## Problem Description

The Pythagorean Triples Coloring Problem^{1} takes in the set of all integers from $1\leq k\leq n$ for some value of $n$ and asks the question, is it possible to assign a color to every value in our set, such that no set of pythagorean triples is all the same color?

For example, the set $\{1,2,3,4,5\}$ can be split into coloring all of $\{1,2,3\}$ being red, and $\{4,5\}$ being blue. This will make it so that no Pythagorean triple is all contained in a single set. Not every number is part of a Pythagorean Triple (for example $1,2$) so those numbers can actually be ignored for this problem as they can go into any set.

There was this question before

Boolean Pythagorean Triples: For a set of integers $n\mathbb{Z}=\{1,2,3,\ldots, n\}$, does there exist two sets $A,B$ with $A\cup B=n\mathbb{Z}$ and $A\cap B=\varnothing$, such that for every Pythagorean Triple $a^2+b^2=c^2$ where $a,b,c\in n\mathbb{Z}$ all of $a,b,c$ at least $1$ of $a,b,c$ is in each of $A,B$?

It actually turns out that the answer to this problem is **no**. In fact this was proven in $2016$ that if $n\geq 7825$ then there does not exist a solution. However, the answer effectively was a brute force^{2} that checked a ridiculous amount of cases. In fact, the computation took $2$ days and spent $4$ years worth of computation power combined across all the CPUs.

This leads us to the following two open questions

### Current Work

Question: Does there exist an algorithm that can verify more quickly that there does not exist a solution for $n=7825$?

Question: Does there exist some amount of colors $k>2$ such that you can in fact color all Pythagorean triples for any value of $n$?^{3}