affine·mess

A solution is two lattice points on one curve

Scale a solution so that x⁶+y⁶+z⁶ = 1. Both triples then sit on the unit superellipsoid, and when the squares also agree (59 of the 74 known small solutions) they sit on one sphere too, hence on the curve where the two surfaces cut. Drag to rotate, scroll to zoom.

M₆: x⁶+y⁶+z⁶ = 1 (superellipsoid)
M₂: x²+y²+z² = r² (sphere)
the curve M₆ ∩ M₂
(|x|,|y|,|z|)   (|u|,|v|,|w|)
drag · scroll

The Brudno pencil on Bremner's surface

Every point of V lies on exactly one fibre Eλ, an elliptic curve. Move λ and watch the curve (drawn in Bremner's Weierstrass model, compactified so that nothing leaves the picture) and the points that the 144 sections of degree ≤ 12 cut out on it. At a rational λ each marked point is an integer solution: hover over it.

Brudno–Kaplansky, 1970 and 1974

In 1970 Brudno reduced a special family of solutions to the quartic z² = x⁴ + 38x²y² + y⁴ and generated new solutions with two classical recursions. In 1974 Brudno and Kaplansky put the same family on the cubic E₁ and found that all of it is the cyclic group generated by one point P.

1970 · the quartic z² = x⁴ + 38x² + 1 (y = 1)

x ↦ (x⁴−y⁴, 2xyz) is multiplication by 2 on the quartic and x ↦ (x(4y⁴z²−g²), y(4x⁴z²−g²)), g = x⁴−y⁴, is multiplication by −3. Each point gives a solution through A = 6xy(x²−y²), B = −(x⁴−2xyz−y⁴), C = x⁴+2xyz−y⁴ and D = z(x²+y²), E = 2xyz, F = (x²−y²)(x²+y²).

1974 · the cubic E₁: W² = (2T+1)(10T²+2T+1)

On the quartic, Lebesgue's recursion is the doubling map [2] and Desboves's is [−3]; the seed (1, 2, 13) corresponds to 2P, so from it the 1970 paper could only reach ±2·2a·3b·P — the solutions 2P, 4P, 6P, 8P, … — and 3P appeared there only through the irrational parameter 17 ± 7√5. The odd multiples, 5P in particular, were out of reach until the 1974 cubic made the whole cyclic group ⟨P⟩ ⊕ ⟨R⟩ visible.