A breakthrough · discrete geometry

AI cracked an 80-year-old conjecture.

based on Sébastien Bubeck on X

An internal OpenAI model resolved the most famous open question in discrete geometry — the unit distance problem. The problem is 80 years old. It is also, on its surface, embarrassingly simple.

This page walks through what the problem is, why everyone thought they knew the answer, and what the model actually did.

The headline

The model did not invent new mathematics. It connected two distant areas — discrete geometry and algebraic number theory — and executed the chain of arguments at expert level. That alone disproved a belief that had stood since 1946.

01 · the problem

How many pairs can sit at the same distance?

Drop n dots on a sheet of paper. Now count: how many pairs of dots are exactly the same distance apart?

By rescaling, we can call that shared distance 1 — hence "unit distance." So the question is just:

The question

Given n points in the plane, how many pairs of them can be at distance exactly 1?

n−1 unit pairs Can we do much better? order n vs. order n²?
Figure 1 the easy baselines. One hub + a unit circle gives n−1 equal distances. The ceiling is n²/2. The truth is somewhere between.

The first easy construction (left): pick one point as a hub and put all the others on a circle of radius 1 around it. That gives n − 1 pairs at distance 1. The ceiling is trivially the total number of pairs, about n²/2. The real question is which end of that gap reality lives on.

02 · the grid guess

Erdős's first idea: just use a grid.

In 1946 Paul Erdős asked the question — and immediately tried the most natural configuration he could think of: a square grid.

Every interior grid point has exactly 4 neighbors at distance 1 (north, south, east, west). Counting all such pairs, and dividing by 2 to avoid double counting, gives about 2n unit-distance pairs.

4 neighbors at distance 1 → about 2n unit-distance pairs total
Figure 2 on a unit grid, each point has 4 partners at distance 1. Total: about 2n equal pairs. Better than the hub, but only by a constant.

So far the grid only gives a constant times n — same order as the hub construction. To get a real improvement, Erdős had to be cleverer about which distance to fix.

03 · the √5 trick

Pick a smarter distance, get more partners.

Instead of distance 1, look at distance $\sqrt{5} = \sqrt{1^2 + 2^2}$. How many grid neighbors does a point have at that distance?

Draw a knight's-move "L" — over 1, up 2 — and rotate / flip it. There are eight different L's, giving 8 grid points at distance √5 from any interior point.

8 points at distance √5
Figure 3 the L-shape can be rotated and reflected eight ways — so distance $\sqrt{5}$ hits 8 grid neighbors instead of 4.

Why does √5 multiply by 2? Because 5 = 1 + 4, and there are essentially two ways to write 5 as a sum of two squares: (±1)² + (±2)² and (±2)² + (±1)². Each sign and swap is a different L. That's not a coincidence — it's the first whisper that prime numbers are doing the work.

04 · primes hidden inside

Gaussian integers: factoring the distance.

Think of a grid point (a, b) as the complex number $z = a + bi$. Its distance from the origin is $|z| = \sqrt{a^2 + b^2}$. So we're asking: how many complex integers $z$ have $|z|^2 = N$?

The key fact: a prime p that is 1 mod 4 (like 5, 13, 17, 29, …) splits in the Gaussian integers as $p = z \cdot \bar{z}$. For example:

5 = (1 + 2i)(1 − 2i)    13 = (3 + 2i)(3 − 2i)    17 = (4 + i)(4 − i)
Re Im 1 + 2i 1 − 2i √5 |1 + 2i|² = 5 |1 − 2i|² = 5 multiply z & z̄ → |z|² · |z̄|² = 25 flip sign: z · z, z · z̄, z̄ · z, z̄ · z̄ 4 distinct points at radius 5
Figure 4 5 splits into two conjugate Gaussian integers z and $\bar{z}$, each at distance √5. Multiplying them together gives two distinct points at radius 5.
The trick

If you have k primes that all split like this, each prime gives you a binary choice: take z, or take its conjugate $\bar{z}$. Multiply them together and you get a Gaussian integer with the same modulus. Different choices → different points → 2k points all at the same distance from the origin.

05 · Erdős's bound

How big can k get?

Take the first k primes that are 1 mod 4: p1, p2, …, pk. Their product R has 2k distinct factorizations as z1$\bar{z}$1 · z2$\bar{z}$2 · …, each picking z or $\bar{z}$. So 2k grid points sit at distance √R from the origin.

p₁ = 5 × 2 p₂ = 13 × 2 p₃ = 17 × 2 pk × 2 R = p₁ · p₂ · ⋯ · pk → 2ᵏ grid points at distance √R (one binary choice per prime: z or z̄)
Figure 5 each split prime doubles the count. k primes give 2k equal-distance points.

How large can k be? We need $\sqrt{R} < \sqrt{n}$ so all 2k points fit in our n-point grid. By the prime number theorem, the first k primes have $\log R \approx k \log k$. Solving $k \log k < \log n$ gives

$$k \approx \frac{\log n}{\log \log n}, \quad 2^k \approx 2^{\log n / \log \log n}$$

Repeating this argument over every interior grid point gives Erdős's bound:

Erdős's 1946 bound

A grid of n points has at least about n · 2log n / log log n unit-distance pairs, which is n1+o(1).

k =
Live the first k primes that are 1 mod 4 (5, 13, 17, 29, …) multiplied together give R. Every orange dot is an integer point (a, b) with a² + b² = R — they all sit on the dashed circle of radius √R.
06 · 80 years of silence

Everyone believed the grid was optimal.

Erdős conjectured that you can never do better than the grid: any configuration of n points should have at most n1+o(1) unit-distance pairs.

For 80 years, zero progress was made on this conjecture. Worse: in the closely related distinct distances problem (introduced in the same 1946 paper), the grid was proven optimal by Guth and Katz in 2015. Everything pointed the same way.

1946 Erdős poses problem — 80 years — no progress on the upper bound grid believed optimal 2025 AI beats the grid
Figure 6 the conjecture sat untouched for eight decades. The closely related distinct-distances problem closed in Erdős's favor, reinforcing the belief.

This was the orthodoxy walking into 2025: the grid is the right shape; the only question left was how to prove a matching upper bound. Everyone was looking up. The breakthrough came from looking down — at the grid itself.

07 · the AI's move

"What if there were more ways to split each prime?"

From the chain-of-thought: the model almost immediately decided to attack the lower bound — to beat the grid. That's the opposite of what mathematicians had been trying.

Its strategy comes from staring at Erdős's argument. The 2 in 2k comes from one fact: each prime splits into two conjugate Gaussian integers, z and $\bar{z}$. What if we worked in a bigger number system where each prime split into many more factors?

Erdős's world: Gaussian integers ℤ[i] p z 2 choices per prime → 2ᵏ equal-distance points AI's move: ring of integers in a higher-degree field K p 𝔭₁ 𝔭₂ 𝔭f f choices per prime (f = degree of K) → fᵏ equal-distance points project the lattice back to ℝ²
Figure 7 use a bigger number field. Each prime splits into f prime ideals instead of 2 conjugates, so the count jumps from 2k to fk. The catch: the lattice now lives in higher dimensions and must be projected back to the plane.

Three obstacles, all of which the model navigated:

obstacle 1 · the wrong fields

First guess: cyclotomic fields. The model tried it in the chain of thought, then quickly saw why it doesn't work and abandoned the idea.

obstacle 2 · unique factorization fails

In higher fields, you can't always factor numbers uniquely — but you can uniquely factor ideals. The model switched to the language of ideals, with the failures of unique factorization tracked by the class group.

obstacle 3 · controlling the projection

A high-dimensional lattice has to be projected back to the plane without too many points colliding. To get a number field with a controlled class number and the right projection properties, the model reached for a sledgehammer from class field theory: the infinite class field towers of Golod–Shafarevich.

The result

The construction gives n1+δ unit-distance pairs for some δ > 0 — strictly more than the grid's n1+o(1). Erdős's conjecture is strongly disproved.

08 · what it means

Knowing everything is itself a kind of creativity.

Step back. The model did not invent class field theory. It did not invent Golod–Shafarevich towers. It did not invent a new branch of mathematics.

What it did do: see a connection between discrete geometry (a combinatorics problem about dots in the plane) and algebraic number theory (rings of integers, class groups, prime splitting). Then it executed a long chain of expert-level calculations end to end, with the right choice of parameters at every step.

discrete geometry grid · distances · counting algebraic number theory primes · ideals · class groups the AI's bridge expert-level execution on: cyclotomic → ideals → class group → G–S towers → projection
Figure 8 the breakthrough is not new mathematics, but a long, correct chain of known mathematics connecting two distant areas.

That is itself the lesson. A lot of scientific progress — not just in mathematics — looks like this: knowing deeply, chaining expertly, choosing parameters right. If a model can do that, a great many open problems are suddenly within reach.

A coda from the OpenAI team: the proof is not being submitted to arXiv. No human author can claim authorship in the traditional sense. Instead, a companion paper by human mathematicians digests, contextualizes, and slightly simplifies the proof. Separating the AI's proof from the human understanding of it may be a useful template going forward.

In one sentence

An AI model used its knowledge of all of mathematics to spot that an 80-year-old problem about dots in a plane was secretly a problem about prime ideals in number fields — and then walked through the resulting argument without a misstep.