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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(5): Solution - flaw? | Comment 6 of 9 |
(In reply to re(4): Solution - flaw? by Jer)

Dear Jer (and Steve),

Jer asked 'Broll, it seems all you did was point out there are _at least_ two solutions for any number of consecutive cubes.'

I don't dissent from this actually - save that I also suggested a simple general approach that demonstrates the truth of the proposition.

As I understood it that's all this (D2) problem called for.

Steve asked 'Are you saying that Jer's solution is complete or incomplete?' I didn't say either; I just thought that Jer might be off-target, in terms of the problem as set. That's why I phrased my comment as I did.

Steve also asked 'And, if it is complete, are you saying that there are exactly two solutions for 5 or 6 or a trillion consecutive cubes?'

I was saying that there are always two. I didn't say that there are always exactly two. However, as I'm now being asked:

The nasty-looking a = (-1/8k((k^2-1)(k-2n-1)+(k-2n-1)^3))^(1/2) clearly has two integer solutions for the substitutions (n=k) and (n=k-1).

I don't for a microsecond believe there could be other solutions. However, I'll happily stand corrected if there are!

Please let me know if there are any other points that need clearing up.


Edited on June 20, 2013, 3:50 am
  Posted by broll on 2013-06-20 03:14:26

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