Given a prime $p$ and by Dirichlet a prime $q = k\cdot p+1$ - minimal of this form -, does then the number $k = (q-1)/p$ have only prime divisors $< p$? What does the research literature say for this question?
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6$\begingroup$ I deleted the non-mathematical part of your question, and I gave an answer to the mathematical part. Please be polite and professional. $\endgroup$GH from MO– GH from MO2025-11-16 06:22:50 +00:00Commented Nov 16 at 6:22
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1$\begingroup$ "Is the question under this title true?" Questions, by their nature, can neither be true nor false. In any event, it would be better if this question had a title that would give other users some clue as to the mathematical content of the question. $\endgroup$Gerry Myerson– Gerry Myerson2025-11-16 06:52:37 +00:00Commented Nov 16 at 6:52
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1$\begingroup$ Since this question is not about mathematical conjectures, but apparently related to a specific conjecture in number theory (as noted by GH in the answer), then it doesn't need the tag [conjectures]. And as others have noted, this is a professional site, and moreover there are terms of use around how people are treated (originally imposed by StackExchange, not MO) which I suspect the original text was skirting the edges of. $\endgroup$David Roberts– David Roberts ♦2025-11-16 08:00:41 +00:00Commented Nov 16 at 8:00
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1$\begingroup$ @GerryMyerson The title has been edited now. $\endgroup$The Amplitwist– The Amplitwist2025-11-16 09:40:18 +00:00Commented Nov 16 at 9:40
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3$\begingroup$ This question is similar to: Reference for a conjecture on the first primes congruent to 1 modulo other primes. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. $\endgroup$Wojowu– Wojowu2025-11-16 10:56:15 +00:00Commented Nov 16 at 10:56
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1 Answer
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For $p$ sufficiently large, $k<p$ follows from a conjecture of Hugh Montgomery recorded as (17.5) in Iwaniec-Kowalski. For a more general version of this implication, see my response here. Of course $k<p$ implies that $k$ only has prime divisors less than $p$.
For further references and related conjectures, see the question and responses here. (Thanks to Wojowu for pointing this out in a comment.)
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1$\begingroup$ Thanks for the reference and clarifying the question. $\endgroup$mathoverflowUser– mathoverflowUser2025-11-16 08:31:28 +00:00Commented Nov 16 at 8:31