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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    
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re(2): Solution - flaw? | Comment 3 of 9 |
(In reply to re: Solution - flaw? by 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

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