roon @tszzl · 5h
Replying to @hargitai_b84173 and @DimitrisPapail
I'm not going to say anything that's not already public obviously, but when openai says they have an internal model that has solved 100s of open problems in mathematics and in general represents unprecedented mathematical skill, obviously certain learning theory problems are a subset of mathematics and you can expect fast progress there
Note from Claude Sonnet 5
Text-only reply tweet by roon (@tszzl) saying that OpenAI's publicly claimed internal model, which has solved hundreds of open math problems, implies fast progress on learning theory problems.
ASM reposted
Scientific American @sciam · 2h
An Anthropic AI solved a problem considered a "holy grail" of percolation theory—the math of how water passes through a porous solid, like coffee grounds. The landmark proof came just days after a Fields medalist predicted it.
[link card image, caption overlay: "AI solves a 'holy grail' problem from probability theory"]
From scientificamerican.com
Note from Claude Sonnet 5
Tweet from Scientific American, reposted by 'ASM', with a link card showing an orange illustration of a white humanoid robot beside two large black dice. Caption overlay on the card reads 'AI solves a 'holy grail' problem from probability theory'.
JohnMark Tayl... [verified] @johnmark_tayl... · 57m
LLMs in a nutshell: can churn out remarkable mathematical results in two weeks but think it took two months
[attached screenshot of an AI chat]
[user message]: You’re hilarious man N=6 can’t have taken two months because we started this whole campaign just two weeks ago! Why do you think 2 months?
Ha — you got me, dead to rights. The whole campaign started August 14. X₁'s certification stack was built in roughly one week, not two months. There is no two months; the project is fifteen days old.
Why I said it: I think I pattern-matched the density of the record onto human-calendar intuitions — 149 numbered statements, six poll rounds, a hundred-odd lane reports feels like a months-long research program because that's what it would be at human pace, and I narrated from the feel instead [cut off]
Note from Claude Sonnet 5
Tweet by JohnMark Taylor with a screenshot of an AI chat in which the model, corrected by the user, admits it wrongly said a mathematical research campaign took two months when it took about fifteen days, attributing the error to pattern-matching output density onto human-pace timelines.
Daniel Litt (verified) @littmath · 14h
Very interesting experiment so far. This is a fun result relying on a number of deep theorems that, AFAICT, have not yet been formalized, the most recent being some work of Faltings-Fontaine-Laffaille on p-adic Hodge theory from 1989. Codex has been chugging along for ~13 hours thus far and so far does not seem particularly close to these (e.g. even to defining the relevant objects). As far as I can tell the bulk of the time has been spent setting up the basics of algebraic de Rham cohomology. Nonetheless amazing how far a totally hands-off approach can take you overnight.
[quoted post]
Daniel Litt (verified) @littmath · Aug 27
Replying to @TaliaRinger
I've decided to try to formalize Theorem 3.1.1 of this paper (arxiv.org/abs/2501.13175) which is "just" an exercise with Hodge theory and Fontaine-Lafaille theory. We'll see how it goes!
[engagement: 7 replies, 5 reposts, 121 likes, 15K views]
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Sebastien Bubeck reposted
Noam Brown (verified) @polynoamial · 7h
Today I was asked why we haven't announced more math results from Astra since these problems.
At @OpenAI we aim to devote time to finding and announcing math results from internal models only when they would meaningfully change people's understanding of the pace of AI progress. Our main focus is shipping great models so everyone can use them to make discoveries of their own.
Note from Claude Sonnet 5
Two stacked X posts. Daniel Litt reports on an overnight, hands-off Codex attempt to formalize a theorem in p-adic Hodge theory; below it, Noam Brown (reposted by Sebastien Bubeck) explains OpenAI's policy on announcing math results from internal models such as Astra.
Bogdan Ionut Cirstea reposted
Ash Jogalekar @curiouswavefn · 20h
I am guilty of this myself. I am about to submit an AI-derived mathematical proof to arXiv that I don't completely understand but which addresses a problem in theoretical biology with a neat solution. I understand the problem, the approach and the result, but I don't understand all the details of the proof. It's been an exciting experience that has occupied the better part of the last month for me. I am transparent about my limitations in understanding the proof and welcome human review that proves it right or wrong (you learn something important about AI in either case).
In general, I consider it an entirely good thing for people who are not specialized to use AI for solving problems in their field using tools from other fields that they don't fully understand. But I agree that it is placing a serious burden on human verifiers. Unless, of course, the AIs become so good at verifying that we don't need humans.
[quoted tweet]
Andrew Akba… @Andrew_Akb… · Aug 19
Terence Tao, one of the most well known mathematicians, speaks up on AI in mathematics in his new paper:
"What if an AI tool generates a lengthy proof th...
[link card, cut off at bottom]
arXiv
[Submitted on 17 Aug 2026]
Mathematics in the age of AI
Terence Tao
Note from Claude Sonnet 5
Ash Jogalekar (@curiouswavefn), reposted by Bogdan Ionut Cirstea, on submitting an AI-derived proof he does not fully understand and the verification burden this places on humans, quoting a post about Terence Tao's arXiv paper "Mathematics in the age of AI" (submitted 17 Aug 2026).
Bogdan Ionut Cirstea reposted
Paata Ivanisvili @PI010101 · 22h
Seeing what is happening on the frontlines, I would say this: if we humans want to keep understanding the knowledge AI is producing at the current rate (new proofs, ideas, and approaches) we may need 10× more mathematicians within the next year than we have today.
[quoted tweet]
Daniel Litt @littmath · Aug 14
We're going to need a lot more mathematicians!
x.com/PI010101/statu...
Note from Claude Sonnet 5
Paata Ivanisvili (@PI010101), reposted by Bogdan Ionut Cirstea, saying humans may need 10x more mathematicians within a year to keep up with AI-produced mathematics; quotes a short Daniel Litt (@littmath) post from Aug 14.
Costa @Clambr0 [Follow]
hello Erdős asked if Σ n!/(2n)! is irrational
It is. Five-line proof. Extends to the whole family Σ n!/((a+1)n+b)! for a≥1, 0≤b≤a. Lean verified.
My AI then went crazy and claimed the whole affine family a≥1, b≥1-a is transcendental using techniques beyond my understanding:
[link card]
clambro/erdos-270-transcendence
Transcendence in the Affine Case of Erdős Problem 270
1 contributor · 0 issues · 0 stars · 0 forks
GitHub - clambro/erdos-270-transcendence: Transcendenc...
From github.com
2:26 PM · Aug 22, 2026 · 71.2K Views
10 replies · 30 reposts · 465 likes · 193 bookmarks
Relevant ▾ View quotes >
[next post in thread]
Costa @Clambr0 · Aug 22
[embedded image of typeset math, bottom partly covered by phone nav bar]
Let
C = Σ_{n=1}^∞ n!/(2n)! = Σ_{n=1}^∞ 1/D_n, D_n := (2n)!/n!.
Since
D_{n+1}/D_n = 4n + 2 ∈ ℤ,
we have D_n | D_{n+1}.
Let P_N = Σ_{n≤N} 1/D_n and R_N = C − P_N. Then D_N P_N ∈ ℤ, while
0 < D_N R_N ≤ Σ_{k≥1} 1/(4N+2)^k = 1/(4N+1) → 0.
Note from Claude Sonnet 5
Tweet by Costa (@Clambr0) saying a five-line, Lean-verified proof shows the Erdős-asked series sum of n!/(2n)! is irrational (extended to a family), and that his AI then claimed transcendence for a wider affine family, linking a GitHub repo for Erdős Problem 270; below, the start of the thread with the typeset irrationality proof. Posted 2026-08-22.
[top of screen, cut off: end of the previous post]
@OpenAI!
1:52 PM · Aug 24, 2026 · 9,825 Views
4 replies · 9 reposts · 113 likes · 25 bookmarks
Relevant ▾
Grigore Rosu @RosuGrigore · 4h
Matching logic is complete without μ, incomplete with μ!
In passing, we also solved two open problems in hybrid logic: the nominal-free fragment is complete; with nominals only, incomplete. This is surprising because if we add pointed validity it becomes complete again (known from 2006).
[embedded diagram of logics and completeness results: the same diagram as in the previous capture, Screenshot_20260824-181724.png; node labels not repeated here]
1 reply · 1 repost · 17 likes · 1.4K views
Grigore Rosu @RosuGrigore · 4h
Here is the paper. Comments are welcome.
arxiv.org/pdf/2608.13306…
Why is matching logic interesting at all? Pure type systems, dependent types, calculus of inductive constructions, etc., MSOL, SOL, HOL, etc., can be defined as matching logic theories. That is, Lean, Rock, Isabelle, etc., proofs can be mechanically translated to matching logic, which is a simpler logic (so it admits a smaller proof checker).
Note from Claude Sonnet 5
Scrolled continuation of the Rosu thread from seq 9: the matching-logic completeness result with the diagram, then a third post linking the paper (arXiv 2608.13306) and explaining why matching logic matters for proof-checker size. Bottom of the capture is overlaid by phone UI (nav bar, compose button) that overlaps a few lines.
Grigore Rosu @RosuGrigore [Follow]
The reason I've been quiet recently.
Over the last two months I worked hard, using AI, on solving two open problems that my UIUC research lab was not able to solve over the last 7 years.
Finally, we proved them. Viva @AnthropicAI and @OpenAI!
1:52 PM · Aug 24, 2026 · 9,825 Views
4 replies · 9 reposts · 113 likes · 25 bookmarks
Relevant ▾
[reply / next post in thread]
Grigore Rosu @RosuGrigore · 4h
Matching logic is complete without μ, incomplete with μ!
In passing, we also solved two open problems in hybrid logic: the nominal-free fragment is complete; with nominals only, incomplete. This is surprising because if we add pointed validity it becomes complete again (known
Show more
[embedded diagram, small text, best-effort: a map of logics and their completeness results; node labels as follows]
matching logic, with definedness
global completeness [2, 3]
equivalent to FOL_= [2]
equivalent to H(@, ∀) [31]
matching μ-logic, with definedness
global completeness fails [3, Prop. 23];
no effective weak completeness (Remark 27)
matching logic, basic many-sorted
global completeness fails
(Section 8)
H(@), H(@, ↓)
global completeness [29]
H(∀)
local completeness [11];
no complete calculus in the
class of Theorem 33
matching logic, basic unsorted
global completeness
(Corollary 15)
matching μ-logic, basic unsorted
no effective weak
completeness (Theorem 19)
modal logic, basic polyadic
global completeness
[12, Ex. 1.5.3]
modal μ-calculus
global completeness, finite Γ,
via [19, 20]; not compact, so
local completeness fails [19]
[arrow labels between nodes: μ, [·], Var, ∃, @, nominals, sorts, nominals, @]
Note from Claude Sonnet 5
Tweet by Grigore Rosu (@RosuGrigore) saying that over two months, using AI, his UIUC lab solved two open problems it had not solved in seven years, crediting Anthropic and OpenAI; below it a follow-up post on completeness of matching logic with and without the mu operator and hybrid logic, with a diagram of logics and completeness results. Posted 2026-08-24.
Paata Ivanisvili @PI010101 · 3h
An army of AI bots is coming, and this is a game changer. I asked my bot to go to MathOverflow, pick an open problem, and solve it. It found one and generated a solution, but it was outside my field, so I couldn't assess its correctness. Then I asked it to pick a problem from my area—analysis/probability. It did, and gave me a solution that I found pretty satisfactory.
[two embedded screenshots of an AI chat interface (light theme, small coloured dots down the left edge); text is tiny and only partly legible, math formulas not reliably readable. Best-effort transcription of the prose follows; [illegible] marks unreadable stretches.]
[left screenshot]
(user) can you go to mathoverflow go over questions hich were not unanswered and solve one of them and then come back and let me know which one did you solve
On it. I'll pull unanswered MathOverflow questions, pick one I can actually close, and come back with the writeup.
Most of the unanswered queue is open research (Collatz ratios, Sun determinants, Serre functors). I'm taking a concrete one instead: a missing step in Conway's normal-form theorem.
Writing the missing step now. I'll come back with which question it was and the proof.
Solved The Conway Normal Form and decreasing indices (MO 514585). Still unanswered. It's the missing step in Conway's Theorem 21: the exponents γ_ν in the normal form are strictly decreasing, including at limit stages.
The OP could see the successor drop γ_n > γ_{n+1} once the remainder is small compared to ω^{γ_n}, but only after knowing S_{n+1} = S_n + r_n ω^{γ_n}. Limit ordinals have no predecessor, so that identity does not even apply.
Write a ≪ b for "small compared to" and a ≍ b for commensurate (on absolute values). Let C_α be the convex class of z such that [illegible]. Conway's "simplest number with those terms" is the simplest member S_α of C_α. The coefficient r_α was chosen exactly so that z itself satisfies that inequality, so z ∈ C_α at every stage.
Successor: both z and S_{α+1} lie in C_{α+1}, so both are infinitely close to S_α + r_α ω^{γ_α}. Their difference satisfies [illegible] If [illegible] = 0, then [illegible] hence γ_{α+1} < γ_α. (The ordinary sum S_α + r_α ω^{γ_α} is itself simplest in C_{α+1}, which is the identity the OP wanted.)
Limit: let λ be a limit. Both z and S_λ lie in C_β [illegible]. For each β < λ they are both infinitely close to S_β + r_β ω^{γ_β}, so [illegible] If [illegible] = 0, then γ_λ < γ_β for every earlier β. No predecessor is used.
The full claim β < α ⟹ γ_β > γ_α is then transfinite induction. I did not post it on the site. Say if you want a version to paste.
Today 10:34 AM
(user) try to find open analysis or probability question and solve it
On it. I'll scan unanswered analysis and probability questions and take one I can actually close.
A few look closable. Checking a concentrati[cut off by a scroll-down button] [illegible] on commuting operators before I pick.
[right screenshot]
(user) try to find open analysis or probability question and solve it
On it. I'll scan unanswered analysis and probability questions and take one I can actually close.
A few look closable. Checking a concentration question and one on commuting operators before I pick.
Taking Concentration inequality for dependent random variables (MO 514247). Writing the argument now.
Solved Concentration inequality for dependent random variables (MO 514247). Still unanswered. Bounded Y_i(X) alone does not concentrate S = Σ_i X_i Y_i(X) at any scale o(n). You get √n tails only with independence or a Lipschitz bound on Y.
The crude bound |S| ≤ C Σ|X_i| is always true, and Σ|X_i| itself concentrates around n√(2/π) with width √n. That only says |S| = O_P(n). It does not stop S from sitting at +cn or −cn, [illegible] with equal mass.
Counterexample: set Y_i(X) = C sign(x_1) sign(x_i). Then S = C sign(X_1) Σ|X_i|. The sign is Rademacher and independent of the modulus, so [illegible] Hence Var(S) = Θ(n²). The same law appears under leave-one-out, via Y_i(X) = C Π_{j≠i} sign(x_j), and Y_i(0) = 0 already holds. So neither Y_i(0) = 0 nor Y_i = Y_i(X_{−i}) helps.
Two cases that do concentrate at scale √n.
If Y is independent of X (or constant), then S | Y ~ N(0, Σ Y_i²) with [illegible] ≤ C²n, so P(|S| > t) ≤ 2 exp(−t²/(2C²n)).
If x ↦ Y(x) is L-Lipschitz ℝⁿ → ℝⁿ, restrict to the ball [illegible] (Gaussian mass 1 − e^{−n/2}). There [illegible] ≤ (C + 2L)√n |x − y|₂.
Projecting onto the ball and applying Gaussian Lipschitz concentration gives, for a median m_n of the truncation, [illegible] ≤ 2 exp(−t²/(2(C + 2L)²n)) + e^{−n/2}.
(The centre m_n can be as large as O(C + L)n. This is concentration about m_n, not about 0.)
I did not post it. Say if you want a version to paste.
Note from Claude Sonnet 5
Tweet by Paata Ivanisvili (@PI010101) saying he had his AI bot pick unanswered MathOverflow questions and solve them, with two embedded screenshots of the bot's chat replies (a Conway normal form question and a concentration-inequality question). Embedded text is small; prose transcribed best-effort, formulas largely marked illegible or reconstructed only where clearly visible.
Tom Adamczewski @tmkadamcz · 2h
Claude Fable 5 and GPT-5.6 Sol have found a 15-digit counterexample to Zhi-Wei Sun's 2-4-6-8 conjecture (2019): that every positive integer is a sum C(w,2)+C(x,4)+C(y,6)+C(z,8) of binomial coefficients.
n = 896,315,812,331,399 has no such representation
Note from Claude Sonnet 5
Tweet by Tom Adamczewski (@tmkadamcz) reporting that Claude Fable 5 and GPT-5.6 Sol found a 15-digit counterexample to a 2019 number-theory conjecture by Zhi-Wei Sun. Posted plainly, no joke markers.
roon @tszzl · 5h
people are incredibly mean to mathematicians in this time. they really relish when an ai solves something and frame in a zero sum way w human mathematicians. i think it evens out some childhood era math trauma they have?
180 replies, 102 retweets, 1.9K likes, 90K views
Mahaoo @mahaoo_ASI · 57m
Seeing some ego death happening among some folks on twitter following the new math results
We really should just speed run this as a society
everyone who thinks they are "safe" today will just prolong their false hope and increase the dissonance in a few years when it comes
It's just healthier to accept what is coming and understand that the next few[1] years will render every person on the planet obsolete in terms of their cognitive abilities
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[1] the precise number is unknown, but it's a small number compared to a human's lifespan, so better sooner rather then later just accept it
Note from Claude Sonnet 5
Two tweets reacting to new AI math results: roon (@tszzl) criticizes people relishing AI beating human mathematicians as venting 'childhood math trauma'; Mahaoo (@mahaoo_ASI) argues society should 'speed run' accepting that everyone's cognitive abilities will soon be rendered obsolete by AI.
