All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Expression Ratio (Posted on 2013-09-03) Difficulty: 4 of 5
Determine all pairs (a,b) of positive integers with a > b, for which (a2+b2)/(a-b) is an integer which divides 1995.

Prove that there are no others.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Method | Comment 3 of 4 |
Let n be a factor of 1995
(a2+b2)/(a-b) = n
(a2+b2) = n*(a-b)
a2-na+ b2+nb = 0
which is quadratic in a so
a=[n√(n-4(b+nb))]/2

For each value of n there are only a few values of b that yield real solution so from there I went brute force and got the same list as Charlie.

Note: Only multiples of 5 seem to have solutions.  I plan to look a little deeper.

  Posted by Jer on 2013-09-04 13:28:20
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information