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Take three consecutive primes p,q,r and find the three remainders for $$p\cdot q\mod r$$ $$p\cdot r\mod q$$ $$q\cdot r\mod p$$ Is it possible for all three remainders to be prime?

Peter
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J. M. Bergot
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    How many examples have you looked at? – Barry Cipra Dec 27 '20 at 22:51
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    Please edit to indicate your efforts. Brute force searching works here. – lulu Dec 27 '20 at 22:58
  • Someone did toss this into a computer and found NO examples. If this is true that never will all three remainders be primes, is there a proof of this? I found example with two remainders being primes and the other an even number and a few with all three remainders being odd. – J. M. Bergot Dec 27 '20 at 23:36
  • @lulu : So you can cite an example with $p,q,r$ less than $10^{10}$? – Eric Towers Dec 27 '20 at 23:53
  • For p=29, remainders are 11,19,16. For p=11, remainders are 1,5,7. For p=47, remainders are 13,17,25. – J. M. Bergot Dec 28 '20 at 00:04
  • @EricTowers I see ${37,41,47}$ as an example. Am I missing something? – lulu Dec 28 '20 at 00:34
  • @EricTowers Ah, I missed the condition that they be consecutive primes. I agree, that's harder. – lulu Dec 28 '20 at 00:35
  • If there are two primes between $p$ and $p+\sqrt p$ then $qr\pmod p=(q-p)(r-p)$ – Empy2 Dec 28 '20 at 00:54
  • If Empy2 speaks truth, then clearly it is not possible to have all three remainders being primes. – J. M. Bergot Dec 28 '20 at 01:32
  • I concede because I didn't study what I already knew about the short cut for to find remainders. I did notice that for smaller primes 4759mod53=17 and 4753mod59=13 giving two consecutive primes (unknown so far?) and that 4353mod47=23 and 4347mod53=7. Could this be the end to the question of finding two prime remainders? – J. M. Bergot Dec 28 '20 at 04:04
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    J.M. as far as anyone can calculate, there are primes between $p$ and $p+ \sqrt p$ for $p$ over some modest threshold, and at least two primes for a slightly less modest threshold. Suggest you run this on computer yourself. There is actually no proof, the best exponent with proof is something like $p + p^{0.58}$ for sufficiently large $p.$ Big gap between computations and what can be proved – Will Jagy Dec 28 '20 at 04:34
  • Even if all three remainders cannot be primes, the problem of finding two remainders from pqmodr and prmodq is still unanswered. If an indefinite number contain two prime remainders can be found, then the question is worthwhile. I am assuming that others DO have computers and enjoy putting their electrons to work to answer unusual questions. – J. M. Bergot Dec 28 '20 at 08:36
  • @J.M.Bergot My program is still running, but I should have displayed the progress. But it seems that beyond $p=47$, even the first two expressions are rarely (or never) both prime. – Peter Dec 28 '20 at 08:39
  • I decided to restart (this time the progress is displayed) – Peter Dec 28 '20 at 08:41
  • I passed $10^{10}$ and found no further triple making the first two expressions simultaneously prime. – Peter Dec 28 '20 at 10:12
  • I stopped at $38\cdot 10^9$ – Peter Dec 28 '20 at 12:21
  • For expression did you use for three consecutive primes p,q,r pr mod q and pq mod r? One is seeking two out of the three and q*r mod p is NOT included. – J. M. Bergot Dec 28 '20 at 23:24

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