# The Boolean Pythagorean triples problem, a $200$-terabyte proof, and $d=163$

I came across this interesting math article,

where one phrase caught my attention and I quote, "... all triples could be multi-coloured in integers up to $7824$". Alternatively, from page 2 of this paper,

Theorem 1. The set {$1,\dots,7824$} can be partitioned into two parts, such that no part contains a Pythagorean triple. This is impossible for {$1,\dots,7825$}.

The number $N=7824$ was awfully familiar. A quick factorization showed that it was in fact,

$$N = 7824 = 2^4 \times 3\times \color{blue}{163}$$

Questions:

1. Does anybody know why the largest Heegner number $163$ figures in the largest $N$ that can be multi-colored in the Boolean Pythagorean triples problem?
2. A272709 is the sequence $2, 4, 8, 16, 24, 48, 96, 192,....0,0,0,0,0\dots$ where the zeros start at $a(7825)$. What is the exact value of $a(7824)$? (In the comments, it just says $a(7824)\geq8$.)
• Sounds like pure coincidence. Do you know of any other relation between these two (seemingly unrelated) issues? BTW, I'd fix it to "Does anybody know..." (pardon my nagging). May 31 '16 at 7:31
• @titopiezasIII may be MO knows?
– Mr.
May 31 '16 at 7:37
• Grammar has been corrected. :) I've done some work investigating the properties of $163$ and I've noticed it sometimes appears in the most intriguing contexts. Do you know, for example, that the moonshine functions span a linear space of exactly $163$ dimensions? Of course, it may be just pure coincidence. But then again.. May 31 '16 at 7:41
• @barakmanos: Well, Euler's polynomial $F(n)=n^2+n+41$ implies the smallest solution to the Pythagorean-like $a^2+b^2=c^2-163$ is $1^2+40^2=42^2-163$. Tangential, but it combines Diophantine equations with $d=163$. May 31 '16 at 7:53
• $a(7824) \ge 8$ because the article says it has a colouring and there are obvious symmetries: you can switch the colours, and $1$ and $2$ can be coloured arbitrarily because they are not in any Pythagorean triples. Better lower bounds shouldn't be hard to find, but I would guess that computing the exact value is several orders of magnitude harder than proving $a(7825)=0$. May 31 '16 at 17:05

It is not difficult at all to show that $a(7824)$ is immensely huge. For example, many numbers do not appear in any pythagorean triple of numbers up to 7824. These can be put in any partition. More precisely, in the article (arXiv:1605.00723, section 6.3) they say they found a solution of 7824 with 1567 free variables. I guess these are boolean variables, so this gives at least $a(7824)\geq 2^{1567}$.
(On a side note, let me share a remark on the appearance of 163. Neither the number 7824 nor the set $\{1,\ldots,7824\}$ look anyhow special to this problem. For instance, the number 7824 is one of the numbers that can be put in any side of the partition. The true special number here is 7825, together with the combinatorial complexity of the Pythagorean triples containing it, etcetera. There is a beautiful system of triples (not involving 7824) that is an obstruction to the problem, and that in fact allows a partition as soon as you remove the 7825. Therefore, I would rather seek for a pattern for $a(163 k+1)$... or better look at the factorization $7825=5^2\times 313$.)