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        riddles >> putnam exam (pure math) >> Sum of subsequent powers (is a power?)
        
(Message started by: JocK on Jan 21st, 2005, 2:21pm)
Title: Sum of subsequent powers (is a power?)
Post by JocK on Jan 21st, 2005, 2:21pm
For K>1 the equation  

[sum]k=1..K kn  =  Nn

has a solution with integer N for n=2: N=70, K=24.

Do integer N solutions also exist for n = 3, 4, 5, .. ?
Title: Re: Sum of subsequent powers (is a power?)
Post by Eigenray on Jan 21st, 2005, 4:15pm
For the case n=3, we're looking at
(K(K+1)/2)2 = N3.
N must be a square, N=m2, so we may write
K(K+1) = 2m3,
(2K+1)2 - 1 = (2m)3.
It's a special case of Catalan's conjecture that m=1, K=1 is the only solution.  I vaguely recall proving the special case
x2 - y3 = 1.
It might even be on the forum somewhere.  Or I could be thinking of another special case, 2x-3y=[pm]1.

[Edit: I think I was thinking of y3-x2=1, which is much easier:
y3 = x2+1 = (x+i)(x-i).
If d=gcd(x+i,x-i), d|2i.  So if d != 1, then 2|y implies 8|x2+1, a contradiction.  Thus x[pm]i have no factor in common, so each factor is a perfect cube, up to a unit:
x+i = ik(a+bi)3,
leading to either
a(a2-3b2) = +/- 1 or
b(3a2-b3) = +/- 1,
so one of a,b is +/- 1, and the other 0; thus x=0, y=1 is the only solution.  But this has nothing to do with the problem at hand]
Title: Re: Sum of subsequent powers (is a power?)
Post by Barukh on Jan 22nd, 2005, 7:34am
Has this problem been settled at last? I thought it is still unsolved...
Title: Re: Sum of subsequent powers (is a power?)
Post by Eigenray on Jan 22nd, 2005, 10:31am
A proof of Catalan's conjecture was given in April 2002 by Preda Mihailescu.  I haven't heard of any holes being found in it since then.
Title: Re: Sum of subsequent powers (is a power?)
Post by JocK on Jan 22nd, 2005, 4:36pm

on 01/21/05 at 16:15:55, Eigenray wrote:
For the case n=3, <..> It's a special case of Catalan's conjecture that m=1, K=1 is the only solution.


Nice. So the proof of Catalan's conjecture also proves that the cannon(hyper)ball conjecture:

For K>1 and n>1 the equation  1n + 2n + .. + Kn = Nn has no integer solutions other than n=2, K=24, N=70

holds for n=3. But what about n>3 ?

Just for clarity: I certainly have no proof for this cannon(hyper)ball conjecture (as you might have guessed!). I just stumbled upon http://mathworld.wolfram.com/CannonballProblem.html and started wondering about the obvious generalisation to higher dimensions. A few limited searches for solutions with n=3, 4, 5 and 6 left me empty handed. Hence the above conjecture.  

Can't imagine that this conjecture is actually new. Anyone who knows a reference?
Title: Re: Sum of subsequent powers (is a power?)
Post by THUDandBLUNDER on Jan 22nd, 2005, 10:24pm

Quote:
Anyone who knows a reference?

Of course (http://www.math.kyushu-u.ac.jp/~mkaneko/kaneko-tachibana.pdf).

 8)
Title: Re: Sum of subsequent powers (is a power?)
Post by JocK on Jan 23rd, 2005, 4:07am
Interesting, thanks! It seems, however, that this paper focusses on the generalisation of the cannonball problem to polygonals other than squares. It doesn't seem to touch the generalisation to higher dimensions (cubes and higher powers).
Title: Re: Sum of subsequent powers (is a power?)
Post by THUDandBLUNDER on Jan 23rd, 2005, 7:05am
This (http://www.math.niu.edu/~rusin/known-math/97/cube.sum) page analyses the more general problem of any sums of consecutive nth powers equalling an nth power.
It seems that there are no solutions for k = m to K when n > 2 and m = 1
However, there are solutions for m [smiley=ne.gif] 1
Title: Re: Sum of subsequent powers (is a power?)
Post by JocK on Jan 23rd, 2005, 8:52am

on 01/23/05 at 07:05:20, THUDandBLUNDER wrote:
It seems that there are no solutions for k = m to K when n > 2 and m = 1

Thanks for the link! Glancing over the text, it indeed seems that the status of the claim that there are no solutions for n>2 is still a conjecture, right?
Title: Re: Sum of subsequent powers (is a power?)
Post by THUDandBLUNDER on Jan 29th, 2005, 1:43am

on 01/23/05 at 08:52:10, JocK wrote:
Thanks for the link! Glancing over the text, it indeed seems that the status of the claim that there are no solutions for n>2 is still a conjecture, right?

I asked around, and nobody asserted that the conjecture is proven.
Title: Re: Sum of subsequent powers (is a power?)
Post by JocK on Jan 29th, 2005, 1:42pm
Hmm.. seems I missed an opportunity here. Should have posted:

----------------

I have discovered a truly remarkable proof which this message window is too small to contain that none of the equations:

1n + 2n = Nn
1n + 2n + 3n = Nn
1n + 2n + 3n + 4n = Nn
...

has integer solutions for n > 2.  

----------------

and then sit back and watch...  ;)
Title: Re: Sum of subsequent powers (is a power?)
Post by Packo on Jan 30th, 2005, 7:11am
Jock, in reply to your signature:

Some people are in a hurry to have sex and don't have the time to solve abstract problems elegantly...

There's this easy to use Minimize-function in Mathematica (Wolfram) to solve xy - y = x5 - y4 - y3 = 20; x>0, y>0.

Well, back to business then  :P
Title: Re: Sum of subsequent powers (is a power?)
Post by packo on Jan 31st, 2005, 1:41pm
aalctuly, oyu gte an estitimaon rhater tnha a sioltuon of urcose  ;)


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