## Generalized version of cannon ball problem

For All p in Natural Number,
Is $$\exists n , n > 1, \sum^{n}_{k=1} k^p = C^2$$ where C is arbitary natural number (not constant) ??

 Quote by kevin0960 For All p in Natural Number, Is $$\exists n , n > 1, \sum^{n}_{k=1} k^p = C^2$$ where C is arbitary natural number (not constant) ??
As far as I'm aware the only solution is:
n = 24
p = 2
C = 70

 Quote by BruceG As far as I'm aware the only solution is: n = 24 p = 2 C = 70
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.. :)

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## Generalized version of cannon ball problem

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

 Quote by CRGreathouse You're trying to solve a sequence of Diophantine equations. For a famous case, search for square triangular number.
Yeah, I know

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

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 Quote by kevin0960 But what I mean was, is it possible to find the solution for arbitary p ?
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

$$3x^8 + 12x^7 + 14x^6 - 7x^4 + 2x^2=24y^2$$

 Quote by 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 $$3x^8 + 12x^7 + 14x^6 - 7x^4 + 2x^2=24y^2$$
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.

 Recognitions: Homework Help Science Advisor 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.

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