OFFSET
1,1
COMMENTS
There are sixteen such triples, namely (2, 2, 2), the three permutations of (2, 2, 3), and the six permutations of each of (2, 6, 11) and (3, 5, 7).
See the proof in the link.
The sixteen triples are (2, 2, 2), (2, 2, 3), (2, 3, 2), (3, 2, 2), (2, 6, 11), (2, 11, 6), (6, 2, 11), (6, 11, 2), (11, 2, 6), (11, 6, 2), (3, 5, 7), (3, 7, 5), (5, 3, 7), (5, 7, 3), (7, 3, 5) and (7, 5, 3).
This sequence is relative to the 2nd problem, proposed by Serbia, during the 56th International Mathematical Olympiad in 2015 at Chiang Mai, Thailand (see links). - Bernard Schott, Mar 17 2021
LINKS
56th International Mathematical Olympiad, Problem N5, 70-72.
Art of Problem Solving, 2015 IMO Problems.
Victor Pambuccian, A problem in Pythagorean Arithmetic, arXiv:1510.01645 [math.LO], 2015.
EXAMPLE
The 3rd primitive triple (x, y, z) = (2, 6, 11) is in the sequence because xy - z = 1, yz - x = 2^6 and zx - y = 2^4.
CROSSREFS
KEYWORD
nonn,fini,full,tabf
AUTHOR
Michel Lagneau, Jan 11 2017
STATUS
approved