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Generalized version of cannon ball problem

  1. Sep 19, 2010 #1
    For All p in Natural Number,
    Is [tex]\exists n , n > 1, \sum^{n}_{k=1} k^p = C^2 [/tex] where C is arbitary natural number (not constant) ??
     
  2. jcsd
  3. Sep 19, 2010 #2
    As far as I'm aware the only solution is:
    n = 24
    p = 2
    C = 70
     
  4. Sep 19, 2010 #3
    No, I checked with mathematica n < 100,000, p < 20

    there are some solutions such as

    p = 5,
    n = 13, 134, etc

    I think there are more solutions.. :)
     
  5. Sep 19, 2010 #4

    CRGreathouse

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    You're trying to solve a sequence of Diophantine equations. For a famous case, search for square triangular number.
     
  6. Sep 19, 2010 #5
    Yeah, I know

    But what I mean was, is it possible to find the solution for arbitary p ??
     
  7. Sep 19, 2010 #6

    CRGreathouse

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    Probably not. Diophantine equations are hard, in the sense of the negative answer to Hilbert's 10th.

    But for any given p it should be possible to at least formulate the problem in that form to see if anything can be discovered. So, for example, with p = 7 you have

    [tex]3x^8 + 12x^7 + 14x^6 - 7x^4 + 2x^2=24y^2[/tex]
     
  8. Sep 19, 2010 #7
    It seems like we cannot find C for arbitary p,

    But can we know the existence of C for arbitary p ??

    I don't need to find the entire solutions, just a single one.
     
  9. Sep 19, 2010 #8

    CRGreathouse

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    It's not clear that a solution exists for a given p. If not, I don't know of an easy way to prove it -- congruence conditions won't be enough, since n = 1 works and so there are always good congruence classes mod any prime power.
     
  10. Sep 20, 2010 #9
    Yeah, It looks almost impossible to use modular to prove...

    Do you know any related article about this??
     
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