Given two integers a and b such that:
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