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Talk:Boolean Pythagorean triples problem

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Explanation Wrong

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I'm no mathematician, but I believe this article is completely wrong. It says "The proof tested all possible colouring of numbers up to 7,825 and found no such colouring was possible."

My understanding is that 7825 is a magical number (that still cannot be explained) where below that number it is possible and above not (or whatever else). But the statement that only up to 7825 was tested is completely wrong. If they only made a test until that number, it would mean that it's still not proven that there's no other solution with bigger numbers and the team solved the problem entirely, with a full proof (which nobody can explain though), so this explanation here is wrong. Can someone who really understands the problem explain it please? --165.222.184.143 (talk) 12:28, 30 May 2016 (UTC)[reply]

As I now understand the problem, I just fixed the main article. It's still difficult to understand - feel free to improve it. --165.222.184.143 (talk) 13:04, 30 May 2016 (UTC)[reply]

There is no need to consider bigger numbers: if one could colour 1,...,n for any n >= 7825 with two colours, such that there is no monochromatic Pythagorean triple, then just restrict that colouring to 1,...,7825, and you obtain one for the considered range (which is impossible, as proven). Oliver Kullmann (talk) 16:42, 5 June 2016 (UTC)[reply]

What number?

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The article says, "Up to the number 7824 it is...".

What number? Is it 'a', 'b', 'c', or the number of triples? Or what?

I'm sure to someone familiar with the field it's clear, but it could stand some clarification. Based on the 2^7824 statement, is it the count of individual integers in the set? — Preceding unsigned comment added by Erikb495 (talkcontribs) 06:44, 1 June 2016 (UTC)[reply]

The ArXiv paper has the clearest statement of the result.
Theorem 1. The set {1, . . . , 7824} can be partitioned into two parts, such that

no part contains a Pythagorean triple, while this is impossible for {1, . . . , 7825}.

It might be an idea to include that. --Salix alba (talk): 08:56, 1 June 2016 (UTC)[reply]
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It seems reasonable to create a "See also" paragraph, and have there a link to https://en.wikipedia.org/wiki/List_of_long_mathematical_proofs

Since I am one of the authors of the underlying paper, I don't know whether I should just go ahead (while the change seems safe to me). Oliver Kullmann (talk) 16:45, 5 June 2016 (UTC)[reply]