下記は、Jamie Condliffe – Gizmodo US[原文]から掲載しています。

If you think you had a hard time filling out pages of algebra at school, spare a thought for the three mathematicians who have just published the world’s largest ever proof. It takes up 200TB of storage space.

Nature reports that the team — from the Universities of Texas at Austin, Kentucky and Swansea — made use of bounteous computing resources to solve the Boolean Pythagorean triples problem. What the hell is that, I hear you scream? Seeing as though you’re interested, it asks:
Natureがそのチームを報告します テキサス大学オースティン、ケンタッキー大学とスウォンジー大学から – ブールピタゴラス3整数問題を解くために豊富なコンピューティングリソースを利用した

Is it possible to colour all the integers either red or blue so that no pythagorean triple of integers a, b, c, satisfying a^2+b^2=c^2 are all the same colour?
整数の全てに赤か青のどちらかを付けて a^2+b^2=c^2を満足するピタゴラス3整数a, b, c, が全て同じ色にならないように色づけることは可能でしょうか?

The puzzle was actually set by mathematician Ronald Graham in the 1980s, and he offered $100 to anyone who could find answer. The trio of researchers have already claimed the reward.
Turns out that the answer to the puzzle is: No. But to reach that simple conclusion, the team had to work through combinations of integers all the way up to 7,825. (The answer was ‘yes’ up until 7,824.) However, by the time you reach 7,825, it turns out there are more than 10^2,300 possible ways to colour all those integers. The team used some mathematical tricks to simplify the situation, but it still left 1 trillion combinations to check.

The team used the University of Texas’s Stampede supercomputer to churn through all the combinations, utilising 800 processors over the course of two days to create 200TB of data. (The previous record was a measly 13GB.)

And, of course, that simple answer: No.
そして もちろん 答えは単純でノーです。

These kinds of computer-aided proofs are increasingly common in mathematics, though there is some debate over whether they’re maths proofs in the truest sense. Still, most mathematicians can probably agree that the quantity of data required to reach this particular solution was simply to large for any human to every generate.


自然数の組 (a, b, c) が原始ピタゴラス数であるためには、ある自然数 m, n が
m と n は互いに素
m > n
m − n は奇数
(a, b, c) = (m^2 − n^2, 2mn, m^2 + n^2) or (2mn, m^2 − n^2, m^2 + n^2)
であることが必要十分である。上記の (m, n) は無数に存在し、2mn は重複しないから、原始ピタゴラス数は無数に存在する。これにより、すべての原始ピタゴラス数を重複なく見つけ出すことができる。
(m, n) = (2, 1) のとき (a, b, c) = (3, 4, 5)
(m, n) = (3, 2) のとき (a, b, c) = (5, 12, 13)
(m, n) = (4, 1) のとき (a, b, c) = (15, 8, 17)
(m, n) = (4, 3) のとき (a, b, c) = (7, 24, 25)
(m, n) = (5, 2) のとき (a, b, c) = (21, 20, 29)

(a, b, c) = (m^2 − n^2, 2mn, m^2 + n^2) or (2mn, m^2 − n^2, m^2 + n^2) m=7825でピタゴラス3整数a, b, c, が全て同じ色になったということでしょうね!


私はリーマン予想のあるプランを持っています。IPv6(Internet Protocol Version 6)は約340澗(340兆の1兆倍の1兆倍)の数の表現が可能で、iTの世界では無限と考えられています。スーパーコンピュータを使ってIPv6の値まで計算すれば、現実社会でのリーマン予想が肯定されるかまたは否定されることになるのではないかと考えています。

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