The discussion revolves around the generalized version of the cannonball problem, specifically exploring whether there exist natural numbers \( n > 1 \) and \( C \) such that the sum of the first \( n \) natural numbers raised to the power \( p \) equals \( C^2 \) for all natural numbers \( p \). The scope includes theoretical exploration of Diophantine equations and potential solutions for various values of \( p \).
Participants express differing views on the existence of solutions for arbitrary \( p \). While some suggest that solutions may exist for specific values of \( p \), others argue that proving the existence of solutions in general is likely very challenging and may not be possible.
There are limitations regarding the assumptions made about the existence of solutions and the applicability of modular arithmetic in proving them. The discussion does not resolve these complexities.
kevin0960 said: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 said:As far as I'm aware the only solution is:
n = 24
p = 2
C = 70
CRGreathouse said:You're trying to solve a sequence of Diophantine equations. For a famous case, search for square triangular number.
kevin0960 said:But what I mean was, is it possible to find the solution for arbitary p ?
CRGreathouse said: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
CRGreathouse said: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.