Month: January 2020

The number 33 has surprising depth

Add three cubed numbers, and what do you get? It is a question that has puzzled mathematicians for centuries.

In 1825, a mathematician known as S. Ryley proved that any fraction could be represented as the sum of three cubes of fractions. In the 1950s, mathematician Louis Mordell asked whether the same could be done for integers, or whole numbers. In other words, are there integers k, x, y and z such that k = x³ + y³ + z³ for each possible value of k?

We still don’t know. “It’s long been clear that there are maths problems that are easy to state, but fiendishly hard to solve,” says Andrew Booker at the University of Bristol, UK – Fermat’s last theorem is a famous example.

Booker has now made another dent in the cube problem by finding a sum for the number 33, previously the lowest unsolved example. He used a computer algorithm to search for a solution:

33 = 8,866,128,975,287,528³ + (-8,778,405,442,862,239)³ + (-2,736,111,468,807,040)³

To cut down calculation time, the program eliminated certain combinations of numbers. “For instance, if x, y and z are all positive and large, then there’s no way that x³ + y³ + z³ is going to be a small number,” says Booker. Even so, it took 15 years of computer-processing time and three weeks of real time to come up with the result.

For some numbers, finding a solution to the equation k = x³ + y³ + z³ is simple, but others involve huge strings of digits. “It’s really easy to find solutions for 29, and we know a solution for 30, but that wasn’t found until 1999, and the numbers were in the millions,” says Booker.

Another example is for the number 3, which has two simple solutions: 1³ + 1³ + 1³  and 4³ + 4³ + (-5) ³ . “But to this day, we still don’t know whether there are more,” he says.

There are certain numbers that we know definitely can’t be the sum of three cubes, including 4, 5, 13, 14 and infinitely many more.

The solution to 74 was only found in 2016, which leaves 42 as the only number less than 100 without a possible solution. There are still 12 unsolved numbers less than 1000.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Donna Lu*

Today I’d like to share an idea. It’s a very simple idea. It’s not fancy and it’s certainly not new. In fact, I’m sure many of you have thought about it already. But if you haven’t—and even if you have!—I hope you’ll take a few minutes to enjoy it with me. Here’s the idea:

So simple! But we can get a lot of mileage out of it.

To start, I’ll be a little more precise: every matrix corresponds to a weighted bipartite graph. By “graph” I mean a collection of vertices (dots) and edges; by “bipartite” I mean that the dots come in two different types/colors; by “weighted” I mean each edge is labeled with a number.

The graph above corresponds to a 3×23×2 matrix MM. You’ll notice I’ve drawn three greengreen dots—one for each row of MM—and two pinkpink dots—one for each column of MM. I’ve also drawn an edge between a green dot and a pink dot if the corresponding entry in MM is non-zero.

For example, there’s an edge between the second green dot and the first pink dot because M21=4M21=4, the entry in the second row, first column of MM, is not zero. Moreover, I’ve labeled that edge by that non-zero number. On the other hand, there is no edge between the first green dot and the second pink dot because M12M12, the entry in the first row, second column of the matrix, is zero.

Allow me to describe the general set-up a little more explicitly.

Any matrix MM is an array of n×mn×m numbers. That’s old news, of course. But such an array can also be viewed as a function M:X×Y→RM:X×Y→R where X={x1,…,xn}X={x1,…,xn} is a set of nn elements and Y={y1,…,ym}Y={y1,…,ym} is a set of mm elements. Indeed, if I want to describe the matrix MM to you, then I need to tell you what each of its ijijth entries are. In other words, for each pair of indices (i,j)(i,j), I need to give you a real number MijMij. But that’s precisely what a function does! A function M:X×Y→RM:X×Y→R associates for every pair (xi,yj)(xi,yj) (if you like, just drop the letters and think of this as (i,j)(i,j)) a real number M(xi,yj)M(xi,yj). So simply write MijMij for M(xi,yj)M(xi,yj).

Et voila. A matrix is a function.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

Lewis Caroll was the pen name for mathematician Charles Dodgson

Curiouser and curiouser! Particle physicists could have the author of Alice’s Adventures in Wonderland to thank for simplifying their calculations.

Lewis Carroll, the 19th century children’s author, was the pen name of mathematician Charles Lutwidge Dodgson. While his mathematical contributions mostly proved unremarkable, one particular innovation may have stood the test of time.

Marcel Golz at Humboldt University, Berlin has built on Dodgson’s work to help simplify the complex equations that arise when physicists try to calculate what happens when particles interact. The hope is that it could allow for speedier and more accurate computations, allowing experimentalists at places like the Large Hadron Collider in Geneva, Switzerland to better design their experiments.

Working out the probabilities of different particle interactions is commonly done using Feynman diagrams, named after the Nobel prize winning physicist Richard Feynman. These diagrams are a handy visual aid for encoding the complex processes at play, allowing them to be converted into mathematical notation.

One early way of representing these diagrams was known as the parametric representation, which has since lost favour among physicists owing to its apparent complexity. To mathematicians, however, patterns in the resulting equations suggest that it might be possible to dramatically simplify them in ways not possible for more popular representations. These simplifications could in turn enable new insights. “A lot of this part of physics is constrained by how much you can compute” says Karen Yeats, a mathematician at the university of Waterloo, Canada.

Golz’s work makes use of the Dodgson identity, a mathematical equivalence noted by Dodgson in an 1866 paper, to perform this exact sort of simplification. While much of the connecting mathematics was done by Francis Brown, one of Golz’s tutors at Oxford University, the intellectual lineage can be traced all the way back to Lewis Carroll. “It’s kind of a nice curiosity,” says Golz. “A nice conversation starter.”

In the past, parametric notation was only useful in calculating simplified forms of quantum theory. Thanks to work like Golz’s, these simplifications could be extended to particle behaviour of real interest to experimentalists. “I can say with confidence that these parametric techniques, applied to the right problems, are game-changing,” says Brown.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Gilead Amit*

Is there an error in there somewhere?

It’s the stuff of Hollywood. Somebody somewhere is surely selling the movie rights to what’s become the biggest spat in maths: a misunderstood genius, a 500-page proof almost nobody can understand and a supporting cast squabbling over what it all means. At stake: nothing less than the future of pure mathematics.

In 2012, Shinichi Mochizuki at Kyoto University in Japan produced a proof of a long-standing problem called the ABC conjecture. Six years later the jury is still out on whether it’s correct. But in a new twist, Peter Scholze at the University of Bonn – who was awarded the Fields Medal, the highest honour in maths, in August – and Jakob Stix at Goethe University Frankfurt – who is an expert in the type of maths used by Mochizuki – claim to have found an error at the heart of Mochizuki’s proof.

Roll credits? Not so fast. The pairs’ reputation means that their claim is a serious blow for Mochizuki. And a handful of other mathematicians claim to have lost the thread of the proof at the same point Scholze and Stix say there is an error. But there is still room for dispute.

a + b = c?

The ABC conjecture was first proposed in the 1980s and concerns a fundamental property of numbers, based around the simple equation a + b = c. For a long time, mathematicians believed that the conjecture was true but nobody had ever been able to prove it.

To tackle the problem, Mochizuki had to invent a fiendish type of maths called Inter-universal Teichmüller (IUT) theory. In an effort to understand IUT better, Scholze and Stix spent a week with Mochizuki in Tokyo in March. By the end of the week, they claim to have found an error.

The alleged flaw comes in Conjecture 3.12, which many see as the crux of the proof. This section involves measuring an equivalence between different mathematical objects. In effect, Scholze and Stix claim that Mochizuki changes the length of the measuring stick in the middle of the process.

No proof

“We came to the conclusion that there is no proof,” they write in their report, which was posted online on 20 September.

But Ivan Fesenko at the University of Nottingham, UK, who says he is one of only 15 people around the world who actually understand Mochizuki’s theory, thinks Scholze and Stix are jumping the gun. “They spent much less time than all of us who have been studying this for many years,” says Fesenko.

Mochizuki has tried to help others understand his work, taking part in seminars and answering questions. Mochizuki was even the one who posted Scholze and Stix’s critical report. “We have this paradoxical situation in which the victim has published the report of the villain,” says Fesenko with a laugh. “This is an unprecedented event in mathematics.”

So is the proof wrong or just badly explained? Fesenko thinks that the six-year dispute exposes something rotten at the heart of pure mathematics. These days mathematicians work in very narrow niches, he says. “People just do not understand what the mathematician in the next office to you is doing.”

This means that mathematicians will increasingly have to accept others’ proofs without actually understanding them – something Fesenko describes as a fundamental problem for the future development of mathematics.

This suggests the story of Mochizuki’s proof may forever lack a satisfactory ending – becoming a war between mathematicians that is doomed to spiral into infinity. “My honest answer is that we will never have consensus about it,” says Fesenko.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Douglas Heaven*