Problems & Puzzles: Puzzles

 Puzzle 486. Reserved for the Gods? JM Bergot sent this nice puzzle: Find couples of prime numbes (p, q) such that: p|Sq & q|Sp where Sr=2+3+5+7+...+r. Bergot reports only one of these couples: (151, 809). Later he added "have been informed that the first solution was found in one second, but NO other solution was found after several hours of execution time... .This puzzle may have just one solution OR to find the 2nd solution is reserved for the Gods" Q1. Can you get the next three couples of reserved primes (p, q)?

Contributions came from J. K. Andersen, Wilfred Whieside

JKA wrote:

A007506 is Primes p with property that p divides the sum of all primes <= p.
If p=q was allowed then A007506 would be solutions: 2, 5, 71, 369119, 415074643.

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WW wrote:

I saw puz486 and gave it a try. But no luck. Below are my non-results:

All distinct prime pairs p,q were tested up to 200,000,000
Side note: n=199,999,991 was the max prime in this range with Sn = 1,075,207,199,997,334

Only the already posted solution was found:
p=151, q=809 with Sp=2427, Sq=50887 found where (Sq%p) = (Sp%q) = 0

The following near misses where found where (Sq%p)<=1 and (Sp%q)<=1
p=3, q=5, Sp=5, Sq=10 yields (Sq%p)=1 and (Sp%q)=0
p=349, q=5443, Sp=10887, Sq=1814801 yields (Sq%p)=1 and (Sp%q)=1
p=19853, q=1042609, Sp=20852181, Sq=40717887557 yields (Sq%p)=0 and (Sp%q)=1
p=27827, q=496901, Sp=39752081, Sq=9794603115 yields (Sq%p)=1 and (Sp%q)=1
p=38803, q=421913, Sp=74678602, Sq=7147046965 yields (Sq%p)=1 and (Sp%q)=1
p=73043, q=116507, Sp=249557995, Sq=607352546 yields (Sq%p)=1 and (Sp%q)=1

I would think that there should be an infinite number of solutions given
that the odds of any given p,q both being prime and satisfying Sq%p = Sp%q = 0 would go like 1/(q*p*ln(q)*ln(p)) which has an infinite double integral as q,p go to infinity. It just climbs very slowly and the solutions would likely be
separated by exponentially increasing orders of magnitude. Anyhow, I doubt
anyone finds another solution, but I'd be happy to be wrong.

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