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).

(In reply to

Thoughts by Brian Smith)

Hints toward formal proof:

Since ( i) (b^2-1)=(a+b)*b-(ab+1)

<ii> b^2-1 =(ab-1)-b(a-b)

(iii) b^2-1 must be divisible by (a+b)*(a-b) (iv)

.show that a and b must be relatively prime ( iv ) i

and a+b & a-b may have only 1 or 2 as common divider

SAY NO MORE.

*Edited on ***November 17, 2016, 5:29 pm**