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

Home > Numbers
Classical Rules 2 (Posted on 2006-04-20) Difficulty: 4 of 5
Find three squares such that each minus the product of the three gives a square.

Classical Rules: Let a "square" be any number that is the square of a rational number.

No Solution Yet Submitted by goFish    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Partial Solution | Comment 2 of 3 |
(In reply to Partial Solution by tomarken)

I first tried to see if there were any solutions where the three squares were equal other than (1,1,1)

If each is rational (say a/b) then

(a/b)^2 - (a/b)^6 = (c/d)^2 where c/d is rational.

This simplifies to (b^4-a^4) = [((b^3)*c)/(ad)]^2 in which case we know from Fermat's Last Theorum that the difference of powers to the fourth cannot equal a square among integers, and since the difference is a non-zero integer (remember I am excluding (1,1,1) its square root cannot be rational without it being a square, which it isn't. Therefore at least there is no other solution with three equal squares.

Edited on April 20, 2006, 10:51 pm
  Posted by Eric on 2006-04-20 22:19:40

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

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information