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Mar 3, 2021 at 10:54 comment added Sylvain JULIEN This may have something to do with Mihailescu theorem (former Catalan conjecture), if we manage to prove $p_{k}^{2}-1$ has to be a perfect power as well as a divisor of $y$ for some $k$.
Dec 27, 2020 at 17:51 comment added JoshuaZ To expand on @FedorPetrov 's comment. the density of Germain primes en.wikipedia.org/wiki/Safe_and_Sophie_Germain_primes strongly suggest that for any such product which is sufficiently large, one will have at least prime p where $p^2 -1$ is divisible by a very big prime tha has not shown up anywhere else in the product.
Dec 27, 2020 at 14:42 answer added მამუკა ჯიბლაძე timeline score: 3
Dec 27, 2020 at 12:39 comment added Fedor Petrov Probably no, but looks hopeless. There should be many primes for which $2p+1$ is prime and $4p-1$ is too large.
Dec 27, 2020 at 11:26 history asked Augusto Santi CC BY-SA 4.0