Talk:Jacobi–Madden equation

Restrictions and Possibilities on the Numbers

Obviously, the equation is true if a = b = c = d = 0. Also, in the set { a, b, c, d }, if three of the four are zero, then the equation is true. If two of the four are zero, then we get $a^{4}+b^{4}=(a+b)^{4}$ , and Pierre de Fermat, himself, showed that this one is impossible for all nonzero numbers { a, b, c }, with $a^{4}+b^{4}=c^{4}$ .

If one of the numbers is zero, then we get $a^{4}+b^{4}+c^{4}=(a+b+c)^{4}$ . This is obviously not true for positive integers, because the right-hand side of the equation would be too large. Possibly,it might be true if one of them is allowed to be negative.

If none of the numbers is zero, then we get $a^{4}+b^{4}+c^{4}+d^{4}=(a+b+c+d)^{4}$ . This is also obviously not true for positive integers, because the right-hand side of the equation would be too large. It has now been shown that there are infinitely-many solutions if some of the numbers are allowed to be negative, and some positive.

This is the new section that I have put in today.74.163.40.58 (talk) 19:41, 29 September 2008 (UTC)[reply]

Hi 74.163.40.58, I removed this section. If you disagree, just put it back and I won't remove it again. I'll explain you why: as you say yourself, the restrictions on the solutions you mention are rather obvious, and don't give any extra insight into the problem. Even the observation that there are no solutions with c and d equal to 0 is very easy, you don't need Fermat's result and can just remark that a solution (a,b) with a and b coprime is equal to a zero of

2a^{2}+3ab+2b^{2}

(which is the difference between the LHS and the RHS divided by

2ab

), which doesn't have nonzero solutions modulo 3. Doetoe (talk) 11:31, 24 October 2008 (UTC)[reply]

I removed the text about Euler's sum of powers conjecture, since that already has its own page. Doetoe (talk) 11:31, 24 October 2008 (UTC)[reply]

Relation of the 88 solutions to the general solution

Does anybody know if the 88 solutions previously found to the problem could be obtained using the general recursive method? In either case it is worth mentioning. Sonoluminesence (talk) 12:24, 25 February 2009 (UTC)[reply]

Firstly, there were more than 88 solutions known. The number 88 comes purely from the MathWorld article [1] that lists just some of the known solutions, however there are other resources, such as Jaroslaw Wroblewski's page [2] with many more solutions.

Secondly, no -- these known solutions cannot be obtained by a recursive method (or at least nobody knows how to do that) -- they were obtained by more or less optimized exhaustive search. Maxal (talk) 03:14, 27 May 2009 (UTC)[reply]

Why Euler's?

I've read the paper, it nowhere says that the equation $a^{4}+b^{4}+c^{4}+d^{4}=(a+b+c+d)^{4}$ was introduced/considered/studied by Euler. Euler conjectured that more general equation $a^{4}+b^{4}+c^{4}+d^{4}=e^{4}$ has integer solutions but that does not mean that every possible special case of this equation should be named after him. Except UPI.com, nobody else (including Jacobi and Madden) calls $a^{4}+b^{4}+c^{4}+d^{4}=(a+b+c+d)^{4}$ "Euler's equation". Therefore, I think "Euler's" should be dropped from the title of this article. It's much more meaningful to name this equation after Jacobi and Madden since they basically introduced and solved it. I propose a new title "Jacobi-Madden equation". Maxal (talk) 00:57, 18 May 2009 (UTC)[reply]

Renamed accordingly. Maxal (talk) 03:05, 27 May 2009 (UTC)[reply]

There is a contradiction point

"first proposed in 1772 by Leonhard Euler who conjectured that four is the minimum number (greater than one) of fourth powers of non-zero integers that can sum up to another fourth power."
That's weird. in the example below,allows negative integers actually. but it said "greater than one"? should it be "Absolute value greater than one"? — Preceding unsigned comment added by Råy kuø (talk • contribs) 12:33, 11 February 2017 (UTC)[reply]

The modifier "greater than one" applies to "four", not to "fourth powers". --JBL (talk) 14:24, 11 February 2017 (UTC)[reply]

Oh,i see. Thank you.--Råy kuø 6:18, 12 February 2017 (UTC)