1$$ (the implication $\geq 1$ is easy but the ratio is also clearly at most $1$) which is already too hard. We can certainly prove

$$\limsup_{n\to\infty} \frac{\log LP(P(n))}{\log |P(n)|} \geq 1/\deg P $$

by finding roots of $P$ modulo large primes $p$ and considering the first lift of each root. I think that's the only statement along these lines that is 'simple'. Analytic number theory methods might be able to increase that lower bound, at least for $P$ of small degree, but not all the way to $1$.

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  • $\begingroup$ Thank you for this clarification! This is interesting! By 'Analytic number theory methods', what do you mean? Could you please explain a bit more, if possible? $\endgroup$
    – James Moriarty
    Commented Nov 20 at 18:19
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    $\begingroup$ @JamesMoriarty See the introduction of Dartyge and Maynard's preprint for a survey of the state of the art on this limsup: arxiv.org/abs/2212.03381 $\endgroup$
    – Ofir Gorodetsky
    Commented Nov 20 at 19:17
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    $\begingroup$ In particular, $1/\deg P$ can be improved when $\deg P \in \{2,3,4\}$, e.g. the largest prime factor of $n^2+1$ is infinitely often $>n^{1.279}$ as shown by Merikoski here: arxiv.org/abs/1908.08816 $\endgroup$
    – Ofir Gorodetsky
    Commented Nov 20 at 19:19
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Yes, in fact for every non-singular quadratic polynomial such a sequence exists. Actually, something much stronger holds: for every nonn-singular quadratic polynomial $f \in \mathbb{Z}[x]$ and for any $\varepsilon > 0$, there exists infinitely many integers $n$ such that the largest prime divisor of $f(n)$ is at most $n^\varepsilon$.

This is proved in the following paper: Smooth values of polynomials.

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    $\begingroup$ $\limsup\frac{|P(a_n)|}{LP(P(a_n))}=\infty$ holds for this sequence $\endgroup$
    – te4
    Commented Nov 20 at 12:56

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