 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Inequality challenge (Posted on 2016-10-25) 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).

 No Solution Yet Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Approaching sqrt(3) Comment 4 of 4 | 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)

 Posted by Brian Smith on 2016-11-20 10:57:55 Please log in:

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