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).

As I noted in my first post, the (a,b) pair (265,153) comes close to sqrt(3). This pair is just one member of a sequence of (a,b) pairs which satisfy the problem and is close to sqrt(3).

The sequence starts with (1,1), (5,3), (19,11), (71,41), (265,153), (989,571), (3691,2131),.... Each member of the sequence is a solution to 3b^2 - a^2 = 2. Both the a and b coordinates follow the recursion t(n) = 4*t(n-1) - t(n-2).

Take a look at the sequence formed by sqrt(3) - a/b: 0.73205, 0.065384, 0.0047781, 0.00034349, 0.000024664, 0.0000017708, 0.00000012714, ....

The ratio between terms is about 13.9282. Using the online Inverse Symbolic Calculator I find 7+4*sqrt(3) = 13.92820323. I suspect this to be the limit of the ratio.

This implies that each successive term in the sequence adds about log(7+4*sqrt(3)) = 1.1439 decimal places of accuracy when each fraction is viewed as a successive approximation of sqrt(3)