wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> general problem-solving / chatting / whatever >> expressing numbers as sum of cubes
        
(Message started by: BenVitale on Jun 18th, 2009, 11:38am)
Title: expressing numbers as sum of cubes
Post by BenVitale on Jun 18th, 2009, 11:38am
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 (http://www.primepuzzles.net/puzzles/puzz_102.htm)

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?
Title: Re: expressing numbers as sum of cubes
Post by Obob on Jun 18th, 2009, 11:50am
Are you requiring distinct cubes, or limiting the number of cubes?  Otherwise there are many, many solutions.
Title: Re: expressing numbers as sum of cubes
Post by towr on Jun 18th, 2009, 12:24pm

on 06/18/09 at 11:50:14, Obob wrote:
Are you requiring distinct cubes, or limiting the number of cubes?  Otherwise there are many, many solutions.
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.
Title: Re: expressing numbers as sum of cubes
Post by pex on Jun 18th, 2009, 1:47pm

on 06/18/09 at 12:24:22, 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.

... and we need to limit the number of cubes to rule out appending +13+(-1)3+23+(-2)3+33+(-3)3+...
Title: Re: expressing numbers as sum of cubes
Post by Eigenray on Jun 18th, 2009, 7:49pm

on 06/18/09 at 11:38:53, BenVitale wrote:
It looks like that the number 100 (=10^2) is the smallest square which is also the sum of 4 consecutive cubes.

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,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif6); adding the torsion point (-3/2,0) gives (1,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mp.gif10).  But these are the only integer points).
Title: Re: expressing numbers as sum of cubes
Post by BenVitale on Jun 18th, 2009, 9:13pm
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]


Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board