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

Home > Numbers
Diophantine And Almost Fermat (Posted on 2007-02-21) Difficulty: 3 of 5
Consider three positive integers x< y< z in arithmetic sequence.

Determine analytically all possible solutions of each of the following equations:

(I) x3 + z3 = y3 + 10yz

(II) x3 + y3 = z3 - 2, with y< 116.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Hints/Tips Hint | Comment 1 of 4

Since x<y<z are three positive integers  in arithmetic sequence (for both the parts I and II) , one may substitute:
x = y-a and z = y+a where a is a positive integer.................

Important
: Remember that each of x, y and z must be positive integers for both the parts.

Edited on February 22, 2007, 1:52 pm

Edited on February 22, 2007, 1:53 pm

Edited on February 22, 2007, 2:05 pm
  Posted by K Sengupta on 2007-02-22 13:51:46

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 (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information