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

Home > Just Math
Quartic Quandary (Posted on 2011-08-23) Difficulty: 3 of 5
Consider three positive integers x < y < z in arithmetic sequence, and determine all possible triplets (x,y,z) that satisfy this equation:

x4 + x2y2 = z4 76

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Comment 1 of 1

I If (x-a)^2*x^2+76 = (x+a)^4 -(x-a)^4, then

II x(8a^3-a^2 x+10ax^2-x^3)=76, with x={1,2,4,19,38,76},

when {a=1,x=2} is the only solution in the positive integers.

  Posted by broll on 2011-08-24 06:39:28
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 (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information