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Aug 20, 2023 at 15:02 history left closed in review Alexey Ustinov
Yemon Choi
Brian Hopkins
Original close reason(s) were not resolved
Aug 17, 2023 at 23:04 review Reopen votes
Aug 20, 2023 at 15:02
May 28, 2023 at 14:51 history closed Denis T
Steven Landsburg
Andrés E. Caicedo
Dave Benson
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May 26, 2023 at 6:02 comment added Philippe Gaucher @BoazTsaban Try platform.openai.com/playground. It is interesting to play with it to understand how it works; keep in mind that for more recent chat bots, the discussion starts with a hidden pre-prompt which is supposed to give some rules of behaviour.
May 24, 2023 at 11:10 comment added Boaz Tsaban @PhilippeGaucher This discussion goes far... This is too subtle an issue to be dealt with in short discussions. Time will tell, I guess.
May 24, 2023 at 7:57 comment added Philippe Gaucher @BoazTsaban chatGPT is nothing else but a text generator generating the most probable sequel of a sequence of words. You can play with the openAI playground to see that. Hence all hallucinations that this so-called AI has: it definitively cannot be trusted. When the sequence of words looks like a question, the most probable sequel is a sequence of words which looks for us, humans, like an answer because of its training. chatGPT does not understand what it's talking about.
May 23, 2023 at 19:49 comment added Boaz Tsaban @YemonChoi This is yet to be discussed for some decades... it is still unclear how far Chat GPT is from human intelligence. I would postpone the decision for at least several years.
May 23, 2023 at 15:31 comment added Yemon Choi @BoazTsaban to me, this is like saying "I rolled some dice and I used the results as coefficients for some polynomial, what is the Galois group of this polynomial". LLMs by their nature cannot understand their output, so saying this is a "conjecture by ChatGPT" seems to be attributing agency/sentience where it doesn't exist.
May 23, 2023 at 15:24 comment added Boaz Tsaban @YemonChoi I disagree. This conjecture was made by Chat GPT, and removing its mention would be plagiarizm.
May 23, 2023 at 9:16 comment added David Roberts @Yemon I agree, it adds nothing to the mathematics.
May 23, 2023 at 2:07 comment added Yemon Choi I would be much happier if all mention of GPT were removed from the original question.
May 22, 2023 at 19:11 history removed from network questions Asaf Karagila
May 22, 2023 at 17:31 history became hot network question
May 22, 2023 at 14:18 review Close votes
May 28, 2023 at 14:51
May 22, 2023 at 14:10 vote accept Boaz Tsaban
May 22, 2023 at 13:50 comment added JRN Maybe we can ask ChatGPT to prove the conjecture?
May 22, 2023 at 12:48 answer added Timothy Chow timeline score: 17
May 22, 2023 at 12:11 comment added Carlo Beenakker in the spirit of the question, I asked chatGPT 4 if this conjecture was known to be true/false or open. It answered: "Up until my last training data in 2021, this conjecture was not resolved. For the most current and accurate information, you should refer to recent mathematical literature or ask a professional mathematician or number theorist."
May 22, 2023 at 11:19 comment added Ofir Gorodetsky @YaakovBaruch Not that I am aware of, but there are partial results about the density of P-P and its structure, see link.springer.com/chapter/10.1007/978-3-319-22240-0_10 and ams.org/journals/proc/2017-145-09/S0002-9939-2017-13533-3
May 22, 2023 at 10:58 comment added Yaakov Baruch Going off the tangent a bit, is there any specific infinite set, not defined using the primes implicitly or explicitly (I know I'm being a bit vague here), known to be contained in $\mathbb{P}-\mathbb{P}$?
May 22, 2023 at 9:53 comment added Fedor Petrov @R.vanDobbendeBruyn you are correct of course. But I am afraid that still this is open for large specific $n$.
May 22, 2023 at 9:52 comment added Ofir Gorodetsky @BoazTsaban Conjecturally P-P contains every even number. More generally, Hardy and Littlewood conjectured that for every 'admissible' tuple $(h_1,\ldots,h_k)$ of positive integers there are infinitely many positive integers $m$ such that $\{m+h_i\}_{i=1}^{k}$ are simultaneously prime (and they predict the count of these $m$ up to $x$, asymptotically). Your conjecture relates to $h_1=0,h_2=n^2$. 'Admissible' means there is no prime $p$ such that $\{ h_i \bmod p\}_{i} = \mathbb{Z}/p\mathbb{Z}$ (see Fedor Petrov's comment). For more information see en.wikipedia.org/wiki/Prime_k-tuple .
May 22, 2023 at 9:51 comment added Fedor Petrov No arithmetic progression of course (look modulo 3). Just infinitely many pairs.
May 22, 2023 at 9:49 answer added Fedor Petrov timeline score: 11
May 22, 2023 at 9:49 comment added Boaz Tsaban @R.vanDobbendeBruyn Does the set P-P have full density? (Density 1)
May 22, 2023 at 9:47 comment added Boaz Tsaban @FedorPetrov Do you mean that there is a conjecture that for each k there is an arithmetic progression of k primes with difference 2?
May 22, 2023 at 9:45 comment added R. van Dobben de Bruyn @FedorPetrov actually it only asks for one prime $p$ so that $p+n^2$ is prime, not infinitely many (which is the case for the twin primes conjecture/bounded gaps theorem). So the $n=2$ case is very much solved!
May 22, 2023 at 9:39 history edited Boaz Tsaban CC BY-SA 4.0
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May 22, 2023 at 9:38 comment added Fedor Petrov It is certainly true and follows from standard conjectures, but widely open even for $n=2$.
May 22, 2023 at 9:31 history asked Boaz Tsaban CC BY-SA 4.0