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Topic: expressing numbers as sum of cubes (Read 676 times) |
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Benny
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expressing numbers as sum of cubes
« on: Jun 18th, 2009, 11:38am » |
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I was wondering how many numbers I could find that could be the sum of cubes. I came across this web page:
Three squares a cube, all primes
3^2 + 3^2 + 3^2 = 3^3
3^2 + 19^2 + 31^2 = 11^3
3^2 + 691^2 + 2293^2 = 179^3
3^2 + 5869^2 + 54959^2 = 1451^3
3^2+ 24967^2 + 60169^2 = 3^2 + 28163^2 + 58741^2 = 1619^3
3^2 + 13127^2 + 121229^2 = 2459^3
I wanted to express the number 100 as the sum of cubes. I found:
(190^3) - (161^3) - (139^3) = 100
and
100 = = 1^3 + 2^3 + 3^3 + 4^3
It looks like that the number 100 (=10^2) is the smallest square which is also the sum of 4 consecutive cubes.
I was told that there are only 3 known answers, 3 ways to express the number 100 as the sum of cubes (we can allow each cube to be positive or negative)
Does anyone know the 3rd way?
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| « Last Edit: Jun 18th, 2009, 2:13pm by Benny » |
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Obob
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Re: expressing numbers as sum of cubes
« Reply #1 on: Jun 18th, 2009, 11:50am » |
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Are you requiring distinct cubes, or limiting the number of cubes? Otherwise there are many, many solutions.
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| « Last Edit: Jun 18th, 2009, 11:50am by Obob » |
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towr
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Re: expressing numbers as sum of cubes
« Reply #2 on: Jun 18th, 2009, 12:24pm » |
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on Jun 18th, 2009, 11:50am, Obob wrote:| Are you requiring distinct cubes, or limiting the number of cubes? Otherwise there are many, many solutions. |
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I think we can safely assume they must be distinct cubes, other wise you can just keep appending +13+(-1)3, which is really uninteresting.
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Wikipedia, Google, Mathworld, Integer sequence DB
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pex
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Re: expressing numbers as sum of cubes
« Reply #3 on: Jun 18th, 2009, 1:47pm » |
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on Jun 18th, 2009, 12:24pm, towr wrote:| I think we can safely assume they must be distinct cubes, other wise you can just keep appending +13+(-1)3, which is really uninteresting. |
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... and we need to limit the number of cubes to rule out appending +13+(-1)3+23+(-2)3+33+(-3) 3+...
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Eigenray
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Re: expressing numbers as sum of cubes
« Reply #4 on: Jun 18th, 2009, 7:49pm » |
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on Jun 18th, 2009, 11:38am, BenVitale wrote:| It looks like that the number 100 (=10^2) is the smallest square which is also the sum of 4 consecutive cubes. |
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Actually
62 = 03 + 13 + 23 + 33,
but there can be only finitely many solutions. Actually, according to Magma, these are the only two (the elliptic curve y2 = x3 + (x+1)3 + (x+2)3 + (x+3)3 has rank 1, generated by (0, 6); adding the torsion point (-3/2,0) gives (1, 10). But these are the only integer points).
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| « Last Edit: Jun 18th, 2009, 7:49pm by Eigenray » |
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Benny
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Re: expressing numbers as sum of cubes
« Reply #5 on: Jun 18th, 2009, 9:13pm » |
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Earlier, I found these sums:
100 = 1903 - 1613 - 1393, and
100 = = 13 + 23 + 33 + 43
And now, I found these sums:
100 = 73 - 33 - 63
100 = 18703 - 9033 - 17973
I was hoping to find sums of 4 cubes, 5 at most.
Since it is known that every integer is a sum of at most 5 signed cubes.
It is believed that 5 can be reduced to 4, so that
N = A3 + B3 + C3 + D3
for any number N, to the exception of numbers of the form (9n + 4) and (9n - 4)
[not proven yet]
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| « Last Edit: Jun 18th, 2009, 9:16pm by Benny » |
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