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Inequality challenge (Posted on 2016-10-25) Difficulty: 4 of 5
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    
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Comments: ( Back to comment list | You must be logged in to post comments.)
Hints/Tips re: Thoughts...hints toward formal proof | Comment 2 of 4 |
(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
  Posted by Ady TZIDON on 2016-11-17 17:14:36

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