Timeline for Are polynomials bounded on the primes possible?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2020 at 5:09 | comment | added | Yaakov Baruch | @AntoineLabelle Nice idea, but wouldn't work. For example the polynomial $P=(x_3-2x_2+x_1)(x_3-2x_2+x_1-1)$ would be $0$ on the pseudo-primes $p_1=2,\dots p_{i+1}=p_i+1+\lfloor \log(i)\rfloor$. | |
| Oct 23, 2020 at 6:59 | comment | added | Yaakov Baruch | @Acccumulation : "are there $n,P$ such that..." | |
| Oct 22, 2020 at 23:15 | comment | added | Acccumulation | It seems to me that there is a quantifier missing for $n$. Is it implied to be the universal one? | |
| Oct 22, 2020 at 21:58 | history | became hot network question | |||
| Oct 22, 2020 at 20:53 | comment | added | Will Sawin | @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 $0$. | |
| Oct 22, 2020 at 19:01 | comment | added | Fedor Petrov | @WillSawin what exactly is Maynard's result, that for at least $c\cdot N^n$ $n$-tuples $(a_1,\ldots,a_n)\in \{1,\ldots,N\}^n$ there exist infinitely many positive integers $x$ such that all $x+a_i$ are prime? | |
| Oct 22, 2020 at 15:24 | comment | added | Antoine Labelle | $P$ 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. | |
| Oct 22, 2020 at 15:13 | comment | added | Will Sawin | @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. | |
| Oct 22, 2020 at 15:11 | comment | added | Yaakov Baruch | @FedorPetrov -- ditto of my comment to Tony's answer! | |
| Oct 22, 2020 at 15:07 | vote | accept | Yaakov Baruch | ||
| Oct 22, 2020 at 14:58 | answer | added | Tony Huynh | timeline score: 12 | |
| Oct 22, 2020 at 14:58 | comment | added | Fedor Petrov | 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. | |
| Oct 22, 2020 at 14:16 | history | edited | Yaakov Baruch | CC BY-SA 4.0 |
added 27 characters in body
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| Oct 22, 2020 at 13:58 | history | asked | Yaakov Baruch | CC BY-SA 4.0 |