# Three Colour Analogue of Boolean Pythagorean Triples Problem

Having read about the Boolean Pythagorean Triples Problem (see here and this question), it occurred to me that a related problem would require the integers to be coloured in three rather than two colours, with each triple containing all three colours. Formally, this problem is to find $N$ such that the following holds:

The set {$1,\dots,N$} can be partitioned into three parts such that every Pythagorean Triple $a^2+b^2=c^2$ with $c\leq N$ contains one integer from each part, and this is impossible for {$1,\dots,N+1$}.

Question: What is $N$?

This problem should be much simpler than the Boolean problem (because the three colour condition is more demanding, limiting the possibilities to be considered), so I expected that it must have been solved long ago. However, my web search did not find any reference to the problem, so I made an attempt to solve it and found the following colouring showing that $N$ is at least 110.

Note that 36 and 105 are both green, so with this colouring the next triple (36,105,111) fails to meet the requirement. It may be that $N$ is 110. But clearly the above does not prove that since there may be another colouring which works beyond 110.

Indeed $N=110$. The triples up to $111$ cannot be coloured. These $60$ triples are far from minimal; there are $65$ subsets of $19$ of them that cannot be coloured. All of these minimal subsets contain the following $14$ triples:

( 5, 12, 13)
(12, 16, 20)
(13, 84, 85)
(15, 20, 25)
(16, 63, 65)
(21, 28, 35)
(25, 60, 65)
(28, 96,100)
(35, 84, 91)
(36,105,111)
(40, 75, 85)
(40, 96,104)
(60, 80,100)
(63, 84,105)


and $5$ from among the following $14$ triples:

( 9, 12, 15)
( 9, 40, 41)
(12, 35, 37)
(15, 36, 39)
(20, 21, 29)
(20, 48, 52)
(21, 72, 75)
(27, 36, 45)
(36, 48, 60)
(36, 77, 85)
(39, 52, 65)
(45, 60, 75)
(60, 63, 87)
(60, 91,109)


Here's one example of an uncolourable subset of $19$ triples:

( 5, 12, 13)
( 9, 12, 15)
(12, 16, 20)
(13, 84, 85)
(15, 20, 25)
(16, 63, 65)
(21, 28, 35)
(25, 60, 65)
(28, 96,100)
(35, 84, 91)
(36, 48, 60)
(36, 77, 85)
(36,105,111)
(40, 75, 85)
(40, 96,104)
(45, 60, 75)
(60, 80,100)
(60, 91,109)
(63, 84,105)


These subsets are minimal in the sense that all subsets of up to $18$ of the $60$ triples up to $c=111$ can be coloured. I don't know whether there are smaller uncolourable sets including triples with higher $c$.

Here's the code I used to obtain these results.

• Thank you. I don't have sufficient programming knowledge to be able to check your code, but was able to confirm 'manually' that your example subset of 19 triples is indeed uncolourable (the details would be rather long to post here), so have accepted your answer on that basis. – Adam Bailey Jun 13 '16 at 14:10