Given two integers a and b such that:
I. a>b>1
ii. a+b divides ab+1
iii. a-b divides ab-1
Prove that a is less than b*sqrt(3).
If a and b are consecutive odd numbers then the two divisibilities are satisfied. a=2k+1, b=2k-1 makes (ab+1)/(a+b) = k and (ab-1)/(a-b) = 2k^2 - 1.
Then a brute force search with a<300 finds nontrivial solutions of (19,11), (41,29), (71,41), (71,55), (109,89), (155,131), (209,181), (239,169), (265,153), (271,239). All these satisfy the criteria a<b*sqrt(3). The pair (265,153) comes very close with 265/153 = 1.7320261 and sqrt(3) = 1.7320508
Thoughts
