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 August 29th, 2013, 12:29 PM #1 Member   Joined: Apr 2013 Posts: 70 Thanks: 0 Express a cube as sum of squares .... Hi there, Is there a way to express n^3 as sum of squares and ns? n is an integer>1 Thank you.
 August 29th, 2013, 01:23 PM #2 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Express a cube as sum of squares .... You need to be more specific with this question. There is only one way of representing $n^3$ in terms of n, although it is possible that $n^3$ can be represented as a sum of squares. Trivially, every cube can be expressed as a sum of squares since 1 is a square. If you were to change this to "sum of distinct squares" then there would be many cubes which could not be so represented, 8 being the first. Another possibility is to restrict the number of squares. For instance, $2^3= 2^2 + 2^2$, so it is representable as a sum of two squares.
 August 29th, 2013, 01:34 PM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Express a cube as sum of squares .... You can express any nonnegative integer as a sum of four squares. Do you have something more specific in mind?
August 29th, 2013, 01:39 PM   #4
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Re: Express a cube as sum of squares ....

Quote:
 Originally Posted by CRGreathouse You can express any nonnegative integer as a sum of four squares.
Do you know how to prove that? I would be interested in seeing the argument.

 August 29th, 2013, 01:40 PM #5 Member   Joined: Apr 2013 Posts: 70 Thanks: 0 Re: Express a cube as sum of squares .... Sorry. Im talking about a general formula. n^3 is expressed by ns (squares and degree 1)
August 29th, 2013, 01:42 PM   #6
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Re: Express a cube as sum of squares ....

Quote:
 Originally Posted by Mouhaha Sorry. Im talking about a general formula. n^3 is expressed by ns (squares and degree 1)
There is no formula in terms of a quadratic function that would work for all n. This is what I meant when I said the only way to represent $n^3$ in terms of n is $n^3$.

August 29th, 2013, 01:44 PM   #7
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Re: Express a cube as sum of squares ....

Quote:
Originally Posted by icemanfan
Quote:
 Originally Posted by Mouhaha Sorry. Im talking about a general formula. n^3 is expressed by ns (squares and degree 1)
There is no formula in terms of a quadratic function that would work for all n. This is what I meant when I said the only way to represent $n^3$ in terms of n is $n^3$.
If so then I have maybe found something very interesting.

 August 29th, 2013, 03:23 PM #8 Member   Joined: Apr 2013 Posts: 70 Thanks: 0 Re: Express a cube as sum of squares .... please read it carefully how to express n^3 as elegant sum of squares and ns? Let n=2k an even integer C(n,3)=(n*(n-1)*(n-2))/6 Here is an identity to check : C(n,3)=sigma(2i^2)) with i=1 to n-1 Example : C(8,3)=2^2+4^2+6^2=56 Let n=2k+1 an odd integer Here is an identity to check : C(n,3)=sigma((2i+1)*^2) with i=0 to n-1 Example : C(11,3)=1^2+3^2+5^2+7^2+9^2=165 Now you can easily deduce n^3 and express it as said above. I will let you do it and write it in Latex.
 August 29th, 2013, 03:52 PM #9 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Express a cube as sum of squares .... If you're allowed unbounded sums there are thousands of identities. You can check this one with Faulhaber's formula, if you like.
August 29th, 2013, 03:54 PM   #10
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Re: Express a cube as sum of squares ....

Quote:
 Originally Posted by CRGreathouse If you're allowed unbounded sums there are thousands of identities. You can check this one with Faulhaber's formula, if you like.
It the identity correct?

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