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 Some steps? | Comment 3 of 8 |
If gcd(p,q)=s, then we can write p=as, q=bs, with gcd(a,b)=1. Replacing, as+b²s²+s³= abs³, so a+b²s+s²= abs², and then (ab-1)s²-b²s-a=0. The positive solution is (b²+√(b^4+4a(ab-1)))/(2ab-2), so necessarily b^4+4a(ab-1) must equal r² for integer r. (This is not a sufficient condition, by the way.)

I checked Charlie's results, and fortunately they agree with this condition, but I couldn't solve it directly, or prove that there are no more possibilities.

Edited on June 21, 2007, 11:18 am
  Posted by Federico Kereki on 2007-06-21 11:17:35

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
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 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information