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If $\{p_i\}$ is the sequence of all primes, is it possible that there exist a non constant $P\in \mathbb{Z}[x_1,\dots x_n]$ such that $P(p_i,p_{i+1},\dots p_{i+n-1})$ is bounded in $i$?

More precisely, can widely believed conjectures, or even heuristic arguments, help make such a claim (even more) unlikely.

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    $\begingroup$ If such polynomial exists, it must depend on differences $p_{i+k}-p_{i+k-1}$. This is not hard to prove unconditionally. The standard conjectures like Hardy — Littlewood then imply that this is not possible. But the fact looks much weaker, possibly it has an unconditional proof. $\endgroup$ – Fedor Petrov Oct 22 at 14:58
  • $\begingroup$ @FedorPetrov -- ditto of my comment to Tony's answer! $\endgroup$ – Yaakov Baruch Oct 22 at 15:11
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    $\begingroup$ @FedorPetrov I was trying to deduce it from Maynard's theorem that a positive proportion of $n$-tuples of natural numbers from $1$ to $N$. I failed because I don't think Maynard rules out that the $n$-tuple could have additional primes between them, which would mess up the polynomials. Still I believe it's pretty likely there's an unconditional proof, maybe just a small modification of this argument. $\endgroup$ – Will Sawin Oct 22 at 15:13
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    $\begingroup$ $P$ takes finitely many values $a_1,\cdots,a_k$. Replacing $P$ by $(P-a_1)\cdots(P-a_k)$ we may assume that $P$ vanish on all tuples $(p_i, \cdots, p_{i+n-1})$. This would give a kind of recurrence relation for the primes. Then maybe we can derive a contradiction from the prime number theorem? I'm not sure. $\endgroup$ – Antoine Labelle Oct 22 at 15:24
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    $\begingroup$ @FedorPetrov Yes, with c depending on $n$. In addition you can replace $\{1,\dots, N\}$ with another set of natural numbers as long as the density of admissible tuples in that subset doesn't go to $0$. $\endgroup$ – Will Sawin Oct 22 at 20:53
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Here is a proof that such a polynomial does not exist assuming that every admissible $n$-tuple occurs infinitely often in the sequence of primes.

To see this let $a:=(0, a_1, \dots, a_{n-1})$ be an admissible $n$-tuple. Suppose $P \in \mathbb{Z}[x_1, \dots, x_n]$ is such that the function $f(i):=P(p_i,p_{i+1},\dots p_{i+n-1})$ is bounded. Replacing $x_i$ by $x_1+a_{i-1}$ for all $i \in \{2, \dots, n\}$ we obtain a polynomial $Q_a \in \mathbb{Z}[x_1]$. Assuming that the $n$-tuple $a$ occurs infinitely often, we have $Q_a(p_i)=f(i)$ for infinitely many $i$. Since $f(i)$ is bounded, $Q_a$ is a constant. Thus $Q_a=p(a)$ where $p$ is a non-constant polynomial only depending on $P$. Since $f$ takes only finitely many values, $p(a)$ only takes on finitely many values over all admissible $n$-tuples $a$. However, this is impossible.

As Terry Tao notes in the comments below:

One can make this argument unconditional by noting that $O(\log^{n−1−o(1)}X)$ of the $O(\log^{n−1}X)$ admissible tuples $a=(0,a_1, \dots ,a_{n−1})$ with $a_1, \dots, a_{n−1}=O(\log X)$ will be associated to consecutive primes $p,p+a_1, \dots,p+a_{n−1}$ for some $p∼X$ (because $∼X/\log X$ primes will generate a tuple by Markov's inequality and each tuple is associated to $O(X/\log^{n−o(1)}X)$ primes by e.g. Selberg sieve). On the other hand, a polynomial constraint on these tuples would instead force at most $O(\log^{n−2}X)$ of these tuples to be admissible (Schwartz-Zippel lemma).

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    $\begingroup$ You cut right through my fog. Such simple and elegant argument! $\endgroup$ – Yaakov Baruch Oct 22 at 15:08
  • $\begingroup$ As Will Sawin says in another comment, Maynard's theorem implies, unconditionally, the existence of one (actually infinitely many, in a strong sense) $n$-tuple occurring infinitely often in a sequence of primes. So provided the proof can be amended, it will be unconditional. $\endgroup$ – Pierre PC Oct 22 at 15:23
  • $\begingroup$ @PierrePC It's easy to amend the proof if you know a positive proportion of admissible tuples occur in the sense that all the members are prime and no numbers between them are prime. The question is entirely to do with the weaker sense of "occur" where we only demand that the members of the tuple are prime. $\endgroup$ – Will Sawin Oct 22 at 15:29
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    $\begingroup$ One can make this argument unconditional by noting that $O(\log^{n-1-o(1)} X)$ of the $O(\log^{n-1} X)$ admissible tuples $a = (0,a_1,\dots,a_{n-1})$ with $a_1,\dots,a_{n-1}=O(\log X)$ will be associated to consecutive primes $p,p+a_1,\dots,p+a_{n-1}$ for some $p \sim X$ (because $\sim X/\log X$ primes will generate a tuple by Markov's inequality and each tuple is associated to $O(X/\log^{n-o(1)} X)$ primes by e.g. Selberg sieve). On the other hand, a polynomial constraint on these tuples would instead force at most $O(\log^{n-2} X)$ of these tuples to be admissible (Schwartz-Zippel lemma). $\endgroup$ – Terry Tao Oct 23 at 2:48
  • $\begingroup$ This proof trivially carries over if $P$ has real coefficients, something that does not seem to me an obvious direct consequence of the result over the rationals/integers. $\endgroup$ – Yaakov Baruch Oct 24 at 20:05

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