Thomas Bloom @thomasfbloom Twitter profile

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Thomas Bloom

@thomasfbloom

4 months

This is a famous picture of Erdős and a young Terence Tao in 1985 discussing mathematics. I recently learnt what they were talking about.

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Thomas Bloom

@thomasfbloom

3 months

One of the most famous anecdotes in mathematics is Fermat's marginal remark that he had 'a truly marvellous proof of <Fermat's last theorem>, which this margin is too narrow to contain'. Few know the actual story behind this - time to correct the record. A thread. 1/18

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Thomas Bloom

@thomasfbloom

4 years

One of my favourite "I can't believe we don't know how to prove that" questions: Is pi^pi^pi^pi an integer?

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Thomas Bloom

@thomasfbloom

1 month

A fun but challenging puzzle: Are there any integer solutions to this equation with a,b,c > 1? (I know the answer, will post the solution in a couple of days.)

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Thomas Bloom

@thomasfbloom

1 year

Erdos asked many beautiful questions, lots of which are still unanswered. I made a website to collect the problems together and keep track of the progress on each. Explore and find a problem to solve!

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Thomas Bloom

@thomasfbloom

1 year

has fallen already! Mattheus and Verstraete have proved that the Ramsey number R(4,k) grows like k^(3-o(1)). Again, first significant progress in 40 years. A year of incredible breakthroughs, and we're only halfway through.

The asymptotics of $r(4,t)$

For integers $s,t \geq 2$, the Ramsey numbers $r(s,t)$ denote the minimum $N$ such that every $N$-vertex graph contains either a clique of order $s$ or an independent set of order $t$. In this...

arxiv.org

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Thomas Bloom

@thomasfbloom

3 years

Happy to report progress on this! The answer is yes: a set of integers of positive density must contain distinct n_1,...,n_k such that 1/n_1+...+1/n_k=1.

On a density conjecture about unit fractions

We prove that any set $A\subset \mathbb{N}$ of positive upper density contains a finite $S\subset A$ such that $\sum_{n\in S}\frac{1}{n}=1$, answering a question of Erdős and Graham.

arxiv.org

Thomas Bloom

@thomasfbloom

4 years

The density version of this is still an open question! If we have some subset A of the integers of positive integers must there by distinct n_1,...,n_k in A such that 1=1/n_1+...+1/n_k? (1/2)

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Thomas Bloom

@thomasfbloom

1 year

Every (positive) rational number is the sum of 1/(p-1) for distinct primes p. For example 1/9 = 1/(13-1) + 1/(37-1)

Unit fractions with shifted prime denominators

We prove that any positive rational number is the sum of distinct unit fractions with denominators in $\{p-1 : p\textrm{ prime}\}$. The same conclusion holds for the set $\{p-h : p\textrm{...

arxiv.org

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Thomas Bloom

@thomasfbloom

4 months

The question that Erdős asked Tao is now on the Erdős problems website at . Tao has kindly shared a letter Erdős sent shortly after this meeting describing the problem:

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Thomas Bloom

@thomasfbloom

1 year

My favourite snappy proof that there are infinitely many primes: For any integer N note that N(N+1) has strictly more prime divisors than N. Iteratively apply with N=1 (say) and we get a forever increasing set of prime divisors.

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Thomas Bloom

@thomasfbloom

1 year

Yet another old Erdos problem falls: Cédric Pilatte, a graduate student at Oxford, has constructed a set which is both a Sidon set AND an asymptotic basis of order three! What does this mean?

A solution to the Erdős-Sárközy-Sós problem on asymptotic Sidon...

A set $S\subset \mathbb{N}$ is a Sidon set if all pairwise sums $s_1+s_2$ (for $s_1, s_2\in S$, $s_1\leq s_2$) are distinct. A set $S\subset \mathbb{N}$ is an asymptotic basis of order 3 if every...

arxiv.org

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Thomas Bloom

@thomasfbloom

3 months

A harder puzzle: There are two strings of letters X and Y such that XY and YX are distinct palindromes: X=ab Y=ba (so XY=abba YX=baab) Are there three strings of letters X,Y,Z such that XYZ, YZX, ZXY are distinct palindromes?

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Thomas Bloom

@thomasfbloom

2 years

The Kahn-Kalai threshold conjecture has been proved! (see also the introduction of for more on why the conjecture is significant)

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Thomas Bloom

@thomasfbloom

3 years

Just heard a lovely puzzle: Choose n>1 subintervals of [0,1] at random (uniformly, independently). What is the probability that one interval intersects all the others? (I'll follow-up with references and the answer later, but don't want to spoil discovery for now!)

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Thomas Bloom

@thomasfbloom

1 year

An incredible breakthrough in additive combinatorics today - Kelley and Meka achieve quasi-polynomial bounds for sets without three term arithmetic progressions!

Strong Bounds for 3-Progressions

We show that for some constant $β> 0$, any subset $A$ of integers $\{1,\ldots,N\}$ of size at least $2^{-O((\log N)^β)} \cdot N$ contains a non-trivial three-term arithmetic progression....

arxiv.org

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Thomas Bloom

@thomasfbloom

3 months

As for the actual proof of FLT, Kevin Buzzard has recently launched a project to formalise it (and therefore a huge amount of modern number theory) in Lean! 18/18

The Fermat's Last Theorem Project | Lean community blog

Kevin Buzzard discusses the project to prove Fermat's Last Theorem in Lean. Introduction Fermat's Last Theorem (FLT) is the claim that some abstract equation has no solutions in positive integers. Th

leanprover-community.github.io

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Thomas Bloom

@thomasfbloom

4 months

It was about sets such that no subset sum is a square number. For example {2,3,8} is such a set because none of 2,3,8,5,10,11,13 are square numbers. But {2,3,6,8} is not because 2+6+8=16 is a square number. Erdős was wondering how large such a set in {1,...,N} can be.

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Thomas Bloom

@thomasfbloom

4 months

Another surprising open problem: Is the sum of 1/(n!-1) for n >1 (so 1+1/5+1/23+...) irrational? The sum of 1/n! (1 + 1/2 + 1/6 + 1/24 +...) is e = 2.71728... which Euler showed almost 300 years ago is irrational. But change the series just by -1 and it's a mystery.

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Thomas Bloom

@thomasfbloom

3 months

Because he didn't know he'd made one! He'd only written about a 'proof' in a doodle of his own personal copy of a book. He had no idea anyone would ever even see this remark. I certainly never bother later correcting all my silly doodles in my private notes. 12/18

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Thomas Bloom

@thomasfbloom

4 months

Let x be a real number such that 2^x and 3^x are both integers. Must x be an integer? Surely yes, but this is not known! (It follows from Schanuel's conjecture, and the weaker four exponentials conjecture, but both of these are also very open.)

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Thomas Bloom

@thomasfbloom

3 months

I am certain that Fermat did not have Wiles' proof, just as am I certain that Plato did not use TikTok. I am also certain that Fermat did not have any other proof (because if he even thought he did he would have told everyone!) 16/18

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Thomas Bloom

@thomasfbloom

3 months

The idea that Fermat had a correct proof, and yet never wrote it down in 30 years, despite the huge amount of other mathematics he wrote in this period which survives, is ludicrous. It would be like da Vinci building a working smartphone and burying it in his garden. 17/18

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Thomas Bloom

@thomasfbloom

4 months

Erdős couldn't find a better construction, and speculated that maybe N^{1/3} is the correct threshold. This is true (at least, up to logarithmic factors), and was proved by Nguyen and Vu in 2010.

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Thomas Bloom

@thomasfbloom

3 months

In fact he did have a proof of the case n=4 (using 'infinite descent') which is valid! Fermat's proof of this case is often still taught in number theory courses today. His valid proof he wrote in several places, and discussed in letters with other mathematicians. 9/18

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Thomas Bloom

@thomasfbloom

3 months

Fermat's last theorem was eventually proved by Andrew Wiles in 1994, using a huge amount of sophisticated machinery, almost none of which was developed until the 20th century. 15/18

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Thomas Bloom

@thomasfbloom

4 months

Explore to find other interesting Erdős problems (both solved and unsolved). Also see for ways you can help the project!

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Thomas Bloom

@thomasfbloom

3 months

(To spoil the punchline: Fermat never had a proof, and aside from a mad moment, of which he quickly corrected himself, never believed that he had one, and never claimed this in public.) 2/18

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Thomas Bloom

@thomasfbloom

3 months

Fermat senior had no idea that his claim would ever be published, and I doubt he would have approved such. He should be remembered for his many proofs of interesting theorems and his *conjecture* of 'Fermat's last theorem', which inspired generations of number theorists. 14/18

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Thomas Bloom

@thomasfbloom

3 months

So what happened? After Fermat died in 1665 his son Clément-Samuel Fermat found his father's copy of Arithmetica and, because of his mathematical fame, published a copy of Arithmetica including his father's notes. The rest is history. 13/18

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Thomas Bloom

@thomasfbloom

4 months

He observed that there is such a set of size about N^{1/3}, because we can take some prime p of size about N^{2/3} and then look at {p,2p,3p,...,Mp} where M has size about N^{1/3}/10 (say).

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Thomas Bloom

@thomasfbloom

1 month

Answer: Yes there are! In fact infinitely many. The smallest I know (as found by a couple of people here) is a = 2^12*3^6 b = 2^8*3^8 c = 2^11*3^7 This is not too hard to find by hand once you start looking for it - a short thread. 1/10

Thomas Bloom

@thomasfbloom

1 month

A fun but challenging puzzle: Are there any integer solutions to this equation with a,b,c > 1? (I know the answer, will post the solution in a couple of days.)

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Thomas Bloom

@thomasfbloom

3 months

What is very likely is that he came up with the n=4 proof and (as many mathematicians do in research) erroneously thought the same trick would just work for all n>2. 10/18

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Thomas Bloom

@thomasfbloom

3 months

Our story begins with Diophantus, a Greek mathematician in the 3rd century AD, one of the early number theorists. His book 'Arithmetica' included discussion of finding integral solutions to polynomial equations (now known as 'Diophantine equations'). 3/18

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Thomas Bloom

@thomasfbloom

4 months

The idea is that all of those subset sums look like p*(some integer at most M^2). Since M^2 is less than p they can't be a square number, because p can only divide them once.

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Thomas Bloom

@thomasfbloom

3 months

"It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvellous proof of this, which this margin is too narrow to contain." 7/18

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Thomas Bloom

@thomasfbloom

3 months

This is not true, it only works for n=4. And presumably Fermat realised this himself as soon as he found some more paper and went to write it out properly! But why didn't he retract his claim then, I hear you cry. 11/18

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Thomas Bloom

@thomasfbloom

3 months

That was in 1637. He lived for almost 30 years after this, yet never wrote about his 'marvellous proof' anywhere else, including in his many letters to other mathematicians. He did mention his conjecture (at least the cases n=3 and n=4), but never said he had a proof. 8/18

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Thomas Bloom

@thomasfbloom

3 months

For example are there integral solutions to x^2+y^2=z^2? Yes, infinitely many, known as 'Pythagorean triples'. e.g. 3^2 + 4^2 = 5^2 Solutions like this were even known to the Babylonians 4000 years ago. 4/18

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Thomas Bloom

@thomasfbloom

3 months

A fun puzzle from Martin Gardner: Can a triangle with one obtuse (>90°) angle be cut into smaller triangles, all acute (all angles <90°)? To show what I mean, here's an attempt that fails (1,2,3 are acute but 4 isn't). Give such a decomposition or prove it's impossible.

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Thomas Bloom

@thomasfbloom

3 months

Arithmetica was translated into Latin in 1621 by Claude Bachet, and was read by the great mathematician (and sometime lawyer) Pierre de Fermat in around 1637. Fermat read about sums of squares being a square, and wondered whether this could be done with cubes? 4th powers? 5/18

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Thomas Bloom

@thomasfbloom

1 year

In making I found ChatGPT invaluable. It was helpful to know some easy programming basics beforehand, but I'd never heard of Flask, SQLite, etc. I just said "I have some latex and I want a website that does such and such" and it said "here's the code".

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Thomas Bloom

@thomasfbloom

3 months

Presumably after a bit of experimentation, Fermat guessed that this wasn't possible. No two cubes sum to a cube, no two 7th powers sum to a 7th power, etc. With the Arithmetica on the desk in front of him, he doodled the infamous remark in the margin (in Latin of course): 6/18

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Thomas Bloom

@thomasfbloom

1 year

Olof Sisask and I have written an exposition of yesterday's breakthrough result by Kelley and Meka (incorporating some simplifications on the technical side), for those interested in an alternative take. (note the KM methods means we only need <20 pages!)

The Kelley--Meka bounds for sets free of three-term arithmetic progressions

We give a self-contained exposition of the recent remarkable result of Kelley and Meka: if $A\subseteq \{1,\ldots,N\}$ has no non-trivial three-term arithmetic progressions then $\lvert A\rvert...

arxiv.org

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Thomas Bloom

@thomasfbloom

1 year

Great new paper out today by Cohen, Pohoata, and Zakharov, improving the (over 40 year old) upper bound for the Heilbronn triangle problem.

A new upper bound for the Heilbronn triangle problem

For sufficiently large $n$, we show that in every configuration of $n$ points chosen inside the unit square there exists a triangle of area less than $n^{-8/7-1/2000}$. This improves upon a result...

arxiv.org

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Thomas Bloom

@thomasfbloom

2 years

@SC_Griffith If pi is algebraic of degree d, so is n*pi for every integer n. These are all roots of sin(x), which is a Taylor series, so a polynomial of 'infinite degree'. It seems unlikely that all the minimal polynomials of the n*pi could divide sin(x) simultaneously.

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Thomas Bloom

@thomasfbloom

1 month

27 new Erdős problems on ! Here's an interesting one (676) - say n is 'small' modulo p^2 (p prime) if it is congruent to some 0<=b<p mod p^2. Is every large enough n small for some p^2 <= n? e.g. 4 = 0 (mod 2^2) 21 = 2 (mod 3^2) 78 = 3 (mod 5^2)

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Thomas Bloom

@thomasfbloom

4 years

@michael_nielsen @wtgowers Yes. It's really to expose our ignorance. Of course, it should be transcendental. But not only can we not show that, we can't show it's irrational. Not only that, we can't show it's not an integer! The only method we know is to calculate its decimal expansion, but too large...

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Thomas Bloom

@thomasfbloom

3 years

The answer is actually 2/3! This surprised me a lot. It's not at all obvious (to me) that the answer is independent of n. I learnt this problem from Bollobas' The Art of Mathematics, along with a beautiful proof due to Graham Brightwell.

Thomas Bloom

@thomasfbloom

3 years

Just heard a lovely puzzle: Choose n>1 subintervals of [0,1] at random (uniformly, independently). What is the probability that one interval intersects all the others? (I'll follow-up with references and the answer later, but don't want to spoil discovery for now!)

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Thomas Bloom

@thomasfbloom

2 years

Inspired by @littmath thought I'd turn to GPT-3 for research advice... Well, it's not wrong.

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Thomas Bloom

@thomasfbloom

4 years

Take some collection G of bijections on a finite set X that is closed under compositions and inverses, all of which (aside from the identity) have at most one fixed point. Then the subset of bijections with NO fixed point (along with the identity) is closed under composition! 1/6

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Thomas Bloom

@thomasfbloom

8 months

New paper! Expository paper with Jared Duker Lichtman about the beautiful determinant method of Bombieri and Pila. If you don't know this method, I recommend you read this right away!

The Bombieri-Pila determinant method

In this expository note, we give a concise and accessible introduction to the real-analytic determinant method for counting integral points on algebraic curves, based on the classic 1989 paper of...

arxiv.org

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Thomas Bloom

@thomasfbloom

8 months

New paper! 'Odd moments and adding fractions', joint with Vivian Kuperberg. This was lots of fun to work on - the big headline result is a proof of an upper bound conjectured in 1986 by Montgomery and Vaughan about odd moments of reduced residues.

Odd moments and adding fractions

We prove near-optimal upper bounds for the odd moments of the distribution of coprime residues in short intervals, confirming a conjecture of Montgomery and Vaughan. As an application we prove...

arxiv.org

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Thomas Bloom

@thomasfbloom

4 years

A thread on Egyptian fractions, which are sums of distinct unit fractions, such as 1/2+1/3+1/6 = 1. (1/10)

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Thomas Bloom

@thomasfbloom

4 years

@SC_Griffith Hot chocolate rather than coffee, but it's that kind of a day!

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Thomas Bloom

@thomasfbloom

4 months

@BlumLenore There is a thread underneath that post explaining, sorry I forgot to indicate that post was the beginning of a thread!

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Thomas Bloom

@thomasfbloom

2 months

I finally managed to get an AI system ( in particular) to predict what a proof of the Riemann Hypothesis might look like. Here's its prediction - mark your calendars for 2054! (More details of the 'proof' in the next tweet.)

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Thomas Bloom

@thomasfbloom

1 year

This problem says how many vertices do we need so that in any red/blue colouring of all the edges between the vertices, there is either a red complete graph on 4 vertices, or a blue complete graph on k. In 1980 Ajtai, Komlos, and Szemeredi showed that k^3 (ish) is enough.

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Thomas Bloom

@thomasfbloom

2 years

This is a very strong conjecture (as Kahn and Kalai wrote, "It would probably be more sensible to conjecture that it is not true"!), and a great achievement. All the more so because the proof is only 4 pages long! Congratulations Park and Pham!

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Thomas Bloom

@thomasfbloom

15 days

@paullee123 @deedydas Your answers to 1 and 2 (at least) are wrong. E.g. for 1 alpha=1 and n=2 is a counterexample. For 2 a=b is not a solution since gcd(a^n+a,a^n+a)=a^n+a which is not constant for all large n.

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Thomas Bloom

@thomasfbloom

2 years

Amazing new tool that converts natural language maths theorem statements into Lean. When this works well for proofs, a real game changer in maths - no need to learn Lean proper, just feed in your Latex paper and it can be converted and verified automatically!

Kevin Buzzard

@XenaProject

2 years

Guest post on my blog by computing+maths Yale undergraduate Zhangir Azerbayev, about a cool Lean toy he's been making.

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Thomas Bloom

@thomasfbloom

3 years

@NoahSD Combinatorial proof: it suffices to show, for all k, n^k < k^k*2^n for all n>k. But 2^n counts the number of subsets of {1,...,n} and n^k/k^k is a lower bound for the number of subsets of size k. So this bound is trivial.

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Thomas Bloom

@thomasfbloom

4 months

This was asked by Erdős, but also by many others.

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Thomas Bloom

@thomasfbloom

1 year

@wtgowers Yes, it's an incredible breakthrough! It's also surprisingly simple (although it makes essential use of pre-existing almost periodicity tools developed by Croot, Sisask, Sanders, and others along with...

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Thomas Bloom

@thomasfbloom

1 year

But do you need k^3? Spencer showed that you need at least k^(5/2) (ish). And now Mattheus and Verstraete have closed the gap!

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Thomas Bloom

@thomasfbloom

3 years

@twarrestrial Yes: let x be the unique solution to x^x=2. One can show this exists and is unique, and also that it must be irrational.

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Thomas Bloom

@thomasfbloom

1 year

Erdos asked many questions about Sidon sets and bases, and I'm sure would have been thrilled with this discovery. Congratulations Cédric!

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Thomas Bloom

@thomasfbloom

1 year

This was conjectured by Sun - see e.g. which includes a table of lots of such representations.

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Thomas Bloom

@thomasfbloom

3 months

Just found this charming student video proof without words of the value of two Ramsey numbers: R(3,3)=6 and R(3,3,3)=17. Probably hard to follow if you don't already know the proofs anyway, but for those who do, it's very relaxing to watch!

A Proof Without Words R(3,3,3)=17

🎵: Stan Forebee - Kensington https://chll.to/bca012a3Made using Blender for creating all the animations, story boards (proof steps), and rendering.This was ...

www.youtube.com

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Thomas Bloom

@thomasfbloom

4 years

The density version of this is still an open question! If we have some subset A of the integers of positive integers must there by distinct n_1,...,n_k in A such that 1=1/n_1+...+1/n_k? (1/2)

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Thomas Bloom

@thomasfbloom

1 year

An infinite Sidon set can never be a basis of order 2, so this is the best one can hope for. So to recap, Pilatte constructs a set which is completely unstructured under 2-fold addition, and under 3-fold addition eventually generates all integers (a very structured set!)

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Thomas Bloom

@thomasfbloom

4 years

@macaronique I am as sure as one can be without a proof that it is not - indeed it should be transcendental, and so not even a rational, let alone an integer. This would follow from Schanuel's conjecture, as others have pointed out.

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Thomas Bloom

@thomasfbloom

4 years

This is now a theorem, proved by Croot in 2003. I think this is probably one of my favourite proofs, using the best of analytic number theory, Fourier analysis, and combinatorial reasoning, and combining them in ingenious ways. (1/3)

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Thomas Bloom

@thomasfbloom

4 years

It's not really clear (to me, at least) whether this should be true or false. Erdős thought it was true. But I don't see much evidence in its favour. It might be worth someone seriously trying to disprove it instead... (1/1)

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Thomas Bloom

@thomasfbloom

4 months

@Moinsdeuxcat Yes! In fact the six exponentials theorem implies that for any three distinct primes p,q,r if p^x,q^x,r^x are all integers then x must be an integer.

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Thomas Bloom

@thomasfbloom

1 year

Of course there was then a long process of tweaking, bug fixing, adding features, but all done in dialogue with gpt. And any time I got an error I just asked gpt what to do!

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Thomas Bloom

@thomasfbloom

1 year

Still a work in progress (170 problems so far added but hundreds still go, and references still being updated) - suggestions and contributions of problems/progress welcome!

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Thomas Bloom

@thomasfbloom

3 years

2/ I should have specified how the intervals are chosen! I meant choosing an interval at random by choosing its endpoints uniformly at random from [0,1].

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Thomas Bloom

@thomasfbloom

1 year

A Sidon set is a set A with no non-trivial additive 'coincidences' - solutions to a+b = c+d aside from the trivial ones where {a,b}={c,d}. A small example is {1,2,4}. So 2-fold addition of things from A is as unstructured as possible.

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Thomas Bloom

@thomasfbloom

2 years

I've just asked a MO question about what happens with the well-known Seymour second neighborhood conjecture for infinite graphs. Tagging @JDHamkins , perhaps the set theory/model theory crowd can shed some light.

Seymour's second neighborhood conjecture for infinite graphs

Let $G$ be a directed graph (say simple, so no loops and each pair of vertices has at most one directed edge between them). Suppose $G$ is 'locally finite', in the sense that each vertex has only

mathoverflow.net

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Thomas Bloom

@thomasfbloom

4 years

Erdős and Graham conjectured (and offered $500 for a proof) that in any finite colouring of the integers, there are some n_1,...,n_k all the same colour such that 1=1/n_1+...1/n_k. (1/4)

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Thomas Bloom

@thomasfbloom

4 years

(Tao recently found a proof that replaces character theory proper with a Fourier analytic proof, which is a step in the right direction, but this proof is still very representation-theoretic!)

A Fourier-analytic proof of Frobenius’ theorem

A finite group $latex {G=(G,\cdot)}&fg=000000$ is said to be a Frobenius group if there is a non-trivial subgroup $latex {H}&fg=000000$ of $latex {G}&fg=000000$ (known as the Frobenius …

terrytao.wordpress.com

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Thomas Bloom

@thomasfbloom

1 year

Infinite Sidon sets are a fascinating area (with big open problems remaining, like how large can they be). Pilatte's construction combines this with another property: being a basis of order three. This just means that every large integer is the sum of three elements of A.

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Thomas Bloom

@thomasfbloom

2 years

Christian Elsholtz and I have written a survey article about Egyptian Fractions (sums of distinct unit fractions), covering some fascinating open problems and recent progress. Comments/questions welcome!

Egyptian Fractions

Any rational number can be written as the sum of distinct unit fractions. In this survey paper we review some of the many interesting questions concerning such 'Egyptian fraction' decompositions,...

arxiv.org

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Thomas Bloom

@thomasfbloom

4 months

@wtgowers As a naive spectator, this looks very impressive - already 20/50 has been achieved after only a couple of weeks. Is this exceeding your expectations so far? I suppose it's not clear whether these are relatively easy early wins, and getting past, say 40/50 will be a bottleneck.

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Thomas Bloom

@thomasfbloom

2 years

@HigherGeometer Not an expert, but I believe that if one knew that L(1,\chi) >> (log q)^{-2022} (as Zhang claims) then one would get a PNT (with good error bounds) for primes up to x, in arithmetic progressions modulo q, for q up to something like exp((log x)^(1/2022)).

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Thomas Bloom

@thomasfbloom

4 years

1/ A thread about some recent developments in additive combinatorics: In 2018 Freddie Manners proved an incredible new bound on the Gowers Uniformity Norm Inverse Theorem. What does this mean, and why is it a big deal?

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Thomas Bloom

@thomasfbloom

4 years

Given positive a/b, take the smallest n such that n > b/a. Then subtract 1/n from a/b and repeat. Key observation: the numerator decreases, so this algorithm terminates in at most a steps with 0, so we can write a/b as the sum of at most a distinct unit fractions. (1/7)

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Thomas Bloom

@thomasfbloom

2 years

@HigherGeometer This is already known for small q with the same error bounds (say q < (log x)^O(1)) - the power of this new bound would be greatly extending the range of permissible q - although still falling short of taking q to be a power of x, which is what L(1,\chi) >> 1/log q would allow.

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Thomas Bloom

@thomasfbloom

4 years

19/ If you want to read more about the inverse theorem and Manners' proof, I wrote an article for the Bourbaki seminar on it, aimed at the non-specialist mathematician:

Quantitative inverse theory of Gowers uniformity norms

(This text is a survey written for the Bourbaki seminar on the work of F. Manners.) Gowers uniformity norms are the central objects of higher order Fourier analysis, one of the cornerstones of...

arxiv.org

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Thomas Bloom

@thomasfbloom

2 years

A new lower bound for the size of cap sets (sets without 3APs in F_3^n) - congratulations Fred! Fred shows that there exist cap sets in F_3^n of size at least 2.218031...^n for large n. This improves on Edel (2002), who had 2.217389...^n.

Fred Tyrrell

@OfficialFredTyr

2 years

Pleased to announce that my paper is now on Arxiv: Feels quite surreal to have my work out there in the real world, alongside professional mathematicians!

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Thomas Bloom

@thomasfbloom

4 years

This awkwardness of notation led to some fascinating questions! The most obvious one is: can we write ALL rationals as the sum of distinct unit fractions? Yes! In fact the greedy approach works here. (1/8)

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Thomas Bloom

@thomasfbloom

4 years

In particular, 4/n is the sum of at most 4 distinct unit fractions. One of the most famous open problems, the Erdős-Straus conjecture, says that in fact it's always the sum of at most 3. Despite many partial results, this is still a conjecture! (1/6)

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Thomas Bloom

@thomasfbloom

4 years

@lang_chris @wtgowers (of course, if I've been too pessimistic about what computers can calculate these days, just add in an extra level of exponent or two...)

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Thomas Bloom

@thomasfbloom

1 month

@mressin The LHS is 2^4 the RHS is 2^8.

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Thomas Bloom

@thomasfbloom

1 month

Ko also proved there are no solutions if a and b are coprime (and I think some proofs given here can be made into correct proofs of this assertion). Later Erdos asked if the infinite families found by Ko were the only ones - I believe this is still open! 10/10

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Thomas Bloom

@thomasfbloom

1 month

I think this is a great lesson in problem solving - most people seemed to assume there was no solution, and (presumably) put all of their effort into constructing a proof. But when you see a question that could go either way, better not to assume! 5/10

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Thomas Bloom

@thomasfbloom

1 year

@wtgowers A dependent random choice argument that has its origins in a method you introduced in your proof of Szemeredi's theorem!)

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Thomas Bloom

@thomasfbloom

4 years

The reason for the name is that in Egyptian hieroglyphics, the general way to write fractions was to add a modifier to the symbol n to get 1/n. So to write non-unit fractions one had to write them as a sum of unit fractions. (1/9)

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Thomas Bloom

@thomasfbloom

4 months

This is my new favourite maths song.

Grant Sanderson

@3blue1brown

4 months

Last week @standupmaths had me join one of his "Evening of Unnecessary Detail" events in New York. I invited @acapellascience to come down and we had a bit too much fun writing a parody of "Ain't No Sunshine" about the twin prime conjecture.

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Thomas Bloom

@thomasfbloom

1 year

Given n points in the unit square, you want to find a triangle between some 3 of the points with as small an area as possible. How well can you do? O(1/n) is trivial. In 1982 Komlos, Pintz and Szemeredi got 1/n^(8/7). The new paper improves the exponent by 1/2000.

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