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

Home > Numbers
Any others? (Posted on 2013-06-18) Difficulty: 2 of 5
100 is a square which is also a sum of 4 consecutive cubes.
Are there any others?

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Solution - flaw? | Comment 3 of 9 |
(In reply to re: Solution - flaw? by Jer)

Jer,

You might not want to spend too much further time investigating this.

Let the square sought be a^2. The sum of the first n cubes is a square, ((n(n+1))/2)^2. So is the sum of the first (n-4) cubes, (((n-4)(n-3))/2)^2, say b^2. The intriguingly Pythagorean look of this is misleading, as a^2+b^2 = c^2 is just as true if a=c and b=0. Here, it's just a matter of zeroing the factors of b to allow the two remaining squares, a and c, to equal each other. Naturally the same is also true if one asks for the sum of 5 cubes, or 6 cubes, or even a trillion cubes.

Edited on June 19, 2013, 7:01 am
  Posted by broll on 2013-06-19 06:06:31

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