Talk:Lagrange's four-square theorem

Problem with Jacobi's Combinations

I can't seem to find information on this elsewhere, but this seems to fail on very simple cases, such as the number 5. Since 5 is odd, and has only itself as a divisor (or do we count the trivial divisor 1?) We should have 8(5) = 40 possible combinations of sums of squares. But there exists only one: 2² + 1² + 2(0²) = 5 Assuming we allow all permutations of order, that only gives us 4!, and this is not unique to this decomposition either. Allowing squares of negative numbers naively seems to be an unnecessary complication, but I still don't see how this works. Explanation someone please? --71.11.128.29 01:36, 4 January 2007 (UTC)[reply]

Actually allowing all permutations of order above gave you 4!/2 = 12. But then allowing pluses and minuses on 1 and 2 multiplies the 12 by 4, giving 48 in all. Since 8*(1+5)=48, Jacobi's theorem is verified for this example. Hope this helps. Thanks for the thoughtful question.Rich 01:50, 4 January 2007 (UTC)[reply]

Question

Why does it matter if the integers are negative?

negative integers can't be represented as the sum of squares. Horndude77 19:02, 16 July 2005 (UTC)[reply]

Removal

I removed the following:

"In 2005, Zhi-Wei Sun proved that any natural number can be represented as the sum of a square, an even square and a triangular number."

This may or may not be true but why does it have to be here? -- Taku 01:11, July 19, 2005 (UTC)