Pythagorean Triples via Rational Points on Conics
The Pythagorean equation x² + y² = z² is intimately connected to the geometry of the unit circle. Finding integer solutions is equivalent to finding rational points on X² + Y² = 1.
The key insight: draw lines through the rational point (−1, 0) with rational slope t = q/p. Each such line intersects the circle at exactly one other rational point, yielding a Pythagorean triple.
Setting t = q/p and clearing denominators gives the classical parametrisation:
The Eisenstein norm form x² − xy + y² looks unfamiliar next to the Gaussian x² + y². But one line of algebra reveals them as the same object — with a single coefficient changed. Complete the square in x:
So x² − xy + y² is a sum of two squares: it's U² + V² where U = x − y/2 and V = √3 · y/2. The Eisenstein norm is the Gaussian norm, viewed through coordinates that stretch one axis by √3.
This is why the unit curve x² − xy + y² = 1 is an ellipse rather than a circle: the √3 compresses the y-direction relative to the x-direction. The eccentricity of the ellipse (≈ 0.816) is determined entirely by this factor.
Now homogenise: set U = X/Z, V = Y/Z, and clear denominators. The unit curve U² + 3V² = 1 becomes
That single 3 is the entire difference between Gaussian and Eisenstein arithmetic, made visible. It is the discriminant of the ring ℤ[ω], the factor that determines which primes split and which remain inert, and geometrically it is the ratio of the eigenvalues of the Eisenstein quadratic form.
The passage from Gaussian to Eisenstein is not a different kind of mathematics — it is the same structure with a different discriminant. The 3 is not a curiosity; it is the organising constant of the entire theory.
The quadratic form X² − XY + Y² = 1 can be written in matrix form as:
The eigenvalues of E are λ₁ = 1/2 and λ₂ = 3/2, giving semi-axes a = √2 (along λ₁) and b = √(2/3) (along λ₂).
The eccentricity is:
This is a fairly eccentric ellipse — closer to a parabola (e = 1) than to a circle (e = 0). Notice that the eigenvalue ratio is exactly 3 — the same 3 from the completed square.
Lines through (−1, 0) with slope t = n/m intersect the Eisenstein ellipse at rational points corresponding to E-P triples.
Setting t = n/m with gcd(m, n) = 1 yields:
| m | n | t = n/m | a | b | c | Classification |
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