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Topic: Sum of subsequent powers (is a power?) (Read 1580 times) |
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JocK
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Sum of subsequent powers (is a power?)
« on: Jan 21st, 2005, 2:21pm » |
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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, .. ?
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Eigenray
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Re: Sum of subsequent powers (is a power?)
« Reply #1 on: Jan 21st, 2005, 4:15pm » |
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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]
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| « Last Edit: Jul 8th, 2008, 4:16am by Eigenray » |
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Barukh
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Re: Sum of subsequent powers (is a power?)
« Reply #2 on: Jan 22nd, 2005, 7:34am » |
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Has this problem been settled at last? I thought it is still unsolved...
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Eigenray
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Re: Sum of subsequent powers (is a power?)
« Reply #3 on: Jan 22nd, 2005, 10:31am » |
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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.
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JocK
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Re: Sum of subsequent powers (is a power?)
« Reply #4 on: Jan 22nd, 2005, 4:36pm » |
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on Jan 21st, 2005, 4:15pm, 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. |
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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?
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| « Last Edit: Jan 22nd, 2005, 4:39pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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ThudnBlunder
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Re: Sum of subsequent powers (is a power?)
« Reply #5 on: Jan 22nd, 2005, 10:24pm » |
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Quote:| Anyone who knows a reference? |
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Of course.
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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JocK
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Re: Sum of subsequent powers (is a power?)
« Reply #6 on: Jan 23rd, 2005, 4:07am » |
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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).
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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ThudnBlunder
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Re: Sum of subsequent powers (is a power?)
« Reply #7 on: Jan 23rd, 2005, 7:05am » |
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This 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
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| « Last Edit: Jan 23rd, 2005, 7:31am by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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JocK
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Re: Sum of subsequent powers (is a power?)
« Reply #8 on: Jan 23rd, 2005, 8:52am » |
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on Jan 23rd, 2005, 7:05am, THUDandBLUNDER wrote:
It seems that there are no solutions for k = m to K when n > 2 and m = 1
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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?
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| « Last Edit: Jan 23rd, 2005, 8:52am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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ThudnBlunder
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Re: Sum of subsequent powers (is a power?)
« Reply #9 on: Jan 29th, 2005, 1:43am » |
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on Jan 23rd, 2005, 8:52am, 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? |
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I asked around, and nobody asserted that the conjecture is proven.
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JocK
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Re: Sum of subsequent powers (is a power?)
« Reply #10 on: Jan 29th, 2005, 1:42pm » |
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Hmm.. seems I missed an opportunity here. Should have posted:
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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.
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and then sit back and watch... 
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Packo
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Re: Sum of subsequent powers (is a power?)
« Reply #11 on: Jan 30th, 2005, 7:11am » |
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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 
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packo
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Re: Sum of subsequent powers (is a power?)
« Reply #12 on: Jan 31st, 2005, 1:41pm » |
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aalctuly, oyu gte an estitimaon rhater tnha a sioltuon of urcose
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