Generalized version of cannon ball problem


by kevin0960
Tags: ball, cannon, generalized, version
kevin0960
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#1
Sep19-10, 10:14 AM
P: 12
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) ??
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BruceG
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#2
Sep19-10, 10:55 AM
P: 40
Quote Quote by kevin0960 View Post
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) ??
As far as I'm aware the only solution is:
n = 24
p = 2
C = 70
kevin0960
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#3
Sep19-10, 06:43 PM
P: 12
Quote Quote by BruceG View Post
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
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#4
Sep19-10, 07:07 PM
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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.
kevin0960
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#5
Sep19-10, 07:15 PM
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Quote Quote by CRGreathouse View Post
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
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#6
Sep19-10, 07:52 PM
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Quote Quote by kevin0960 View Post
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

[tex]3x^8 + 12x^7 + 14x^6 - 7x^4 + 2x^2=24y^2[/tex]
kevin0960
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#7
Sep19-10, 08:47 PM
P: 12
Quote Quote by CRGreathouse View Post
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]
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
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#8
Sep19-10, 10:01 PM
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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.
kevin0960
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#9
Sep20-10, 04:37 AM
P: 12
Quote Quote by CRGreathouse View Post
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??


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