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

Home > Just Math
Cubic Equation (Posted on 2013-06-09) Difficulty: 4 of 5
Determine all triplets (a,b,c) of nonzero integers satisfying:
987654321*a + 123456789*b + c = (a + b + c)3

Prove that there are no others.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
A simplification? | Comment 10 of 13 |
Let x = (a+b+c)

Then 987654321*a + 123456789*b + (x-a-b) = x^3

Then 987654320*a + 123456788*b = x^3 - x = (x-1)x(x+1)

The GCD of 987654320 and 123456788 is 4, so it appears that the equation has integer solutions whenever (x-1)x(x+1) is a multiple of 4. Three consecutive integers are a multiple of 4 whenever the middle integer is odd or a multiple of 4.  In other words, the equation has a solution whenever x <> 2 (mod 4).  There are an infinite number of solutions, even after subtracting out that small fraction of them where a or b or c = 0. 

Edited on June 13, 2013, 10:49 pm
  Posted by Steve Herman on 2013-06-11 09:56:22

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 (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

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