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Computer Cracks 200 Terabyte Math Proof

Here’s a cute problem, called the Boolean Pythagorean Theorem problem. Here are the first few Pythagorean triples:

3^2 + 4^2 = 5^2

6^2 + 8^2 = 10^2

5^2 + 12^2 = 13^2

9^2 + 12^2 = 15^2

8^2 + 15^2 = 17^2

12^2 + 16^2 = 20^2

7^2 + 24^2 = 25^2

15^2 + 20^2 = 25^2

10^2 + 24^2 = 26^2

20^2 + 21^2 = 29^2

18^2 + 24^2 = 30^2

16^2 + 30^2 = 34^2

21^2 + 28^2 = 35^2

12^2 + 35^2 = 37^2

15^2 + 36^2 = 39^2

27^2 + 36^2 = 45^2

9^2 + 40^2 = 41^2

27^2 + 36^2 = 45^2

OK, let’s have some fun with this. Let’s write every multiple of 5 (5, 10, 15, 20, 25, 30, 35, 40, 45) in boldface:

3^2 + 4^2 = {\bf 5}^2

6^2 + 8^2 = {\bf 10}^2

{\bf 5}^2 + 12^2 = 13^2

9^2 + 12^2 = {\bf 15}^2

8^2 + {\bf 15}^2 = 17^2

12^2 + 16^2 = {\bf 20}^2

7^2 + 24^2 = {\bf 25}^2

{\bf 15}^2 + {\bf 20}^2 = {\bf 25}^2

{\bf 10}^2 + 24^2 = 26^2

{\bf 20}^2 + 21^2 = 29^2

18^2 + 24^2 = {\bf 30}^2

16^2 + {\bf 30}^2 = 34^2

21^2 + 28^2 = {\bf 35}^2

12^2 + {\bf 35}^2 = 37^2

{\bf 15}^2 + 36^2 = 39^2

27^2 + 36^2 = {\bf 45}^2

9^2 + {\bf 40}^2 = 41^2

27^2 + 36^2 = {\bf 45}^2

For nearly all of these equations, there is one number that’s in boldface and one that’s not. However, there’s one that is all in one typeface: {\bf 15}^2 + {\bf 20}^2 = {\bf 25}^2.

So here’s a question: is it possible to divide the integers so that every Pythagorean triple (not just the small ones listed above) has at least one number in boldface and another that’s not?

This May, it was proved that it’s impossible. The proof is very brute-force (from https://cosmosmagazine.com/mathematics/computer-cracks-200-terabyte-maths-proof):

The team found all triples could be multi-coloured in integers up to 7,824. As soon as they hit 7,825, it became impossible.

But to prove a solution doesn’t exist, you need to try all possibilities. There are more than 10^{2300} ways to colour all those integers, so the scientists used a few mathematical tricks to reduce the number of combinations to trial to just under one trillion.

Two days later, with 800 processors at the University of Texas Stampede supercomputer crunching all possibilities in parallel, the team had their answer – no.

There is no way to colour the integers 1 to 7,825 in a way that leaves all Pythagorean triples multi-coloured, the team reported in arXiv.

I had to read this news article a couple of times to appreciate this: a supercomputer ran for two days on a supercomputer (without parallelization, computation time was 51,000 hours), producing an output file of 200 terabytes, comparable “to the size of the entire digitized text held by the US Library of Congress.” Wow.

What a Quantitative Literacy Course Should Look Like

A recent opinion piece in the New York Times discussed what a good quantitative literacy course should look like. Unlike the algebra-precalculus-calculus sequence or even a class in AP statistics, quantitative literacy is specifically designed for students who are not interested in a STEM major, teaching them how to view numbers to make sense of the world and be an informed citizen.

Another Reasoning Puzzle From ChefMongoose

I enjoyed this challenge.

Joseph Nebus's avatarnebusresearch

My friend ChefMongoose had another reasoning problem come to him, and I’m happy to share it further. It’s rather like that famous Singapore Birthday Problem that drove people crazy a couple of months ago. Here’s the problem:

I have a combination lock at work. There are three digits, all in the range 1 – 40; they’re all prime numbers. They’re X+Y, X+2Y, X+3Y — where X and Y are positive integers.

If I told you what X was but not Y, you wouldn’t be able to tell me the combination. If I told you what Y was but not X, you wouldn’t be able to tell me the combination. Now, what’s the combination?

I did work out the puzzle. It did make me notice a couple of strings of uniformly-spaced prime numbers I hadn’t done before, too, such as 3-13-23. (However, 3-13-23 isn’t one of the possible answers, because of…

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Useless Numerology for 2016: Part 5

The following entertaining (but useless) facts about the number 2,016 appeared in a recent Facebook post (and subsequent comments) by the American Mathematical Monthly.

2016 = \displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2

2016 = 2^{11} - 2^5

Establishing that these two expressions are equal takes a little bit of work. We begin by dividing the sum into even and odd values of n. The odd values of n from n = 1 to n = 63 have the form n = 2k+1, where k varies between k = 0 and k = 31. The even values of n from n = 0 to n = 62 have the form n = 2k, where again k varies between k = 0 and k = 31.

\displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2 = \displaystyle \sum_{n=0 \atop n =2k+1}^{63} (-1)^{n+1} n^2 + \displaystyle \sum_{n=0 \atop n =2k}^{63} (-1)^{n+1} n^2

= \displaystyle \sum_{k=0}^{31} (-1)^{(2k+1)+1} (2k+1)^2 + \displaystyle \sum_{k=0}^{31} (-1)^{2k+1} (2k)^2

= \displaystyle \sum_{k=0}^{31} \left[ (-1)^{2k+2} (2k+1)^2 + (-1)^{2k+1} (2k)^2 \right]

= \displaystyle \sum_{k=0}^{31} \left[(2k+1)^2 - (2k)^2 \right]

= \displaystyle \sum_{k=0}^{31} \left[4k^2 + 4k + 1 - 4k^2 \right]

= \displaystyle \sum_{k=0}^{31} \left[4k + 1 \right]

This last sum is an arithmetic series. The first (k = 0) term is 1, the last term is 4(31) + 1, and there are 32 terms. Using the formula for an arithmetic series, we find

\displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2= \displaystyle \frac{32 [1 + 4(31) + 1]}{2}

= \displaystyle \frac{32[4(31) + 2]}{2}

= 32[2(31) + 1]

= 32[2(32-1) + 1]

= 32[2(32) - 2 + 1]

= 32[64 - 1]

= 2^5[2^6 -1]

= 2^{11} - 2^5.

Useless Numerology for 2016: Part 3

The following entertaining (but useless) facts about the number 2,016 appeared in a recent Facebook post (and subsequent comments) by the American Mathematical Monthly.

2016 = 1+2+3 + \dots + 62 + 63

2016 = 2^{11} - 2^5

In this post, we’ll explore why these two expressions have to be equal.

The sum 1 + 2 + 3 + \dots + 62 + 63 is an arithmetic series. The first term is 1, the last term is 63, and there are 63 terms in the series. Using the formula for an arithmetic series, we find

1 + 2 + 3 + \dots + 62 + 63 = \displaystyle \frac{(63)(1 + 63)}{2}

= \displaystyle \frac{63 \times 64}{2}

= 63 \times 32

= (64-1) \times 32

= (2^6 - 1) \times 2^5

= 2^{11} - 2^5.

Useless Numerology for 2016: Part 2

The following entertaining (but useless) facts about the number 2,016 appeared in a recent Facebook post (and subsequent comments) by the American Mathematical Monthly.

2016 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5

2016 = 2^{11} - 2^5

Not surprisingly, there’s a natural reason why these two expressions are equal. (However, there isn’t a natural reason why the answer happens to match the current year other than coincidence.)

To begin, 2^5 + 2^6 + 2^7 + 2^8 + 2^9 + 2^{10} is a finite geometric series. The first term is 2^5 = 32, the common ratio is 2, and there are 6 terms in the series. Using the formula for a finite geometric series,

2^5 + 2^6 + 2^7 + 2^8 + 2^9 + 2^{10} = \displaystyle \frac{2^5 (1-2^6)}{1-2} = \frac{2^5-2^{11}}{-1} = 2^{11} - 2^5,

thus establishing that these two expressions are equal.