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

Home > Just Math
A 2009 problem (Posted on 2006-08-27) Difficulty: 3 of 5
Find all integers p and q that satisfy p³+27pq+2009=q³.

See The Solution Submitted by K Sengupta    
Rating: 2.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Factors and restrictions | Comment 4 of 5 |

The prime factorization of 2009 is 7x7x41, and none of these are factors of 27. This means that they may be the only factors that p and q share. because otherwise, you can factor it out from all three terms except 2009 which would be impossible to solve.

Looking at the cases p=|q|, it is clear that p doesn't equal q. if p=-q, then p³-27p²+2009=-p³ or -p³-27p²+2009=p³. The first form equals p²(2p-27)=-2009 and the second form equals p²(2p-27)=2009. It is clear that (7,-7) is the only possible solution there.

Edited on August 28, 2006, 12:02 am
  Posted by Gamer on 2006-08-28 00:00:41

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 (3)
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