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

Home > Numbers
Multiply Together, Get Cubic (Posted on 2007-06-21) Difficulty: 4 of 5
Determine all possible integer pairs (p,q) such that p+q²+s³=pqs, where s=gcd(p,q) and gcd denotes the greatest common divisor.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts If s = 2 | Comment 5 of 8 |
If s = 2,    p + q^2 + 8 = 2pq

p = (q^2 + 8)/(2q - 1)

4p = (4q^2 + 32)/(2q - 1) = (4Q^2 - 1 + 33)/(2q - 1) =
      (2q + 1)  + (33/(2q - 1))

A necessary (but not sufficient) condition for p integral is that 4p is integral, and this can only happen if (2q - 1) is 3 or 11 or 1 or -1 or -3 or -11.  i.e, if q = 2 or 6 or 1 or 0 or -1 or -5.  But if s = 2, only 2 or 6 are possible values of q, because 2 cannot be a gcd of 1, 0 , -1 , or -5.  Both q = 2 and 6 lead to solutions:

(p,q) = (4,2) and (4,6)

Next step: s > 2



  Posted by Steve Herman on 2007-06-21 19:36:00
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 (12)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information