Problems & Puzzles: Puzzles

Puzzle 5.- Find Pn =prime >61 such that Pn divides Pn+1*Pn+2 + 1

G. L. Honaker, Jr. - a science/mathematics teacher from Bristol, Virginia - observes that three consecutive primes {p1,p2,p3} satisfy the

following relation (p2 x p3 + 1)mod p1 = 0 in the following cases: {2,3,5}, {3,5,7}, & {61,67,71} and asks if there are other triplet solutions.

He offers $61 to those who obtain a solution or prove the nonexistence of any others.


We have a new collaborator, Trevor Green. He worked the first problem sent by G.L. Honaker, Jr. This are his results :

"Well, I found no more triplets up to the 7 500 000th prime, which is about 98,5% of 2^28. I have it in a file which is too large (74 MB) for my text editor to open, so I can't give you the exact number. (I didn't allocate enough memory )-:) I, too, think that we will find no more triplets" (7/July/98)


Jud McCranie wrote this (Fri, 17 Jul 1998) :

"There are no other solutions < 6.5 x 10^15, and it isn't likely that there
are any larger solutions.  Here's a heuristic argument as to why.

I'll use consecutive primes p, q, r; p<q<r.  We're looking for p a divisor
of qr+1, or (qr+1) == 0 mod p.  But write this as q=p+x and r = p+y.  The
equation becomes (p+x)(p+y) + 1 == 0 mod p.  Which is the same as xy+1 == 0 mod p.  So the least solution is xy=p-1.  (As in the 61,67,71 example, x=6, y=10, xy=61-1.)  But x is the gap between primes, and y is the sum of 2 gaps.  If g is the largest possible gap (between primes of that size) then x <= g and y <= 2g.  So 2g^2 approximately = p, so the gap g would have to be at least approximately sqrt(p/2) for a solution to be possible.

So the question boils down to - can the gap g be as large as sqrt(p/2)?  If the Extended Riemann Hypothesis (ERH) is true, then there is a proof that the gap can't be any larger than slightly more than sqrt(p), but that bound isn't good enough to prove that there are no solutions.  However, there is a conjecture that the gap is actually <= (log(p))^2, and the tabulation of prime data agrees with it so far.  (log(p))^2 is much smaller than sqrt(p) as p gets large. 

The gap record I know of is a gap of 863 at 6,505,941,701,960,039 (but
there is probably a newer one known), so there is no other solution less
than that number.  There's no realistic chance of a gap being big enough to allow another solution.  Even if it isn't proven that the gap < (log(p))^2, anything less than sqrt(p) will suffice to prove that no more solutions are possible."


Puzzle totally solved  in 2005 by Caldwell & Cheng in 2005.



Records   |  Conjectures  |  Problems  |  Puzzles