Thomas Bloom
@thomasfbloom
Followers
2K
Following
370
Media
19
Statuses
322
Royal Society University Research Fellow at the University of Oxford. Mathematician. Tries to count interesting things in interesting ways. He/him/his.
Oxford, England
Joined December 2020
In particular, despite what you may hear, something being an "Erdos problem" is no guarantee of being a significant or hard open problem. He asked thousands of questions, they can't all be deep!
0
0
17
This was a lot of my motivation in setting up https://t.co/7CRT0qI5Gg. I suspected that there were many old questions that could be easily solved if only the right person looked at it. I wanted to clear away such 'noise' to see what genuinely interesting/hard problems remain.
I expect we'll see successes in attention-bottlenecked conjectures in mathematics, especially false ones. There are a lot of open conjectures where no one has ever spent a few hours/days just searching for a counterexample; to some extent this can likely be automated.
1
4
64
One step closer to defeating Erdos's aliens! The Ramsey number R(5) is now at most 46. (So is between 43 and 46 now - the previous upper bound was 48.) Congratulations Brendan and Vigleik! https://t.co/viWSyXwZr6
arxiv.org
We prove that the Ramsey number $R(5,5)$ is less than or equal to~$46$. The proof uses a combination of linear programming and checking a large number of cases by computer. All of the computations...
4
46
227
(Of course, with a little liberty interpreting spaces as full stops - I wonder if this is possible without any full stops involved...)
0
0
5
Amazing fact I learnt today from the 1917 book Amusements in Mathematics book by Dudeney: You can cut a chessboard up into letters that form a complete sentence! The poetic and nonsensical "Cut thy life".
2
0
25
Open Erdos problem: are there any x,y such that x,y and x+1,y+1 and x+2,y+2 are all compatible? https://t.co/BdW1yFT0bC 3/3
3
2
31
Another example is 14,224 (both with primes {2,7}) and 15,225 (both with primes {3,5}). In general, you can take x = 2(2^r-1) and y = x(x+2). 2/3
1
0
13
Call two numbers x,y compatible if they have the same set of prime factors - e.g. 18 and 12 are compatible because both have prime divisors {2,3}. Are there any compatible x,y such that x+1,y+1 are also compatible? Yes! For example, 2,8 and 3,9 are both compatible pairs. 1/3
1
5
52
I wrote about an election stat that has been annoying me: the concept of the “crossover age”, at which you become more likely to vote Conservative than Labour.
mpaldridge.github.io
There’s an election statistic going around that annoys me. It’s some version of this:
0
3
4
27 new Erdős problems on https://t.co/7CRT0qI5Gg! 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)
4
7
41
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
0
0
7
Erdos originally thought that there were no solutions also; I think the first to construct the solution above was Chao Ko in 1940. In fact Ko constructed an infinite family of solutions (where a,b,c have the form 2^(something)*(2^n-1)^(something)). 9/10
2
0
4
The psychology here is interesting; we want to think that in maths there is always a 'clean' answer - either a trivial solution or a proof there are no solutions (even if it's a hard proof, like FLT). But sometimes the answer is just a bit messy. 8/10
1
0
4
(e.g. if you have a wrong proof that there are no solutions then seeing why it goes wrong gives a clue as to what form the actual solutions might take). 7/10
1
0
3
In general in mathematics it's wise to put equal effort into both proving and constructing a counterexample; in fact often the efforts in one direction help you in another. 6/10
1
0
5
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
2
0
6
You only need to plug in the first few powers of 2 and 3 before you find a combination that works. Not too difficult by hand, and by computer this is very quick. 4/10
1
0
3
What about powers of 2 and 3? This seems a lot better - more variables to play around with, more flexibility. In particular after making a couple of simplifying assumptions you get a system of two equations you need to satisfy (since the powers of 2 and 3 need to agree) 3/10
1
0
3
It's an equation in integers, involving powers, so the natural thing to do is to think about the prime factorisations. You want to find a 'simple' solution, so first try just one prime: a,b,c being powers of 2. Try small values, don't get anywhere. Not enough variables. 2/10
1
0
3