
#1
Sep1910, 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) ?? 



#2
Sep1910, 10:55 AM

P: 40

n = 24 p = 2 C = 70 



#3
Sep1910, 06:43 PM

P: 12

there are some solutions such as p = 5, n = 13, 134, etc I think there are more solutions.. :) 



#4
Sep1910, 07:07 PM

Sci Advisor
HW Helper
P: 3,680

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.




#5
Sep1910, 07:15 PM

P: 12

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



#6
Sep1910, 07:52 PM

Sci Advisor
HW Helper
P: 3,680

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] 



#7
Sep1910, 08:47 PM

P: 12

But can we know the existence of C for arbitary p ?? I don't need to find the entire solutions, just a single one. 



#8
Sep1910, 10:01 PM

Sci Advisor
HW Helper
P: 3,680

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.




#9
Sep2010, 04:37 AM

P: 12

Do you know any related article about this?? 


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