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Continue in portrait anywayOn May 20, 2026, an AI built by OpenAI disproved a problem from 1946. The problem, set by Paul Erdős, asks how many pairs of dots on a sheet of paper can sit exactly the same distance apart. For 78 years, everyone believed a regular grid was the best you could do. The AI showed something better — and the engine of its construction is what's spinning on your screen right now.
Pick three whole numbers — call them a, b, c. Plug them into a recipe:
point = a + b·θ + c·θ²
where θ is the awkward irrational number 2·cos(2π/7). Do it three times — once each for the three different ways of placing seven equally-spaced angles on a circle — and you get three numbers. Those three numbers are one point in 3D space.
Run every triple (a, b, c) you can fit and you get the cyan forest of dots above. It looks like a grid because the recipe is linear, but it's a sheared grid — twisted by the irrational θ in three different directions at once.
Press your face against the floor and look up. Every 3D point casts a shadow straight down. The shadows are not on a grid. They scatter densely, irregularly — and the more points you let in, the messier the shadow becomes.
That's a tidy 3D pattern, smashed flat into a 2D mess. It's already non-grid. Good — but this isn't the dot pattern the proof actually counts. The real construction uses twice as many integers.
For the actual proof, take two elements of the cubic lattice. Call them u and v. Together they need six integers: a, b, c for u and d, e, f for v. Think of the pair as a single complex number, z = u + i·v.
Keep only the pairs where σ_k(u)² + σ_k(v)² ≤ R² for all three embeddings k = 1, 2, 3 — that's the K-polydisc constraint, the condition that z sits inside a product of three discs. Plot the survivors as 2D points (σ₁(u), σ₁(v)).
What appears in the right panel is the true planar set the proof analyzes: a rank-6 lattice in ℝ⁶ projected to ℝ². Compare it to the floor shadow on the left — the K-set is denser, more clustered, and the dashed boundary is a circle, not a square, because the constraint is genuinely a disc.
The magenta dots on the inner unit-circle in the right panel are the heart of the proof. Each one is a specific complex number of length exactly 1 — built by picking a prime that splits in three different ways (we use q = 13, the smallest that works), then for each of the 8 choices of which prime-pair to take, dividing one ideal generator by its complex conjugate. The result is a unit-modulus number.
In the proof, these 8 numbers act as translations on a much finer lattice (1/13)·O_K — too fine to draw here without millions of points. Each translation moves a planar point by exactly distance 1. The proof shows you get so many of these unit-distance pairs that the count beats any grid.
Notice the magenta dots come in conjugate pairs across the real axis. That's the symmetry from ε and its bitwise complement: flipping every choice swaps each prime with its conjugate, which complex-conjugates the result.
a, b, c are allowed to roam.The math here is exact. θ₁ ≈ 1.247, θ₂ ≈ −0.445, θ₃ ≈ −1.802. They're the three real roots of x³ + x² − 2x − 1 = 0 — the smallest cyclic cubic field, and the smallest case where this construction works.
(u, v) of L-lattice elements gives a point inside the dashed cyan disc. 0 points. The magenta dots on the inner circle are σ₁(u_ε) — the 8 unit-distance translation directions the proof constructs from q = 13.