Talk:Lagrange's four-square theorem

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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" gave you 4!/2 = 12, because 2 of the 4 numbers being permuted are identical(0 and 0). But then allowing pluses and minuses on 1 and 2 multiplies the 12 by 4, giving 48 in all. Since 8 times (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]

The article should very clearly explain that Jacobi's formula counts integer (as opposed to natural) solutions. To the best of my knowledge, there is no useful formula for the number of the natural solutions, although Emil Grosswald wrote a book on the subject of counting natural solutions for the case of representations of an integer as a sum of $k$ squares. 129.15.11.123 06:56, 7 March 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]

He was referring to "More formally, for every positive integer n there exist non-negative integers x1, x2, x3, x4". Since negatives square always square up to positives I don't see any reason why the should be non-negative. I've edited accordingly. --poorsod^talk 16:53, 23 May 2007 (UTC)[reply]

Hurwitz quaternions is a non commutative ring. unique factorization domain is used in case of commutative rings. Why that is used here? I want to point out that unique factorization domain, called UFD for short, is any integral domain in which every nonzero noninvertible element has a unique factorization. Hurwitz quaternions is not UFD. — Preceding unsigned comment added by 110.234.113.82 (talk) 08:36, 4 May 2012 (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)

Word kevinbocking 22:18, 3 July 2007 (UTC)[reply]

Proof

The proof appears in ja article. I might translate it to here if I have time. Anyone interested to see it? -- Taku 07:40, 18 July 2007 (UTC)[reply]

Fermat's proof?

The article says (uncited) "An earlier proof by Fermat was never published." Does this proof definitely exist, or is it another of those marginal notes like the one about his last theorem? AndrewWTaylor 17:20, 24 October 2007 (UTC)[reply]

It does indeed exist. According to Wolfram's MathWorld: "Although the theorem was proved by Fermat using infinite descent, the proof was suppressed. Euler was unable to prove the theorem. " —Preceding unsigned comment added by 71.88.108.163 (talk) 03:18, 8 July 2008 (UTC)[reply]

Ramanujan's General Solution

By symmetry, the statement "if we assume, without loss of generality, that a ≤ b ≤ c ≤ d then there are exactly 54 possible choices for a, b, c and d such that (*) is solvable in integers x1, x2, x3, x4 for all n" is meaningless. Each x_n is a square.

Borat fan (talk) 23:36, 17 July 2008 (UTC)[reply]

Jacobi's Four Square Theorem

In the article currently, it says that the total number of ways a given positive integer n can be represented as the sum of four squares is eight times the sum of the divisors of n if n is odd and 24 times the sum of the odd divisors of n if n is even. Other sources, e.g. this one say it is eight times the sum of all its divisors which are not divisible by 4. These are equivalent (although not immediately obvious in some cases). It might be helpful to give both formulas. MSGJ (talk) 13:38, 11 August 2008 (UTC)[reply]

On reflection I have moved the stuff on Jacobi's theorem to a new article Jacobi's four-square theorem because although related to the Lagrange theorem, it seems not to belong here. MSGJ (talk) 13:38, 11 August 2008 (UTC)[reply]

There are some gaps in this 'proof' !

I think I have found some gaps in this pseudo-proof. First of all: where do we use the fact that 'The generalized Euclidean algorithm' works ? The second one: I am not sure but I think that Hurwitz integers haven't got a property of unique factorization. Because: 2 = (1+i)(1-i) = (1-j)(1+j) = (1-k)(1+k) The last one: even if this proof is correct, we only show that 'p' is a sum of four square of integers or four square of half-integer. — Preceding unsigned comment added by 89.231.20.14 (talk) 08:29, 9 July 2011 (UTC)[reply]

Also the norm link is wrong, it goes to the metric space definition instead of the group theory one. — Preceding unsigned comment added by 46.64.84.54 (talk) 12:12, 31 October 2011 (UTC)[reply]

The proof is flawed. Hurwitz integers do have a kind of unique factorization, but it's not the kind associated with commutative rings. In particular, it's unique up to unit migration, meta-commutation, and recombination. More information can be found here [1]. Conway and Smith also covered this here [2].