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.



Are polynomials bounded on the primes possible?
Here is a proof that such a polynomial does not exist assuming that every admissible [$n$-tuple](https://en.wikipedia.org/wiki/Prime_k-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:



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> 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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