Three Colour Analogue of Boolean Pythagorean Triples Problem

Question

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.

Answer 1

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.