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. Solution 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 I'll use consecutive primes p, q, r; p<q<r. We're
looking for p a divisor 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 *** Puzzle totally solved in 2005 by Caldwell & Cheng in 2005. ***





