Generalized version of cannon ball problem

[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) ??

 

[BruceG]

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

 

[kevin0960]

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.. :)

 

[CRGreathouse]

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

 

[kevin0960]

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 ??

 

[CRGreathouse]

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

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

 

[kevin0960]

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.

 

[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]

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...

Do you know any related article about this??