Generalized version of cannon ball problem

kevin0960

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) ??

BruceG

As far as I'm aware the only solution is:
n = 24
p = 2
C = 70

kevin0960

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.. :)

CRGreathouse

You're trying to solve a sequence of Diophantine equations. For a famous case, search for square triangular number.

kevin0960

Yeah, I know

But what I mean was, is it possible to find the solution for arbitary p ??

CRGreathouse

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]

kevin0960

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.

CRGreathouse

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.

kevin0960

Yeah, It looks almost impossible to use modular to prove...

Do you know any related article about this??