Talk:Lander, Parkin, and Selfridge conjecture

New page
Hi, I've created this page; a small lead section has been added. This is quite a technical topic, feel free to suggest how to make it more accessible. Cheers, BRE [forgot to sign originally] Breggen (talk) 01:24, 3 August 2013 (UTC)

Current status
The sentence It is not known if ... solutions exist that would be counterexamples ... (other than the solutions that follow from the commutative and reflexive properties; there are solution for n = 2, 3, 4). It's not clear what this means exactly? And what is the reference? Spectral sequence (talk) 20:46, 1 August 2013 (UTC)


 * This was added by a 3rd party, I can elaborate. I think in my version I simply had "non-trivial". Breggen (talk) 01:23, 3 August 2013 (UTC)


 * The passage other than the solutions that follow from the commutative and reflexive properties was added by Georgia Guy on 23 July. I reverted it today (albeit a tad belatedly), and he put it back in again. Georgia Guy, I ask again that you please remember not to put original research into Wikipedia articles. Also, please remember WP:BRD -- Bold (which you were when you put it in), Revert (which I just did), and then Discuss, which you should do next here on the talk page rather than undoing a reversion.


 * Perhaps you could give an example of a solution that follows from the commutative and reflexive properties, requiring the caveat.


 * I'm taking it out again, as per WP:BRD and because there is a consensus in this section of the talk page that the passage is incomprehensible and unreferenced. Duoduoduo (talk) 16:18, 16 August 2013 (UTC)


 * Commutative property: 2^5 + 3^5 = 3^5 + 2^5 (a+b = b+a)
 * Reflexive property: 2^5 + 3^5 = 2^5 + 3^5 (a = a)


 * Georgia guy (talk) 16:20, 16 August 2013 (UTC)


 * These are not counterexamples to the conjecture, which is that


 * if $$\sum_{i=1}^{n} a_i^k = \sum_{j=1}^{m} b_j^k$$, where ai ≠ bj are positive integers for all 1 ≤ i ≤ n and 1 ≤ j ≤ m, then m+n ≥ ''k


 * since they violate ai ≠ bj. Duoduoduo (talk) 16:59, 16 August 2013 (UTC)

An similar conjecture
Conjecture: if ±(a_1)^n±(a_2)^n±(a_3)^n±...±(a_m)^n=0 (all a_i are positive integers) and m > 2, then n ≤ 2m-2. Is it right? That is, if $$\sum_{i=1}^{n} a_i^k = \sum_{j=1}^{m} b_j^k$$, where ai ≠ bj are positive integers for all 1 ≤ i ≤ n and 1 ≤ j ≤ m, then 2m+n-2 ≥ k. — Preceding unsigned comment added by 49.214.6.73 (talk) 08:43, 16 April 2015 (UTC)