Introduction
Over the past few years, large language models have become increasingly useful for tasks such as writing, programming, and planning. More recently, they have started to be capable of contributing intellectually to scientific research. With GPT-5, we see early signs that, guided by an expert, the model can propose helpful ideas, perform deep literature searches, and even produce complete new proofs. This report documents internal and external cases in which GPT-5 contributed to scientific progress, and how it fell short. Our aim is to highlight what is possible today, what is still out of reach, and implications for the future of scientific research.
We focus on concrete examples spanning mathematics, physics, astronomy, computer science, biology, and materials science. Since the subject matter varies widely by section, each begins with an explanation of the background and motivation before discussing the contribution made by GPT-5. We also provide ChatGPT conversation transcripts when possible, and describe the human input that was needed, in order to make clear where the model currently adds value and where expert oversight remains essential. Several related accounts in a similar spirit have also surfaced in recent months, including [32 , 25 , 41 , 2 , 42 , 58 , 35 ] . We note that the latter paper, on AlphaEvolve by Google, is slightly different in flavor, as it focuses on search problems with a well-defined objective function that can be hill-climbed. By contrast, we (and the other cited papers) are focused on a general-purpose system that can answer any type of query. The two approaches are complementary, each providing unique advantages to scientists.
We are also careful to point out the limits of current AI models. GPT-5 is imperfect: it can confidently make mistakes, ardently defend them, and confuse itself (and us) in the process. Results may depend on fine details of both the initial prompt and follow-up responses, and thus can be challenging to reproduce. Despite these limitations, we see real progress. GPT-5 can search broad conceptual spaces, integrate diverse information sources, and iterate quickly. It can tirelessly propose new ideas, help turn an imprecise idea into a concrete result, and sanity-check or extend a given line of thought. In some areas, such as literature search, it is often uniquely effective.
Our aim is not to claim more than the evidence allows. It is to show, with specific examples, what GPT-5 can and cannot do today, and to give a clear path for how researchers can use it to accelerate scientific discovery while keeping standards high. We believe GPT-5 already provides substantial value for scientific researchers today, and will become an even more powerful tool tomorrow. The rest of the paper is organized as follows:
Chapter I collects examples in which GPT-5 independently rediscovers known results at the research frontier in math, physics, and biology.
Chapter II highlights the ability of GPT-5 to perform “deep literature search”. Its ability to focus on the core concepts rather than the words used to describe them surmounts barriers of language between scientific disciplines to uncover seemingly forgotten or hard-to-find connections.
Chapter III displays examples of human researchers working in tandem with GPT-5 to accelerate their research workflow.
Chapter IV presents examples of GPT-5 obtaining novel research-level results.
Chapter I
Independent rediscovery of known results at the scientific frontier
Improved step-size condition in a recent convex optimization result – Sébastien Bubeck
This section can be viewed as a warm-up for the rest of the paper. The experiment was conducted on August 20th 2025, at a time when it was not yet clear whether GPT-5 could be used to push the frontier in research level questions. The goal of this experiment was more modest than what is described in several of the other sections, namely it was merely to see if GPT-5 could reproduce the main result of a recent paper, specifically the main theorem in [10 ] . This particular paper was chosen for several reasons:
First, obviously, the paper is not in the training data for GPT-5 as it is too recent. But even more importantly, the main theorem is about a genuinely new question for a very classical field (convex optimization), and in particular no trace of this question can be found in the training data either.
The paper has three versions on arxiv (v1, v2, v3), and crucially v1 presents a suboptimal result which is then refined (and made optimal) in v2. The challenge I was interested in is: given v1, can GPT-5 rederive v2?
In particular, if GPT-5 could rederive v2 from v1, it would indicate that it could have counterfactually accelerated the scientific discovery process. As we will see, GPT-5 did not manage to fully rederive the v2 result, but it basically went half-way between v1 and v2. (See also the comment at the end of this section on deriving v2 from scratch with models that can reason for longer.)
Remarkably, this was my first attempt at an experiment of this type (i.e., I did not try any other paper than [10 ] ).
Description of the problem
The most basic algorithm to optimize an objective function f : R d → R f : \mathbb{R}^d \to\mathbb{R} f : R d → R η \eta η
x n + 1 = x n − η ∇ f ( x ) . x_{n+1} = x_n - \eta\nabla f(x). x n + 1 = x n − η ∇ f ( x ) . A classical result in convex optimization is that, for smooth convex functions, convergence happens if and only if the step size is smaller than 2 / L 2/L 2/ L L L L x ↦ ∇ f ( x ) x \mapsto\nabla f(x) x ↦ ∇ f ( x ) [10 ] the authors study a new variant of the convergence question, which as far as we know (both the authors of that paper and the present author) was never studied before: instead of merely asking for convergence, under what condition is the curve traced by the visited objective function values itself convex, i.e., when is the piecewise linear function going through the points { ( n , f ( x n ) ) } n ∈ N \{(n, f(x_n))\}_{n\in\mathbb{N}} {( n , f ( x n )) } n ∈ N
In v1 of the paper, the authors prove that η ≤ 1 / L \eta\le1/L η ≤ 1/ L η ≤ 1.75 / L \eta\le1.75/L η ≤ 1.75/ L η ≤ 1.75 / L \eta\le1.75/L η ≤ 1.75/ L
GPT-5 acting as a collaborator: from v1 getting close to v2
Attachment: [2503.10138v1.pdf]
Can you improve the condition on the step-size in Theorem 1? I don’t want to add any more hypotheses, I just want you to work to improve the step-size condition under the same assumptions as Theorem 1.
[Reasoned for 17m 35s]
Yes. Under the same assumptions as Theorem 1 (convex L L L f f f η ≤ 1 / L \eta\le1/L η ≤ 1/ L
η ≤ 3 2 L \eta\le\frac{3}{2L} η ≤ 2 L 3 i.e., the optimization curve n ↦ f ( x n ) n \mapsto f(x_n) n ↦ f ( x n ) η ∈ ( 0 , 3 / ( 2 L ) ] \eta\in(0, 3/(2L)] η ∈ ( 0 , 3/ ( 2 L )]
As discussed above, the experiment I was interested in is given v1, can GPT-5 rederive v2? To check this, I uploaded the PDF of the v1 paper, and simply asked GPT-5 (Pro) to try to improve the result. The entire interaction is captured in Figure I.1 above. In particular, one can see that GPT-5 claims to have improved the condition from η ≤ 1 / L \eta\le1/L η ≤ 1/ L η ≤ 1.5 / L \eta\le1.5/L η ≤ 1.5/ L η ≤ 1.75 / L \eta\le1.75/L η ≤ 1.75/ L fig:I.2 , which the present author has verified to be correct. Moreover, I inspected the reasoning traces to make sure that no web search was performed (as this could have revealed the existence of v2), and furthermore the proof given by GPT-5 is quite different from the one in v2. Indeed, the GPT-5 proof can be viewed as a more canonical variant of the v1 proof, whereas the (human) v2 proof requires a clever weighting of different inequalities for certain cancellations to happen.
Figure I.1. The interaction with GPT-5 Pro to derive the improved bound.
To say it plainly, such a result (improving from 1 / L 1/L 1/ L 1.5 / L 1.5/L 1.5/ L
We end this section by noting that our internal models, which can think for a few hours, were able to derive the optimal bound 1.75 / L 1.75/L 1.75/ L
Theorem I.1.1 (Improved step-size condition). Let f : R d → R f:\mathbb{R}^{d}\to\mathbb{R} f : R d → R L L L
x k + 1 = x k − η ∇ f ( x k ) , k ≥ 0. x_{k+1}=x_k-\eta\nabla f(x_k),\qquad k\ge0. x k + 1 = x k − η ∇ f ( x k ) , k ≥ 0. If η ∈ ( 0 , 3 2 L ] \eta\in\left(0,\frac{3}{2L}\right] η ∈ ( 0 , 2 L 3 ] { f ( x k ) } k ≥ 0 \{f(x_k)\}_{k\ge0} { f ( x k ) } k ≥ 0
D k : = f ( x k ) − f ( x k + 1 ) D_k:=f(x_k)-f(x_{k+1}) D k := f ( x k ) − f ( x k + 1 ) form a nonincreasing sequence: D k + 1 ≤ D k D_{k+1}\le D_k D k + 1 ≤ D k k ≥ 0 k\ge0 k ≥ 0
Proof. Write g k : = ∇ f ( x k ) g_k:=\nabla f(x_k) g k := ∇ f ( x k ) Δ k : = g k + 1 − g k \Delta_k:=g_{k+1}-g_k Δ k := g k + 1 − g k D k D_k D k D k + 1 D_{k+1} D k + 1
(1) A lower bound for D k D_k D k For convex L L L f f f
1 2 L ∥ ∇ f ( x ) − ∇ f ( y ) ∥ 2 ≤ f ( x ) − f ( y ) − ⟨ ∇ f ( y ) , x − y ⟩ . \frac{1}{2L}\lVert\nabla f(x)-\nabla f(y)\rVert^2\le f(x)-f(y)-\langle\nabla f(y),x-y\rangle. 2 L 1 ∥ ∇ f ( x ) − ∇ f ( y ) ∥ 2 ≤ f ( x ) − f ( y ) − ⟨ ∇ f ( y ) , x − y ⟩ . With x = x k x=x_k x = x k y = x k + 1 y=x_{k+1} y = x k + 1 x k − x k + 1 = η g k x_k-x_{k+1}=\eta g_k x k − x k + 1 = η g k
D k ≥ η ⟨ g k + 1 , g k ⟩ + 1 2 L ∥ Δ k ∥ 2 . (1) D_k\ge\eta\langle g_{k+1},g_k\rangle+\frac{1}{2L}\lVert\Delta_k\rVert^2. \tag*{(1)} D k ≥ η ⟨ g k + 1 , g k ⟩ + 2 L 1 ∥ Δ k ∥ 2 . (1) (2) An upper bound for D k + 1 D_{k+1} D k + 1 By convexity, f ( x ) − f ( y ) ≤ ⟨ ∇ f ( x ) , x − y ⟩ f(x)-f(y)\le\langle\nabla f(x),x-y\rangle f ( x ) − f ( y ) ≤ ⟨ ∇ f ( x ) , x − y ⟩ x = x k + 1 x=x_{k+1} x = x k + 1 y = x k + 2 y=x_{k+2} y = x k + 2 x k + 1 − x k + 2 = η g k + 1 x_{k+1}-x_{k+2}=\eta g_{k+1} x k + 1 − x k + 2 = η g k + 1
D k + 1 ≤ η ∥ g k + 1 ∥ 2 . (2) D_{k+1}\le\eta\lVert g_{k+1}\rVert^2. \tag*{(2)} D k + 1 ≤ η ∥ g k + 1 ∥ 2 . (2) (3) Subtract and use cocoercivity. Subtracting (B) from (A) yields
D k − D k + 1 ≥ η ⟨ g k + 1 , g k − g k + 1 ⟩ + 1 2 L ∥ Δ k ∥ 2 = − η ⟨ g k + 1 , Δ k ⟩ + 1 2 L ∥ Δ k ∥ 2 . D_k-D_{k+1}\ge\eta\langle g_{k+1},g_k-g_{k+1}\rangle+\frac{1}{2L}\lVert\Delta_k\rVert^2=-\eta\langle g_{k+1},\Delta_k\rangle+\frac{1}{2L}\lVert\Delta_k\rVert^2. D k − D k + 1 ≥ η ⟨ g k + 1 , g k − g k + 1 ⟩ + 2 L 1 ∥ Δ k ∥ 2 = − η ⟨ g k + 1 , Δ k ⟩ + 2 L 1 ∥ Δ k ∥ 2 . Since g k + 1 = g k + Δ k g_{k+1}=g_k+\Delta_k g k + 1 = g k + Δ k ⟨ g k + 1 , Δ k ⟩ = ⟨ g k , Δ k ⟩ + ∥ Δ k ∥ 2 \langle g_{k+1},\Delta_k\rangle=\langle g_k,\Delta_k\rangle+\lVert\Delta_k\rVert^2 ⟨ g k + 1 , Δ k ⟩ = ⟨ g k , Δ k ⟩ + ∥ Δ k ∥ 2
D k − D k + 1 ≥ − η ⟨ g k , Δ k ⟩ + ( 1 2 L − η ) ∥ Δ k ∥ 2 . D_k-D_{k+1}\ge-\eta\langle g_k,\Delta_k\rangle+\left(\frac{1}{2L}-\eta\right)\lVert\Delta_k\rVert^2. D k − D k + 1 ≥ − η ⟨ g k , Δ k ⟩ + ( 2 L 1 − η ) ∥ Δ k ∥ 2 . For convex L L L f f f 1 / L 1/L 1/ L
1 L ∥ Δ k ∥ 2 ≤ ⟨ Δ k , x k + 1 − x k ⟩ = ⟨ Δ k , − η g k ⟩ = − η ⟨ g k , Δ k ⟩ . \frac{1}{L}\lVert\Delta_k\rVert^2\le\langle\Delta_k,x_{k+1}-x_k\rangle=\langle\Delta_k,-\eta g_k\rangle=-\eta\langle g_k,\Delta_k\rangle. L 1 ∥ Δ k ∥ 2 ≤ ⟨ Δ k , x k + 1 − x k ⟩ = ⟨ Δ k , − η g k ⟩ = − η ⟨ g k , Δ k ⟩ . Therefore,
D k − D k + 1 ≥ ( 1 L + 1 2 L − η ) ∥ Δ k ∥ 2 = ( 3 2 L − η ) ∥ Δ k ∥ 2 . D_k-D_{k+1}\ge\left(\frac{1}{L}+\frac{1}{2L}-\eta\right)\lVert\Delta_k\rVert^2=\left(\frac{3}{2L}-\eta\right)\lVert\Delta_k\rVert^2. D k − D k + 1 ≥ ( L 1 + 2 L 1 − η ) ∥ Δ k ∥ 2 = ( 2 L 3 − η ) ∥ Δ k ∥ 2 . If η ≤ 3 2 L \eta\le\frac{3}{2L} η ≤ 2 L 3 D k + 1 ≤ D k D_{k+1}\le D_k D k + 1 ≤ D k k k k { f ( x k ) } k ≥ 0 \{f(x_k)\}_{k\ge0} { f ( x k ) } k ≥ 0
Figure 2. GPT-5 Pro’s proof of the sufficient condition η ≤ 1.5 / L \eta\le1.5/L η ≤ 1.5/ L
Discovering new black hole symmetries with GPT-5 – Alex Lupsasca
We document, in a standardized format, a theoretical calculation in black hole physics performed by an AI. GPT-5 Pro (re)derived nontrivial Lie point symmetries—including an SL ( 2 , R ) \operatorname{SL}(2,\mathbb{R}) SL ( 2 , R ) [47 ] (to which the model did not have access).
The problem in context
Astrophysical black holes are characterized by their mass M M M J = a M J=aM J = a M ( M , a ) (M,a) ( M , a )
We study massless, stationary, axisymmetric waves on a rotating (Kerr) black hole. In Boyer-Lindquist coordinates ( t , r , x , ϕ ) (t,r,x,\phi) ( t , r , x , ϕ ) x = cos θ x=\cos\theta x = cos θ
∂ r [ Δ ( r ) ∂ r ψ ( r , x ) ] + ∂ x [ ( 1 − x 2 ) ∂ x ψ ( r , x ) ] = 0 , Δ ( r ) = r 2 − 2 M r + a 2 . (3) \partial_r\left[\Delta(r)\,\partial_r\psi(r,x)\right]+\partial_x\left[(1-x^2)\,\partial_x\psi(r,x)\right]=0,\qquad\Delta(r)=r^2-2Mr+a^2. \tag*{(3)} ∂ r [ Δ ( r ) ∂ r ψ ( r , x ) ] + ∂ x [ ( 1 − x 2 ) ∂ x ψ ( r , x ) ] = 0 , Δ ( r ) = r 2 − 2 M r + a 2 . (3) This is a linear second-order partial differential equation in two variables. Its solutions encode the static tidal response of black holes; their asymptotics determine horizon deformability (“Love numbers”). In general relativity, black holes exhibit vanishing (static) Love numbers, a surprising rigidity often traced to hidden symmetries of the field equations.
A key question is thus: what are the symmetries of Eq. (31) ?
Human route. Classically, one would apply Lie’s algorithm for reduction of order to Eq. (31) . In this procedure, one first posits a symmetry generator X = ξ r ∂ r + ξ x ∂ x + η ∂ ψ X=\xi^r\partial_r+\xi^x\partial_x+\eta\partial_\psi X = ξ r ∂ r + ξ x ∂ x + η ∂ ψ X X X (31) , and finally solves the resulting overdetermined system for the unknown functions ( ξ r , ξ x , η ) (\xi^r,\xi^x,\eta) ( ξ r , ξ x , η ) r r r x x x Δ \Delta Δ 1 − x 2 1-x^2 1 − x 2 [47 ] , I carried out this analysis and was surprised to find three nontrivial Lie point symmetries that generate an SL ( 2 , R ) \operatorname{SL}(2,\mathbb{R}) SL ( 2 , R ) H 0 H_0 H 0
Additional references. For background on Kerr and wave equations: [43 , 66 , 67 ] . For black-hole tidal response and Love numbers, our main technical reference is [47 ] and citations therein.
The task given to GPT-5 Pro: goal, prompts, and model behavior
Goal. Elicit from GPT-5 Pro the Lie point symmetries of (31) , ideally reproducing the SL ( 2 , R ) \operatorname{SL}(2,\mathbb{R}) SL ( 2 , R ) [47 ] .
Outcome. After ∼ 5 \sim5 ∼ 5
What are the Lie point symmetries of the PDE eq:I.1 ?
[Reasoned for 5m 13s] [Incorrect answer: GPT-5 Pro finds no symmetries and wrongly asserts that none exist.]
Warm-up prompt and outcome (success on flat space). To scaffold the task, we initialized a new instance of the model and first posed the flat-space limit M , a → 0 M,a \to0 M , a → 0 ( ρ , z ) (\rho,z) ( ρ , z ) ρ = r 1 − x 2 \rho= r\sqrt{1-x^2} ρ = r 1 − x 2 z = r x z=rx z = r x
( ∂ ρ 2 + ∂ z 2 + 1 ρ ∂ ρ ) ψ ( ρ , z ) = 0. (4) \left(\partial_{\rho}^{2}+\partial_{z}^{2}+\frac{1}{\rho}\partial_{\rho}\right)\psi(\rho,z)=0. \tag*{(4)} ( ∂ ρ 2 + ∂ z 2 + ρ 1 ∂ ρ ) ψ ( ρ , z ) = 0. (4) Outcome. After 10 minutes and 27 seconds, the model produced all the symmetries, including three that generate SL ( 2 , R ) \operatorname{SL}(2,\mathbb{R}) SL ( 2 , R )
H + = ∂ z , H 0 = ρ ∂ ρ + z ∂ z + 1 2 , H − = 2 ρ z ∂ ρ − ( ρ 2 − z 2 ) ∂ z + z . (5) H_{+}=\partial_{z}, \qquad H_{0}=\rho\,\partial_{\rho}+z\,\partial_{z}+\frac{1}{2}, \qquad H_{-}=2\rho z\,\partial_{\rho}-(\rho^{2}-z^{2})\,\partial_{z}+z. \tag*{(5)} H + = ∂ z , H 0 = ρ ∂ ρ + z ∂ z + 2 1 , H − = 2 ρ z ∂ ρ − ( ρ 2 − z 2 ) ∂ z + z . (5) Here H − H_{-} H −
What are the Lie point symmetries of (4) ?
[Reasoned for 10m 27s] [Correct answer including generators (5) .]
What are the Lie point symmetries of eq:I.1 ?
[Reasoned for 18m 9s] [Correct answer including generators eq:I.4 .]
Second curved-space attempt and outcome (success). We then gave the same instance of GPT-5 the same prompt as before: “What are the Lie point symmetries of eq:I.1 ?”
Outcome. Within ∼ 18 \sim18 ∼ 18 S L ( 2 , R ) \mathrm{SL}(2,\mathbb{R}) SL ( 2 , R )
H + = x Δ ∂ r + ( r − M ) ( 1 − x 2 ) ∂ x ( r − M ) 2 − ( M 2 − a 2 ) x 2 , (6) H_{+} = \frac{x \Delta\, \partial_{r} + (r-M)(1-x^{2}) \, \partial_{x}}{(r-M)^{2}-(M^{2}-a^{2})x^{2}}, \tag*{(6)} H + = ( r − M ) 2 − ( M 2 − a 2 ) x 2 x Δ ∂ r + ( r − M ) ( 1 − x 2 ) ∂ x , (6) H 0 = ( r − M ) Δ ∂ r + ( M 2 − a 2 ) x ( 1 − x 2 ) ∂ x ( r − M ) 2 − ( M 2 − a 2 ) x 2 + 1 2 , (7) H_{0} = \frac{(r-M)\Delta\, \partial_{r} + (M^{2}-a^{2})x(1-x^{2}) \, \partial_{x}}{(r-M)^{2}-(M^{2}-a^{2})x^{2}}+\frac{1}{2}, \tag*{(7)} H 0 = ( r − M ) 2 − ( M 2 − a 2 ) x 2 ( r − M ) Δ ∂ r + ( M 2 − a 2 ) x ( 1 − x 2 ) ∂ x + 2 1 , (7) H − = ( M 2 − a 2 ) x Δ ∂ r − ( r − M ) ( 1 − x 2 ) [ Δ − ( M 2 − a 2 ) x 2 ] ∂ x ( r − M ) 2 − ( M 2 − a 2 ) x 2 + x Δ ∂ r + ( r − M ) x . (8) H_{-} = \frac{(M^{2}-a^{2})x\Delta\, \partial_{r}-(r-M)(1-x^{2})[\Delta-(M^{2}-a^{2})x^{2}] \, \partial_{x}}{(r-M)^{2}-(M^{2}-a^{2})x^{2}}+x\Delta\, \partial_{r}+(r-M)x. \tag*{(8)} H − = ( r − M ) 2 − ( M 2 − a 2 ) x 2 ( M 2 − a 2 ) x Δ ∂ r − ( r − M ) ( 1 − x 2 ) [ Δ − ( M 2 − a 2 ) x 2 ] ∂ x + x Δ ∂ r + ( r − M ) x . (8) Analysis. In summary, here is
What GPT-5 got right: The relatively simple flat-space symmetries; the full nontrivial curved-space coefficients; the S L ( 2 , R ) \mathrm{SL}(2,\mathbb{R}) SL ( 2 , R )
What GPT-5 got wrong (along the way). The cold start on Eq. eq:I.1 incorrectly concluded “no symmetries.” The model appears to have needed to “warm up” via the simpler flat-space problem (4) sharing the same symmetry structure.
Log. For verification and reproducibility purposes, this conversation with GPT-5 can be accessed here: [46 ] .
Result, implications, and next steps
Result. GPT-5 Pro (re)discovered the curved-space S L ( 2 , R ) \mathrm{SL}(2,\mathbb{R}) SL ( 2 , R ) eq:I.4 of Eq. eq:I.1 . This matches the key structural insight of [47 ] . Practically, once the symmetry is known, downstream results (e.g., constraints on tidal response and the vanishing of static Love numbers in this sector) follow with comparatively modest analysis.
Reflection on the interaction. Two observations:
Scaffolding mattered. The model failed “cold” but succeeded rapidly after a closely related warm-up. This suggests retrieval or internal pattern activation can be primed by presenting a simpler member of the same symmetry class.
Algorithmic plausibility. The final generators are too structured to be a lucky guess. The model likely executed (implicitly) a mix of: recognizing conformal invariance in the flat equation, hypothesizing a curved analogue, and/or exploiting a coordinate map that simplifies Eq. eq:I.1 toward Eq. (4) .
Implications
AI as a symmetry engine. With minimal domain scaffolding, current models can carry out nontrivial Lie-symmetry discovery for PDEs with non-constant coefficients.
Research velocity. Given such capabilities, the time from idea to publishable result can compress from months to days once the right prompts and scaffolds are in place.
Takeaway. GPT-5 Pro, when properly scaffolded, uncovered the S L ( 2 , R ) \mathrm{SL}(2,\mathbb{R}) SL ( 2 , R )
Mechanistic analysis and outcome prediction for in vitro immune system experiments using GPT-5 Pro – Derya Unutmaz, M.D.
Here, I demonstrate that GPT-5 Pro successfully analyzed a figure from an experiment with human T cells cultured with 2-deoxy-D-glucose (2-DG) that showed an increased proinflammatory Th17 cell subset. This experiment was performed in our lab several years ago, but the mechanism remained unclear. GPT-5 Pro provided the key mechanism that could explain these findings and, in addition, made highly relevant experimental suggestions. The mechanistic insight and further hypothesis to dissect these findings were highly valuable and not immediately obvious, despite our deep expertise in this field. In a subsequent unpublished figure, GPT-5 Pro interpreted flow cytometry data of the checkpoint inhibitors PD-1 and LAG-3 on cytotoxic T cells after transient glycolysis inhibition, inferring that 2-DG was reprogramming inhibitory receptor expression through combined effects on glycosylation and through attenuated T cell receptor signaling. It further correctly predicted that a brief 2-DG pulse during CAR-T cell generation from these cells would enhance their cytotoxicity towards target cancer cell lines, which we had internally validated in unpublished results. Together, these examples illustrate how GPT-5 Pro can function as a true mechanistic co-investigator in biomedical research, compressing months of reasoning into minutes, uncovering non-obvious hypotheses, and directly shaping experimentally testable strategies.
Problem Context
We had been studying how subsets of human immune cells called T cells respond to modifications in their glucose metabolism. There was significant evidence that glucose uptake and energy metabolism influence the differentiation of T cells into effector subsets, which can enhance protection against infections or drive pathologies such as chronic inflammation and autoimmunity, and could have implications during cancer immunotherapy. To dissect this mechanism, we used a competitive analog of glucose, 2-deoxy-D-glucose (2-DG), as an inhibitor of glucose uptake in T cells. Because T cells are highly heterogeneous, we first sorted them into C D 4 + CD4^+ C D 4 + C D 8 + CD8^+ C D 8 + C D 4 + CD4^+ C D 4 +
The task given to GPT-5: goal, prompts, and model behavior.
The figure in the prompt below is derived from flow cytometry data on cells cultured with 2-DG. I uploaded it as part of the prompt, which led to the following conversation with GPT-5. While we had published a similar findings as a preprint, this specific figure or experiment had not been previously published.
Figure 1.
Context and question: We tested how transient glycolysis inhibition during human T cell priming affects later Th17 function. Sorted C D 4 + CD4^{+} C D 4 + C D 4 + CD4^{+} C D 4 +
A quantitative readout of dose responses for IL-17A, CCR6, and CD161, including percent positive, median fluorescence intensity, and effect sizes with confidence intervals across donors.
A comparison of expected outcomes if the experiment started with naive C D 4 + CD4^{+} C D 4 + C D 4 + CD4^{+} C D 4 + C D 4 + CD4^{+} C D 4 +
[Reasoned for 17m]
Briefly exposing human C D 4 + \mathrm{CD4}^{+} CD4 + I L - 17 A + \mathrm{IL\text{-}17A}^{+} IL - 17A + C C R 6 + \mathrm{CCR6}^{+} CCR6 + C D 161 + \mathrm{CD161}^{+} CD161 + I L - 17 A \mathrm{IL\text{-}17A} IL - 17A C C R 6 + / C D 161 + \mathrm{CCR6}^{+}/\mathrm{CD161}^{+} CCR6 + / CD161 + I L - 17 A \mathrm{IL\text{-}17A} IL - 17A
After 17 minutes of reasoning, GPT-5 Pro produced multiple histogram figures like these two examples, that perfectly analyzed the data, along with an outstanding executive summary that captured the key mechanistic points.
Figure 1.
It also generated a quantitative dose–response readout for the figure (output not shown) and a CSV file containing all underlying data used to generate the histograms. GPT-5 Pro then provided an interpretation of how 2-DG treatment altered Th17 biology. The mechanistic inferences from GPT-5 Pro were pivotal, and some were ideas we had not previously considered. In retrospect, in particular, the proposed mechanism of reduced IL-2 signaling via interference with N-linked glycosylation made clear biological sense, because it could directly explain the disinhibition of Th17 cell differentiation under 2-DG treatment. However, this mechanistic hypothesis had not occurred to us previously, as making this connection was not immediately obvious. Other mechanistic hypotheses were also highly relevant and plausible.
What the data say:
A single 2DG exposure at priming led, two weeks later, to (i) a 13 × 13\times 13 × + ^{+} + 16 16 16 20 × 20\times 20 × + ^{+} + + ^{+} + + ^{+} + enrichment and/or imprinting of Th17 lineage memory rather than a uniform gain in acute effector cytokine output across all cells. Many CCR6+ ^{+} + + ^{+} + − ^{-} − poised Th17 memory that can produce IL-17A under the right cues (for example, IL-1β \beta β
Mechanistic inferences. Several nonexclusive mechanisms could generate this durable skew:
Reduced IL2 signaling via N-linked glycosylation interference. 2-DG at mM levels perturbs Nglycan synthesis, which in activated T cells reduces proper surface expression of glycoproteins—including IL2Rα \alpha α disinhibits Th17 differentiation and limits conversion of Th17 cells to Th1-like states. The stronger effect of 2DG than low glucose supports a role for glycosylation beyond mere ATP/pyruvate restriction.
Acute metabolic checkpointing that locks in a lineage program. Transient glycolytic inhibition can shift activation metabolism toward mitochondrial oxidation , altering mTORC1/HIF1α \alpha α α \alpha α epigenetic memory at Th17-relevant loci (RORC, IL23R, CCR6). The persistence after washout is most consistent with such a program.
Selective survival/expansion. 2DG may preferentially limit proliferation or survival of non-Th17 subsets (e.g., Th1), effectively enriching preexisting CCR6+ ^{+} + + ^{+} + present in the sorted total CD4+ ^{+} +
Altered TCR/CD28 signal strength. Perturbed glycosylation also modulates TCR and co-stimulatory receptor function during priming. Weaker IL-2-rich effector programs and preserved RORγ \gamma γ
High level conclusions and potential impact: A onetime metabolic perturbation during priming imprints a durable Th17-leaning memory phenotype in human CD4+ ^{+} +
GPT-5 Pro then proposed key follow-up experiments. We had already performed the first of these experiments, the mannose rescue experiment, weeks after the first experiment. The result matched GPT-5 Pro’s prediction exactly. However, although we have not published this specific figure, we did publish a preprint in BioRxiv that included the mannose rescue experiment in a similar story. Therefore, while impressive, a caveat for this suggestion is that GPT-5 Pro may have known about this finding and made the connection with this result.
That said, GPT-5 Pro also suggested additional, even more important experiments targeting N-linked glycosylation using inhibitors, as well as experiments to interrogate the IL-2 axis during priming, neither of which we have performed nor published in the context of this specific finding. We are also not aware of any other publication performing these experiments for Th17 cell differentiation status after 2-DG treatment. In addition, the suggestions by GPT-5 Pro to map the functional plasticity of Th17 cells and their epigenetic imprinting are striking, because they require an extremely deep understanding of Th17 cell biology.
To discriminate among mechanisms and establish causality, the most informative next steps are: Partition “glycolysis block” vs “glycosylation block.” Repeat the priming step with:
2 2 2 ± \pm ±
specific glycolysis inhibitors that do not inhibit N-glycosylation (e.g., PFKFB3 inhibitor 3PO or iodoacetate at carefully titrated, nontoxic doses ), and
N-glycosylation inhibitors (e.g., tunicamycin) at low doses. If mannose reverses the 2-DG effect while glycolysis-only inhibitors do not phenocopy it, glycosylation is the driver.
Interrogate the IL2 axis during priming. Quantify surface CD25 maturation (PNGase F shift), pSTAT5, and IL2 consumption. Rescue experiments with exogenous IL2 , or IL2/antiIL2 immune complexes that enforce STAT5 signaling, should counteract Th17 bias if IL2 pathway suppression is causal.
Track proliferation/survival and composition. Use CellTrace/Ki67 and Annexin V at 24–96 h after priming to determine whether 2-DG is enriching CCR6+/CD161+ cells through selective growth/survival. Include absolute cell counts so percentages can be interpreted.
Define the epigenetic imprint. Perform ATACseq and targeted CUT&Tag for H3K27ac/H3K4me3 at RORC, IL-17A/F, IL23R, CCR6, and PRDM1 after priming ± \pm ±
Map functional plasticity. After the two-week washout, restimulate with IL-1β \beta β (Th17 stabilizing) or IL12 (Th1 skewing) to see whether 2-DG-primed cells are preferentially locked into or resistant to Th1 conversion.
Rule out MAIT contamination. Because CD161 is also expressed by MAIT cells, add TRAV12 (Vα \alpha α
Metabolism readouts. Seahorse ECAR/OCR at 24–48h post-priming to verify the intended glycolytic restriction and compensatory mitochondrial changes; measure mTORC1 readouts (pS6) and HIF1α \alpha α
In the follow-up prompt, I also uploaded the results of the mannose rescue experiment we had published. GPT-5 Pro confirmed previous suggestions and further refined the mechanistic model to explain these results together with the similar follow-up experiments suggested (output not shown).
Integrating both datasets and the naïve vs memory result
Your original titration showed a monotonic 2-DG dose–response and persistent increases in CCR6/CD161 well after washout; your new experiment shows that mannose fully normalizes or nearly normalizes these increases. Together with your subset sort, three strong inferences follow.
The 2-DG imprint is driven primarily by impaired N-linked glycosylation during priming. Mannose bypasses the 2-DG block in the dolichol-linked oligosaccharide pathway and rescues normal glycoprotein maturation. The near-complete rescue at 3mM mannose means glycolysis inhibition is not the principal cause of the Th17 skew.
The responding cells are memory, not naïve, CD4+ T cells. The absence of effect in naïve cells rules out a simple “2DG instructs naïve Th17 differentiation” model under your conditions. Memory cells likely contain preexisting CCR6+/CD161+ Th17-lineage clones; 2-DG preserves or expands them during priming by attenuating IL-2/STAT5 and/or by dampening TCR/CD28 glycoprotein function, both of which otherwise push cells away from Th17 toward Th1/CTL-like fates.
Programming plus selection best explains the magnitude and durability. Short term glycosylation stress at priming can both select for Th17-lineage memory cells (differential proliferation/survival) and program chromatin at Th17 loci. The persistence after two weeks without drug argues that at least part of the effect is epigenetically stabilized.
Mechanistic model that fits all observations. During the first 24–48h of activation, 3mM 2DG impairs N-glycosylation of multiple surface receptors (notably IL2Rα/CD25, TCR, CD28, potentially IL-6R and IL-23R). Hypoglycosylated receptors are retained or function less efficiently at the membrane, which reduces IL-2/STAT5 signaling and blunts strong co-stimulation. Because IL-2/STAT5 antagonizes Th17 and promotes Th1/Treg fates, this temporary reduction biases activated memory cells toward retaining or consolidating a Th17 program. Mannose restores N-glycan synthesis, normalizing receptor glycosylation and thereby reversing the imprint. Naïve cells, which depend heavily on IL-2 for survival and lack precommitted Th17 chromatin, do not show the same skew; any reduction in IL-2 signaling at priming in naïve cells tends to reduce expansion rather than drive Th17 features.
This model predicts that, in 2-DG-primed memory cultures: surface CD25 will show lower expression and altered glycoform size; pSTAT5 will be reduced; ATAC-seq will show enhanced accessibility at RORC, IL-23R, CCR6; and proliferation/survival curves will show relative sparing or enrichment of CCR6+/CD161+ cells.
This final section on impact, implications, and high-level interpretation offered particularly remarkable insight into these hypothesis and proposed experiments.
Because naïve cells did not respond, the 2DG effect does not instruct de novo Th17 programming from naïve precursors under these conditions. Instead, it selects and stabilizes Th17-lineage features in memory cells already carrying partial RORC/IL-23R/CCR6 programs. This distinction has practical consequences. If your aim is to manufacture Th17-leaning T cell products for therapy, enriching memory CD4⁺ (and potentially CD8⁺) compartments before a short 2-DG priming pulse will be far more effective than applying 2-DG to naïve enriched starting material. Conversely, if you are attempting to avoid Th17 skew in a culture process, maintaining normal glycosylation (adequate mannose/glucose; avoiding 2-DG exposure during early activation) is essential.
Impact and applications. For immunotherapy, a brief, controlled glycosylation perturbation at priming could be used to generate Th17/Tc17-leaning memory with improved persistence and trafficking, then returned to normal glycosylation by washout or mannose supplementation. For autoimmunity, the data warn that 2-DG –or any intervention that transiently depresses N-glycosylation during T cell activation—could exacerbate Th17 immunity, especially within existing memory pools. For vaccine design, manipulating glycosylation during priming may tune helper subset balance without permanent genetic changes.”
I then uploaded an unpublished flow cytometry figure showing the effect of 2-DG on the checkpoint inhibitors PD-1 and LAG-3 on activated CD8⁺ naïve and memory T cells (Figure I.3 shown below) with similar prompt to the original. GPT-5 Pro perfectly analyzed and described the data in the figure:
A brief, transient exposure to 3 mM 2-DG during anti-CD3/CD28 activation profoundly remodels early checkpoint receptor expression on human CD8⁺ T cells measured on day 4. In naïve CD8⁺ cells, 2-DG collapses the PD-1⁺ LAG-3⁺ population from ∼ 94% control to ∼ 0.8%, leaving most cells double-negative. In memory CD8⁺ cells, 2-DG reduces LAG-3 strongly (from 92 percent control to ∼ 57% LAG-3⁺) and decreases the double-positive subset to ∼ 55%, with total PD-1⁺ only modestly reduced (from ∼ 99% to ∼ 89%). In glucose-free media (glycolysis restriction without glycosylation block), naïve cells show only a partial reduction of LAG-3 (to ∼ 38%) and little change in PD-1; memory cells are largely unchanged. Thus, 2-DG’s effects are far stronger than glucose deprivation and are consistent with impaired N-linked glycosylation rather than energy shortage. The pattern dovetails with your earlier findings in CD4⁺ T cells: a brief, early glycosylation perturbation leaves a durable imprint that shifts fate and checkpoint expression.
In the follow-up, GPT-5 Pro provided another key mechanistic interpretation of why LAG-3 and PD-1 could be lower on CD8⁺ T cells due to glycosylation impairment, given that both are heavily N-glycosylated. It also provided a second and very insightful mechanistic explanation in which reduced signal strength could attenuate TCR signaling during 2-DG exposure during the activation phased of these cells and thus this modulation in signaling could have prevented upregulation of PD-1 and LAG-3, especially in naive T cells. Remarkably, we have direct and unpublished experiments showing this exact reduction in signal strength, which is remarkably much more pronounced in naive T cells compared to memory subset. Both PD-1 and LAG-3 are key checkpoint inhibitors that dampen responses against cancer cells. Indeed, the major cancer immunotherapy treatments target PD-1 or its ligand to enhance the responses of CD8+ ^{+} +
Figure I.3. Naive and memory CD8+ ^{+} +
Mechanistic interpretation. 2-DG is a glucose analog that both limits glycolysis and interferes with N-linked glycosylation, a post-translational modification required for correct folding, trafficking, and surface stability of many receptors. LAG-3 and PD-1 are heavily N-glycosylated. The drastic, mannose-rescuable effects you already demonstrated in CD4+ ^{+} +
Direct receptor effect. Hypoglycosylation reduces proper surface expression and stability of LAG-3 (and to a lesser extent PD-1, especially in naïve cells), rapidly lowering measured positivity.
Signal-strength effect. By hypoglycosylating TCR/CD28 and IL-2Ralpha (CD25) during priming, 2-DG attenuates early TCR/co-stimulatory and IL-2/STAT5 signals that otherwise drive robust PD-1/LAG-3 induction. This is more evident in naïve cells, which rely on strong primary signals; memory cells, with lower activation thresholds, still upregulate PD-1 but lose much of their LAG-3.
High-level conclusions and impact
A short, early 2-DG pulse is a powerful dial that reduces inhibitory checkpoint display (especially LAG-3) on CD8+ T cells, with a stronger effect in naïve than in memory cells. The biology is glycosylation-dependent rather than energy-deficit-dependent. Because PD-1/LAG-3 co-expression correlates with early dysfunction/exhaustion, transient 2-DG during priming may produce CD8+ products that are less prone to early exhaustion and better poised for sustained activity after drug washout.
Immunotherapy implications. For CAR-T or TCR-engineered products derived from memory CD8+ cells, a brief 2-DG exposure during the first 24–48 hour could generate cells with lower LAG-3 and slightly lower PD-1, potentially enhancing resistance to exhaustion under antigen pressure and improving serial killing over days. This matches your CD4+ memory dataset, where early 2-DG favored type-17-like, memory-leaning states that persist after washout. Conversely, for autoimmunity the data caution that interfering with glycosylation during priming could produce effector-competent, checkpoint-low CD8+ responses that exacerbate pathology.
Importantly, I asked GPT-5 Pro to simulate an experiment in which we would genetically engineer these CD8+ T cells with anti-CD19 CAR molecules to target CD19+ lymphomas, and to predict how prior 2-DG treatment would impact the cytotoxicity of these cells toward the cancer cells. Because we had already performed this experiment, also unpublished, I was able to validate the prediction from GPT-5 Pro, and it was nearly perfect in predicting that cytotoxicity would be enhanced. It also proposed mechanistic insights and follow-up experiments that we would need to perform before publishing these results. In this regard, GPT-5 Pro made sufficient contributions to this work to the extent that it would warrant its inclusion as a co-author in this new study.
What this predicts for anti-CD19 CAR memory CD8+ cells (simulation)
Memory CD8+ cells are activated, transduced with an anti-CD19 CAR, expanded for 2 weeks in IL-2, washed free of drug, then co-cultured with CD19+ lymphoma cells for 3 days. Outcomes are read as specific lysis and serial-killing capacity. To isolate functional differences, analyses gate on CAR+ cells, to control for any 2-DG effects on transduction efficiency. Mechanistic expectations applied to CAR-T context. The 2-DG pulse yields lower LAG-3 and modestly lower PD-1 on day 4, which typically translates to less activation-induced dysfunction over repeated target encounters. Although acute glycolysis supports immediate cytotoxic degranulation, your exposure is transient and occurs two weeks before the assay; metabolic capacity should recover, while the checkpoint/lineage program imprinted early is retained. Therefore, initial killing at 24 h may be similar or slightly lower, but cumulative 72 h killing should be higher with 2-DG priming, especially at low E:T ratios.
Key takeaways and implications from this case study of GPT-5 Pro
In summary, GPT-5 Pro provided remarkable key insights and hypothesis based on our experimental data. If we had had these interpretations and the recommended next experimental plan from GPT-5 Pro, we would have resolved or hypothesized the mechanistic insights within minutes upon data analysis. Importantly, we still missed key insights provided by GPT-5 Pro, despite our deep expertise of this field. In addition, GPT-5 Pro suggested several experiments that we will need to perform to publish these findings, that’s a co-investigator level contribution. More specific takeaways and implications are:
Precision interpretation of complex biology. GPT-5 Pro rapidly connected the observed phenotypes to a mechanistic hypothesis that distinguished glycolysis blockade from impaired N-linked glycosylation, proposed the IL-2 receptor pathway as the driver, and predicted that memory rather than naïve T cells carry the effect.
Testable next steps that held up in the lab. GPT-5 Pro produced a clean decision tree of experiments, including the mannose rescue to restore glycosylation, metabolic readouts, and epigenetic assays. This shows the model can generate highly relevant hypotheses that are testable in wet-labs.
Mechanism-first thinking that avoids false trails. By separating selection effects from programming effects, and by proposing controls that disentangle glycosylation from energy restriction, GPT-5 Pro reduced the risk of chasing attractive but potentially unnecessary experiments, which would have wasted many months of testing.
AI-guided bioengineering for cell therapies. By correctly predicting that a brief 2-DG exposure during priming would lower PD-1/LAG-3, preserve cytotoxic potential, and enhance serial killing in anti-CD19 CAR memory CD8+ T cells, GPT-5 Pro illustrates how foundation models can propose concrete, testable tweaks to CAR-T cell development protocols that enhance therapeutic performance against cancer or autoimmune diseases by rapidly iterating a variety of conditions in silico, before wet-lab validation.
Long-term impact on biomedicine. As models like GPT-5 Pro become native to lab operating systems, we should expect: faster mechanism discovery across immunology, oncology, and metabolism; cheaper negative results because failed branches are pruned in silico; more reproducible science due to selecting better hypothesis and well designed experimental approaches. The net effect will be a much higher discovery rate per experiment and a shorter route from observation to discovery to intervention, thus profoundly accelerating the biomedical scientific process.
Chapter II
Deep literature search
From density estimation and convex geometry to multi-objective optimization – Nikita Zhivotovskiy
The experiment discussed here originally dates to late August 2025, before a number of subsequent applications circulated more broadly. The goal was to apply GPT-5 not to obtain a direct literature answer, but to see whether a newly proved geometric statement would immediately surface adjacent literature and applications. This was useful to the authors because plausible extensions were unclear and would otherwise require asking a number of experts (and luck) to find the right connections.
Our preliminary conclusion from this experiment is that, given only a core mathematical statement, GPT-5 can rapidly surface nontrivial and technically aligned links across areas (here, multiobjective optimization and approximate Pareto sets), providing context for new applications.
First, we record the statement that motivated the experiment. Let K K K R + d \mathbb{R}_{+}^{d} R + d α ≥ 1 \alpha\ge1 α ≥ 1 A ⊂ K A \subset K A ⊂ K α \alpha α K K K θ = ( θ 1 , … , θ d ) ∈ K \theta= (\theta_1,\ldots,\theta_d) \in K θ = ( θ 1 , … , θ d ) ∈ K ϕ = ( ϕ 1 , … , ϕ d ) ∈ A \phi= (\phi_1,\ldots,\phi_d) \in A ϕ = ( ϕ 1 , … , ϕ d ) ∈ A θ j ≤ α ϕ j \theta_j \le\alpha\phi_j θ j ≤ α ϕ j j = 1 , … , d j = 1,\ldots,d j = 1 , … , d
Theorem II.1.1 (Compton, Lugosi, Mourtada, Qian, Zhivotovskiy, 2025). For every d ≥ 1 d \ge1 d ≥ 1 K ⊂ R + d K \subset\mathbb{R}_{+}^{d} K ⊂ R + d A ⊂ K A \subset K A ⊂ K 2 8 d 2^{8d} 2 8 d 32 32 32 K K K [23 ]
Note that GPT-5 was not involved in the proof of the result above. The original motivation for Theorem II.1.1 is the statistical problem of density estimation for mixtures. Let { f 1 , … , f d } \{f_1,\ldots,f_d\} { f 1 , … , f d } R p \mathbb{R}^{p} R p θ ∈ K ⊂ Δ d − 1 \theta\in K \subset\Delta^{d-1} θ ∈ K ⊂ Δ d − 1 Δ d − 1 = { θ ∈ R + d : ∑ i = 1 d θ i = 1 } \Delta^{d-1} = \{\theta\in\mathbb{R}_{+}^{d} : \sum_{i=1}^{d}\theta_i = 1\} Δ d − 1 = { θ ∈ R + d : ∑ i = 1 d θ i = 1 } p θ ( x ) = ∑ i = 1 d θ i f i ( x ) p_{\theta}(x) = \sum_{i=1}^{d}\theta_i f_i(x) p θ ( x ) = ∑ i = 1 d θ i f i ( x ) A A A α \alpha α K K K θ ∈ K \theta\in K θ ∈ K ϕ ∈ A \phi\in A ϕ ∈ A θ ≤ α ϕ \theta\le\alpha\phi θ ≤ α ϕ p θ ( x ) ≤ α p ϕ ( x ) p_{\theta}(x) \le\alpha p_{\phi}(x) p θ ( x ) ≤ α p ϕ ( x ) x x x p θ ( x ) p ϕ ( x ) ≤ α \frac{p_{\theta}(x)}{p_{\phi}(x)} \le\alpha p ϕ ( x ) p θ ( x ) ≤ α KL ( p θ ∥ p ϕ ) ≤ log α \operatorname{KL}(p_{\theta}\|p_{\phi}) \le\log\alpha KL ( p θ ∥ p ϕ ) ≤ log α { p θ : θ ∈ K } \{p_{\theta} : \theta\in K\} { p θ : θ ∈ K }
When obtaining a result of the form of Theorem II.1.1, one naturally asks whether it is known and where it applies. What GPT-5 was helpful to do, once the authors had Theorem II.1.1, was to surface related results that broaden the scope into theoretical computer science. To keep the experiment simple and clean:
the work [23 ] was not public at that time, so its statements and possible connections were not available to the model;
for this note, the queries were rerun in a temporary, new session to avoid contamination from prior searches/interactions.
As can be seen in Figure II.1 , among a series of other relevant results, the August 2025 exchange prominently highlighted the seminal work of [51 ] on multiobjective optimization. At first sight (see Figure II.2 ), this seems unrelated to Theorem II.1.1 and could be mistaken for a hallucination. However, unpacking their proof shows that their result can be phrased as a coordinatewise ( 1 + ε ) (1+\varepsilon) ( 1 + ε ) K ⊆ [ 1 / R , R ] d K \subseteq[1/R,R]^d K ⊆ [ 1/ R , R ] d O ( ( log R ε ) d − 1 ) O\left(\left(\frac{\log R}{\varepsilon}\right)^{d-1}\right) O ( ( ε l o g R ) d − 1 ) K K K 2 8 d 2^{8d} 2 8 d R R R K K K
Figure II.1. An excerpt from GPT-5’s reply. The full response can be found in [68 ] .
Figure II.2. A result of [51 ] closely related to Theorem II.1.1.
A substantial part of [51 ] is devoted to constructing near-optimal ε \varepsilon ε ( 1 + ε ) (1+\varepsilon) ( 1 + ε ) O ( ( log ( 1 / ε ) ε ) d − 1 ) O\left(\left(\frac{\log(1/\varepsilon)}{\varepsilon}\right)^{d-1}\right) O ( ( ε l o g ( 1/ ε ) ) d − 1 ) K K K R R R log R \log R log R O ( ( log R ε ) d − 1 ) O\left(\left(\frac{\log R}{\varepsilon}\right)^{d-1}\right) O ( ( ε l o g R ) d − 1 )
Erdős problems (part 1/2) – Mehtaab Sawhney and Mark Sellke
Introduction
Paul Erdős was a prolific mathematician whose mathematical impact ranges across diverse areas of mathematics including combinatorics, number theory and analysis. Along with publishing more than 1500 mathematical papers, Erdős posed a substantial number of mathematical conjectures with several of these conjectures becoming central problems in mathematics.
Recently Thomas Bloom [14 ] created an online index of Erdős problems, which has greatly facilitated the process of identifying and solving the problems and opened the door to numerous online collaborations. The role of large language models in our story already begins here with the creation of the website in early 2024, which relied on a previous version of ChatGPT (https://www.erdosproblems.com/faq ) to help write the code underlying the website. Each listed problem has its own webpage, with a status (roughly speaking, either “open” or “solved”) as well as a discussion thread for potential collaborations. The list is continually growing, with 685 “open” problems out of 1105 total as of October 31, 2025; the problems posed by Erdős over the years are scattered across many disorganized sources. Unfortunately the solutions are also often similarly scattered, and so it can be very challenging to determine for sure whether a particular problem is actually open, or if it was instead solved decades ago and forgotten to most mathematicians. This has resulted in a non-trivial amount of futile effort by mathematicians trying to solve problems that turned out to already be known. For instance as a graduate student, the first author spent a bit of time thinking about [Bloom, Problem #490] and even mentioned it an informal meeting with other graduate students as a beautiful problem. Eventually the first author was able to locate a reference by [65 ] solving the problem, with the vague title “On a problem of P. Erdős”. Thus, in practice a problem being listed as “open” roughly indicates that at least 1 professional mathematician attempted and failed to find a previously published solution by searching the internet. We discuss our successful attempts to improve the status of this database using GPT-5, which led to the following progress:
Locating previously published solutions to 10 problems not previously marked as known: 223, 339, 494, 515, 621, 822, 883 (part 2/2), 903, 1043, 1079.
Reporting noteworthy partial progress in the existing literature for 10 other problems: 32, 167, 188, 750, 788, 811, 827, 829, 1017, 1011.
Correcting a misprint in the statement of problem 1041.
Generating a new idea for Problem #848, which, together with previous suggestions by online commenters van Doorn, Weisenberg, and Cambie, enabled us to settle the problem.
In this section we focus on several examples from the first two bullets above. Problem #848 and its solution are presented later in Section IV.1.
Literature search
The database of Erdős problems serves as a testbed for the abilities of GPT-5. For every problem listed as “open”, we asked GPT-5 to locate a solution if it existed, and otherwise to report significant partial progress that merits being mentioned on the website. This was done using a combination of the ChatGPT interface, the OpenAI API, and internal OpenAI tools; the latter two allow automated queries of many problems in parallel, and automated filtering of responses also using GPT-5; aside from the automated aspect, we expect similar results for any particular problem could have been obtained purely using the ChatGPT interface. The prompts used were completely problem agnostic, consisting of a link to or a screenshot of the problem, and an instruction to search the internet for a solution or partial progress, despite the “open” designation on the website.1
The references provided were generally accurate and easy to verify manually. We did not observe any cases in which GPT-5 pretended to have found a correct reference (though in some cases it was overly enthusiastic about partial progress it had found). We also emphasize that the problem numbers used correspond to their labeling on the database, beginning in early 2024. In particular these numbers cannot be used in websearches to locate pre-2024 literature on a given problem.
Typical practice for a mathematician attempting to find literature is either following long chains of references or attempting to “guess the correct search phrase” hoping to find relevant literature. With GPT-5, one can skip this artificial middle step and simply plug in the statement! Furthermore in mathematical practice, one is often tasked with locating literature in a separate field “which surely exists”; GPT-5 can do so provided often vague descriptions or candidate “theorem” statements.
We highlight a few representative cases. The first is problem [14 ] , the surprising success of which was the impetus for our efforts.
Problem #339 ([28 ] ): Let A ⊆ N A \subseteq\mathbb{N} A ⊆ N r r r r r r A A A r r r A A A (I.e., must this set occupy a non-vanishing fraction of the integers { 1 , 2 , … , N } \{1,2,\ldots,N\} { 1 , 2 , … , N } N N N N N N
A difficulty in searching for references is that Problem #339 is raised in a 100-page paper of Erdős–Graham [28 ] which lists many other problem and has roughly 700 citations. Sorting through to find which of these are relevant for the specific problem above would be extremely time-consuming for a human. However, GPT-5 Pro was able to find the solution in the first query we tried, given only a screenshot of the problem webpage. The discussion in which GPT-5 Pro found the solution is here: [59 ] . We note that before the solution was found, the problem attracted some discussion among participants on the online forum https://www.erdosproblems.com/forum/thread/339 .
Further Examples
Here we present three more examples, which further illustrate the challenges present in mathematical literature search. We reiterate that the numbers below correspond only to the recently created online database of Erdős problems, and thus are not useful for searching the pre-2024 literature.
Problem #515 ([29 ] ): Let f ( z ) f(z) f ( z ) γ \gamma γ λ > 0 \lambda> 0 λ > 0 ∫ γ ∣ f ( z ) ∣ − λ d z \int_{\gamma} |f(z)|^{-\lambda} \, dz ∫ γ ∣ f ( z ) ∣ − λ d z
Partial result toward Problem #750 ([26 ] ): Fix ε > 0 \varepsilon> 0 ε > 0 G G G m m m ( 1 2 − ε ) m \left(\frac{1}{2}-\varepsilon\right)m ( 2 1 − ε ) m
Problem #1043 ([27 ] ): Let f ∈ C [ x ] f \in\mathbb{C}[x] f ∈ C [ x ] ℓ \ell ℓ { z : ∣ f ( z ) ∣ ≤ 1 } \{z : |f(z)| \le1\} { z : ∣ f ( z ) ∣ ≤ 1 } ℓ \ell ℓ
GPT-5 located references solving the above problems, respectively in [45 ] , [31 ] , [53 ] . Identifying these decades-old papers required GPT-5 to go far beyond the functionality of a search engine, and indeed to read each of these papers in detail and apply a genuine understanding of mathematics.
In the first case, the work [45 ] actually proves something more general than the statement of Problem #515. Namely, the authors study integrals of e − F ( z ) e^{-F(z)} e − F ( z ) F : C → R F : \mathbb{C} \to\mathbb{R} F : C → R subharmonic function, meaning its average value on any circle is smaller than its value at the center. It is a classical fact that F ( z ) = log ∣ f ( z ) ∣ F(z) = \log|f(z)| F ( z ) = log ∣ f ( z ) ∣ f f f [45 ] do imply Problem #515. However this connection is not the focus of the paper, and indeed neither Erdős nor his above problem are mentioned. This use of different vocabulary for similar underlying concepts is a common challenge in searching the mathematical literature, and would have made [45 ] quite hard to find using a standard websearch. Indeed the main feature of Problem #515 is the integral expression, which would be quite challenging to search for directly. GPT-5 in fact later located a corroborating reference, a (256 page!) survey [39 ] that mentions the connection between [45 ] and Problem #515 on page 27.
Additionally, there is a subtle technical point in connection between [45 ] and the problem it solves, which is that the function log ∣ f ( z ) ∣ \log|f(z)| log ∣ f ( z ) ∣ f ( z ) = 0 f(z) = 0 f ( z ) = 0 [45 ] to Problem 515, it is crucial that their definition of a “subharmonic” function allows such singularities. The paper requires a careful reading to confirm that their definition of subharmonic is sufficiently general for the required implication; here as well, GPT-5 assisted us in pointing out parts of the paper that made the intended definition clear.
For Problem #750, the result from [31 ] actually concerns the slightly different problem of finding a bipartite subgraph, and shows such subgraphs exist with ( 1 − ε ) m (1-\varepsilon)m ( 1 − ε ) m
Problem #1043 is stated in [27 ] together with 15 other problems; several of these including 1043 were solved in [53 ] . Unlike the others, Problem #1043 is addressed in a brief side comment between two theorems, and was consequently overlooked by many including the MathSciNet review of [53 ] . Additionally it turns out that [52 ] (displayed as reference [52 ] in the screenshot of [53 ] ) actually contains the main work in solving Problem #1043, and is written in German. GPT-5 translated and explained the proofs from [52 ] to us so that we could verify them ourselves.
Figure II.3. Pommerenke’s solution to Problem #1043, appearing on page 7 of [53 ] . The proof follows by combining the “approximation theorem” from [53 ] and Pommerenke’s previous work [52 ] (cited as [53 ] , p. 73 in the red block above). See https://www.erdosproblems.com/forum/thread/1043 for a more detailed outline.
Looking forward
Motivated in part by our efforts, other researchers including Thomas Bloom and Terence Tao soon started posting negative literature searches to the Erdős problem website, reporting that GPT-5 and other large language models were unable to locate an existing solution to various problems. This provides a convenient, crowd-sourceable “soft certificate” that the solution is unlikely to appear in the published literature. See e.g. https://www.erdosproblems.com/forum/thread/689 , https://www.erdosproblems.com/forum/thread/915 , https://www.erdosproblems.com/forum/thread/990 .
The landscape of mathematical results is substantially broader than any one mathematician plausibly has available at their fingertips. GPT-5 therefore provides the practicing mathematician a new mechanism to access the collective breadth of the mathematical literature.
Clique-avoiding codes: a cautionary tale
The following was communicated by Venkatesan Guruswami and Parikshit Gopalan, edited by us and included with their permission
A binary linear code is a subspace of F 2 N ≃ { 0 , 1 } N \mathbb{F}_2^N \simeq\{0,1\}^N F 2 N ≃ { 0 , 1 } N n n n [36 ] , motivated by the study of maximally recoverable codes, although it was never published.
Let us define the problem formally. Let [ n ] = { 1 , … , n } [n]=\{1,\ldots,n\} [ n ] = { 1 , … , n } ( n 2 ) \binom{n}{2} ( 2 n ) F 2 ( n 2 ) \mathbb{F}_2^{\binom{n}{2}} F 2 ( 2 n ) { i , j } \{i,j\} { i , j } 1 ≤ i < j ≤ n 1\le i<j\le n 1 ≤ i < j ≤ n S ⊆ [ n ] S\subseteq[n] S ⊆ [ n ] χ S ∈ F 2 ( n 2 ) \chi_S\in\mathbb{F}_2^{\binom{n}{2}} χ S ∈ F 2 ( 2 n ) S S S χ S ( { i , j } ) = 1 [ i ∈ S and j ∈ S ] \chi_S(\{i,j\})=\mathbf{1}[i\in S\text{ and }j\in S] χ S ({ i , j }) = 1 [ i ∈ S and j ∈ S ] C ⊆ F 2 ( n 2 ) C\subseteq\mathbb{F}_2^{\binom{n}{2}} C ⊆ F 2 ( 2 n ) clique-avoiding if χ S ∉ C \chi_S\notin C χ S ∈ / C S ⊆ [ n ] S\subseteq[n] S ⊆ [ n ] ∣ S ∣ ≥ 2 |S|\ge2 ∣ S ∣ ≥ 2
The question posed to GPT-5 was the following:
What is the minimum number of parity checks (i.e., the minimum co-dimension) denoted r = r ( n ) r=r(n) r = r ( n ) n n n
A simple probabilistic (random coding) argument, which GPT-5 also found right away (with some additive O ( 1 ) O(1) O ( 1 )
Lemma II.3.1. We have r ( n ) ≤ n r(n)\le n r ( n ) ≤ n
It is tempting to think that the random coding bound is tight, but there is no obvious lower bound. This was the question we posed to GPT-5 (without really thinking about it ourselves). Initially, it was convinced that this bound was tight (up to an additive constant), and tried to convince us of this using a sequence of buggy arguments, resorting to linear algebra and proof by authority:
Another incorrect argument nevertheless contained the germ of the main idea. It was to consider quadratic forms Q 1 , … , Q r Q_1,\ldots,Q_r Q 1 , … , Q r syndrome S ( x ) = ( Q 1 ( x ) , … , Q r ( x ) ) S(x)=(Q_1(x),\ldots,Q_r(x)) S ( x ) = ( Q 1 ( x ) , … , Q r ( x )) x ↦ S ( x ) x\mapsto S(x) x ↦ S ( x ) r ≥ n r\ge n r ≥ n
When challenged about this, GPT-5 came up with an elegant (and correct) algebraic argument based on the classical Chevalley-Warning theorem, which established Theorem II.3.1 below. (On other attempts, it would come up with the correct formulation using quadratic polynomials modulo 2 2 2 r ≥ ⌈ ( n − log ( n ) ) / 2 ⌉ r\ge\lceil(n-\log(n))/2\rceil r ≥ ⌈( n − log ( n )) /2 ⌉
Theorem II.3.1. For all n ≥ 2 n \ge2 n ≥ 2 r ( n ) ≥ ⌊ n / 2 ⌋ r(n) \ge\lfloor n/2 \rfloor r ( n ) ≥ ⌊ n /2 ⌋
We later discovered a construction (without using GPT-5) showing that the lower bound is exactly tight : the simple probabilistic bound above is suboptimal, and in fact r ( n ) = ⌊ n / 2 ⌋ r(n) = \lfloor n/2 \rfloor r ( n ) = ⌊ n /2 ⌋ n n n
An unexpected twist
We were very excited to have an exact solution to a question which we had been curious about for years, especially because one direction of the proof was due to an AI. However, we soon learned that the same tight bound (as well as a different proof of the matching upper bound) had appeared on Arxiv nearly 3 years previously in the short paper [3 ] . In fact, GPT-5’s proof of the lower bound was exactly the same as Alon’s! Thus, it appears that GPT-5 reproduced Alon’s proof and passed it along to us without realizing its source. As it turns out, a fresh attempt at querying GPT-5 Pro was able to recover this source, as depicted in Figure II.4 .
Figure II.4. An interaction in which GPT-5 Pro was able to locate the reference [3 ] solving our problem. This query was made after we first became aware of [3 ] .
Our experience illustrates a pitfall in using AI: although GPT-5 possesses enormous internal knowledge and the capability to locate even more using the internet, it may not always report the original information sources accurately. This has the potential to deceive even seasoned researchers into thinking their findings are novel. We expect that our experience is not unique, and urge others to take special care in attribution when working with LLM-assisted proofs.
Do you know anything about the following question? This concerns the minimum co-dimension of a binary linear code C ⊂ F 2 ( n 2 ) C \subset\mathbb{F}_2^{\binom{n}{2}} C ⊂ F 2 ( 2 n ) n n n
Answer. Yes. The minimum possible co-dimension of a binary linear code C ⊂ F 2 ( n 2 ) C \subset\mathbb{F}_2^{\binom{n}{2}} C ⊂ F 2 ( 2 n )
⌊ n / 2 ⌋ . \left\lfloor n/2 \right\rfloor. ⌊ n /2 ⌋ . Explanation (sketch). A linear code C C C r r r r r r G 1 , … , G r G_1,\ldots,G_r G 1 , … , G r [ n ] [n] [ n ] x ∈ F 2 ( n 2 ) x \in\mathbb{F}_2^{\binom{n}{2}} x ∈ F 2 ( 2 n ) C C C
⟨ x , 1 E ( G i ) ⟩ = 0 for all i . \langle x,1_{E(G_i)}\rangle=0 \quad\text{for all } i. ⟨ x , 1 E ( G i ) ⟩ = 0 for all i . Avoiding all clique indicators means that for every nonempty vertex set T ⊆ [ n ] T \subseteq[n] T ⊆ [ n ] K T K_T K T
∣ E ( K T ) ∩ E ( G i ) ∣ = 1 ( m o d 2 ) for at least one i . \lvert E(K_T) \cap E(G_i)\rvert= 1 \pmod2 \quad\text{for at least one } i. ∣ E ( K T ) ∩ E ( G i )∣ = 1 ( mod 2 ) for at least one i . Upper bound: codim ( C ) ≤ ⌊ n / 2 ⌋ \operatorname{codim}(C) \le\left\lfloor n/2 \right\rfloor codim ( C ) ≤ ⌊ n /2 ⌋ n n n [ n − 1 ] [n-1] [ n − 1 ] B i = { 2 i − 1 , 2 i } B_i=\{2i-1,2i\} B i = { 2 i − 1 , 2 i } ( n − 1 ) / 2 (n-1)/2 ( n − 1 ) /2 G i G_i G i B i B_i B i j > 2 i j>2i j > 2 i K T K_T K T G i G_i G i C C C ( n − 1 ) / 2 (n-1)/2 ( n − 1 ) /2 n n n
codim ( C ) ≤ ⌊ n / 2 ⌋ . \operatorname{codim}(C) \le\left\lfloor n/2 \right\rfloor. codim ( C ) ≤ ⌊ n /2 ⌋ . Lower bound: codim ( C ) ≥ ⌊ n / 2 ⌋ \operatorname{codim}(C) \ge\left\lfloor n/2 \right\rfloor codim ( C ) ≥ ⌊ n /2 ⌋ r < n / 2 r<n/2 r < n /2 T ⊆ [ n ] T \subseteq[n] T ⊆ [ n ] ∣ E ( K T ) ∩ E ( G i ) ∣ \lvert E(K_T)\cap E(G_i)\rvert ∣ E ( K T ) ∩ E ( G i )∣ i i i K T K_T K T C C C r ≥ n / 2 r\ge n/2 r ≥ n /2 n n n
codim ( C ) ≥ ⌊ n / 2 ⌋ . \operatorname{codim}(C) \ge\left\lfloor n/2 \right\rfloor. codim ( C ) ≥ ⌊ n /2 ⌋ . Conclusion. Both bounds match, so the exact minimum co-dimension is
⌊ n / 2 ⌋ . \left\lfloor n/2 \right\rfloor. ⌊ n /2 ⌋ . For a full, elegant presentation, see N. Alon, Graph-Codes , Theorem 1.6.
Chapter III
Working in tandem with AI
Recent experiences of using LLMs as research partners – Timothy Gowers
This is a brief report on one or two recent experiences of using LLMs while actually doing mathematics, by which I mean thinking about research-level problems at an earlyish stage where I have some ideas but have not yet found an approach that feels likely to work. I can summarize the situation so far by saying that I have not (yet) had the experience of an LLM making a decisive contribution, but I have found them useful in the following ways.
If I suspect that a definition or mathematical observation I have come up with is in the literature, an LLM can often help me find it. GPT-5 seems to be significantly better at this than GPT-4. In particular, with GPT-5 my experience has been that the references are rarely hallucinated, and even the hallucinations can turn out to be pointers to references that exist and are useful.
I have sometimes formulated well-defined subproblems that look as though they shouldn’t be too hard but also look as though they would take a bit of time to solve, and GPT-5 has solved them for me in a matter of seconds.
I have had reasonably precise ideas for solving problems that I have run past GPT-5 Pro, which has explained to me why they cannot work.
On the negative side, if I ask more open-ended questions, or offer more sketchy ideas for proof attempts, then that seems to encourage the more annoying characteristics of LLMs to come to the fore: they will tell me that my ideas do indeed work, and will write something that supposedly fleshes out the details but that does not withstand close scrutiny. This applies even to GPT-5 Pro, to which I have had access for the last few days (at the time of writing).
Taking this together, my current assessment of LLMs is that they are just beginning to be useful as research collaborators: one can bounce ideas off them in the way that one can with a human collaborator and one gets a response extremely quickly. As with human collaborators, even the less good ideas of an LLM can sometimes stimulate me to make progress. (I think of this as the “That clearly doesn’t work ... but wait a minute!" phenomenon.) So we have reached the stage where LLMs can speed up the process of thinking about a problem, especially if that problem is a little outside one’s primary domain of expertise, but we have not yet reached the stage where an LLM is likely to have the main idea for solving a difficult problem.
(That last statement has to be interpreted carefully. I may find a problem difficult because I am unfamiliar with the domain-specific tricks needed to solve it, but if the problem would be found easy by somebody with the relevant expertise, then it may well be easy for an LLM as well. Thus, by “difficult” I mean a problem for which the standard bag of tricks in the relevant area does not suffice.)
I shall give two examples here of recent interactions with GPT-5 and GPT-5 Pro. I stress that in each case, the main contribution of the LLM was to save me a small amount of time: the observations it made and proofs it provided are ones that I would have expected to find for myself before all that long, but with the help of an LLM I did not to have to make any effort at all and I got the answers within seconds.
It is interesting to think about whether I should have made more effort before turning to an LLM for help. Everybody will have to make their own decision about this. My current feeling is that this issue is a bit like the issue of whether one should use calculators to do arithmetic. One is clearly a much better user of calculators if one knows how to do the calculations oneself and uses the calculators only as a time-saving device. Similarly, in these early days, I feel much more comfortable asking an LLM to help me if I think it is carrying out a task for which I have a good idea of how I would set about it and that I could almost certainly do without help, given a bit of time. I would be less comfortable simply feeding it a difficult problem and hoping for the answer (and also, for now, much less confident that I would get a useful answer if I did so).
Example 1: a compact subset of L 2 L_2 L 2
I was trying to solve a problem where I had a somewhat complicated non-linear map T : L 2 → L 2 T : L_2 \to L_2 T : L 2 → L 2 f , T f , T 2 f , … f, Tf, T^2f, \ldots f , T f , T 2 f , … f f f T T T K ⊂ L 2 K \subset L_2 K ⊂ L 2
I knew that in an arbitrary Banach space, one source of compact sets is convex hulls of null sequences, and after a little thought I realized that I could prove that in ℓ 2 \ell_2 ℓ 2 s s s { a ∈ ℓ 2 : ∣ a ∣ ≤ s } \{a \in\ell_2 : |a| \leq s\} { a ∈ ℓ 2 : ∣ a ∣ ≤ s } f f f
I soon realized that that could not be the case, because one can take a sequence of more and more rapidly oscillating functions, and those will have no convergent subsequence.
However, I had reason to believe that the iterates of f f f T T T T T T f f f f f f f ^ \hat{f} f ^
This felt plausible to me, so I asked GPT-5 about it. After 22 seconds, it provided me with a proof, which relied on a lemma that I had not heard of called the Kolmogorov–Riesz lemma. The proof looked correct, given the lemma, and the lemma looked plausible. I checked online and found that it did indeed exist. I don’t know how long it would have taken me to realize that the lemma was the right intermediate step (after which I could have used GPT-5 for its semantic search capabilities) but I estimate that this saved me anything between an hour and three hours of work.
There is an epilogue to this story. I am collaborating on the project with two research students, who were also good at pointing out drawbacks with my proposals for proof approaches. In particular, one of them pointed out that the condition that f f f f ^ \hat{f} f ^ f f f f f f f ^ \hat{f} f ^ e − c x 2 e^{-cx^{2}} e − c x 2 c = 1 / 2 c = 1/2 c = 1/2
I was fooled by this, because each Hermite function is a polynomial multiplied by e − x 2 / 2 e^{-x^{2}/2} e − x 2 /2 e − c x 2 e^{-cx^{2}} e − c x 2 c < 1 / 2 c < 1/2 c < 1/2 C e − c x 2 Ce^{-cx^{2}} C e − c x 2 C C C c c c e − c x 2 e^{-cx^{2}} e − c x 2 c < 1 / 2 c < 1/2 c < 1/2 C e − c x 2 Ce^{-cx^{2}} C e − c x 2 c > 0 c > 0 c > 0
x h n = n + 1 2 h n + 1 + n 2 h n − 1 . xh_n = \sqrt{\frac{n+1}{2}}h_{n+1} + \sqrt{\frac{n}{2}}h_{n-1}. x h n = 2 n + 1 h n + 1 + 2 n h n − 1 . Since the Hermite functions are orthonormal it follows that
∫ − ∞ ∞ x 2 h n ( x ) 2 d x = n + 1 2 + n 2 = n + 1 2 . \int_{-\infty}^{\infty} x^{2}h_n(x)^{2}\,\mathrm{d}x = \frac{n+1}{2} + \frac{n}{2} = n + \frac{1}{2}. ∫ − ∞ ∞ x 2 h n ( x ) 2 d x = 2 n + 1 + 2 n = n + 2 1 . It follows that for sufficiently large n n n C 2 ∫ x 2 e − c x 2 d x C^{2}\int x^{2}e^{-cx^{2}}\,\mathrm{d}x C 2 ∫ x 2 e − c x 2 d x n n n n n n ∣ h n ∣ |h_n| ∣ h n ∣ C e − c x 2 Ce^{-cx^{2}} C e − c x 2
I then asked GPT-5 Pro whether if 0 < c < 1 / 2 0 < c < 1/2 0 < c < 1/2 f f f ∣ f ∣ |f| ∣ f ∣ ∣ f ^ ∣ |\hat{f}| ∣ f ^ ∣ C e − c x 2 Ce^{-cx^{2}} C e − c x 2 f f f f f f
The final twist in the story is another remark by my human collaborators. I had been slightly careless in assuming that the set of all f f f ∣ f ∣ |f| ∣ f ∣ ∣ f ^ ∣ |\hat{f}| ∣ f ^ ∣ C e − c x 2 Ce^{-cx^2} C e − c x 2 T T T
Example 2: graphs with uncharacteristically small clique number
A very nice problem I recently heard about asks whether there is an algorithm with the following properties.
It takes as its input a graph with n n n
If the graph contains no clique of size ( 3 / 2 ) log 2 n (3/2)\log_2 n ( 3/2 ) log 2 n
For almost all graphs (meaning a 1 − o ( 1 ) 1-o(1) 1 − o ( 1 )
It runs in polynomial time.
The reason one might hope for such an algorithm to exist is that random graphs have clique number approximately 2 log 2 n 2\log_2 n 2 log 2 n ( 3 / 2 ) log 2 n (3/2)\log_2 n ( 3/2 ) log 2 n ( 3 / 2 ) log 2 n (3/2)\log_2 n ( 3/2 ) log 2 n n / log n \sqrt{n}/\log n n / log n
I decided to try to think of ways of creating a class of graphs that did not have cliques of size ( 3 / 2 ) log 2 n (3/2)\log_2 n ( 3/2 ) log 2 n
First idea: use linear algebra
A well known result in combinatorics, and a surprising one when you first see it, is that the number of subsets of { 1 , 2 , … , n } \{1,2,\ldots,n\} { 1 , 2 , … , n } n n n F 2 n \mathbb{F}_2^n F 2 n F 2 \mathbb{F}_2 F 2 n n n
If we form a graph where the vertices are all sets of odd size and two vertices are joined by an edge if their intersection has even size, then the above result tells us that there can be no clique of size greater than n n n N = 2 n − 1 N=2^{n-1} N = 2 n − 1 ( 3 / 2 ) log 2 N (3/2)\log_2 N ( 3/2 ) log 2 N N N N
So far, it is easily distinguishable from a random graph: for example, it is much more regular than a random graph will typically be. I proposed to deal with that problem by passing to a random subset of the vertices of size 2 α n 2^{\alpha n} 2 α n α \alpha α
GPT-5 Pro pointed out to me that the rank over F 2 \mathbb{F}_2 F 2 n n n 2 α n 2^{\alpha n} 2 α n 2 α n 2^{\alpha n} 2 α n
Second idea: use angles between unit vectors
My first idea had failed because it was in some sense too algebraic, but it occurred to me that I knew another result about vectors that implies the existence of a graph without large cliques. It states that the largest number of vectors you can find in R d \mathbb{R}^d R d d + 1 d+1 d + 1 d d d d + 1 < ( 3 / 2 ) log 2 n d+1 < (3/2)\log_2 n d + 1 < ( 3/2 ) log 2 n n n n u 1 , … , u n u_1,\ldots,u_n u 1 , … , u n R d \mathbb{R}^d R d ⟨ u i , u j ⟩ \langle u_i,u_j\rangle ⟨ u i , u j ⟩ d d d
However, GPT-5 Pro almost immediately gave me a test for detecting matrices created this way, based on the fact that the larger ⟨ u i , u j ⟩ \langle u_i,u_j\rangle ⟨ u i , u j ⟩ u i u_i u i u j u_j u j θ \theta θ 1 / 2 − θ / 2 π 1/2-\theta/2\pi 1/2 − θ /2 π
I then noticed that a test of this form also applies to the F 2 n \mathbb{F}_2^n F 2 n x + y + z + w = 0 x+y+z+w=0 x + y + z + w = 0 v v v x x x y y y z z z w w w 2 α n 2^{\alpha n} 2 α n α \alpha α x , y , z , w x,y,z,w x , y , z , w
Third idea: replace linear algebra by polynomials
I wondered rather vaguely whether one could rescue the more algebraic approach by devising some polynomial condition for two elements x , y x,y x , y F 2 n \mathbb{F}_2^n F 2 n P : F 2 n × F 2 n → F 2 P:\mathbb{F}_2^n \times\mathbb{F}_2^n \to\mathbb{F}_2 P : F 2 n × F 2 n → F 2 x x x y y y x x x y y y P ( x , y ) = 1 P(x,y)=1 P ( x , y ) = 1
This time GPT-5 Pro gave me a longish answer explaining that unless the degree of P P P P P P d d d x x x P ( x , y ) P(x,y) P ( x , y ) y y y F 2 n \mathbb{F}_2^n F 2 n x x x P ( x , y ) P(x,y) P ( x , y ) y y y d d d x i 1 ⋯ x i s x_{i_1}\cdots x_{i_s} x i 1 ⋯ x i s s ≤ d s\le d s ≤ d F 2 n \mathbb{F}_2^n F 2 n F 2 \mathbb{F}_2 F 2 ∑ s = 0 d ( n s ) \sum_{s=0}^{d}\binom{n}{s} ∑ s = 0 d ( s n ) 2 α n 2^{\alpha n} 2 α n 2 α n 2^{\alpha n} 2 α n d d d n n n
But GPT-5 Pro also told me that if the degree of P P P ( 3 / 2 ) log 2 N (3/2)\log_2 N ( 3/2 ) log 2 N N N N
Attempting to go in the other direction
Given that I had a uniform explanation for the failure of my first two attempted examples, I then decided to have a conversation about whether a graph that passes certain strong tests for randomness based on the sizes of the intersections of neighbourhoods could be shown to contain a clique of size ( 3 / 2 ) log 2 ( n ) (3/2)\log_2(n) ( 3/2 ) log 2 ( n ) ( 3 / 2 ) log 2 n (3/2)\log_2 n ( 3/2 ) log 2 n
Conclusion
As a research supervisor I have a rule of thumb for when a contribution I make to the research of one of my PhD students is at the level where I should be a joint author. The rule is that if the student comes to discuss the problem with me and I have, in the course of that discussion, an idea that comes more naturally to me than to them and that turns out to be helpful, then that is not enough for joint authorship. But if I spend time struggling with the problem (of course, I will do this only if the project is officially a joint one) and during the course of that struggle I come up with an idea that required more than just standard expertise that I happened to have, then I have made a genuine contribution to the work.
My experience so far with LLMs is that they are capable of playing this knowledgeable-research-supervisor role with me, which can be extremely useful given just how much knowledge they have, but that they are not yet at the level (or at least have not yet exhibited that level in my own interactions with them) at which a human mathematician who follows my convention above would ask for joint authorship.
Power spectra of gravitational radiation from cosmic strings – Robert Scherrer
The problem in context
Cosmic strings are hypothetical one-dimensional topological defects that might have formed at a cosmological phase transition in the early universe (see, e.g., [40 ] for a review). After a burst of interest in the 1980s, interest in cosmic strings declined sharply. However, there has been a recent revival of interest in these objects because they can serve as copious emitters of gravitational radiation such as the stochastic background recently observed by the Pulsar Timing Array.
In a previous paper [64 ] , a graduate student (David Storm) and I examined the spectrum of radiation that might arise from the class of cosmic strings first examined by Garfinkle and Vachaspati [34 ] . While the total power emitted per unit solid angle of the gravitational radiation for these cosmic strings was investigated in [34 ] , the quantity relevant for the stochastic background is the total power P n P_n P n n n n ω n = 4 π n / L \omega_n = 4\pi n/L ω n = 4 π n / L L L L μ \mu μ r ^ \hat{r} r ^ [34 ]
P n = 32 G μ 2 π 3 n 2 ∫ d Ω [ 1 − ( − 1 ) n cos ( n π e 1 ) ] [ 1 − ( − 1 ) n cos ( n π e 2 ) ] ( 1 − e 1 2 ) ( 1 − e 2 2 ) , (9) P_n = \frac{32G\mu^2}{\pi^3 n^2} \int d\Omega\frac{[1 - (-1)^n \cos(n\pi e_1)][1 - (-1)^n \cos(n\pi e_2)]}{(1-e_1^2)(1-e_2^2)}, \tag*{(9)} P n = π 3 n 2 32 G μ 2 ∫ d Ω ( 1 − e 1 2 ) ( 1 − e 2 2 ) [ 1 − ( − 1 ) n cos ( nπ e 1 )] [ 1 − ( − 1 ) n cos ( nπ e 2 )] , (9) where e 1 = r ^ ⋅ a ^ e_1 = \hat{r}\cdot\hat{a} e 1 = r ^ ⋅ a ^ e 2 = r ^ ⋅ b ^ e_2 = \hat{r}\cdot\hat{b} e 2 = r ^ ⋅ b ^ a ^ \hat{a} a ^ b ^ \hat{b} b ^ a ^ \hat{a} a ^ b ^ \hat{b} b ^ α \alpha α a ^ \hat{a} a ^ b ^ \hat{b} b ^ α = π / 2 \alpha= \pi/2 α = π /2 α \alpha α
In our earlier paper [64 ] , we examined a range of values for α \alpha α d P n / d Ω dP_n/d\Omega d P n / d Ω n − 2 n^{-2} n − 2 n n n [24 ] , but seemed to diverge from the predicted result. As our results disagreed with received wisdom, it was important to confirm them analytically.
Human route. After roughly six months attacking the problem, I was able to derive a partial result, but only for the simplest case: α = π / 2 \alpha= \pi/2 α = π /2 n n n
P n = [ 128 G μ 2 π 2 n 2 ] [ γ + ln π + ln ( n ) ] . (10) P_n = \left[\frac{128G\mu^2}{\pi^2 n^2}\right][\gamma+ \ln\pi+ \ln(n)]. \tag*{(10)} P n = [ π 2 n 2 128 G μ 2 ] [ γ + ln π + ln ( n )] . (10) where γ \gamma γ [64 ] .
The task given to GPT-5 Pro: goal, prompts, and model behavior
Goal. While my eventual goal was to determine analytically the power spectrum for all α \alpha α P n P_n P n (9) —for odd values of n n n n n n (10) , but the methods I used to derive this result could not be applied to the odd-n n n
Prompt. In my initial prompt, I asked GPT-5 Pro to evaluate the integral in (9) for odd n n n n n n
I would like to analytically integrate the following integral over the sphere, so that theta goes from 0 to π \pi π 2 π 2\pi 2 π [ 1 + cos ( n π e 1 ) ] [ 1 + cos ( n π e 2 ) ] / [ ( 1 − e 1 2 ) ( 1 − e 2 2 ) ] [1 + \cos(n\pi e_1)][1 + \cos(n\pi e_2)]/[(1 - e_1^2)(1 - e_2^2)] [ 1 + cos ( nπ e 1 )] [ 1 + cos ( nπ e 2 )] / [( 1 − e 1 2 ) ( 1 − e 2 2 )] n n n e 1 = cos ( θ ) e_1 = \cos(\theta) e 1 = cos ( θ ) e 2 = sin ( θ ) sin ( ϕ ) e_2 = \sin(\theta)\sin(\phi) e 2 = sin ( θ ) sin ( ϕ ) θ \theta θ ϕ \phi ϕ θ \theta θ z z z ϕ \phi ϕ n n n n n n n n n
I was particularly pedantic about the definitions of θ \theta θ ϕ \phi ϕ n n n (9) and dropped the ( − 1 ) n (-1)^n ( − 1 ) n
Outcome. After reasoning for 40 minutes, GPT-5 Pro produced a result for large odd n n n n n n (10) . It used a completely different method of solution from my own. While my own solution involved splitting the integral into three pieces and taking an ad hoc approach to each piece, GPT-5 used an expansion in Legendre polynomials, and then extended this to a Bessel function expansion. I had suspected that some sort of solution in Bessel functions was possible but had never found it myself. GPT-5 Pro also gave the leading order correction term to this formula, so that it becomes
4 π [ ln ( n π ) + γ − Ci ( n π ) ] , (11) 4\pi[\ln(n\pi) + \gamma- \operatorname{Ci}(n\pi)], \tag*{(11)} 4 π [ ln ( nπ ) + γ − Ci ( nπ )] , (11) where Ci is the cosine integral function. I was not aware of this correction term, which is relevant for small values of n n n n n n n n n n n n n n n ( − 1 ) n (-1)^n ( − 1 ) n (9) . For reference, the full interaction is available here: [61 ] .
Result, implications, and next steps
Results. GPT-5 Pro derived a new analytical result that confirmed our earlier numerical simulations for the gravitational wave power spectrum from a class of cosmic strings. Had this result been available to me earlier, I would have incorporated it into Ref. [64 ] .
Implications. GPT-5 Pro is capable of solving complex analytic integrations that are beyond the reach of symbolic manipulation programs such as Mathematica.
Next Steps. The obvious next step will be to solve the general case for arbitrary α \alpha α
AI-assisted reduced-physics modeling of thermonuclear burn propagation – Brian Keith Spears
Premise
GPT-5 has evolved to be a quite capable research assistant. It is right now remarkable for its moderate depth of knowledge (relative to, say, a 2-decade veteran PhD researcher) striped across an incredibly broad range of topics. We test and exploit this capability here by recruiting its assistance to formulate an inertial confinement fusion problem that probes this breadth.
We aim to develop an improved understanding of the local physics of fusion ignition and burn propagation at the interface between hot and cold nuclear fuel – a critical complex of processes that lead to the production of more energy by fusion than the energy absorbed by the ICF target [1 ] . To this end, we set up a reduced-physics model that isolates the generation of a thermonuclear burn wave from the complex implosion dynamics of a full ICF implosion simulation [49 ] .
We are interested in AI-driven assistance with several problem phases:
The setup of a simplified physics model with which we can develop intuition;
The numerical simulation and solution of such a model;
The design of numerical experiments to help us elicit the key physics sensitivities;
Development of a theory that explains the numerical findings.
We have developed a simplified, static (no fluid motion) reaction–diffusion model for an ICF deuterium–tritium (DT) capsule to investigate the ignition and propagation of a thermonuclear burn wave into cold, dense fuel. This framework balances fidelity with transparency: it isolates the relative roles of thermal conduction, alpha-particle energy deposition, radiation losses, and density-profile shaping on wave propagation.
Introduction
To get going, I prompted GPT-5 to help me set up a model environment that I could use to explore thermonuclear burn propagation in inertial confinement fusion (ICF). This physics couples complex microphysics with steep thermodynamic gradients, and I would like to intentionally remove the dynamics that continually and crucially changes the power balance. The goal is simply to find the ideal fusion interface in the static case; later we can figure out how to map it into a real implosion and control it (via shock timing, drive preheating, etc.).
I start in my usual way, with what is probably (perhaps not affectionately) known as a “boomer prompt”:
I want to explore the dynamics of a thermonuclear burn wave propagating into cold deuterium-tritium fuel in an imploding ICF capsule. I want to build a simplified, static (no material motion) model of the ignition and propagation of such a wave in spherical geometry.
Let’s build a reaction diffusion equation to model it. I want the model to have 3 regions. The first region is a hot, low-density DT plasma (hot spot) that will initiate the burn wave. It should have properties similar to the central hot spot of a NIF implosion at peak ion temperature and should have similar radius. The next region will be relatively cool, dense DT fuel with a quadratic density profile (one parameter for average slope and one for curvature), ramping from the density of the hot spot to high-density cold fuel and a similar temperature ramp from hot to cold. The third and final region is uniform density (relatively) cold fuel, also at the conditions and thickness of a well shock-timed NIF implosion at peak shell compression (after rebound shock passage).
The reaction diffusion equation dynamics should capture clearly three effects: first, thermal energy should diffuse according to a good approximation of the density- and temperature-dependent thermal diffusivity; second, the reaction term should capture energy production by alpha particle generation and the non-local deposition of that alpha energy consistent with the condition-dependent stopping power of alpha particles in the nearby material; finally, there should be radiation loss which is transported far away from the local heat front.
With this model, we want to ask a key question. What density profile in region 2 most easily allows us to propagate a burn wave? We presume that for very shallow ramps, the hot spot can initially heat region two easily, but wave propagation “flames out.” For very steep ramps, we presume it requires more energy in the hot spot to get the wave going, but that it propagates more easily. The intuition, then, is that there is an optimal profile (average slope and curvature).
Set up this model with appropriate physical constants and assumptions. Embed it in an optimization framework that delivers the profile slope that provides the most robust initiation and propagation of a burn wave. Demonstrate the results with a few plots showing temperature wave propagation under a few scenarios, and add a figure-of-merit plot with axes difference in propagation depth (or speed) vs. fuel profile (slope and curvature).
This initial prompt led to an extended conversation with GPT-5 Pro, with a lot of back and forth which I omit here. The remainder of this section summarizes the work that I was able to carry out in conjunction with GPT-5 Pro. Most of the results given here were produced by GPT-5; I merely provided it with some feedback and guidance. Interested readers may consult this long exchange with GPT-5 Pro here: [63 ] .
The result was a pretty good start at a static model – the formulation is good, the terms are aptly formed and named, and the physics property choices are relevant. After some dialogue, we settled on familiar and more detailed physics data, all of which were offered by GPT-5 in advance of my requesting them. This is how the conversation would go with a national lab colleague and expert.
Together, we consider spherical symmetry with no material motion. The dependent variable is the local DT temperature T ( r , t ) T(r,t) T ( r , t ) P d V P\,dV P d V
ρ C v ∂ T ∂ t = 1 r 2 ∂ ∂ r ( r 2 ρ χ ( T , ρ ) ∂ T ∂ r ) + Q α ( T , ρ ) − Q r a d ( T , ρ ) , (12) \rho C_v \frac{\partial T}{\partial t} = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\rho\chi(T,\rho)\frac{\partial T}{\partial r}\right)+Q_\alpha(T,\rho)-Q_{\rm rad}(T,\rho), \tag*{(12)} ρ C v ∂ t ∂ T = r 2 1 ∂ r ∂ ( r 2 ρ χ ( T , ρ ) ∂ r ∂ T ) + Q α ( T , ρ ) − Q rad ( T , ρ ) , (12) where C v = 3 2 k B / m i C_v=\frac{3}{2}k_B/m_i C v = 2 3 k B / m i χ \chi χ Q α Q_\alpha Q α Q r a d Q_{\rm rad} Q rad
For thermal transport, we adopt a Spitzer–Härm scaling for electronic heat conduction [? , 37 ]:
χ ( T , ρ ) = χ 0 T 5 / 2 ρ , (13) \chi(T,\rho)=\chi_0\frac{T^{5/2}}{\rho}, \tag*{(13)} χ ( T , ρ ) = χ 0 ρ T 5/2 , (13) with χ 0 = 0.12 \chi_0=0.12 χ 0 = 0.12 ρ ≈ 1 g c m − 3 \rho\approx1\ {\rm g\,cm^{-3}} ρ ≈ 1 g c m − 3
The fusion power density follows from the Bosch–Hale [15 ] fit for DT reactivity ⟨ σ v ⟩ \langle\sigma v\rangle ⟨ σ v ⟩
P f u s = n D n T ⟨ σ v ⟩ E f u s , E f u s = 17.6 M e V , (14) P_{\rm fus}=n_Dn_T\langle\sigma v\rangle E_{\rm fus},\qquad E_{\rm fus}=17.6\ {\rm MeV}, \tag*{(14)} P fus = n D n T ⟨ σ v ⟩ E fus , E fus = 17.6 MeV , (14) where n D = n T = ρ / ( 2 m i ) n_D=n_T=\rho/(2m_i) n D = n T = ρ / ( 2 m i )
Q α = f α P f u s , (15) Q_\alpha=f_\alpha P_{\rm fus}, \tag*{(15)} Q α = f α P fus , (15) with f α = E α / E f u s = 3.5 / 17.6 ≈ 0.199 f_\alpha=E_\alpha/E_{\rm fus}=3.5/17.6\approx0.199 f α = E α / E fus = 3.5/17.6 ≈ 0.199
We model finite-range alpha transport using a nonlocal Helmholtz operator [6 ] :
( I − λ α 2 ∇ 2 ) q α = f d e p P α , (16) \left(I-\lambda_\alpha^2\nabla^2\right)q_\alpha=f_{\rm dep}P_\alpha, \tag*{(16)} ( I − λ α 2 ∇ 2 ) q α = f dep P α , (16) where λ α ( T , ρ ) = R m ( T ) / ρ \lambda_\alpha(T,\rho)=R_m(T)/\rho λ α ( T , ρ ) = R m ( T ) / ρ R m ( T ) R_m(T) R m ( T ) g c m − 2 {\rm g\,cm^{-2}} g c m − 2
R m ( T ) = 0.30 ( T 3 k e V ) 1 / 2 g c m − 2 , (17) R_m(T)=0.30\left(\frac{T}{3\ {\rm keV}}\right)^{1/2}\ {\rm g\,cm^{-2}}, \tag*{(17)} R m ( T ) = 0.30 ( 3 keV T ) 1/2 g c m − 2 , (17) consistent with 3.5 MeV alpha stopping data. Deposition was limited to 90% of total energy (f d e p ≤ 0.9 f_{\rm dep}\leq0.9 f dep ≤ 0.9
Finally, for radiation losses, we use an approximate gray diffusion with an escape probability [? , 16 ]:
Q r a d = C r a d ρ 2 T 1 1 + κ 0 ρ R o u t , (18) Q_{\rm rad}=C_{\rm rad}\rho^2\sqrt{T}\frac{1}{1+\kappa_0\rho R_{\rm out}}, \tag*{(18)} Q rad = C rad ρ 2 T 1 + κ 0 ρ R out 1 , (18) where C r a d = 3 × 10 − 36 W m 3 k g − 2 k e V − 1 / 2 C_{\rm rad}=3\times10^{-36}\ {\rm W\,m^3\,kg^{-2}\,keV^{-1/2}} C rad = 3 × 1 0 − 36 W m 3 k g − 2 ke V − 1/2 κ 0 = 5 × 10 − 3 \kappa_0=5\times10^{-3} κ 0 = 5 × 1 0 − 3
GPT-5 is quite good at this kind of model development and setup. It required little intervention on my part to make this plausible and complete. That would change in the numerics.
Numerical Implementation
GPT-5, without explicit prompting, selected for a partial differential equation (PDE) solving a second-order Crank–Nicolson implicit scheme on a uniform radial grid (N ≈ 150 N \approx150 N ≈ 150
∂ T ∂ r = 0 at r = 0 , ∂ T ∂ r = 0 at r = R o u t e r . (19) \frac{\partial T}{\partial r}=0 \quad\text{at } r=0, \qquad\frac{\partial T}{\partial r}=0 \quad\text{at } r=R_{\mathrm{outer}}. \tag*{(19)} ∂ r ∂ T = 0 at r = 0 , ∂ r ∂ T = 0 at r = R outer . (19) Reaction, diffusion, and radiative source terms were split via operator splitting. Alpha deposition was solved by a tridiagonal Helmholtz inversion at each step.
These are reasonable choices for the kind of testbed we seek to develop. It should be noted that this took a few minutes to code up. While I could have done this, I do not do it every day, and I suspect it would have taken me a few work days to really get solid code working. The point is that I can now deliver in minutes as if I were at the highest level I have ever been at for this kind of work. There are those who will say this implementation is not remarkable, and it is not. But, to have it in minutes is remarkable.
It was at this point that my human labor and intelligence were required. I needed to suggest some parameterization for the density and temperature profiles at the fusion boundary. GPT-5 also needed me to help it find the region of parameter space that exhibited the kinds of threshold behaviors we were looking for. Make the hot spot too strong, and the whole thing singularly explodes. Make it too weak, and no burn wave will initiate and propagate. To understand the effect of the profile we are imagining, we need a delicately balanced marginal system. Given the nonlinearities in ICF, this takes not only intuition, but some numerical empiricism. GPT-5 made me work here.
For the density profile, three regions were specified:
ρ ( r ) = ρ h s + s ( r − R h s ) + c ( r − R h s ) 2 , (20) \rho(r) = \rho_{hs} + s(r-R_{hs}) + c(r-R_{hs})^2, \tag*{(20)} ρ ( r ) = ρ h s + s ( r − R h s ) + c ( r − R h s ) 2 , (20) scaled to ensure ρ ( R h s + L r a m p ) = ρ c o l d \rho(R_{hs} + L_{ramp}) = \rho_{cold} ρ ( R h s + L r am p ) = ρ co l d
The ramp is controlled by the slope s s s c c c
s ∈ [ 1.5 , 6.5 ] × 10 9 k g m − 4 and c ∈ [ − 6.5 , 0 ] × 10 13 k g m − 5 . s \in[1.5,6.5] \times10^9\ \mathrm{kg\ m}^{-4}
\quad\text{and} \quad
c \in[-6.5,0] \times10^{13}\ \mathrm{kg\ m}^{-5}. s ∈ [ 1.5 , 6.5 ] × 1 0 9 kg m − 4 and c ∈ [ − 6.5 , 0 ] × 1 0 13 kg m − 5 . I spent a few hours of my time adjusting hot spot conditions, cold fuel conditions, conductivities, alpha particle stopping powers, and profile designs to get a physically reasonable result. This activity is what we in ICF might call “design work” – getting to the place where the physics are interesting and useful. GPT-5 was a bad designer, offering results that were null, noisy, or invalid (NaNs), while claiming that glory had been achieved. However, when re-prompted to examine a pathological result or null signal, GPT-5 offered quite sophisticated solutions, including different implementations of FFTs to prevent aliasing, improved resolutions to track burn fronts, altered performance metrics to amplify signals. If a user gave up at the first wacky response, they would miss out on some rather powerful capability of the model. It needs to be highlighted that this requires you to be a very confident physicist. You have to know that an assertion is wrong, you have to confidently push for a better solution, and you have to be good enough to know when a real solution has been reached. I often find that GPT-5 offers solutions that I have to think deeply about to determine if they are actually correct, or just convenient.
We settled on simulations that were performed at a fixed hot-spot temperature T h s = 1.5 T_{\mathrm{hs}} = 1.5 T hs = 1.5
v m e a n = ∫ t 0 t 1 d r f d t d t t 1 − t 0 . (21) v_{\mathrm{mean}} = \frac{\int_{t_0}^{t_1} \frac{d r_f}{d t}\,\mathrm{d}t}{t_1-t_0}. \tag*{(21)} v mean = t 1 − t 0 ∫ t 0 t 1 d t d r f d t . (21) The resulting v m e a n ( s , c ) v_{\mathrm{mean}}(s,c) v mean ( s , c ) − 1 ^{-1} − 1
s ≈ 3.0 × 10 9 k g m − 4 , c ≈ − 4.5 × 10 13 k g m − 5 , (22) s \approx3.0 \times10^{9}\ \mathrm{kg\,m^{-4}}, \qquad c \approx-4.5 \times10^{13}\ \mathrm{kg\,m^{-5}}, \tag*{(22)} s ≈ 3.0 × 1 0 9 kg m − 4 , c ≈ − 4.5 × 1 0 13 kg m − 5 , (22) with local maxima up to v m e a n ≈ 100 v_{\mathrm{mean}} \approx100 v mean ≈ 100 − 1 ^{-1} − 1
We spent some time examining the density profiles along the ridge. Five equally spaced points along this ridge were selected; their parameters are listed in Table III.1 .
Table III.1. Five equally spaced points along the propagation ridge in ( s , c ) (s,c) ( s , c )
Figure III.1 illustrates that as curvature becomes less negative, the density near the hot-spot boundary rises faster – stiffening the thermal coupling and slightly reducing the front speed. Optimal profiles exhibit a gentle initial density rise, allowing alphas to preheat the ramp before losses dominate deeper in the cold fuel.
Figure III.1. The color map represents the mean propagation speed of a thermonuclear burn wave propagating into the cold ICF fuel. That wave speed depends on the profile of the interface between hot and cold DT fuel. More specifically, both the average slope and the curvature play a key role. Together they set an optimum profile for delivering the highest speed wave under the given physics assumptions. These optimal profiles lie along the yellow ridge.
Within a few hours of noodling, we had a decent numerical picture of a relationship between the parameters chosen to describe the interface and their optimal values for propagating rapid burn fronts.
Theoretical Explanation
While the numerics show a trend, a good design physicist will aim to explain that trend using the theoretical tools at his disposal. I asked GPT-5 to help do this here for two reasons. First, it is useful to have this theoretical picture for forming the intuition we are looking for. Second, it provides an alternative check that GPT-5 has not hallucinated the physics and that I have not been fooled. This is a much-needed verification process.
GPT-5 quite quickly settled (following my nudging prompt) on a traveling-wave power balance to understand what sets the ridge.
We settled on a quasi–steady burn front, written in a co-moving coordinate ξ = r − v t \xi= r - vt ξ = r − v t ρ = ρ ( ξ ) \rho= \rho(\xi) ρ = ρ ( ξ )
− v ρ C v d T d ξ = d d ξ ( ρ χ ( T , ρ ) d T d ξ ) + Q α [ T , ρ ; ρ ( ⋅ ) ] − Q r a d ( T , ρ , ρ R o u t ) . (23) -v \rho C_v \frac{dT}{d\xi}
=
\frac{d}{d\xi}\left(\rho\chi(T,\rho)\frac{dT}{d\xi}\right)
+ Q_\alpha[T,\rho;\rho(\cdot)]
- Q_{\mathrm{rad}}(T,\rho,\rho R_{\mathrm{out}}). \tag*{(23)} − v ρ C v d ξ d T = d ξ d ( ρ χ ( T , ρ ) d ξ d T ) + Q α [ T , ρ ; ρ ( ⋅ )] − Q rad ( T , ρ , ρ R out ) . (23) It identified, I believe correctly, key scalings:
Fusion source: Q α ∼ f α E f u s ( n D n T ) ⟨ σ v ⟩ ( T ) D [ ρ ] Q_\alpha\sim f_\alpha E_{\mathrm{fus}}(n_D n_T)\langle\sigma v\rangle(T)D[\rho] Q α ∼ f α E fus ( n D n T ) ⟨ σ v ⟩ ( T ) D [ ρ ] n ∝ ρ n \propto\rho n ∝ ρ Q α ∝ ρ 2 ⟨ σ v ⟩ ( T ) Q_\alpha\propto\rho^2\langle\sigma v\rangle(T) Q α ∝ ρ 2 ⟨ σ v ⟩ ( T ) D [ ρ ] D[\rho] D [ ρ ]
The ridge (maximal mean front speed) consists of ( s , c ) (s,c) ( s , c )
( conduction from hot side ) + ( nonlocal alpha deposition ) ⏟ effective drive is maximized against ( radiative loss ) + ( downstream conduction ) + ρ C v v Δ T ⏟ sinks + enthalpy (24) \underbrace{\left(\text{conduction from hot side}\right) + \left(\text{nonlocal alpha deposition}\right)}_{\text{effective drive}}
\text{ is maximized against }
\underbrace{\left(\text{radiative loss}\right) + \left(\text{downstream conduction}\right) + \rho C_v v \Delta T}_{\text{sinks + enthalpy}} \tag*{(24)} effective drive ( conduction from hot side ) + ( nonlocal alpha deposition ) is maximized against sinks + enthalpy ( radiative loss ) + ( downstream conduction ) + ρ C v v Δ T (24) It argued for an optimal slope s s s
ρ ( ξ ) ≈ ρ h s + s Δ + c Δ 2 , Δ ≡ ( ξ − ξ h s ) ∈ [ 0 , L r a m p ] . (25) \rho(\xi) \approx\rho_{\mathrm{hs}} + s\Delta+ c\Delta^2,\qquad\Delta\equiv(\xi-\xi_{\mathrm{hs}}) \in[0,L_{\mathrm{ramp}}]. \tag*{(25)} ρ ( ξ ) ≈ ρ hs + s Δ + c Δ 2 , Δ ≡ ( ξ − ξ hs ) ∈ [ 0 , L ramp ] . (25) Make s s s ρ \rho ρ
Make s s s ρ \rho ρ
Pros: stronger Q α ∝ ρ 2 Q_\alpha\propto\rho^2 Q α ∝ ρ 2 ρ C v \rho C_v ρ C v ρ R \rho R ρR f d e p f_{\mathrm{dep}} f dep
Net: there is a sweet spot in s s s ρ \rho ρ
It then moved on to optimal curvature c c c
Negative curvature (c < 0 c < 0 c < 0 ρ \rho ρ χ \chi χ λ α \lambda_\alpha λ α
Positive curvature (c > 0 c > 0 c > 0 ρ \rho ρ χ \chi χ λ α \lambda_\alpha λ α T T T
This is why the data favor moderately negative curvature at nearly any slope that lies on the ridge: it positions the right ρ \rho ρ
Ultimately, a simple power-balance criterion predicts an “ideal” c c c s s s δ \delta δ Δ ⋆ \Delta^\star Δ ⋆ Δ ⋆ ∼ \Delta^\star\sim Δ ⋆ ∼ T i g ≈ 5 k e V T_{\mathrm{ig}} \approx5\ \mathrm{keV} T ig ≈ 5 keV
Q α i n ( Δ ⋆ ) ⏟ nonlocal from hot side + Q c o n d i n ( Δ ⋆ ) ⏟ ∝ χ Δ T / ℓ ≳ Q r a d ( ρ ( Δ ⋆ ) , T ) ⏟ ∝ ρ ( Δ ⋆ ) 2 T / ( 1 + κ 0 ρ R o u t ) + ρ ( Δ ⋆ ) C v T i g − T 0 τ ⏟ enthalpy over timescale τ . (26) \underbrace{Q_{\alpha}^{\mathrm{in}}(\Delta^\star)}_{\text{nonlocal from hot side}}+\underbrace{Q_{\mathrm{cond}}^{\mathrm{in}}(\Delta^\star)}_{\propto\chi\Delta T/\ell}\gtrsim\underbrace{Q_{\mathrm{rad}}(\rho(\Delta^\star),T)}_{\propto\rho(\Delta^\star)^2\sqrt{T}/(1+\kappa_0\rho R_{\mathrm{out}})}+\underbrace{\rho(\Delta^\star)C_v\frac{T_{\mathrm{ig}}-T_0}{\tau}}_{\text{enthalpy over timescale }\tau}. \tag*{(26)} nonlocal from hot side Q α in ( Δ ⋆ ) + ∝ χ Δ T / ℓ Q cond in ( Δ ⋆ ) ≳ ∝ ρ ( Δ ⋆ ) 2 T / ( 1 + κ 0 ρ R out ) Q rad ( ρ ( Δ ⋆ ) , T ) + enthalpy over timescale τ ρ ( Δ ⋆ ) C v τ T ig − T 0 . (26) Two closure relations connect ρ ( Δ ⋆ ) \rho(\Delta^\star) ρ ( Δ ⋆ ) ( s , c ) (s,c) ( s , c )
ρ ( Δ ⋆ ) Δ ⋆ ∼ R m (order unity in areal density) . (27) \rho(\Delta^\star)\Delta^\star\sim R_m \qquad\text{(order unity in areal density)}. \tag*{(27)} ρ ( Δ ⋆ ) Δ ⋆ ∼ R m (order unity in areal density) . (27) For a fixed s s s ρ ( Δ ⋆ ) \rho(\Delta^\star) ρ ( Δ ⋆ ) c c c ρ ( Δ ⋆ ) = ρ h s + s Δ ⋆ + c Δ ⋆ 2 \rho(\Delta^\star)=\rho_{\mathrm{hs}}+s\Delta^\star+c\Delta^{\star2} ρ ( Δ ⋆ ) = ρ hs + s Δ ⋆ + c Δ ⋆ 2
χ 0 T 5 / 2 ρ ( Δ ⋆ ) ∼ Δ ⋆ 2 τ . (28) \frac{\chi_0T^{5/2}}{\rho(\Delta^\star)}\sim\frac{\Delta^{\star2}}{\tau}. \tag*{(28)} ρ ( Δ ⋆ ) χ 0 T 5/2 ∼ τ Δ ⋆ 2 . (28) This again fixes ρ ( Δ ⋆ ) \rho(\Delta^\star) ρ ( Δ ⋆ ) c c c s s s τ \tau τ
Put together, for each slope s s s ρ ( Δ ⋆ ) \rho(\Delta^\star) ρ ( Δ ⋆ ) ρ Δ ⋆ ∼ R m \rho\Delta^\star\sim R_m ρ Δ ⋆ ∼ R m ℓ c ∼ Δ ⋆ \ell_c\sim\Delta^\star ℓ c ∼ Δ ⋆ c c c ρ ( Δ ) \rho(\Delta) ρ ( Δ ) Δ ⋆ \Delta^\star Δ ⋆ c ⋆ ( s ) c^\star(s) c ⋆ ( s ) c ⋆ ( s ) c^\star(s) c ⋆ ( s )
Intuition from that criterion:
The ridge curvature is thus the locus where ρ ( Δ ⋆ ) \rho(\Delta^\star) ρ ( Δ ⋆ ) Q α Q_\alpha Q α
We also developed tests to be sure we were clear about how the ridge should shift with physics knobs.
This picture immediately predicts how the ridge moves if you change microphysics:
Finally, we ended with a compact “engineering rule.” Pick a target preheat distance Δ ⋆ \Delta^{\star} Δ ⋆ ( s , c ) (s,c) ( s , c )
ρ ( Δ ⋆ ) Δ ⋆ ≈ R m and χ 0 T i 5 / 2 ρ ( Δ ⋆ ) ≈ Δ ⋆ 2 τ . (29) \rho(\Delta^{\star})\Delta^{\star}\approx R_m \quad\text{and}\quad\frac{\chi_0 T_i^{5/2}}{\rho(\Delta^{\star})}\approx\frac{\Delta^{\star2}}{\tau}. \tag*{(29)} ρ ( Δ ⋆ ) Δ ⋆ ≈ R m and ρ ( Δ ⋆ ) χ 0 T i 5/2 ≈ τ Δ ⋆ 2 . (29) This reduces to a single condition on ρ ( Δ ⋆ ) \rho(\Delta^{\star}) ρ ( Δ ⋆ ) c ⋆ c^{\star} c ⋆ s s s
This formula offers a check on our results, using microphysics and power balance to estimate the expected c ⋆ ( s ) c^{\star}(s) c ⋆ ( s ) Figure III.2 , the red engineering rule and the numerically obtained ridge are pretty close. I invested about 6 hours in this project (it is quite interesting, but the exercise was to be quick), and I ended with both a numerical result and a theoretical explanation that give me confidence to proceed. As I mentioned earlier, this process itself is not novel. It is the workflow of a design physicist operating in the cutting-edge field of ICF. However, executing this workflow from concept, to numerical exploration, to theoretical supporting statement in hours is rather amazing.
Figure III.2. The red line represents the theoretical ridge slope developed by our theoretical exposition. It fits reasonably well to the numerical ridge. The combination of a numerical result verified by a theoretical power balance argument increases confidence in results.
Physics Conclusions and Study Outcomes
At the outset, I was looking for a set of physics goals, and we delivered:
This is a representative use of GPT-5 for my experience. The model took the seed of an idea that I had – a static model to get a feel for key physics at the initiation and propagation of a thermonuclear burn wave – and it helped me very quickly build a partial differential equation representation of the model. On command, it incorporated detailed stopping power and familiar (to the expert) physical data known to the ICF community. It discretized the problem, set up the optimization routine I imagined, and delivered the result I thought we would expect. This took me about 6 hours from start to finish – idea to writing. There is not too much expertise that I could not find by querying my colleagues at the national labs, doing internet search, deriving some things by hand, recalling some undergraduate and graduate numerical tricks, and writing a few hundred lines of code. And, that is amazing. It made me a one-person army of experts all rolled up, as if I were the best I had ever been at everything I have ever thought about ICF. I can name the world experts it felt like I was talking to or the ones I would have had to reach out to to help me with some small step. But, they were right there in my laptop (more or less).
The theoretical explanation that I worked up with it is probably the best part. This explanation directly ties known physics relations together to give a closed-form predictor of what we had already found. This is a good way of ensuring that the results built over hours of elicitation with the model are not wrong and that the model had not seduced me into buying its arguments.
The results, I must say, do not come without work. The model, in its eagerness to please, often introduces numerical duct tape to smooth over a thorny issue, silently swaps out detailed numerical solves for approximations with trends it knows I want, and confidently declares victory when numerical signals are still obviously noise. When pointed to these errors and unwanted outcomes, the model is very good at actually fixing the issue. But, users must be expert enough to catch the oversimplifications, must persist to get the model to reconsider, and must be vigilant to ensure that the refined results are driving in the right direction.
Overall, as I have noted many times, I feel like my 6 hours of work here yielded something I could have done over a month or two with a very good pair of postdocs – one for theory and one for numerics. So, say 6 person-months reduced to 6 person-hours. That is a compression of about a factor of 1000. I can only imagine what two postdocs and I could do if we collaborated with each of us working in this AI-accelerated mode.
Part II — Final Ridge Map Code
Below is the complete Python code used to generate the Gaussian-smoothed ridge map and extract the five density profiles.
import numpy as np
import matplotlib.pyplot as plt
from numpy.linalg import lstsq
# ------------------ Physical Constants ------------------
MeV_to_J = 1.602176634e-13
m_u = 1.66053906660e-27
m_ion = 2.5 * m_u
k_B = 1.380649e-23
K_per_keV = 1.1604518e7
Cv = 1.5 * k_B / m_ion
E_fus = 17.6 * MeV_to_J
E_alpha = 3.5 * MeV_to_J
alpha_fraction = E_alpha / E_fus
# ------------------ Regions & Profiles ------------------
class Regions:
def __init__(self):
self.R_hs = 50e-6
self.L_ramp = 45e-6
self.L_cold = 90e-6
self.rho_hs = 1200.0
self.rho_cold = 1.5e5
self.T_cold_keV = 0.3
def make_regions(): return Regions()
def density_profile_norm(r, regs, slope, curvature):
R1, R2 = regs.R_hs, regs.R_hs + regs.L_ramp
rho = np.empty_like(r)
rho[r <= R1] = regs.rho_hs
mask = (r > R1) & (r <= R2)
d = r[mask] - R1
ramp = regs.rho_hs + slope*d + curvature*(d**2)
end0 = regs.rho_hs + slope*(R2-R1) + curvature*(R2-R1)**2
scale = (regs.rho_cold - regs.rho_hs)/(end0 - regs.rho_hs)
ramp = regs.rho_hs + scale*(ramp - regs.rho_hs)
rho[mask] = np.clip(ramp, regs.rho_hs, regs.rho_cold)
rho[r > R2] = regs.rho_cold
return rho
# ------------------ Transport / Sources ------------------
def thermal_diffusivity(T, rho, chi0=0.12):
return chi0 * np.maximum(1e-6, T)**2.5 / np.maximum(1e-3, rho)
def bosch_hale_DT_cm3s(T_keV):
C1 = 1.17302e-9
C = [0.0, C1, 1.51361e-2, 7.51886e-2, 4.60643e-3, 1.35e-2,
-1.0675e-4, 1.366e-5]
B_G = 34.3827; mr_c2 = 1124656.0
T = np.maximum(1e-6, T_keV)
th = T/(1 - (T*(C[2]+T*(C[4]+T*C[6])))/(1+T*(C[3]+T*(C[5]+T*C[7]))))
eta = (B_G**2/(4.0*th))**(1/3)return C[1]*th*np.sqrt(eta/(mr_c2*T**3))*np.exp(-3*eta)
def fusion_power_density(T, rho, burn):
sv = bosch_hale_DT_cm3s(T)*1e-6
n_i = rho/m_ion
return 0.25*n_i**2*(1-burn)**2*sv*E_fus
def radiation_loss_gray(T, rho, rhoR, C_rad=3e-36, kappa0=5e-3):
emiss = C_rad*rho**2*np.sqrt(np.maximum(1e-6, T))
Pesc = 1/(1 + kappa0*np.maximum(0, rhoR))
return emiss*Pesc
# realistic alpha mass range
def mass_range_Rm(T): return 0.30*(T/3)**0.5 # g/cm^2
def alpha_range_length(T, rho):
return (mass_range_Rm(T)/np.maximum(1e-6, rho/1000))*1e-2
def rhoR_outward(r, rho):
dr = r[1]-r[0]; rho_gcc = rho/1000
return np.cumsum(rho_gcc[::-1])*dr*100.0[::-1]
# ------------------ Finite-Hotspot Solver ------------------
def step_CN_finite(T, B, r, rho, dt, chi0=0.12):
N = len(T); dr = r[1]-r[0]
chi = thermal_diffusivity(T, rho, chi0)
chi_i = 0.5*(chi[:-1]+chi[1:])
a=np.zeros(N-1); b=np.zeros(N); c=np.zeros(N-1)
for i in range(1,N-1):
Aip=((r[i+1]+r[i])/2)**2*chi_i[i]/dr
Aim=((r[i]+r[i-1])/2)**2*chi_i[i-1]/dr
vol=r[i]**2; a[i-1]=Aim/vol; c[i]=Aip/vol; b[i]=-(Aim+Aip)/vol
b[0]=-2*chi_i[0]/dr**2; b[-1]=-2*chi_i[-1]/dr**2
rhs = T + dt*((fusion_power_density(T,rho,B)*alpha_fraction
- radiation_loss_gray(T,rho,0)) / (Cv*rho*K_per_keV))
# simple implicit diffusion
M=np.zeros((N,N)); np.fill_diagonal(M,1-dt*b)
for i in range(N-1): M[i,i+1]=-dt*c[i]; M[i+1,i]=-dt*a[i]
Tnew=np.linalg.solve(M,rhs)
return np.clip(Tnew,0,400),B
def run_sim_finite(regs,slope,curv,Ths,t_end=3.5e-9,N=150,Nt=140):
Rtot=regs.R_hs+regs.L_ramp+regs.L_cold
r=np.linspace(0,Rtot,N)
rho=density_profile_norm(r,regs,slope,curv)
T=np.full_like(r,regs.T_cold_keV); T[r<=regs.R_hs]=Ths
B=np.zeros_like(T)
times=np.linspace(0,t_end,Nt+1); dt=t_end/Nt
Th=np.zeros((Nt+1,N)); Th[0]=T
for n in range(1,Nt+1):
T,B=step_CN_finite(T,B,r,rho,dt)
Th[n]=T
class Result: pass
res=Result(); res.r=r; res.times=times; res.T_hist=Th; res.rho=rho; res.
meta={'regs':regs}
return res
# ------------------ Speed Metric ------------------def interp_front(r,T,Tthr=5.0):
idx=np.where(T>=Tthr)[0]
if len(idx)==0: return np.nan
i2=idx[-1]; i1=max(i2-1,0)
T1,T2=T[i1],T[i2]; r1,r2=r[i1],r[i2]
return r1+(Tthr-T1)/(T2-T1+1e-12)*(r2-r1)
def mean_speed(res,Tthr=5.0,t0=1,t1=3):
t=res.times*1e9; r=res.r
mask=(t>=t0)&(t<=t1)
rf=[interp_front(r,res.T_hist[i],Tthr) for i in np.where(mask)[0]]
rf=np.array(rf); ok=~np.isnan(rf)
if ok.sum()<3: return 0
A=np.vstack([t[mask][ok]*1e-9,np.ones(ok.sum())]).T
v,_=lstsq(A,rf[ok],rcond=None)[0]
return max(0,v)
# ------------------ Ridge Map Generation ------------------
regs=make_regions(); T_hs=1.5
slopes=np.linspace(1.5e9,6.5e9,11)
curvs=np.linspace(-6.5e13,0,11)
V=np.zeros((len(slopes),len(curvs)))
for i,s in enumerate(slopes):
for j,c in enumerate(curvs):
res=run_sim_finite(regs,s,c,T_hs)
V[i,j]=mean_speed(res)
# Gaussian kernel smoother
def gauss_smooth(V,sigma=1):
from scipy.ndimage import gaussian_filter
return gaussian_filter(V,sigma=sigma,mode='nearest')
Vg=gauss_smooth(V,1)
# Ridge extraction
ridge_s=[]; ridge_c=[]
for j in range(Vg.shape[1]):
i=np.argmax(Vg[:,j]); ridge_s.append(slopes[i]); ridge_c.append(curvs[j])
ridge_s=np.array(ridge_s); ridge_c=np.array(ridge_c)
# Ridge plot
plt.figure(figsize=(8,5))
im = plt.imshow(Vg.T, origin='lower',
extent=[slopes.min(), slopes.max(), curvs.min(), curvs.max()],
aspect='auto')
plt.colorbar(im, label='mean front speed (m/s)')
plt.plot(ridge_s, ridge_c, 'k--', linewidth=2, label='ridge')
plt.xlabel('slope (kg/m^4)'); plt.ylabel('curvature (kg/m^5)')
plt.legend(); plt.tight_layout()
plt.show()Listing III.1: Ridge map generation and profile extraction.
New scientific results obtained with AI
Erdős problems (part 2/2) – Mehtaab Sawhney and Mark Sellke
We continue from where we left off in Section II.2 and explain how GPT-5 Pro assisted us in solving Erdős Problem #848.
AI–assisted solution to Problem #848
Erdős Problem #848 is in combinatorial number theory, and was posted as Problem 23 in [30 ] . This problem was genuinely open, and has now been solved by the first author and GPT-5, in combination with online comments of van Doorn, Weisenberg and Cambie. The statement is:
Problem #848: Let A A A { 1 , 2 , … , N } \{1,2,\ldots,N\} { 1 , 2 , … , N } a b + 1 ab+1 ab + 1 a , b ∈ A a,b\in A a , b ∈ A a b + 1 ab+1 ab + 1 p 2 p^2 p 2 p p p N N N A A A A = { 1 ≤ a ≤ N : a ≡ 7 ( m o d 25 ) } A=\{1\leq a\leq N:a\equiv7\pmod{25}\} A = { 1 ≤ a ≤ N : a ≡ 7 ( mod 25 )}
Of course, the given set A A A 7 2 + 1 ≡ 0 ( m o d 25 ) 7^2+1\equiv0\pmod{25} 7 2 + 1 ≡ 0 ( mod 25 ) A A A { 1 ≤ a ≤ N : a ≡ 18 ( m o d 25 ) } \{1\leq a\leq N:a\equiv18\pmod{25}\} { 1 ≤ a ≤ N : a ≡ 18 ( mod 25 )} ε > 0 \varepsilon>0 ε > 0 A A A ( 1 − ε ) N 25 \frac{(1-\varepsilon)N}{25} 25 ( 1 − ε ) N
This question is in a line of extremal questions raised by Erdős at the intersection of elementary number theory and combinatorics. For instance, Erdős and Silverman asked what is the densest subset of [ N ] [N] [ N ] A + A A+A A + A [44 ] who showed that ∣ A ∣ ≤ ( 11 / 32 + o ( 1 ) ) N |A|\leq(11/32+o(1))N ∣ A ∣ ≤ ( 11/32 + o ( 1 )) N [30 ] ) about the largest collection A ⊆ [ N ] A\subseteq[N] A ⊆ [ N ] a b ab ab
The starting point of the solution is to observe that the diagonal constraint that a 2 + 1 a^2+1 a 2 + 1 A A A p ≡ 1 ( m o d 4 ) p\equiv1\pmod{4} p ≡ 1 ( mod 4 ) a 2 + 1 a^2+1 a 2 + 1 2 / p 2 2/p^2 2/ p 2 a a a a 2 + 1 a^2+1 a 2 + 1 p p p
1 − ∏ p prime p ≡ 1 ( m o d 4 ) ( 1 − 2 p 2 ) ≈ 0.105 1-\prod_{\substack{p\ \text{prime}\\p\equiv1\ (\mathrm{mod}\ 4)}}\left(1-\frac{2}{p^2}\right)\approx0.105 1 − p prime p ≡ 1 ( mod 4 ) ∏ ( 1 − p 2 2 ) ≈ 0.105 for the density of A A A
What is more challenging is to use the off-diagonal constraints on a b + 1 ab+1 ab + 1 Figure IV.1 below (which itself built on online comments of van Doorn, Weisenberg and Cambie).
Figure IV.1. The start of the interaction with GPT-5 Pro to derive the improved bound for Problem #848. In the above, it has just summarized the problem and the argument on https://www.erdosproblems.com . Immediately after this, GPT-5 put forward the new idea that led to our solution (see next page). The full transcript can be found here: [60 ] .
The idea suggested by GPT-5’s reply in Figure IV.1 gives a method to use any single number b ∈ A b\in A b ∈ A a ∈ A a\in A a ∈ A A ∗ A^* A ∗ A A A A ∗ A^* A ∗ b ∈ A ∗ b\in A^* b ∈ A ∗ a ∈ A a\in A a ∈ A a 2 + 1 a^2+1 a 2 + 1 a b + 1 ab+1 ab + 1 p 2 p^2 p 2 p p p b ∈ A ∗ b\in A^* b ∈ A ∗ a ∈ A a\in A a ∈ A
23 25 ( 1 − ∏ p ≡ 1 ( m o d 4 ) p ≥ 13 ( 1 − 2 p 2 ) ) + 2 25 ( 1 − ∏ p ≠ 5 ( 1 − 1 p 2 ) ) ≤ .02517 + .0294 ≈ .05457. \begin{aligned}
\frac{23}{25}\left(1-\prod_{\substack{p\equiv1\ (\mathrm{mod}\ 4)\\p\geq13}}\left(1-\frac{2}{p^2}\right)\right)+\frac{2}{25}\left(1-\prod_{p\ne5}\left(1-\frac{1}{p^2}\right)\right) \\
&\leq.02517+.0294\approx.05457.
\end{aligned} 25 23 1 − p ≡ 1 ( mod 4 ) p ≥ 13 ∏ ( 1 − p 2 2 ) + 25 2 1 − p = 5 ∏ ( 1 − p 2 1 ) ≤ .02517 + .0294 ≈ .05457. Indeed for the ( 23 / 25 ) (23/25) ( 23/25 ) a ≡ 7 , 18 ( m o d 25 ) a\equiv7,18\pmod{25} a ≡ 7 , 18 ( mod 25 ) a b + 1 ab+1 ab + 1 p 2 p^2 p 2 p ≠ 5 p\ne5 p = 5
A first step of the eventual solution was to note that if b ∈ A ∗ b\in A^* b ∈ A ∗ a b + 1 ab+1 ab + 1
23 25 ( 1 − ∏ p ≡ 1 ( m o d 4 ) p ≥ 13 ( 1 − 2 p 2 ) ) + 2 25 ( 1 − ∏ p ∉ { 2 , 5 } ( 1 − 1 p 2 ) ) ≤ .02517 + .0125 ≈ .03767. \begin{aligned}
\frac{23}{25}\left(1-\prod_{\substack{p\equiv1\ (\mathrm{mod}\ 4)\\p\geq13}}\left(1-\frac{2}{p^2}\right)\right)+\frac{2}{25}\left(1-\prod_{p\notin\{2,5\}}\left(1-\frac{1}{p^2}\right)\right) \\
&\leq.02517+.0125\approx.03767.
\end{aligned} 25 23 1 − p ≡ 1 ( mod 4 ) p ≥ 13 ∏ ( 1 − p 2 2 ) + 25 2 1 − p ∈ / { 2 , 5 } ∏ ( 1 − p 2 1 ) ≤ .02517 + .0125 ≈ .03767. Since .03767 < 1 / 25 .03767<1/25 .03767 < 1/25 A A A A ∗ A^* A ∗ b ∈ A ∗ b\in A^* b ∈ A ∗ [60 ] .
We emphasize that the first observation mentioned is due to Wouter van Doorn and the particular prompt given to GPT-5 was based on a comment of Stijn Cambie; these were already present in the comments of the official website https://www.erdosproblems.com/forum/thread/848 . Thus, the final proof consists of a key solution step due to GPT-5 sandwiched between two layers of human mathematics!
Fix a ≥ 1 a \ge1 a ≥ 1 t ∈ { 0 , … , 24 } t \in\{0,\ldots,24\} t ∈ { 0 , … , 24 } t ≠ − a − 1 ( m o d 25 ) t \ne-a^{-1} \pmod{25} t = − a − 1 ( mod 25 ) 25 ∤ a b + 1 25 \nmid ab+1 25 ∤ ab + 1 b ≡ t ( m o d 25 ) b \equiv t \pmod{25} b ≡ t ( mod 25 )
B a ( X ) : = # { b ≤ X : b ≡ t ( m o d 25 ) , a b + 1 is squarefree } . B_a(X) := \#\{b \le X : b \equiv t \pmod{25},\ ab+1\text{ is squarefree}\}. B a ( X ) := # { b ≤ X : b ≡ t ( mod 25 ) , ab + 1 is squarefree } . For any odd prime p ≠ 5 p \ne5 p = 5 p ∤ a p \nmid a p ∤ a p 2 ∣ a b + 1 p^2 \mid ab+1 p 2 ∣ ab + 1 p 2 p^2 p 2 p ∣ a p \mid a p ∣ a S S S
# { b ≤ X : b ≡ t ( m o d 25 ) , p 2 ∤ a b + 1 ∀ p ∈ S } X / 25 = ∏ p ∈ S p ≠ 5 ( 1 − 1 p 2 ) + O ( 1 X ) . \frac{\#\{b \le X : b \equiv t \pmod{25},\ p^2 \nmid ab+1\ \forall p \in S\}}{X/25}
= \prod_{\substack{p \in S\\p\ne5}}\left(1-\frac{1}{p^2}\right)+O\left(\frac{1}{X}\right). X /25 # { b ≤ X : b ≡ t ( mod 25 ) , p 2 ∤ ab + 1 ∀ p ∈ S } = p ∈ S p = 5 ∏ ( 1 − p 2 1 ) + O ( X 1 ) . Letting S S S ∏ p ( 1 − 1 / p 2 ) \prod_p(1-1/p^2) ∏ p ( 1 − 1/ p 2 ) a a a
Lemma 1 (squarefree density in a fixed 25-class).
For any fixed a a a t ≠ − a − 1 ( m o d 25 ) t \ne-a^{-1} \pmod{25} t = − a − 1 ( mod 25 )
B a ( X ) X / 25 = ∏ p ≠ 5 ( 1 − 1 p 2 ) + o ( 1 ) ≥ ∏ p ≠ 5 ( 1 − 1 p 2 ) + o ( 1 ) , \frac{B_a(X)}{X/25}
= \prod_{p \ne5}\left(1-\frac{1}{p^2}\right)+o(1)
\ge\prod_{p \ne5}\left(1-\frac{1}{p^2}\right)+o(1), X /25 B a ( X ) = p = 5 ∏ ( 1 − p 2 1 ) + o ( 1 ) ≥ p = 5 ∏ ( 1 − p 2 1 ) + o ( 1 ) , as X → ∞ X \to\infty X → ∞
δ : = ∏ p ≠ 5 ( 1 − 1 p 2 ) = 1 ζ ( 2 ) ⋅ 1 1 − 1 / 25 = 25 4 π 2 = 0.633257 … \delta:= \prod_{p \ne5}\left(1-\frac{1}{p^2}\right)
= \frac{1}{\zeta(2)}\cdot\frac{1}{1-1/25}
= \frac{25}{4\pi^2}=0.633257\ldots δ := p = 5 ∏ ( 1 − p 2 1 ) = ζ ( 2 ) 1 ⋅ 1 − 1/25 1 = 4 π 2 25 = 0.633257 … Thus, for every fixed a a a t t t 25 ∣ a b + 1 25 \mid ab+1 25 ∣ ab + 1 δ − o ( 1 ) \delta-o(1) δ − o ( 1 ) b b b a b + 1 ab+1 ab + 1
1 − δ = 1 − 25 4 π 2 = 0.366742 … 1-\delta=1-\frac{25}{4\pi^2}=0.366742\ldots 1 − δ = 1 − 4 π 2 25 = 0.366742 … of those b b b
Figure IV.2. A continuation of the interaction with GPT-5 Pro to derive the improved bound. Here it proposes the main new idea that led to the solution presented in Subsection IV.1.3. Namely if S S S all other numbers in S S S a a a b b b [60 ] .
Looking Forward
Our investigation of Erdős problem #848 highlights that GPT-5 has the ability to serve as an effective mathematical assistant, capable of recalling relevant lemmas, identifying analogies and locating relevant results from vague, ill-specified prompts. However, current models remain limited in perceiving the “negative space” of mathematics. While models are able to suggest plausible proof strategies, they often do not realize certain “obvious” examples which block progress, and are overly confident in the power of existing methods. This is unsurprising as such discussion is largely absent in mathematical literature itself; mathematicians don’t systematically record why problems are out reach, or why a more obvious strategy doesn’t work, or why certain techniques are inherently unable to solve certain classes of problems.
Proposition IV.1.1. There exists an integer N 0 N_0 N 0 N ≥ N 0 N \ge N_0 N ≥ N 0 A ⊆ [ N ] A \subseteq[N] A ⊆ [ N ] a , b ∈ A a,b \in A a , b ∈ A a b + 1 ab+1 ab + 1
∣ A ∣ ≤ ∣ { n ∈ [ N ] : n ≡ 7 ( m o d 25 ) } ∣ , |A| \le\left|\{n \in[N] : n \equiv7 \pmod{25}\}\right|, ∣ A ∣ ≤ ∣ { n ∈ [ N ] : n ≡ 7 ( mod 25 )} ∣ , i.e. { n ∈ [ N ] : n ≡ 7 ( m o d 25 ) } \{n \in[N] : n \equiv7 \pmod{25}\} { n ∈ [ N ] : n ≡ 7 ( mod 25 )} ∣ A ∣ |A| ∣ A ∣
Remark IV.1.1. Our proof actually allows one to go further and characterize all approximately maximal sets A A A A = { n ∈ [ N ] : n ≡ 7 ( m o d 25 ) } A = \{n \in[N] : n \equiv7 \pmod{25}\} A = { n ∈ [ N ] : n ≡ 7 ( mod 25 )} A = { n ∈ [ N ] : n ≡ 18 ( m o d 25 ) } A = \{n \in[N] : n \equiv18 \pmod{25}\} A = { n ∈ [ N ] : n ≡ 18 ( mod 25 )} n n n ∣ A ∣ ≥ 0.038 ⋅ N |A| \ge0.038 \cdot N ∣ A ∣ ≥ 0.038 ⋅ N N N N A A A { n ∈ [ N ] : n ≡ 7 ( m o d 25 ) } \{n \in[N] : n \equiv7 \pmod{25}\} { n ∈ [ N ] : n ≡ 7 ( mod 25 )} { n ∈ [ N ] : n ≡ 18 ( m o d 25 ) } \{n \in[N] : n \equiv18 \pmod{25}\} { n ∈ [ N ] : n ≡ 18 ( mod 25 )} A A A
We first require the following pair of preliminary lemmas; the first was already used by van Doorn (see [14 ] ). In order to be self-contained, we provide proofs with suboptimal error terms.
Lemma IV.1.1. Let P \mathcal{P} P max P ≤ N 1 / 2 \max\mathcal{P} \le N^{1/2} max P ≤ N 1/2 q q q p ∈ P p \in\mathcal{P} p ∈ P R p \mathcal{R}_p R p p 2 p^2 p 2 ∣ R p ∣ ≤ 2 |\mathcal{R}_p| \le2 ∣ R p ∣ ≤ 2 R p = ∅ \mathcal{R}_p = \varnothing R p = ∅ ( p , q ) ≠ 1 (p,q) \ne1 ( p , q ) = 1
∣ ( { n ∈ [ N ] : n ≡ t ( m o d q ) } ∩ ⋃ p ∈ P { n ( m o d p 2 ) ∈ R p } ) − N q ( 1 − ∏ p ∈ P ( 1 − ∣ R p ∣ p 2 ) ) ∣ ≤ O ( N ( log N ) − 1 / 2 ) . \left|\left(\{n \in[N] : n \equiv t \pmod q\} \cap\bigcup_{p \in\mathcal{P}}\{n \pmod{p^2} \in\mathcal{R}_p\}\right)-\frac{N}{q}\left(1-\prod_{p \in\mathcal{P}}\left(1-\frac{|\mathcal{R}_p|}{p^2}\right)\right)\right| \le O\left(N(\log N)^{-1/2}\right). { n ∈ [ N ] : n ≡ t ( mod q )} ∩ p ∈ P ⋃ { n ( mod p 2 ) ∈ R p } − q N 1 − p ∈ P ∏ ( 1 − p 2 ∣ R p ∣ ) ≤ O ( N ( log N ) − 1/2 ) . Proof. Let T = ⌊ log N ⌋ T = \lfloor\sqrt{\log N}\rfloor T = ⌊ log N ⌋
∣ [ N ] ∩ ⋃ T ≤ p ≤ N 1 / 2 { n ( m o d p 2 ) ∈ R p } ∣ ≤ O ( ∑ T ≤ p ≤ N 1 / 2 N p 2 ) ≤ O ( N T ) . \left|[N]\cap\bigcup_{T\le p\le N^{1/2}}\{n \pmod{p^2}\in\mathcal{R}_p\}\right|\le O\left(\sum_{T\le p\le N^{1/2}}\frac{N}{p^2}\right)\le O\left(\frac{N}{T}\right). [ N ] ∩ T ≤ p ≤ N 1/2 ⋃ { n ( mod p 2 ) ∈ R p } ≤ O T ≤ p ≤ N 1/2 ∑ p 2 N ≤ O ( T N ) . Therefore it suffices to estimate
∣ ( { n ∈ [ N ] : n ≡ t ( m o d q ) } ∩ ⋃ p ∈ P p ≤ T { n ( m o d p 2 ) ∈ R p } ) ∣ . (30) \left|\left(\{n \in[N] : n \equiv t \pmod q\}\cap\bigcup_{\substack{p\in\mathcal{P}\\p\le T}}\{n \pmod{p^2}\in\mathcal{R}_p\}\right)\right|. \tag*{(30)} { n ∈ [ N ] : n ≡ t ( mod q )} ∩ p ∈ P p ≤ T ⋃ { n ( mod p 2 ) ∈ R p } . (30) Note that as T ≤ log N T \le\sqrt{\log N} T ≤ log N ∏ p ≤ T p 2 ≤ N o ( 1 ) \prod_{p\le T}p^2 \le N^{o(1)} ∏ p ≤ T p 2 ≤ N o ( 1 )
∣ ( { n ∈ [ N ] : n ≡ t ( m o d q ) } ∩ ⋃ p ∈ P p ≤ T { n ( m o d p 2 ) ∈ R p } ) ∣ = ∑ S ⊆ P ∩ [ T ] S ≠ ∅ ( − 1 ) ∣ S ∣ − 1 ⋅ ∣ ⋂ p ∈ S { n ( m o d p 2 ) ∈ R p } ∩ { n ∈ [ N ] : n ≡ t ( m o d q ) } ∣ = N o ( 1 ) + ∑ S ⊆ P ∩ [ T ] S ≠ ∅ ( − 1 ) ∣ S ∣ − 1 ⋅ N q ⋅ ∏ p ∈ S ∣ R p ∣ ∏ p ∈ S p 2 = N q ⋅ ( 1 − ∏ p ∈ P ∩ [ T ] ( 1 − ∣ R p ∣ p 2 ) ) + N o ( 1 ) . \begin{aligned}
\left|\left(\{n\in[N]:n\equiv t\pmod q\}\cap\bigcup_{\substack{p\in\mathcal{P}\\p\le T}}\{n\pmod{p^2}\in\mathcal{R}_p\}\right)\right| \\
&=\sum_{\substack{S\subseteq\mathcal{P}\cap[T]\\S\ne\varnothing}}(-1)^{|S|-1}\cdot\left|\bigcap_{p\in S}\{n\pmod{p^2}\in\mathcal{R}_p\}\cap\{n\in[N]:n\equiv t\pmod q\}\right| \\
&=N^{o(1)}+\sum_{\substack{S\subseteq\mathcal{P}\cap[T]\\S\ne\varnothing}}(-1)^{|S|-1}\cdot\frac{N}{q}\cdot\frac{\prod_{p\in S}|\mathcal{R}_p|}{\prod_{p\in S}p^2} \\
&=\frac{N}{q}\cdot\left(1-\prod_{p\in\mathcal{P}\cap[T]}\left(1-\frac{|\mathcal{R}_p|}{p^2}\right)\right)+N^{o(1)}.
\end{aligned} { n ∈ [ N ] : n ≡ t ( mod q )} ∩ p ∈ P p ≤ T ⋃ { n ( mod p 2 ) ∈ R p } = S ⊆ P ∩ [ T ] S = ∅ ∑ ( − 1 ) ∣ S ∣ − 1 ⋅ p ∈ S ⋂ { n ( mod p 2 ) ∈ R p } ∩ { n ∈ [ N ] : n ≡ t ( mod q )} = N o ( 1 ) + S ⊆ P ∩ [ T ] S = ∅ ∑ ( − 1 ) ∣ S ∣ − 1 ⋅ q N ⋅ ∏ p ∈ S p 2 ∏ p ∈ S ∣ R p ∣ = q N ⋅ 1 − p ∈ P ∩ [ T ] ∏ ( 1 − p 2 ∣ R p ∣ ) + N o ( 1 ) . Completing the product, we have the desired result. □
We next require a variant of the previous lemma which allows us to incorporate the “off-diagonal” constraints that a b + 1 ab+1 ab + 1
Lemma IV.1.2. Fix a positive integer N N N t ( m o d q ) t\pmod q t ( mod q ) q q q b b b 1 ≤ b ≤ N 1\le b\le N 1 ≤ b ≤ N p p p p 2 ∣ q p^2\mid q p 2 ∣ q p 2 ∣ ( b t + 1 ) p^2\mid(bt+1) p 2 ∣ ( b t + 1 ) Then we have that
∣ { a ∈ [ N ] : a ≡ t ( m o d q ) ∧ μ ( a b + 1 ) = 0 } − N q ⋅ ( 1 − ∏ ( p , q b ) = 1 ( 1 − 1 p 2 ) ) ∣ ≤ O ( N log N ) . \left|\left\{a\in[N]:a\equiv t\pmod q\land\mu(ab+1)=0\right\}-\frac{N}{q}\cdot\left(1-\prod_{(p,qb)=1}\left(1-\frac{1}{p^2}\right)\right)\right|\le O\left(\frac{N}{\sqrt{\log N}}\right). { a ∈ [ N ] : a ≡ t ( mod q ) ∧ μ ( ab + 1 ) = 0 } − q N ⋅ 1 − ( p , q b ) = 1 ∏ ( 1 − p 2 1 ) ≤ O ( log N N ) . Proof. Note that a b + 1 ≤ N 2 + 1 ab+1\le N^2+1 ab + 1 ≤ N 2 + 1 μ ( a b + 1 ) = 0 \mu(ab+1)=0 μ ( ab + 1 ) = 0 p p p p 2 ∣ ( a b + 1 ) p^2\mid(ab+1) p 2 ∣ ( ab + 1 ) ( p , q b ) = 1 (p,qb)=1 ( p , q b ) = 1 p ∣ b p\mid b p ∣ b p ∤ a b + 1 p\nmid ab+1 p ∤ ab + 1 p ∣ q p\mid q p ∣ q a b + 1 ≡ b t + 1 ≢ 0 ( m o d p 2 ) ab+1\equiv bt+1\not\equiv0\pmod{p^2} ab + 1 ≡ b t + 1 ≡ 0 ( mod p 2 )
We now let T = ⌊ log N ⌋ T=\lfloor\sqrt{\log N}\rfloor T = ⌊ log N ⌋
∑ T ≤ p ≤ N ∣ { a ∈ [ N ] : p 2 ∣ ( a b + 1 ) } ∣ ≤ O ( ∑ T ≤ p ≤ N ( N p 2 + 1 ) ) ≤ O ( N T ) . \sum_{T\le p\le N}\left|\{a\in[N]:p^2\mid(ab+1)\}\right|\le O\left(\sum_{T\le p\le N}\left(\frac{N}{p^2}+1\right)\right)\le O\left(\frac{N}{T}\right). T ≤ p ≤ N ∑ { a ∈ [ N ] : p 2 ∣ ( ab + 1 )} ≤ O T ≤ p ≤ N ∑ ( p 2 N + 1 ) ≤ O ( T N ) . Therefore it suffices to handle p ≤ T p\le T p ≤ T ( p , q b ) = 1 (p,qb)=1 ( p , q b ) = 1 p 2 ∣ a b + 1 p^2\mid ab+1 p 2 ∣ ab + 1 p p p
We now give the remainder of the analysis to conclude the proof. The proof consists of breaking the set into the parts which are 7 ( m o d 25 ) 7\pmod{25} 7 ( mod 25 ) 18 ( m o d 25 ) 18\pmod{25} 18 ( mod 25 ) ∏ ( p , q b ) = 1 ( 1 − 1 / p 2 ) \prod_{(p,qb)=1}(1-1/p^2) ∏ ( p , q b ) = 1 ( 1 − 1/ p 2 ) 169 169 169 289 289 289 [60 ] in particular uses the auxiliary prime 3 3 3
Proof of Proposition IV.1.1. Let η > 0 \eta> 0 η > 0 η = 0.002 \eta= 0.002 η = 0.002 N N N ∣ A ∣ ≥ ( 1 / 25 − η ) ⋅ N |A| \ge(1/25-\eta)\cdot N ∣ A ∣ ≥ ( 1/25 − η ) ⋅ N
A 7 = { a ∈ A : a ≡ 7 ( m o d 25 ) } A_7=\{a\in A:a\equiv7\pmod{25}\} A 7 = { a ∈ A : a ≡ 7 ( mod 25 )} A 18 = { a ∈ A : a ≡ 18 ( m o d 25 ) } A_{18}=\{a\in A:a\equiv18\pmod{25}\} A 18 = { a ∈ A : a ≡ 18 ( mod 25 )} A ∗ = A ∖ ( A 7 ∪ A 18 ) . A^*=A\setminus(A_7\cup A_{18}). A ∗ = A ∖ ( A 7 ∪ A 18 ) . We first prove that A ∗ A^* A ∗
Suppose that there exists b ∈ A ∗ b\in A^* b ∈ A ∗ 2 ∣ b 2\mid b 2 ∣ b x ∈ A ∗ x\in A^* x ∈ A ∗ p ≡ 1 ( m o d 4 ) p\equiv1\pmod{4} p ≡ 1 ( mod 4 ) p ≥ 13 p\ge13 p ≥ 13 p 2 ∣ x 2 + 1 p^2\mid x^2+1 p 2 ∣ x 2 + 1 A ∗ A^* A ∗ 25 25 25
∣ A ∗ ∣ N ≤ ( 23 25 ) ( 1 − ∏ p ≡ 1 ( m o d 4 ) p ≥ 13 ( 1 − 2 p 2 ) ) + o ( 1 ) . \frac{|A^*|}{N}\le\left(\frac{23}{25}\right)\left(1-\prod_{\substack{p\equiv1\pmod{4}\\p\ge13}}\left(1-\frac{2}{p^2}\right)\right)+o(1). N ∣ A ∗ ∣ ≤ ( 25 23 ) 1 − p ≡ 1 ( mod 4 ) p ≥ 13 ∏ ( 1 − p 2 2 ) + o ( 1 ) . For a ∈ A 7 ∪ A 18 a\in A_7\cup A_{18} a ∈ A 7 ∪ A 18 p ≠ 2 , 5 p\ne2,5 p = 2 , 5 p 2 ∣ a b + 1 p^2\mid ab+1 p 2 ∣ ab + 1
∣ A 7 ∪ A 18 ∣ N ≤ ( 2 25 ) ( 1 − ∏ p ≠ 2 , 5 ( 1 − 1 p 2 ) ) + o ( 1 ) . \frac{|A_7\cup A_{18}|}{N}\le\left(\frac{2}{25}\right)\left(1-\prod_{p\ne2,5}\left(1-\frac{1}{p^2}\right)\right)+o(1). N ∣ A 7 ∪ A 18 ∣ ≤ ( 25 2 ) 1 − p = 2 , 5 ∏ ( 1 − p 2 1 ) + o ( 1 ) . Combining these bounds we have
∣ A ∣ N ≤ ( 23 25 ) ( 1 − ∏ p ≡ 1 ( m o d 4 ) p ≥ 13 ( 1 − 2 p 2 ) ) + 2 25 ( 1 − ∏ p ≠ 2 , 5 ( 1 − 1 p 2 ) ) + o ( 1 ) \frac{|A|}{N}\le\left(\frac{23}{25}\right)\left(1-\prod_{\substack{p\equiv1\pmod{4}\\p\ge13}}\left(1-\frac{2}{p^2}\right)\right)+\frac{2}{25}\left(1-\prod_{p\ne2,5}\left(1-\frac{1}{p^2}\right)\right)+o(1) N ∣ A ∣ ≤ ( 25 23 ) 1 − p ≡ 1 ( mod 4 ) p ≥ 13 ∏ ( 1 − p 2 2 ) + 25 2 1 − p = 2 , 5 ∏ ( 1 − p 2 1 ) + o ( 1 ) ≤ 0.0252 + 0.0125 = 0.0377 < 0.04 − η ; \le0.0252+0.0125=0.0377<0.04-\eta; ≤ 0.0252 + 0.0125 = 0.0377 < 0.04 − η ; here we have absorbed the o ( 1 ) o(1) o ( 1 ) N N N ∣ A ∣ ≥ ( 1 / 25 − η ) N |A|\ge(1/25-\eta)N ∣ A ∣ ≥ ( 1/25 − η ) N
We have now reduced to the case that A ∗ A^* A ∗ A ∗ A^* A ∗
∣ A ∗ ∣ N ≤ ( 1 2 ) ( 23 25 ) ( 1 − ∏ p ≡ 1 ( m o d 4 ) p ≥ 13 ( 1 − 2 p 2 ) ) + o ( 1 ) . \frac{|A^*|}{N}\le\left(\frac{1}{2}\right)\left(\frac{23}{25}\right)\left(1-\prod_{\substack{p\equiv1\pmod{4}\\p\ge13}}\left(1-\frac{2}{p^2}\right)\right)+o(1). N ∣ A ∗ ∣ ≤ ( 2 1 ) ( 25 23 ) 1 − p ≡ 1 ( mod 4 ) p ≥ 13 ∏ ( 1 − p 2 2 ) + o ( 1 ) . We now split further into cases based on whether or not A 7 ∪ A 18 A_7\cup A_{18} A 7 ∪ A 18 A 7 ∪ A 18 A_7\cup A_{18} A 7 ∪ A 18 A 7 ⊆ { 7 , 57 } ( m o d 100 ) A_7\subseteq\{7,57\}\pmod{100} A 7 ⊆ { 7 , 57 } ( mod 100 ) A 18 ⊆ { 43 , 93 } ( m o d 100 ) A_{18}\subseteq\{43,93\}\pmod{100} A 18 ⊆ { 43 , 93 } ( mod 100 ) b ∈ A ∗ b\in A^* b ∈ A ∗ A 7 A_7 A 7 A 18 A_{18} A 18 2 2 2^2 2 2
∣ A 7 ∪ A 18 ∣ N ≤ 1 50 + ( 1 50 ) ( 1 − ∏ p ≠ 2 , 5 ( 1 − 1 p 2 ) ) + o ( 1 ) . \frac{|A_7\cup A_{18}|}{N}\le\frac{1}{50}+\left(\frac{1}{50}\right)\left(1-\prod_{p\ne2,5}\left(1-\frac{1}{p^2}\right)\right)+o(1). N ∣ A 7 ∪ A 18 ∣ ≤ 50 1 + ( 50 1 ) 1 − p = 2 , 5 ∏ ( 1 − p 2 1 ) + o ( 1 ) . This gives another contradiction as
∣ A ∣ N ≤ ( 1 2 ) ( 23 25 ) ( 1 − ∏ p ≡ 1 ( m o d 4 ) p ≥ 13 ( 1 − 2 p 2 ) ) + 1 50 + ( 1 50 ) ( 1 − ∏ p ≠ 2 , 5 ( 1 − 1 p 2 ) ) + o ( 1 ) \frac{|A|}{N}\le\left(\frac{1}{2}\right)\left(\frac{23}{25}\right)\left(1-\prod_{\substack{p\equiv1\pmod{4}\\p\ge13}}\left(1-\frac{2}{p^2}\right)\right)+\frac{1}{50}+\left(\frac{1}{50}\right)\left(1-\prod_{p\ne2,5}\left(1-\frac{1}{p^2}\right)\right)+o(1) N ∣ A ∣ ≤ ( 2 1 ) ( 25 23 ) 1 − p ≡ 1 ( mod 4 ) p ≥ 13 ∏ ( 1 − p 2 2 ) + 50 1 + ( 50 1 ) 1 − p = 2 , 5 ∏ ( 1 − p 2 1 ) + o ( 1 ) ≤ 0.0126 + 0.0200 + 0.0032 = 0.0358 < 0.04 − η . \le0.0126+0.0200+0.0032=0.0358<0.04-\eta. ≤ 0.0126 + 0.0200 + 0.0032 = 0.0358 < 0.04 − η . Therefore we have reduced to the case where A ∗ A^* A ∗ A 7 A_7 A 7 A 18 A_{18} A 18 b ∈ A ∗ b \in A^* b ∈ A ∗ b ′ ∈ A 7 b' \in A_7 b ′ ∈ A 7 2 ∣ b ′ 2 \mid b' 2 ∣ b ′ A 18 A_{18} A 18 ∣ A ∗ ∣ |A^*| ∣ A ∗ ∣ b b b
∣ A 7 ∣ N ≤ 1 25 ( 1 − ∏ p ≠ 5 ( 1 − 1 p 2 ) ) + o ( 1 ) . \frac{|A_7|}{N} \le\frac{1}{25}\left(1-\prod_{p\ne5}\left(1-\frac{1}{p^2}\right)\right)+o(1). N ∣ A 7 ∣ ≤ 25 1 1 − p = 5 ∏ ( 1 − p 2 1 ) + o ( 1 ) . Additionally, using b ′ ∈ A 7 b' \in A_7 b ′ ∈ A 7
∣ A 18 ∣ N ≤ 1 25 ( 1 − ∏ p ≠ 2 , 5 ( 1 − 1 p 2 ) ) + o ( 1 ) . \frac{|A_{18}|}{N} \le\frac{1}{25}\left(1-\prod_{p\ne2,5}\left(1-\frac{1}{p^2}\right)\right)+o(1). N ∣ A 18 ∣ ≤ 25 1 1 − p = 2 , 5 ∏ ( 1 − p 2 1 ) + o ( 1 ) . Therefore we have that
∣ A ∣ N ≤ ( 1 2 ) ( 23 25 ) ( 1 − ∏ p ≡ 1 ( m o d 4 ) p ≥ 13 ( 1 − 2 p 2 ) ) + 1 25 ( 1 − ∏ p ≠ 5 ( 1 − 1 p 2 ) ) + 1 25 ( 1 − ∏ p ≠ 2 , 5 ( 1 − 1 p 2 ) ) + o ( 1 ) ≤ 0.0126 + 0.0147 + 0.0063 = 0.0336 < 0.04 − η . \begin{aligned}
\frac{|A|}{N} &\le\left(\frac{1}{2}\right)\left(\frac{23}{25}\right)\left(1-\prod_{\substack{p\equiv1\pmod{4}\\p\ge13}}\left(1-\frac{2}{p^2}\right)\right)+\frac{1}{25}\left(1-\prod_{p\ne5}\left(1-\frac{1}{p^2}\right)\right)\\
&\quad+\frac{1}{25}\left(1-\prod_{p\ne2,5}\left(1-\frac{1}{p^2}\right)\right)+o(1)\\
&\le0.0126+0.0147+0.0063=0.0336<0.04-\eta.
\end{aligned} N ∣ A ∣ ≤ ( 2 1 ) ( 25 23 ) 1 − p ≡ 1 ( mod 4 ) p ≥ 13 ∏ ( 1 − p 2 2 ) + 25 1 1 − p = 5 ∏ ( 1 − p 2 1 ) + 25 1 1 − p = 2 , 5 ∏ ( 1 − p 2 1 ) + o ( 1 ) ≤ 0.0126 + 0.0147 + 0.0063 = 0.0336 < 0.04 − η . Therefore we have now reduced to the case where A ∗ = ∅ A^*=\varnothing A ∗ = ∅ A 7 A_7 A 7 A 18 A_{18} A 18 b ∈ A 7 b\in A_7 b ∈ A 7 b ′ ∈ A 18 b'\in A_{18} b ′ ∈ A 18
∣ A ∣ N ≤ 2 25 ( 1 − ∏ p ≠ 5 ( 1 − 1 p 2 ) ) + o ( 1 ) ≤ 0.0294 < 0.04 − η . \frac{|A|}{N}\le\frac{2}{25}\left(1-\prod_{p\ne5}\left(1-\frac{1}{p^2}\right)\right)+o(1)\le0.0294<0.04-\eta. N ∣ A ∣ ≤ 25 2 1 − p = 5 ∏ ( 1 − p 2 1 ) + o ( 1 ) ≤ 0.0294 < 0.04 − η . Therefore A A A A 7 A_7 A 7 A 18 A_{18} A 18
We record various numerical constants to 4 digits, rounding to preserve the stated inequalities.
∏ p ( 1 − 1 p 2 ) = 6 π 2 ≥ 0.6079 \prod_p\left(1-\frac{1}{p^2}\right)=\frac{6}{\pi^2}\ge0.6079 ∏ p ( 1 − p 2 1 ) = π 2 6 ≥ 0.6079 1 − 6 π 2 ≤ 0.3921 1-\frac{6}{\pi^2}\le0.3921 1 − π 2 6 ≤ 0.3921
∏ p ≠ 5 ( 1 − 1 p 2 ) = 25 4 π 2 ≥ 0.6332 \prod_{p\ne5}\left(1-\frac{1}{p^2}\right)=\frac{25}{4\pi^2}\ge0.6332 ∏ p = 5 ( 1 − p 2 1 ) = 4 π 2 25 ≥ 0.6332 1 − 25 4 π 2 ≤ 0.3668 1-\frac{25}{4\pi^2}\le0.3668 1 − 4 π 2 25 ≤ 0.3668
∏ p ≠ 2 , 5 ( 1 − 1 p 2 ) = 25 3 π 2 ≥ 0.8443 \prod_{p\ne2,5}\left(1-\frac{1}{p^2}\right)=\frac{25}{3\pi^2}\ge0.8443 ∏ p = 2 , 5 ( 1 − p 2 1 ) = 3 π 2 25 ≥ 0.8443 1 − 25 3 π 2 ≤ 0.1557 1-\frac{25}{3\pi^2}\le0.1557 1 − 3 π 2 25 ≤ 0.1557
1 − ∏ p ≡ 1 ( m o d 4 ) p ≥ 13 ( 1 − 2 p 2 ) ≤ 0.0274 1-\prod_{\substack{p\equiv1\pmod{4}\\p\ge13}}\left(1-\frac{2}{p^2}\right)\le0.0274 1 − ∏ p ≡ 1 ( mod 4 ) p ≥ 13 ( 1 − p 2 2 ) ≤ 0.0274
New online algorithms lower bounds – Christian Coester
This section describes how GPT-5 assisted in proving new bounds on the performance of algorithms for a fundamental problem in online optimization called convex body chasing . The problem was introduced by [45 ] and captures how to make sequential decisions under uncertainty when each step restricts feasible options in a geometric way. Formally, one is given a sequence of convex sets K 1 , K 2 , … , K T ⊂ R d K_1, K_2, \ldots, K_T \subset\mathbb{R}^d K 1 , K 2 , … , K T ⊂ R d K t K_t K t p t ∈ K t p_t \in K_t p t ∈ K t ∑ t = 1 T ∥ p t − p t − 1 ∥ \sum_{t=1}^{T} \lVert p_t-p_{t-1}\rVert ∑ t = 1 T ∥ p t − p t − 1 ∥ online as it has to make decisions on the fly, without knowledge of the future. In contrast, an offline algorithm is one that knows the entire input upfront, allowing it to move optimally. The standard measure of performance of an online algorithm is its competitive ratio , defined as the worst-case (i.e., maximum) ratio between the online algorithm’s cost and the optimum (offline) cost on any instance.
Intuitively, convex body chasing formalizes the tension between adaptivity and stability: each new convex body represents a changing feasible region (e.g., updated constraints, new data, or evolving system state), and the online algorithm must adjust with minimal movement. It arises naturally in contexts like control theory (tracking feasible states), machine learning (online convex optimization with movement costs), and operations research (dynamic resource allocation). In theoretical computer science, it serves as a clean abstraction for understanding how well an online decision process can perform relative to an omniscient optimum, with close connections to other central problems in online optimization [18 ] .
When introducing the problem, [45 ] showed that the competitive ratio of any online algorithm is at least d \sqrt{d} d d d d [19 , 5 , 62 ] ), it is now established that the optimal competitive ratio lies in the interval [ d , d ] [\sqrt{d},d] [ d , d ] nested convex bodies, where K 1 ⊃ K 2 ⊃ ⋯ ⊃ K T K_1 \supset K_2 \supset\cdots\supset K_T K 1 ⊃ K 2 ⊃ ⋯ ⊃ K T [7 , 4 , 18 ] ).
The following sections show how GPT-5 refuted competitiveness of a natural algorithm in response to a single prompt, and how it assisted in raising the lower bound on the competitive ratio of arbitrary online algorithms from d \sqrt{d} d π 2 ⌊ d 2 ⌋ ≈ 1.11 d \frac{\pi}{2}\sqrt{\left\lfloor\frac{d}{2}\right\rfloor} \approx1.11\sqrt{d} 2 π ⌊ 2 d ⌋ ≈ 1.11 d
Lower bound for Follow-the-Leader
When the initial location is p 0 = 0 p_0=0 p 0 = 0 nested convex body chasing, the optimal offline algorithm will move to arg min p ∈ K T ∥ p ∥ \arg\min_{p\in K_T}\lVert p\rVert arg min p ∈ K T ∥ p ∥ p t = arg min p ∈ K t ∥ p ∥ p_t=\arg\min_{p\in K_t}\lVert p\rVert p t = arg min p ∈ K t ∥ p ∥ t t t π / 2 \pi/2 π /2 π / 2 \pi/2 π /2
Consider nested convex body chasing in 2-dimensional Euclidean space, where the initial location is 0 0 0 0 0 0 π / 2 \pi/2 π /2
[Reasoned for 16m 19s]
Short answer: the statement is false .
[...]
Two basic facts about the greedy minimizer
Let K 0 ⊇ K 1 ⊇ K 2 ⊇ ⋯ ⊂ R 2 K_0 \supseteq K_1 \supseteq K_2 \supseteq\cdots\subset\mathbb{R}^2 K 0 ⊇ K 1 ⊇ K 2 ⊇ ⋯ ⊂ R 2 K 0 K_0 K 0 0 0 0
Let p t ∈ K t p_t \in K_t p t ∈ K t ∥ x ∥ \lVert x\rVert ∥ x ∥ K t K_t K t K t K_t K t
Write p t = r t u t p_t = r_tu_t p t = r t u t r t = ∥ p t ∥ ∈ [ 0 , 1 ] r_t = \lVert p_t\rVert\in[0,1] r t = ∥ p t ∥ ∈ [ 0 , 1 ] u t ∈ S 1 u_t \in S^1 u t ∈ S 1
Two standard properties hold:
Half-space support: for every t t t
K t ⊂ H ( p t ) : = { x : ⟨ p t , x ⟩ ≥ r t 2 } . K_t \subset H(p_t) := \{x : \langle p_t,x\rangle\ge r_t^2\}. K t ⊂ H ( p t ) := { x : ⟨ p t , x ⟩ ≥ r t 2 } . Equivalently, for all s ≥ t s \ge t s ≥ t
⟨ p t , p s ⟩ ≥ r t 2 ⟺ cos ( ∠ ( u t , u s ) ) ≥ r t r s . (31) \langle p_t,p_s\rangle\ge r_t^2 \quad\Longleftrightarrow\quad\cos(\angle(u_t,u_s)) \ge\frac{r_t}{r_s}. \tag*{(31)} ⟨ p t , p s ⟩ ≥ r t 2 ⟺ cos ( ∠ ( u t , u s )) ≥ r s r t . (31) (This is just the normal cone condition at a Euclidean projection.)
Realizability: conversely, if a sequence ( p t ) (p_t) ( p t ) r t r_t r t s ≥ t s \ge t s ≥ t
K t : = B ( 0 , 1 ) ‾ ∩ ⋂ i ≤ t H ( p i ) K_t := \overline{B(0,1)} \cap\bigcap_{i\le t} H(p_i) K t := B ( 0 , 1 ) ∩ i ≤ t ⋂ H ( p i ) yields a nested sequence for which the unique minimum-norm point in K t K_t K t p t p_t p t p t ∈ K t p_t \in K_t p t ∈ K t H ( p t ) H(p_t) H ( p t ) < r t < r_t < r t
Thus, to disprove any universal upper bound on the greedy cost, it suffices to exhibit a sequence ( p t ) ⊂ B ( 0 , 1 ) (p_t) \subset B(0,1) ( p t ) ⊂ B ( 0 , 1 ) ∑ t ∥ p t + 1 − p t ∥ \sum_t \lVert p_{t+1}-p_t\rVert ∑ t ∥ p t + 1 − p t ∥
An alternating construction
Pick a nonincreasing sequence of angles { φ m } m ≥ 1 \{\varphi_m\}_{m\ge1} { φ m } m ≥ 1
0 < φ m + 1 ≤ φ m < π 8 , ∑ m = 1 ∞ φ m = ∞ , ∑ m = 1 ∞ φ m 2 < ∞ . 0 < \varphi_{m+1} \le\varphi_m < \frac{\pi}{8}, \qquad\sum_{m=1}^{\infty} \varphi_m = \infty, \qquad\sum_{m=1}^{\infty} \varphi_m^2 < \infty. 0 < φ m + 1 ≤ φ m < 8 π , m = 1 ∑ ∞ φ m = ∞ , m = 1 ∑ ∞ φ m 2 < ∞. (For instance, φ m = c m \varphi_m = \frac{c}{m} φ m = m c c ∈ ( 0 , π / 8 ) c \in(0,\pi/8) c ∈ ( 0 , π /8 ) r 1 > 0 r_1 > 0 r 1 > 0
p 2 m − 1 = r 2 m − 1 ( cos ( − φ m ) , sin ( − φ m ) ) , p_{2m-1} = r_{2m-1}\left(\cos(-\varphi_m),\sin(-\varphi_m)\right), p 2 m − 1 = r 2 m − 1 ( cos ( − φ m ) , sin ( − φ m ) ) , p 2 m = r 2 m ( cos ( φ m ) , sin ( φ m ) ) , p_{2m} = r_{2m}\left(\cos(\varphi_m),\sin(\varphi_m)\right), p 2 m = r 2 m ( cos ( φ m ) , sin ( φ m ) ) , with radii chosen recursively so that the immediately preceding constraint in (31) is tight:
r 2 m = r 2 m − 1 cos ( φ m + φ m ) = r 2 m − 1 cos ( 2 φ m ) , r 2 m + 1 = r 2 m cos ( φ m + φ m + 1 ) . (32) \begin{aligned}
r_{2m} &= \frac{r_{2m-1}}{\cos(\varphi_m+\varphi_m)} = \frac{r_{2m-1}}{\cos(2\varphi_m)},\\
r_{2m+1} &= \frac{r_{2m}}{\cos(\varphi_m+\varphi_{m+1})}. \tag*{(32)}
\end{aligned} r 2 m r 2 m + 1 = cos ( φ m + φ m ) r 2 m − 1 = cos ( 2 φ m ) r 2 m − 1 , = cos ( φ m + φ m + 1 ) r 2 m . (32) (Geometrically, each new point lies on the boundary of the supporting half-space defined by the previous minimizer, with the largest possible angular change allowed by that one constraint.)
[...]
Figure IV.3 contains an illustration of the construction. GPT goes on to justify the feasibility of the construction: “We must verify that (1) holds not only for the immediate predecessor but for all earlier indices.” The argument it gives for this is actually incorrect, but a correct argument is easy to see: For s ≥ t + 1 s \ge t+1 s ≥ t + 1
Figure IV.3. Beginning of a hard instance for the follow-the-leader algorithm. Figure made by GPT-5.
cos ( ∠ ( u t , u s ) ) ≥ cos ( ∠ ( u t , u t + 1 ) ) = r t r t + 1 ≥ r t r s \cos(\angle(u_t,u_s)) \ge\cos(\angle(u_t,u_{t+1})) = \frac{r_t}{r_{t+1}} \ge\frac{r_t}{r_s} cos ( ∠ ( u t , u s )) ≥ cos ( ∠ ( u t , u t + 1 )) = r t + 1 r t ≥ r s r t GPT completes the proof (within the same response) as shown in the next chat box. The first line in it contains a slight inaccuracy, since we need the inequality not only for ∣ x ∣ ≤ π / 8 \lvert x\rvert\le\pi/8 ∣ x ∣ ≤ π /8 ∣ x ∣ ≤ π / 4 \lvert x\rvert\le\pi/4 ∣ x ∣ ≤ π /4
Finally, because ∑ φ m 2 < ∞ \sum\varphi_m^2 < \infty ∑ φ m 2 < ∞ 1 / cos x ≤ exp ( x 2 ) 1/\cos x \le\exp(x^2) 1/ cos x ≤ exp ( x 2 ) ∣ x ∣ ≤ π / 8 \lvert x\rvert\le\pi/8 ∣ x ∣ ≤ π /8
sup t r t ≤ r 1 ⋅ exp ( C ∑ m ≥ 1 φ m 2 ) < ∞ . \sup_t r_t \le r_1 \cdot\exp\left(C\sum_{m\ge1}\varphi_m^2\right) < \infty. t sup r t ≤ r 1 ⋅ exp ( C m ≥ 1 ∑ φ m 2 ) < ∞. Choosing r 1 > 0 r_1 > 0 r 1 > 0 r t ≤ 1 r_t \le1 r t ≤ 1 t t t p t p_t p t
The cost explodes
The distance traveled in the two types of steps is explicit, because in (2) the immediately previous constraint is tight:
∥ p 2 m − p 2 m − 1 ∥ = r 2 m 2 − r 2 m − 1 2 = r 2 m − 1 tan ( 2 φ m ) . \lVert p_{2m} - p_{2m-1}\rVert= \sqrt{r_{2m}^2-r_{2m-1}^2} = r_{2m-1}\tan(2\varphi_m). ∥ p 2 m − p 2 m − 1 ∥ = r 2 m 2 − r 2 m − 1 2 = r 2 m − 1 tan ( 2 φ m ) . ∥ p 2 m + 1 − p 2 m ∥ = r 2 m + 1 2 − r 2 m 2 = r 2 m tan ( φ m + φ m + 1 ) . \lVert p_{2m+1} - p_{2m}\rVert= \sqrt{r_{2m+1}^2-r_{2m}^2} = r_{2m}\tan(\varphi_m+\varphi_{m+1}). ∥ p 2 m + 1 − p 2 m ∥ = r 2 m + 1 2 − r 2 m 2 = r 2 m tan ( φ m + φ m + 1 ) . Therefore, after N N N
L N = ∑ m = 1 N ( r 2 m − 1 tan ( 2 φ m ) + r 2 m tan ( φ m + φ m + 1 ) ) ≥ ∑ m = 1 N ( 2 r 2 m − 1 φ m + r 2 m ( φ m + φ m + 1 ) ) ( since tan x ≥ x for x ∈ [ 0 , π / 2 ] ) ≥ c ∑ m = 1 N φ m \begin{aligned}
L_N = \sum_{m=1}^{N}\left(r_{2m-1}\tan(2\varphi_m)+r_{2m}\tan(\varphi_m+\varphi_{m+1})\right) \\
&\ge\sum_{m=1}^{N}\left(2r_{2m-1}\varphi_m+r_{2m}(\varphi_m+\varphi_{m+1})\right) \qquad(\text{since } \tan x \ge x \text{ for } x \in[0,\pi/2]) \\
&\ge c\sum_{m=1}^{N}\varphi_m
\end{aligned} L N = m = 1 ∑ N ( r 2 m − 1 tan ( 2 φ m ) + r 2 m tan ( φ m + φ m + 1 ) ) ≥ m = 1 ∑ N ( 2 r 2 m − 1 φ m + r 2 m ( φ m + φ m + 1 ) ) ( since tan x ≥ x for x ∈ [ 0 , π /2 ]) ≥ c m = 1 ∑ N φ m for some constant c > 0 c>0 c > 0 r 1 r_1 r 1 { φ m } \{\varphi_m\} { φ m } ∑ m φ m = ∞ \sum_m\varphi_m=\infty ∑ m φ m = ∞ L N → ∞ L_N\to\infty L N → ∞ N → ∞ N\to\infty N → ∞
Thus the “follow-the-leader” cost can be made arbitrarily large while all requests stay nested inside the unit disk. The original interaction is available at [21 ] .
Improved lower bounds for general algorithms
We now discuss how GPT-5’s assistance helped in proving a better lower bound on the competitive ratio of arbitrary online algorithms. The idea for this lower bound is based on a previous 1.5358 lower bound due to [13 ] for chasing lines (i.e., where each K t K_t K t d \sqrt{d} d d = 2 d = 2 d = 2 [13 ] is rather ad-hoc, giving a small number of request lines issued in three steps, using six rational constants to specify those lines. It was acknowledged by [13 ] that 1.5358 does not seem to be tight, since a generalization of their idea, using more steps and request lines, should achieve a better bound. However, this would make their analysis even more involved.
My goal was to derive a better and cleaner lower bound using a generalization of their approach where the request line moves continuously. My initial attempts to prompt GPT-5 to help with this construction were unsuccessful: It only reproved the known 1.5358 lower bound, failing to understand why my suggested continuous generalization would be helpful. Curiously though, the reasoning trace preview mentioned at one point that it was aiming for a lower bound of π / 2 ≈ 1.5708 \pi/2 \approx1.5708 π /2 ≈ 1.5708
Lower bound idea (see Figure IV.4 ). Consider an instance with initial location ( 0 , 0 ) (0,0) ( 0 , 0 ) y = 0 y = 0 y = 0 B = ( 1 , 0 ) B = (1,0) B = ( 1 , 0 ) p t p_t p t K t K_t K t K t K_t K t 1 / 2 1/2 1/2 ( 1 / 2 , 0 ) (1/2,0) ( 1/2 , 0 ) π / 2 \pi/2 π /2 1 1 1 B B B t t t B ′ B' B ′ B B B K t K_t K t 1 1 1 p t p_t p t B B ′ BB' B B ′ x ∈ B B ′ ‾ x \in\overline{BB'} x ∈ B B ′ ∥ x ∥ \lVert x\rVert ∥ x ∥ p t p_t p t B B ′ ‾ \overline{BB'} B B ′ B ′ B' B ′ B B B p t p_t p t p t p_t p t 1 1 1 ≥ π / 2 \geq\pi/2 ≥ π /2
Figure IV.4. Illustration of π / 2 \pi/2 π /2
The task for GPT-5. A subtle difficulty in the preceding argument lies in defining the precise moment of pivot switching. If the switch occurs exactly when the algorithm crosses below the semicircle, then at that moment the algorithm is still located on the semicircle, at the location where it intersects the new semicircle. But then it might also try to go below the new semicircle immediately, leading to a possibly infinite sequence of pivot switches at the same instant, effectively freezing the process.
My initial plan for resolving this was to argue that the algorithm must move at a temporarily faster speed in order to go below the semicircle, allowing the pivot switch to be delayed until its speed decreases again; but formalizing this approach seemed technically cumbersome. GPT-5 suggested a much simpler and cleaner solution: trigger the switch once the algorithm is below the semicircle at distance ≥ ϵ \ge\epsilon ≥ ϵ p t p_t p t ϵ → 0 \epsilon\to0 ϵ → 0 π / 2 \pi/2 π /2 Figure IV.5 .
Figure IV.5. An excerpt from GPT-5 Pro’s response, in which it suggested the introduction of a small parameter ε \varepsilon ε
Extension to higher dimension. Finally, I was hoping to extend the lower bound to arbitrary dimension. My own approach (so far unsuccessful) was to first aim for an extension of the above idea to three dimensions, but it does not seem obvious how to do this. I posed the problem to GPT-5, with a prompt that included the lower bound idea above (without the ϵ \epsilon ϵ
[...] Try to generalize this idea for d = 2 d = 2 d = 2 d \sqrt{d} d d d d
Within its first response, GPT-5 proposed an approach yielding a lower bound of π 2 ⌊ d 2 ⌋ ≈ 1.11 d \frac{\pi}{2}\sqrt{\left\lfloor\frac{d}{2}\right\rfloor} \approx1.11\sqrt{d} 2 π ⌊ 2 d ⌋ ≈ 1.11 d R d \mathbb{R}^{d} R d ⌊ d / 2 ⌋ \left\lfloor d/2\right\rfloor ⌊ d /2 ⌋ R 2 \mathbb{R}^{2} R 2 π / 2 \pi/2 π /2 [22 ] (containing also subsequent unsuccessful attempts to push the ratio further). Notably, my initial focus on three dimensions might have caused me to miss this simple decomposition idea.
Reflection
GPT-5 assisted the research process on this problem at several stages. Perhaps the most impressive part is its proof refuting the follow-the-leader algorithm, produced from a single [Reasoned for 10m 38s]
[...]
A robust fix: ε \varepsilon ε
Pick any ε ∈ ( 0 , 1 ) \varepsilon\in(0,1) ε ∈ ( 0 , 1 )
u P ( ϕ ) = t ( ϕ ) − cos ϕ ≥ − ε . u_P(\phi) = t(\phi) - \cos\phi\ge-\varepsilon. u P ( ϕ ) = t ( ϕ ) − cos ϕ ≥ − ε . When the first time τ \tau τ u P ( τ ) = − ε u_P(\tau) = -\varepsilon u P ( τ ) = − ε switch pivot to Q : = Q := Q := L τ L_\tau L τ , and reverse the rotation direction.
u Q ( ψ ) : = s ( ψ ) − cos ψ ≥ − ε u_Q(\psi) := s(\psi) - \cos\psi\ge-\varepsilon u Q ( ψ ) := s ( ψ ) − cos ψ ≥ − ε (here s ( ψ ) s(\psi) s ( ψ ) Q Q Q u Q = − ε u_Q = -\varepsilon u Q = − ε
Key geometry at a switch
At the switch time τ \tau τ P P P Q Q Q M M M [ P , Q ] [P,Q] [ P , Q ] p ∗ ( τ ) p^*(\tau) p ∗ ( τ ) L τ L_\tau L τ u P ( τ ) = − ε u_P(\tau) = -\varepsilon u P ( τ ) = − ε ε \varepsilon ε closer to P P P M M M L τ L_\tau L τ Q Q Q ε \varepsilon ε farther than M M M
u Q ( τ ) = + ε . u_Q(\tau) = +\varepsilon. u Q ( τ ) = + ε . So after every switch we begin strictly outside the new Thales circle, with a uniform gap ε \varepsilon ε
[...]
prompt without any guidance on how to approach the task. This result saved significant time I might have otherwise spent unsuccessfully searching for an upper bound, and it is unclear if I would have been able to come up with the lower bound, which contained several non-trivial steps. The aid it provided for the lower bound on arbitrary algorithms was on the one hand the inspiring appearance of the number π / 2 \pi/2 π /2 some solution to the freezing myself, it would have taken longer to formalize and likely been less elegant. GPT-5’s extension to arbitrary dimension was simple, but beyond the scope of my approach of focusing on three dimensions first.
Naturally, several limitations are also still apparent: it initially failed to understand how viewing the problem in continuous time could yield stronger bounds, and upon later attempts to push the lower bounds further, it presented arguments containing serious flaws. In the proofs presented in this section, GPT-5’s responses also contained some errors, but these were easy to fix for a human, overall accelerating the research process.
Inequalities on subgraph counts in trees – Sébastien Bubeck, Mark Sellke, and Steven Yin
We conduct here another experiment, this time in graph theory, on a problem formulated and studied by the first author of this section, together with Nati Linial, in a 2013 preprint that was later published in 2016 in the Journal of Graph Theory ([20 ] ). The paper lists 7 open problems, and the one we consider here is the first open problem, arguably the one we were most interested in solving back in 2013:
The follow-up paper [17 ] shows that, except possibly for the first two, none of these inequalities are actually true. Moreover the main result of [17 ] is a proof that the first inequality is indeed true. The second inequality is left as an open problem.
Aided by somewhat sophisticated scaffolding (aimed at doing mathematics research), GPT-5 was able to reprove the first inequality, and then build on this to also prove the second (open) inequality. Both of GPT-5’s proofs are quite different from any of the arguments in [20 , 17 ] . We present the re-proof of the first inequality in Section IV.3.2 and the new proof of the second in Section IV.3.3.
Both proofs below are entirely AI-generated, aside from minor editing for clarity. No human input was needed beyond first asking to prove the first inequality, and then (with the solution to the first inequality in context) asking to prove the second inequality. (A few incorrect proofs were also generated and rejected by human checking.) See Figure IV.7 and Figure IV.8 for the precise prompts.
Figure IV.7. Question posed to (a scaffolded version of) GPT-5 to reprove Theorem IV.3.1. The answer it provided is written in Section IV.3.2, with minor edits for clarity.
Figure IV.8. Question posed to (a scaffolded version of) GPT-5 to prove Conjecture 1. The answer it provided is written in Section IV.3.3, with minor edits for clarity.
We moreover note that GPT-5 did not have access to web search in this experiment, and did not seem to be aware of the existing proof in [17 ] (indeed the latter paper has remained quite niche, with only 3 citations in 10 years). The pre-existing (human-generated) proof rests on a somewhat cumbersome analysis, that analyzes several different cases across 4 pages of calculation. In contrast GPT-5’s proof is short and elegant, and based on a somewhat miraculous identity. The entire argument in Section IV.3.2 fits in a page and a half.
We note that we first attempted to use this research scaffolding on top of GPT-5 to solve a few Erdős problems, but with no success. The experiment described here was our first attempt at a problem outside of Erdős’ problems.
Paths, stars and wyes in trees
Let G G G Figure IV.6 ):
Figure IV.6. The three isomorphism types of trees with 5 vertices. TikZ figure produced by GPT-5.
Theorem IV.3.1 ([17 ] ). For any finite tree G G G one has
Y − 9 S − P ≤ 6. (33) Y - 9S - P \le6. \tag*{(33)} Y − 9 S − P ≤ 6. (33) Conjecture 1 ([20 , 17 ] ). There exists some constant K > 0 K > 0 K > 0 such that for any finite tree G G G one has
29 Y − 42 P − 144 S ≤ K . (34) 29Y - 42P - 144S \le K. \tag*{(34)} 29 Y − 42 P − 144 S ≤ K . (34) The motivation for the above problems is that the set of achievable asymptotic proportions ( Y Y + S + P , S Y + S + P , P Y + S + P ) \left(\frac{Y}{Y+S+P},\frac{S}{Y+S+P},\frac{P}{Y+S+P}\right) ( Y + S + P Y , Y + S + P S , Y + S + P P ) Y + S + P → ∞ Y+S+P \to\infty Y + S + P → ∞ flag algebra semidefinite program of [55 ] generates all such inequalities, but that linear inequalities of the form above can still be computationally undecidable [38 ] .
In some sense Theorem IV.3.1 gives the “first face” of this convex set, while Conjecture 1 gives the “second face”. We note that Conjecture 1 was also made in [20 ] , together with an infinite set of conjectured inequalities, one for each face, and those conjectured inequalities were all disproven in [17 ] except for Conjecture 1 which we will show here to be actually true.
Aside from being checked and rewritten for clarity, the new proofs below are entirely AI-generated.
New proof of Theorem IV.3.1
For v ∈ G v \in G v ∈ G d v d_v d v v v v
x v = d v − 1 , T v = ∑ u ∼ v x u , A v = A d v = ( d v − 1 2 ) . x_v=d_v-1,\qquad T_v=\sum_{u\sim v}x_u,\qquad A_v=A_{d_v}=\binom{d_v-1}{2}. x v = d v − 1 , T v = u ∼ v ∑ x u , A v = A d v = ( 2 d v − 1 ) . It is not hard to see that:
S = ∑ v ∈ G ( d v 4 ) , P = 1 2 ∑ v ∈ G [ T v 2 − ∑ u ∼ v x u 2 ] , Y = ∑ v ∈ G A v T v . S=\sum_{v\in G}\binom{d_v}{4},\qquad P=\frac{1}{2}\sum_{v\in G}\left[T_v^2-\sum_{u\sim v}x_u^2\right],\qquad Y=\sum_{v\in G}A_vT_v. S = v ∈ G ∑ ( 4 d v ) , P = 2 1 v ∈ G ∑ [ T v 2 − u ∼ v ∑ x u 2 ] , Y = v ∈ G ∑ A v T v . Indeed the value of S S S P P P v v v T v 2 T_v^2 T v 2 v v v ∑ u ∼ v x u 2 \sum_{u\sim v}x_u^2 ∑ u ∼ v x u 2 u ∼ v u\sim v u ∼ v T v 2 − ∑ u ∼ v x u 2 T_v^2-\sum_{u\sim v}x_u^2 T v 2 − ∑ u ∼ v x u 2 P P P Y Y Y v v v T v T_v T v v v v Y Y Y d v − 1 d_v-1 d v − 1 v v v A d v A_{d_v} A d v
The first key claim is the identity
Y − 9 S − P = ∑ v ∈ G [ F ( d v ) − ( T v − A v ) 2 2 ] (35) Y-9S-P=\sum_{v\in G}\left[F(d_v)-\frac{(T_v-A_v)^2}{2}\right] \tag*{(35)} Y − 9 S − P = v ∈ G ∑ [ F ( d v ) − 2 ( T v − A v ) 2 ] (35) where F F F
F ( k ) = ( k − 1 2 ) 2 2 + k ( k − 1 ) 2 2 − 9 ( k 4 ) . F(k)=\frac{\binom{k-1}{2}^2}{2}+\frac{k(k-1)^2}{2}-9\binom{k}{4}. F ( k ) = 2 ( 2 k − 1 ) 2 + 2 k ( k − 1 ) 2 − 9 ( 4 k ) . To prove (35) , firstly note that 9 ( d v 4 ) = 9 S v 9\binom{d_v}{4}=9S_v 9 ( 4 d v ) = 9 S v T v 2 / 2 T_v^2/2 T v 2 /2 A v T v A_vT_v A v T v
1 2 ∑ v ∑ u ∼ v x u 2 = ? ( x v 2 ) 2 2 + d v x v 2 2 − A v 2 2 . \frac{1}{2}\sum_v\sum_{u\sim v}x_u^2\stackrel{?}{=}\frac{\binom{x_v}{2}^2}{2}+\frac{d_vx_v^2}{2}-\frac{A_v^2}{2}. 2 1 v ∑ u ∼ v ∑ x u 2 = ? 2 ( 2 x v ) 2 + 2 d v x v 2 − 2 A v 2 . In fact, by definition ( x v 2 ) 2 = A v 2 \binom{x_v}{2}^2=A_v^2 ( 2 x v ) 2 = A v 2
∑ v ∑ u ∼ v x u 2 = ∑ u ∑ v ∼ u x u 2 = ∑ u d u x u 2 = ∑ v d v x v 2 . \sum_v\sum_{u\sim v}x_u^2=\sum_u\sum_{v\sim u}x_u^2=\sum_u d_ux_u^2=\sum_v d_vx_v^2. v ∑ u ∼ v ∑ x u 2 = u ∑ v ∼ u ∑ x u 2 = u ∑ d u x u 2 = v ∑ d v x v 2 . This establishes (35) .
Next, we have the miraculous identity for each v v v
− ( T v − A v ) 2 2 + F ( d v ) = 5 2 ∑ u ∼ v ( d v − d u ) − 3 ( d v − 2 ) − d v ( d v − 1 ) ( d v − 3 ) ( d v − 4 ) 4 ( T v − A v − 2 2 ) . (36) -\frac{(T_v-A_v)^2}{2}+F(d_v)=\frac{5}{2}\sum_{u\sim v}(d_v-d_u)-3(d_v-2)-\frac{d_v(d_v-1)(d_v-3)(d_v-4)}{4}\binom{T_v-A_v-2}{2}. \tag*{(36)} − 2 ( T v − A v ) 2 + F ( d v ) = 2 5 u ∼ v ∑ ( d v − d u ) − 3 ( d v − 2 ) − 4 d v ( d v − 1 ) ( d v − 3 ) ( d v − 4 ) ( 2 T v − A v − 2 ) . (36) Given (36) the rest of the proof is immediate. Indeed the first term on the right-hand side telescopes when summing over v ∈ G v\in G v ∈ G ( d v − d u ) (d_v-d_u) ( d v − d u ) − 1 -1 − 1 1 (36) implies that
Y − 9 S − P ≤ 6 ∣ V ( G ) ∣ − 3 ∑ v ∈ G d v = 6 Y - 9S - P \le6|V(G)| - 3\sum_{v\in G} d_v = 6 Y − 9 S − P ≤ 6∣ V ( G ) ∣ − 3 v ∈ G ∑ d v = 6 where the latter equality holds for all trees.
To verify (36) , we first compute (see e.g. Wolfram Alpha ) that
F ( k ) + k ( k − 1 ) ( k − 3 ) ( k − 4 ) 4 = 5 k 2 4 − 7 k 4 + 1 2 F(k) + \frac{k(k-1)(k-3)(k-4)}{4} = \frac{5k^2}{4} - \frac{7k}{4} + \frac{1}{2} F ( k ) + 4 k ( k − 1 ) ( k − 3 ) ( k − 4 ) = 4 5 k 2 − 4 7 k + 2 1 Meanwhile
( T v − A v ) 2 2 − ( T v − A v − 2 2 ) = 5 2 ( T v − A v ) − 3. \frac{(T_v-A_v)^2}{2} - \left(\frac{T_v-A_v-2}{2}\right) = \frac{5}{2}(T_v-A_v)-3. 2 ( T v − A v ) 2 − ( 2 T v − A v − 2 ) = 2 5 ( T v − A v ) − 3. Finally
∑ u ∼ v ( d v − d u ) = d v 2 − ∑ u ∼ v ( x u + 1 ) = d v 2 − T v − d v . \sum_{u\sim v}(d_v-d_u)=d_v^2-\sum_{u\sim v}(x_u+1)=d_v^2-T_v-d_v. u ∼ v ∑ ( d v − d u ) = d v 2 − u ∼ v ∑ ( x u + 1 ) = d v 2 − T v − d v . Thus (36) is equivalent to:
5 d v 2 4 − 7 d v 4 + 1 2 ? = 5 2 ( d v 2 − T v − d v ) − 3 d v + 6 + 5 2 ( T v − A v ) − 3. \frac{5d_v^2}{4}-\frac{7d_v}{4}+\frac{1}{2} \mathrel{?=} \frac{5}{2}(d_v^2-T_v-d_v)-3d_v+6+\frac{5}{2}(T_v-A_v)-3. 4 5 d v 2 − 4 7 d v + 2 1 ? = 2 5 ( d v 2 − T v − d v ) − 3 d v + 6 + 2 5 ( T v − A v ) − 3. The T v T_v T v A v A_v A v
− 5 d v 2 4 + 15 d v 4 − 5 2 ? = − 5 A v 2 . (37) -\frac{5d_v^2}{4}+\frac{15d_v}{4}-\frac{5}{2} \mathrel{?=} -\frac{5A_v}{2}. \tag*{(37)} − 4 5 d v 2 + 4 15 d v − 2 5 ? = − 2 5 A v . (37) Finally, expanding
A v = ( d v − 1 2 ) = d v 2 − 3 d v + 2 2 A_v=\binom{d_v-1}{2}=\frac{d_v^2-3d_v+2}{2} A v = ( 2 d v − 1 ) = 2 d v 2 − 3 d v + 2 one finds that all terms in (37) match. This completes the proof.
Proof of Conjecture 1
As mentioned above, the work [20 ] actually made an infinite sequence of conjectures parametrized by a positive integer C ≥ 2 C\ge2 C ≥ 2
C ( 2 C 2 + 7 C + 7 ) Y − C ( C + 1 ) ( C 2 + 3 C + 4 ) P − 24 ( C + 1 ) ( C + 2 ) S ≤ O ( 1 ) C(2C^2+7C+7)Y-C(C+1)(C^2+3C+4)P-24(C+1)(C+2)S\le O(1) C ( 2 C 2 + 7 C + 7 ) Y − C ( C + 1 ) ( C 2 + 3 C + 4 ) P − 24 ( C + 1 ) ( C + 2 ) S ≤ O ( 1 ) for any fixed C C C C = 1 C=1 C = 1 [20 ] actually disproved this for C ≥ 3 C\ge3 C ≥ 3 C = 2 C=2 C = 2
Theorem IV.3.2. For any tree G G G
29 Y − 42 P − 144 S ≤ 504. 29Y-42P-144S\le504. 29 Y − 42 P − 144 S ≤ 504. [Theorem IV.3.1 and Section IV.3.2 are put in context.]
More generally, we can state the inequality above like so
C ( C + 1 ) ( C 2 + 3 C + 4 ) P + 24 ( C + 1 ) ( C + 2 ) S − C ( 2 C 2 + 7 C + 7 ) Y ≥ cst C(C+1)(C^2+3C+4)P+24(C+1)(C+2)S-C(2C^2+7C+7)Y \ge\text{cst} C ( C + 1 ) ( C 2 + 3 C + 4 ) P + 24 ( C + 1 ) ( C + 2 ) S − C ( 2 C 2 + 7 C + 7 ) Y ≥ cst where the implied constant does not depend on G G G C = 1 C=1 C = 1 C = 2 C=2 C = 2
Proof. We will double the inequality to be proved to conveniently reduce the number of non-integer terms. Using the same formulas for S S S P P P Y Y Y
E : = 58 Y − 84 P − 288 S = ∑ v ( 58 A v T v − 42 T v 2 ) + ∑ v ( 42 d v ( d v − 1 ) 2 − 288 ( d v 4 ) ) . E := 58Y - 84P - 288S = \sum_v \left(58A_vT_v - 42T_v^2\right) + \sum_v \left(42d_v(d_v-1)^2 - 288\binom{d_v}{4}\right). E := 58 Y − 84 P − 288 S = v ∑ ( 58 A v T v − 42 T v 2 ) + v ∑ ( 42 d v ( d v − 1 ) 2 − 288 ( 4 d v ) ) . We will rewrite this using the identities:
58 A v T v − 42 T v 2 = − 29 ( T v − A v ) 2 − 13 T v 2 + 29 A v 2 , 58A_vT_v - 42T_v^2 = -29(T_v-A_v)^2 - 13T_v^2 + 29A_v^2, 58 A v T v − 42 T v 2 = − 29 ( T v − A v ) 2 − 13 T v 2 + 29 A v 2 , − 29 ( T v − A v ) 2 = − 58 ( T v − A v − 3 2 ) − 203 ( T v − A v ) + 348 , -29(T_v-A_v)^2 = -58\binom{T_v-A_v-3}{2} - 203(T_v-A_v) + 348, − 29 ( T v − A v ) 2 = − 58 ( 2 T v − A v − 3 ) − 203 ( T v − A v ) + 348 , − 13 T v 2 = − 26 ( T v − 3 2 ) − 91 T v + 156. -13T_v^2 = -26\binom{T_v-3}{2} - 91T_v + 156. − 13 T v 2 = − 26 ( 2 T v − 3 ) − 91 T v + 156. Substituting gives:
E = ∑ v ( − 58 ( T v − A v − 3 2 ) − 26 ( T v − 3 2 ) − 294 T v + L ( d v ) ) . E = \sum_v \left(-58\binom{T_v-A_v-3}{2} - 26\binom{T_v-3}{2} - 294T_v + L(d_v)\right). E = v ∑ ( − 58 ( 2 T v − A v − 3 ) − 26 ( 2 T v − 3 ) − 294 T v + L ( d v ) ) . where
L ( d ) = 42 d ( d − 1 ) 2 − 288 ( d 4 ) + 29 ( d − 1 2 ) 2 + 203 ( d − 1 2 ) + 504. L(d) = 42d(d-1)^2 - 288\binom{d}{4} + 29\binom{d-1}{2}^2 + 203\binom{d-1}{2} + 504. L ( d ) = 42 d ( d − 1 ) 2 − 288 ( 4 d ) + 29 ( 2 d − 1 ) 2 + 203 ( 2 d − 1 ) + 504. Here we used 203 + 91 = 294 203+91=294 203 + 91 = 294 348 + 156 = 504 348+156=504 348 + 156 = 504
∑ u ∼ v ( d v − d u ) = d v 2 − T v − d v \sum_{u\sim v}(d_v-d_u)=d_v^2-T_v-d_v u ∼ v ∑ ( d v − d u ) = d v 2 − T v − d v and so
− 294 T v = 294 ∑ u ∼ v ( d v − d u ) − 294 ( d v 2 − d v ) . -294T_v = 294\sum_{u\sim v}(d_v-d_u)-294(d_v^2-d_v). − 294 T v = 294 u ∼ v ∑ ( d v − d u ) − 294 ( d v 2 − d v ) . The first term on the right-hand side telescopes to 0 from summing over v ∈ G v \in G v ∈ G
E = ∑ v ( − 58 ( T v − A v − 3 2 ) − 26 ( T v − 3 2 ) + J ( d v ) ) E = \sum_v \left(-58\binom{T_v-A_v-3}{2}-26\binom{T_v-3}{2}+J(d_v)\right) E = v ∑ ( − 58 ( 2 T v − A v − 3 ) − 26 ( 2 T v − 3 ) + J ( d v ) ) where we have (again see Wolfram Alpha) that
J ( d ) = L ( d ) − 294 ( d 2 − d ) = − 19 4 d 2 + 141 2 d 3 − 1257 4 d 2 + 33 2 d + 736. J(d)=L(d)-294(d^2-d)=-\frac{19}{4}d^2+\frac{141}{2}d^3-\frac{1257}{4}d^2+\frac{33}{2}d+736. J ( d ) = L ( d ) − 294 ( d 2 − d ) = − 4 19 d 2 + 2 141 d 3 − 4 1257 d 2 + 2 33 d + 736. The next step is to maximize each binomial coefficient binomial coefficient above, holding A v A_v A v T v T_v T v E E E d d d A = ( d − 1 2 ) A=\binom{d-1}{2} A = ( 2 d − 1 ) Y v = T v − 3 Y_v=T_v-3 Y v = T v − 3
58 ( T v − A v − 3 2 ) + 26 ( T v − 3 2 ) = 42 Y v 2 − ( 58 A v + 42 ) Y v + 29 A v ( A v + 1 ) . 58\binom{T_v-A_v-3}{2}+26\binom{T_v-3}{2}=42Y_v^2-(58A_v+42)Y_v+29A_v(A_v+1). 58 ( 2 T v − A v − 3 ) + 26 ( 2 T v − 3 ) = 42 Y v 2 − ( 58 A v + 42 ) Y v + 29 A v ( A v + 1 ) . (We are trying to upper bound E E E E E E Y v = 29 A v + 21 42 Y_v=\frac{29A_v+21}{42} Y v = 42 29 A v + 21
29 A v ( A v + 1 ) − ( 29 A v + 21 2 ) 42 . 29A_v(A_v+1)-\frac{(29A_v+21^2)}{42}. 29 A v ( A v + 1 ) − 42 ( 29 A v + 2 1 2 ) . However since T v T_v T v Y v Y_v Y v
58 ( T v − A v − 3 2 ) + 26 ( T v − 3 2 ) ≥ 29 A v ( A v + 1 ) − ( 29 A v + 21 2 ) 42 + s 2 42 58\binom{T_v-A_v-3}{2}+26\binom{T_v-3}{2}\ge29A_v(A_v+1)-\frac{(29A_v+21^2)}{42}+\frac{s^2}{42} 58 ( 2 T v − A v − 3 ) + 26 ( 2 T v − 3 ) ≥ 29 A v ( A v + 1 ) − 42 ( 29 A v + 2 1 2 ) + 42 s 2 where s = min ( r , 42 − r ) s=\min(r,42-r) s = min ( r , 42 − r ) r = 29 A + 21 ( m o d 42 ) r=29A+21\pmod{42} r = 29 A + 21 ( mod 42 ) s / 42 s/42 s /42 Y v = 29 A v + 21 42 Y_v=\frac{29A_v+21}{42} Y v = 42 29 A v + 21 A A A r r r s s s d v d_v d v
Combining the above, we have bounded E E E d v d_v d v
E ≤ ∑ v ∈ G J ( d v ) − m ( d v ) E\le\sum_{v\in G}J(d_v)-m(d_v) E ≤ v ∈ G ∑ J ( d v ) − m ( d v ) where J J J m ( d v ) = 29 A v ( A v + 1 ) − ( 29 A v + 21 2 ) 42 + s 2 42 m(d_v)=29A_v(A_v+1)-\frac{(29A_v+21^2)}{42}+\frac{s^2}{42} m ( d v ) = 29 A v ( A v + 1 ) − 42 ( 29 A v + 2 1 2 ) + 42 s 2 ∑ v ( d v − 2 ) = − 2 \sum_v(d_v-2)=-2 ∑ v ( d v − 2 ) = − 2
E − 1008 ≤ ∑ v ∈ G ( J ( d v ) − m ( d v ) + 504 ( d v − 2 ) ) . E-1008\le\sum_{v\in G}\left(J(d_v)-m(d_v)+504(d_v-2)\right). E − 1008 ≤ v ∈ G ∑ ( J ( d v ) − m ( d v ) + 504 ( d v − 2 ) ) . It remains to show each term on the right-hand side is at most 0 for any value of d v d_v d v d v ∈ { 1 , 2 , 4 , 5 } d_v\in\{1,2,4,5\} d v ∈ { 1 , 2 , 4 , 5 } 2
COLT problem on dynamic networks – Sébastien Bubeck, Mark Sellke, and Steven Yin
We consider a random tree ( G t ) t ≥ 1 (G_t)_{t \ge1} ( G t ) t ≥ 1 t = 1 t = 1 t = 1 t + 1 t+1 t + 1 Y t + 1 Y_{t+1} Y t + 1 X t + 1 ∈ { Y 1 , … , Y t } X_{t+1} \in\{Y_1,\ldots,Y_t\} X t + 1 ∈ { Y 1 , … , Y t }
Each vertex v v v a ( v ) ∈ { 1 , w } a(v) \in\{1,w\} a ( v ) ∈ { 1 , w } attractiveness-weighted degree of x ∈ V t x \in V_t x ∈ V t
A t ( x ) : = ∑ y ∼ t x a ( y ) . A_t(x) := \sum_{y \sim_t x} a(y). A t ( x ) := y ∼ t x ∑ a ( y ) . (Here ∼ t \sim_t ∼ t t t t X t + 1 X_{t+1} X t + 1 Y t + 1 Y_{t+1} Y t + 1
P ( X t + 1 = x ∣ F t ) = A t ( x ) S t , S t : = ∑ x ∈ V t A t ( x ) . \mathbb{P}(X_{t+1}=x \mid\mathcal{F}_t) = \frac{A_t(x)}{S_t}, \qquad S_t := \sum_{x \in V_t} A_t(x). P ( X t + 1 = x ∣ F t ) = S t A t ( x ) , S t := x ∈ V t ∑ A t ( x ) . Here F t = σ ( G 1 , … , G t ) \mathcal{F}_t = \sigma(G_1,\ldots,G_t) F t = σ ( G 1 , … , G t )
Figure IV.9. Illustration of the process at t = 3 t = 3 t = 3 w w w A t A_t A t
Our goal will be to approximately recover w w w G t G_t G t t t t without observing the vertex labels . This was posed as an open problem3 [54 ] , who observed that single-edge label statistics do not help determine w w w [9 ] (in the case w = 1 w = 1 w = 1 w w w dynamical behavior of an evolving network while only observing the network at just 1 time. Note that there is a redundancy in the model above: reciprocating w w w ( G t ) t ≥ 1 (G_t)_{t\ge1} ( G t ) t ≥ 1 w ≥ 1 w \ge1 w ≥ 1 Figure IV.10 ).
Figure IV.10. Question posed to (a scaffolded version of) GPT-5. It solved this problem by coming up with and proving Theorem IV.4.1, and observing that this implies Corollary IV.4.1. (In particular the decision to study the quantity L ( t ) L(t) L ( t )
GPT-5 was able to prove that w w w f ( w ) f(w) f ( w ) L 1 ( t ) L_1(t) L 1 ( t ) L w ( t ) L_w(t) L w ( t ) t t t 1 1 1 w w w L ( t ) = L 1 ( t ) + L w ( t ) L(t) = L_1(t) + L_w(t) L ( t ) = L 1 ( t ) + L w ( t ) t t t G t G_t G t L 1 ( t ) L_1(t) L 1 ( t ) L w ( t ) L_w(t) L w ( t )
Theorem IV.4.1. For any w > 0 w > 0 w > 0
lim t → ∞ L ( t ) t = f ( w ) ≡ 3 4 − 1 2 ( w + 2 ) + 1 4 ( 2 w + 1 ) = 3 ( w + 1 ) 2 2 ( w + 2 ) ( 2 w + 1 ) . (38) \lim_{t \to\infty} \frac{L(t)}{t} = f(w) \equiv\frac{3}{4} - \frac{1}{2(w+2)} + \frac{1}{4(2w+1)} = \frac{3(w+1)^2}{2(w+2)(2w+1)}. \tag*{(38)} t → ∞ lim t L ( t ) = f ( w ) ≡ 4 3 − 2 ( w + 2 ) 1 + 4 ( 2 w + 1 ) 1 = 2 ( w + 2 ) ( 2 w + 1 ) 3 ( w + 1 ) 2 . (38) In fact, one has
lim t → ∞ L ( 1 ) ( t ) t = w + 1 2 ( w + 2 ) , lim t → ∞ L ( w ) ( t ) t = w + 1 2 ( 2 w + 1 ) . \lim_{t \to\infty} \frac{L^{(1)}(t)}{t} = \frac{w+1}{2(w+2)}, \qquad\lim_{t \to\infty} \frac{L^{(w)}(t)}{t} = \frac{w+1}{2(2w+1)}. t → ∞ lim t L ( 1 ) ( t ) = 2 ( w + 2 ) w + 1 , t → ∞ lim t L ( w ) ( t ) = 2 ( 2 w + 1 ) w + 1 . The function f f f (38) is easily seen to be strictly increasing, with
f ′ ( w ) = 1 2 ( 1 ( w + 2 ) 2 − 1 ( 2 w + 1 ) 2 ) > 0 , ∀ w > 1. f'(w) = \frac{1}{2}\left(\frac{1}{(w+2)^2} - \frac{1}{(2w+1)^2}\right) > 0,\qquad\forall w > 1. f ′ ( w ) = 2 1 ( ( w + 2 ) 2 1 − ( 2 w + 1 ) 2 1 ) > 0 , ∀ w > 1. Identifiability of w w w
Corollary IV.4.1. For any w ≥ 1 w \ge1 w ≥ 1 w ^ ( t ) = f − 1 ( L ( t ) / t ) \widehat{w}(t) = f^{-1}(L(t)/t) w ( t ) = f − 1 ( L ( t ) / t ) w w w t → ∞ t \to\infty t → ∞ f − 1 : [ 2 / 3 , 3 / 4 ] → [ 1 , ∞ ] f^{-1}: [2/3, 3/4] \to[1,\infty] f − 1 : [ 2/3 , 3/4 ] → [ 1 , ∞ ] f ∣ [ 1 , ∞ ] f|_{[1,\infty]} f ∣ [ 1 , ∞ ] f − 1 ( x ) = 1 f^{-1}(x) = 1 f − 1 ( x ) = 1 x ≤ 2 / 3 x \le2/3 x ≤ 2/3 f − 1 ( x ) = ∞ f^{-1}(x) = \infty f − 1 ( x ) = ∞ x ≥ 3 / 4 x \ge3/4 x ≥ 3/4 *
Figure IV.11. A simulation of the asymptotic result (38) in Theorem IV.4.1. The orange plot shows the average (and empirical standard deviation) of the fraction of leaves in G 10 5 G_{10^5} G 1 0 5 w w w f ( w ) f(w) f ( w ) w ∈ [ 1 , ∞ ) w \in[1,\infty) w ∈ [ 1 , ∞ ) f ( 1 ) = 2 / 3 f(1)=2/3 f ( 1 ) = 2/3 lim w → ∞ f ( w ) = 3 / 4 \lim_{w\to\infty} f(w)=3/4 lim w → ∞ f ( w ) = 3/4
It is worth noting that the above type of algorithm provably fails in a related model of evolving trees studied in [18 ] (in which attachment probabilities depend only on label agreement and not on degree). In their model, the fraction of degree k k k 2 − k 2^{-k} 2 − k L ( t ) L(t) L ( t )
Remark IV.4.1. [54 ] in fact asks for a polynomial rate of convergence, i.e. an estimator w ^ ( t ) \widehat{w}(t) w ( t ) P [ ∣ w ^ ( t ) − w ∣ ≤ ε ] ≥ 1 − δ \mathbb{P}[|\widehat{w}(t)-w|\le\varepsilon]\ge1-\delta P [ ∣ w ( t ) − w ∣ ≤ ε ] ≥ 1 − δ t t t ( ε , δ ) (\varepsilon,\delta) ( ε , δ ) [54 ] also considers a slightly more general setting in which incoming vertices have probability p p p w w w w ↔ 1 / w w\leftrightarrow1/w w ↔ 1/ w
Remark IV.4.2. While all major proof ideas below are due to GPT-5, a few details of proof writing are human-supplied. Our scaffolded version of GPT-5 autonomously produced exactly the high-level argument below and the same main references, but omitted the proof details for certain intermediate assertions, stating (correctly) that they followed from routine stochastic approximation arguments. When we asked (an unscaffolded) GPT-5 to provide more detail for these latter arguments, it made several false starts, stating for example that [57 ] directly implies V t → 0 V_t\to0 V t → 0 V t V_t V t E [ V t ] → 0 \mathbb{E}[V_t]\to0 E [ V t ] → 0
Proof Overview
We assume w ≥ 1 w \ge1 w ≥ 1 w w w O ( ⋅ ) O(\cdot) O ( ⋅ )
S t ( 1 ) : = ∑ x : a ( x ) = 1 A t ( x ) , S t ( w ) : = ∑ x : a ( x ) = w A t ( x ) . S_t^{(1)} := \sum_{x:a(x)=1} A_t(x), \qquad S_t^{(w)} := \sum_{x:a(x)=w} A_t(x). S t ( 1 ) := x : a ( x ) = 1 ∑ A t ( x ) , S t ( w ) := x : a ( x ) = w ∑ A t ( x ) . Then S t ( 1 ) + S t ( w ) = S t S_t^{(1)} + S_t^{(w)} = S_t S t ( 1 ) + S t ( w ) = S t
R t ≡ S t ( w ) S t = P [ a ( X t + 1 ) = w ] R_t \equiv\frac{S_t^{(w)}}{S_t} = \mathbb{P}[a(X_{t+1}) = w] R t ≡ S t S t ( w ) = P [ a ( X t + 1 ) = w ] is the chance for the next vertex to attach to a w w w G t G_t G t
D t : = S t ( w ) − 1 2 S t = S t ( w ) − S t ( 1 ) 2 = S t ⋅ ( R t − 1 2 ) , μ = E [ a ( new vertex ) ] = 1 + w 2 . (39) D_t := S_t^{(w)} - \frac{1}{2}S_t = \frac{S_t^{(w)} - S_t^{(1)}}{2} = S_t \cdot\left(R_t - \frac{1}{2}\right), \qquad\mu= \mathbb{E}[a(\text{new vertex})] = \frac{1+w}{2}. \tag*{(39)} D t := S t ( w ) − 2 1 S t = 2 S t ( w ) − S t ( 1 ) = S t ⋅ ( R t − 2 1 ) , μ = E [ a ( new vertex )] = 2 1 + w . (39) and set (with E t = E ( G t ) E_t = E(G_t) E t = E ( G t ) t t t
T t : = ∑ { u , v } ∈ E t a ( u ) a ( v ) = S t ( 1 ) + w S t ( w ) 2 . T_t := \sum_{\{u,v\}\in E_t} a(u)a(v) = \frac{S_t^{(1)} + wS_t^{(w)}}{2}. T t := { u , v } ∈ E t ∑ a ( u ) a ( v ) = 2 S t ( 1 ) + w S t ( w ) . Then
E [ a ( X t + 1 ) ∣ F t ] = 2 T t S t . (40) \mathbb{E}[a(X_{t+1}) \mid\mathcal{F}_t] = \frac{2T_t}{S_t}. \tag*{(40)} E [ a ( X t + 1 ) ∣ F t ] = S t 2 T t . (40) For notational convenience, given a discrete-time stochastic process ( Z t ) t ≥ 1 (Z_t)_{t\ge1} ( Z t ) t ≥ 1 Δ Z t = Z t + 1 − Z t \Delta Z_t = Z_{t+1} - Z_t Δ Z t = Z t + 1 − Z t
The proof consists of two main steps. In the first we show R t → 1 / 2 R_t \to1/2 R t → 1/2 S t / t → w + 1 S_t/t \to w+1 S t / t → w + 1 eq:IV.4.1 and eq:IV.4.2 ). In the second we use this to analyze the growth of L ( 1 ) ( t ) L^{(1)}(t) L ( 1 ) ( t ) L ( w ) ( t ) L^{(w)}(t) L ( w ) ( t ) [56 , 57 ] .
One-Step Dynamics
When the ( t + 1 ) (t+1) ( t + 1 ) y = Y t + 1 y = Y_{t+1} y = Y t + 1 x = X t + 1 ∈ { Y 1 , … , Y t } x = X_{t+1} \in\{Y_1,\ldots,Y_t\} x = X t + 1 ∈ { Y 1 , … , Y t }
Δ A t ( x ) ≡ A t + 1 ( x ) − A t ( x ) = a ( y ) , Δ A t ( y ) = a ( x ) , \Delta A_t(x) \equiv A_{t+1}(x) - A_t(x) = a(y), \qquad\Delta A_t(y) = a(x), Δ A t ( x ) ≡ A t + 1 ( x ) − A t ( x ) = a ( y ) , Δ A t ( y ) = a ( x ) , and A t ( z ) A_t(z) A t ( z ) z ∈ { Y 1 , … , Y t } ∖ { x } z \in\{Y_1,\ldots,Y_t\}\setminus\{x\} z ∈ { Y 1 , … , Y t } ∖ { x }
Δ S t = a ( x ) + a ( y ) . \Delta S_t = a(x) + a(y). Δ S t = a ( x ) + a ( y ) . Using (40) one obtains with some algebra that
E [ Δ S t ∣ F t ] = μ + 2 T t S t = D t ( w − 1 ) S t + ( w + 1 ) . (41) \mathbb{E}[\Delta S_t \mid\mathcal{F}_t] = \mu+ \frac{2T_t}{S_t} = \frac{D_t(w-1)}{S_t} + (w+1). \tag*{(41)} E [ Δ S t ∣ F t ] = μ + S t 2 T t = S t D t ( w − 1 ) + ( w + 1 ) . (41) Likewise,
Δ S t ( w ) = 1 { a ( x ) = w } a ( y ) + 1 { a ( y ) = w } a ( x ) , \Delta S_t^{(w)} = \mathbf{1}\{a(x)=w\}a(y) + \mathbf{1}\{a(y)=w\}a(x), Δ S t ( w ) = 1 { a ( x ) = w } a ( y ) + 1 { a ( y ) = w } a ( x ) , and one can easily compute
E [ Δ D t ∣ F t ] = μ ( R t − 1 2 ) = μ D t S t . (42) \mathbb{E}[\Delta D_t \mid\mathcal{F}_t] = \mu\left(R_t-\frac{1}{2}\right) = \frac{\mu D_t}{S_t}. \tag*{(42)} E [ Δ D t ∣ F t ] = μ ( R t − 2 1 ) = S t μ D t . (42) We also trivially have
2 t ≤ S t ≤ 2 w t (43) 2t \le S_t \le2wt \tag*{(43)} 2 t ≤ S t ≤ 2 w t (43) and ∣ Δ R t ∣ ≤ O ( 1 / S t ) |\Delta R_t| \le O(1/S_t) ∣Δ R t ∣ ≤ O ( 1/ S t )
Convergence of R t R_t R t S t S_t S t
Lemma IV.4.1. lim t → ∞ R t = 1 2 \lim_{t\to\infty} R_t = \frac{1}{2} lim t → ∞ R t = 2 1 almost surely.
Proof. Define
V t : = ( R t − 1 2 ) 2 = ( D t / S t ) 2 . V_t := \left(R_t-\frac{1}{2}\right)^2 = \left(D_t/S_t\right)^2. V t := ( R t − 2 1 ) 2 = ( D t / S t ) 2 . We will show V t → 0 V_t \to0 V t → 0 Q t = D t / S t Q_t = D_t/S_t Q t = D t / S t
Δ V t = 2 Q t Δ Q t + ( Δ Q t ) 2 . \Delta V_t = 2Q_t\Delta Q_t + (\Delta Q_t)^2. Δ V t = 2 Q t Δ Q t + ( Δ Q t ) 2 . Next it is easy to see by Taylor’s approximation theorem that the quotient rule holds approximately for discrete differences; since ∣ Δ S t ( w ) ∣ , ∣ Δ S t ∣ ≤ O ( 1 ) |\Delta S_t^{(w)}|, |\Delta S_t| \le O(1) ∣Δ S t ( w ) ∣ , ∣Δ S t ∣ ≤ O ( 1 )
Δ Q t = S t Δ D t − D t Δ S t S t 2 + O ( S t − 2 ) . \Delta Q_t = \frac{S_t\Delta D_t-D_t\Delta S_t}{S_t^2} + O(S_t^{-2}). Δ Q t = S t 2 S t Δ D t − D t Δ S t + O ( S t − 2 ) . Combining the previous two displays, we find:
Δ V t = 2 D t S t 2 ( Δ D t − D t Δ S t S t ) + O ( S t − 2 ) . \Delta V_t = \frac{2D_t}{S_t^2}\left(\Delta D_t-\frac{D_t\Delta S_t}{S_t}\right) + O(S_t^{-2}). Δ V t = S t 2 2 D t ( Δ D t − S t D t Δ S t ) + O ( S t − 2 ) . Using (IV.9)–(IV.10) yields
E [ Δ V t ∣ F t ] = − 4 D t 2 T t S t 4 + O ( S t − 2 ) = − V t ( 4 T t S t 2 ) + O ( S t − 2 ) . \mathbb{E}[\Delta V_t\mid\mathcal{F}_t] = -\frac{4D_t^2T_t}{S_t^4}+O(S_t^{-2})=-V_t\left(\frac{4T_t}{S_t^2}\right)+O(S_t^{-2}). E [ Δ V t ∣ F t ] = − S t 4 4 D t 2 T t + O ( S t − 2 ) = − V t ( S t 2 4 T t ) + O ( S t − 2 ) . Since T t ≍ S t ≍ t T_t \asymp S_t \asymp t T t ≍ S t ≍ t
E [ Δ V t ∣ F t ] ≤ − Ω ( S t − 1 ) V t + O ( S t − 2 ) . \mathbb{E}[\Delta V_t\mid\mathcal{F}_t] \le-\Omega(S_t^{-1})V_t+O(S_t^{-2}). E [ Δ V t ∣ F t ] ≤ − Ω ( S t − 1 ) V t + O ( S t − 2 ) . By (IV.11), we have
∑ t ≥ 1 S t − 1 = ∞ , ∑ t ≥ 1 S t − 2 = S ‾ ( w ) < ∞ . \sum_{t\ge1} S_t^{-1}=\infty,\qquad\sum_{t\ge1}S_t^{-2}=\overline{S}(w)<\infty. t ≥ 1 ∑ S t − 1 = ∞ , t ≥ 1 ∑ S t − 2 = S ( w ) < ∞. The almost-supermartingale theorem of [57 ] implies that lim t → ∞ V t \lim_{t\to\infty}V_t lim t → ∞ V t E t = E [ V t ] E_t=\mathbb{E}[V_t] E t = E [ V t ]
E t + 1 ≤ ( 1 − c S t − 1 ) E t + O ( S t − 2 ) . E_{t+1}\le(1-cS_t^{-1})E_t+O(S_t^{-2}). E t + 1 ≤ ( 1 − c S t − 1 ) E t + O ( S t − 2 ) . for a uniform constant c = c ( w ) > 0 c=c(w)>0 c = c ( w ) > 0
E T ≤ ( E 1 + S ‾ ( w ) ) ⋅ ∏ 1 ≤ t < T ( 1 − c S t − 1 ) . E_T\le(E_1+\overline{S}(w))\cdot\prod_{1\le t<T}(1-cS_t^{-1}). E T ≤ ( E 1 + S ( w )) ⋅ 1 ≤ t < T ∏ ( 1 − c S t − 1 ) . This tends to 0 0 0 T → ∞ T\to\infty T → ∞ ∑ S t − 1 = ∞ \sum S_t^{-1}=\infty ∑ S t − 1 = ∞ E t → 0 E_t\to0 E t → 0 V t ∈ [ 0 , 1 ] V_t\in[0,1] V t ∈ [ 0 , 1 ] t t t t → ∞ t\to\infty t → ∞ 0 0 0 □ \square □
Lemma IV.4.2. S t / t → w + 1 S_t/t\to w+1 S t / t → w + 1 almost surely.
Proof. Recalling (IV.9), we have
Δ S t = D t ( w − 1 ) S t + ( w + 1 ) + ξ t + 1 \Delta S_t=\frac{D_t(w-1)}{S_t}+(w+1)+\xi_{t+1} Δ S t = S t D t ( w − 1 ) + ( w + 1 ) + ξ t + 1 for a bounded martingale difference ξ t + 1 \xi_{t+1} ξ t + 1 D t / S t → 0 D_t/S_t\to0 D t / S t → 0
1 t ∑ i = 1 t ( D i ( w − 1 ) S i + ( w + 1 ) ) → w + 1. \frac{1}{t}\sum_{i=1}^{t}\left(\frac{D_i(w-1)}{S_i}+(w+1)\right)\to w+1. t 1 i = 1 ∑ t ( S i D i ( w − 1 ) + ( w + 1 ) ) → w + 1. The martingale SLLN gives 1 t ∑ i = 1 t ξ i → 0 \frac{1}{t}\sum_{i=1}^{t}\xi_i\to0 t 1 ∑ i = 1 t ξ i → 0 S t / t → w + 1 S_t/t\to w+1 S t / t → w + 1 □ \square □
Stochastic Approximation for Leaf Fractions
We use the following stochastic approximation result, which gives an ODE limit for certain stochastic processes. It follows by combining [12 ] , Propositions 4.1 and 4.2 (using q = 2 q = 2 q = 2
Theorem IV.4.2. Consider a scalar recursion
Z t + 1 = Z t + γ t + 1 ( F ( Z t ) + ε t + 1 + δ t + 1 ) Z_{t+1} = Z_t + \gamma_{t+1}\left(F(Z_t) + \varepsilon_{t+1} + \delta_{t+1}\right) Z t + 1 = Z t + γ t + 1 ( F ( Z t ) + ε t + 1 + δ t + 1 ) driven by:
γ t → 0 , ∑ t ≥ 1 γ t = ∞ , ∑ t ≥ 1 γ t 2 < ∞ . \gamma_t \to0,\qquad\sum_{t \ge1} \gamma_t = \infty,\qquad\sum_{t \ge1} \gamma_t^2 < \infty. γ t → 0 , t ≥ 1 ∑ γ t = ∞ , t ≥ 1 ∑ γ t 2 < ∞. Suppose the ODE z ′ ( T ) = F ( z ( T ) ) z'(T) = F(z(T)) z ′ ( T ) = F ( z ( T )) z ⋆ z^\star z ⋆ z ( 0 ) ∈ R z(0) \in\mathbb{R} z ( 0 ) ∈ R lim T → ∞ z ( T ) = z ⋆ \lim_{T \to\infty} z(T) = z^\star lim T → ∞ z ( T ) = z ⋆
lim t → ∞ Z t = z ⋆ . \lim_{t \to\infty} Z_t = z^\star. t → ∞ lim Z t = z ⋆ . Recall the definitions of L 1 ( t ) , L w ( t ) L_1(t), L_w(t) L 1 ( t ) , L w ( t )
E [ Δ L 1 ∣ F t ] = S t ( 1 ) − L 1 ( t ) S t , E [ Δ L w ∣ F t ] = S t ( w ) − w L w ( t ) S t . (44) \mathbb{E}[\Delta L_1 \mid\mathcal{F}_t] = \frac{S_t^{(1)} - L_1(t)}{S_t},\qquad\mathbb{E}[\Delta L_w \mid\mathcal{F}_t] = \frac{S_t^{(w)} - wL_w(t)}{S_t}. \tag*{(44)} E [ Δ L 1 ∣ F t ] = S t S t ( 1 ) − L 1 ( t ) , E [ Δ L w ∣ F t ] = S t S t ( w ) − w L w ( t ) . (44) Indeed L 1 ( t ) / S t L_1(t)/S_t L 1 ( t ) / S t L w ( t ) / S t L_w(t)/S_t L w ( t ) / S t F t \mathcal{F}_t F t X t + 1 X_{t+1} X t + 1 w w w X t + 1 X_{t+1} X t + 1 G t + 1 G_{t+1} G t + 1
The main quantities of interest will be the leaf fractions
x t = L 1 ( t ) / t , y t = L w ( t ) / t . x_t = L_1(t)/t,\qquad y_t = L_w(t)/t. x t = L 1 ( t ) / t , y t = L w ( t ) / t . Since e.g. x t + 1 = t x t t + 1 x_{t+1} = \frac{t x_t}{t+1} x t + 1 = t + 1 t x t Δ L 1 ( t ) = 0 \Delta L_1(t) = 0 Δ L 1 ( t ) = 0
Δ x t = Δ L 1 ( t ) − x t t + 1 , Δ y t = Δ L w ( t ) − y t t + 1 . (45) \Delta x_t = \frac{\Delta L_1(t) - x_t}{t+1},\qquad\Delta y_t = \frac{\Delta L_w(t) - y_t}{t+1}. \tag*{(45)} Δ x t = t + 1 Δ L 1 ( t ) − x t , Δ y t = t + 1 Δ L w ( t ) − y t . (45) We showed in Lemma IV.4.1 and IV.4.2 above that almost surely
( S t ( 1 ) t , S t ( w ) t , S t t ) → ( w + 1 2 , w + 1 2 , w + 1 ) . (46) \left(\frac{S_t^{(1)}}{t},\frac{S_t^{(w)}}{t},\frac{S_t}{t}\right) \to\left(\frac{w+1}{2},\frac{w+1}{2},w+1\right). \tag*{(46)} ( t S t ( 1 ) , t S t ( w ) , t S t ) → ( 2 w + 1 , 2 w + 1 , w + 1 ) . (46) In light of (IV.12) and (IV.13), we will apply Theorem IV.4.2 to x t x_t x t y t y_t y t γ t + 1 = 1 / ( t + 1 ) \gamma_{t+1}=1/(t+1) γ t + 1 = 1/ ( t + 1 )
F 1 ( x ) = 1 2 − x − x 1 + w , F w ( y ) = 1 2 − y − w y 1 + w . F_1(x)=\frac{1}{2}-x-\frac{x}{1+w},\qquad F_w(y)=\frac{1}{2}-y-\frac{wy}{1+w}. F 1 ( x ) = 2 1 − x − 1 + w x , F w ( y ) = 2 1 − y − 1 + w w y . and the martingale noises
ε t + 1 X : = Δ L 1 ( t ) − E [ Δ L 1 ( t ) ∣ F t ] , ε t + 1 Y : = Δ L w ( t ) − E [ Δ L w ( t ) ∣ F t ] . \varepsilon^X_{t+1}:=\Delta L_1(t)-\mathbb{E}[\Delta L_1(t)\mid\mathcal{F}_t],\qquad\varepsilon^Y_{t+1}:=\Delta L_w(t)-\mathbb{E}[\Delta L_w(t)\mid\mathcal{F}_t]. ε t + 1 X := Δ L 1 ( t ) − E [ Δ L 1 ( t ) ∣ F t ] , ε t + 1 Y := Δ L w ( t ) − E [ Δ L w ( t ) ∣ F t ] . Focusing without loss of generality on the x t x_t x t
x t + 1 = x t + γ t + 1 ( F 1 ( x t ) + ε t + 1 X + δ t + 1 X ) . x_{t+1}=x_t+\gamma_{t+1}\left(F_1(x_t)+\varepsilon^X_{t+1}+\delta^X_{t+1}\right). x t + 1 = x t + γ t + 1 ( F 1 ( x t ) + ε t + 1 X + δ t + 1 X ) . where δ t X \delta^X_t δ t X δ t X → 0 \delta^X_t\to0 δ t X → 0 ∣ Δ L 1 ∣ ≤ 1 |\Delta L_1|\le1 ∣Δ L 1 ∣ ≤ 1 ε t X \varepsilon^X_t ε t X x ′ ( T ) = F 1 ( x ( T ) ) x'(T)=F_1(x(T)) x ′ ( T ) = F 1 ( x ( T )) F 1 F_1 F 1
x ∗ = w + 1 2 ( w + 2 ) . x^*=\frac{w+1}{2(w+2)}. x ∗ = 2 ( w + 2 ) w + 1 . Thus we may apply Theorem IV.4.2 to conclude the almost sure limit:
x t / t → w + 1 2 ( w + 2 ) . x_t/t\to\frac{w+1}{2(w+2)}. x t / t → 2 ( w + 2 ) w + 1 . Identical reasoning shows that y t y_t y t F w F_w F w
y t / t → y ∗ = w + 1 2 ( 2 w + 1 ) . y_t/t\to y^*=\frac{w+1}{2(2w+1)}. y t / t → y ∗ = 2 ( 2 w + 1 ) w + 1 . This completes the proof of Theorem IV.4.1.
Conclusion
AI models are well-known for assisting with routine tasks; what is less well-known is that frontier models like GPT-5 are increasingly able to accelerate novel research. We highlighted examples across a range of scientific disciplines showing GPT-5 assisting in ideation, powering deep literature search, and even proposing and implementing proofs of (appropriately-sized) open problems. None of this would have been possible just twelve months ago. The fact that GPT-5 is already showing it can accelerate scientific research, combined with the rate of improvement of frontier AI, suggests that the coming years will be an exhilarating time in science and mathematics.
Acknowledgments
A. Lupsasca is supported in part by the National Science Foundation through the CAREER award 2340457 and grant 2307888, and by the Simons Foundation through the grant SFI-MPS-BH-00012593-09. This research was conducted during the period M. Sawhney served as a Clay Fellow. C. Coester is funded by the European Union (ERC, CCOO, 101165139). Views and opinions expressed are however those of the author(s) only and do not necessarily reflect those of the European Union or the European Research Council. Neither the European Union nor the granting authority can be held responsible for them. D. Unutmaz thanks Oral Alpan, M.D. for critical reading of Section 3 of this report.
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