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Elliptic geometry of square classes · 4.2

F4+: An Elliptic Surface for Pairs of Pythagorean Areas

Equality of rational square classes of two Pythagorean-triangle areas reduces to one cubic with a common parameter. Its Weierstrass model forms a rational elliptic surface of Mordell–Weil rank 2; a quadratic base change produces an elliptic K3 surface of the same geometric rank. The two original parametrizations also determine two independent points on the universal congruent-number curve, so that curve has rank at least 2 over the function field of F4+.

1. Pythagorean areas and congruent-number curves

For nonzero rational u,v, put

f(u,v)=uv(u^2-v^2).

The classical Pythagorean parametrization gives the right triangle

(u^2-v^2)^2+(2uv)^2=(u^2+v^2)^2, \qquad \frac{(u^2-v^2)(2uv)}2=f(u,v).

Thus f(u,v) is its oriented area. The pair is nondegenerate when uv(u−v)(u+v)≠0. Simultaneous scaling of the pair does not change the square class of the area:

f(\lambda u,\lambda v)=\lambda^4f(u,v), \qquad \lambda\in\mathbb Q^\times.

Suppose the positive square class of f(u,v) is represented by the squarefree integer T and f(u,v)=Tq². Dividing the triangle sides by q gives a rational right triangle of area T. Equivalently, the pair determines a nontrivial rational point on the congruent-number curve

C_T:\quad V^2=U^3-T^2U, \qquad (U,V)= \left( \frac{Tu}{v}, \frac{T^2q}{v^2} \right).

Therefore two parameter pairs whose areas have the same positive square class determine two rational points on the same curve C_T. What follows classifies the relation between such pairs; the curve C_T and the auxiliary F4+ surface are distinct elliptic objects.

2. Complete normalization to a common parameter

Normalization theorem

Let (u,v) and (r,s) be two nondegenerate rational pairs. Their areas belong to the same oriented square class if and only if, after projectively scaling the second pair, there are ρ∈ℚ× and a nondegenerate triple a,b,d∈ℚ such that

[(u,v)]=[(a,b)],\qquad [(r,s)]=[(a,d)],\qquad f(a,b)=\rho^2f(a,d).

Here nondegeneracy of the triple means abd(a²−b²)(a²−d²)≠0.

Proof

If the square classes agree, then f(u,v)=σ²f(r,s) for some σ∈ℚ×. Scale the second pair by u/r and put

a=u,\qquad b=v,\qquad d=\frac{us}{r},\qquad \rho=\sigma\left(\frac ru\right)^2.

Fourth-degree homogeneity gives

f(a,d) = f\!\left(\frac ur r,\frac ur s\right) = \left(\frac ur\right)^4f(r,s), \qquad \rho^2f(a,d)=\sigma^2f(r,s)=f(a,b).

The converse follows immediately from the last equality. Thus the common first parameter is not an additional restriction: it is a coordinate normalization for every pair of equal square classes.

3. The normalized cubic

Now consider the complete equation

ab(a^2-b^2)=\rho^2ad(a^2-d^2).

For nondegenerate triples a,b,d, divide by ab³ and introduce the ratios

x=\frac ab,\qquad y=\frac db,\qquad \tau=\rho^2.

This gives the family of plane cubics

S_\tau:\qquad x^2-1=\tau y(x^2-y^2).

In the original arithmetic problem, τ must be a nonzero rational square. To study the surface, it is useful to regard τ temporarily as an independent parameter over ℚ.

Weierstrass-model theorem

For τ≠0 and τ²≠1, the cubic S_τ is birationally equivalent to the elliptic curve

\mathcal E_\tau:\qquad Y^2=X^3-3\tau^2X+\tau^2(\tau^2+1).

The forward map is

X=\frac{\tau(\tau-y)}{1-\tau y}, \qquad Y=\frac{\tau(\tau^2-1)x}{1-\tau y}.

The inverse map is

x=\frac{Y}{\tau(X-1)}, \qquad y=\frac{X-\tau^2}{\tau(X-1)}.

4. Proof of birationality

Denote the left-hand side of the equation of S_τ by

R_\tau(x,y)=x^2-1-\tau y(x^2-y^2).

Direct substitution into the Weierstrass model gives the exact factor identity

\begin{aligned} Y^2-&\left(X^3-3\tau^2X+\tau^2(\tau^2+1)\right)\\ &= \frac{\tau^2(\tau^2-1)^2} {(1-\tau y)^3} R_\tau(x,y). \end{aligned}

Hence every point of S_τ in the domain of the map is sent to 𝓔_τ. Conversely,

R_\tau\!\left( \frac{Y}{\tau(X-1)}, \frac{X-\tau^2}{\tau(X-1)} \right) = \frac{\tau^2-1}{\tau^2(X-1)^3} \left[ Y^2-X^3+3\tau^2X-\tau^2(\tau^2+1) \right].

Both compositions are identities. For the first one, this follows immediately from

X-1=\frac{\tau^2-1}{1-\tau y}, \qquad X-\tau^2= \frac{\tau y(\tau^2-1)}{1-\tau y}.

Substituting y=1/τ into R_τ gives (1−τ²)/τ². Thus the denominator of the forward map cannot vanish on S_τ when τ²≠1. The inverse map excludes the two points

(X,Y)=\left(1,\pm(\tau^2-1)\right),

which form the boundary of the chosen affine normalization b≠0. For the original problem one must additionally remove points whose recovered triple is degenerate:

xy(x^2-1)(x^2-y^2)=0.

5. Singular fibers and the known conic

For the model 𝓔_τ, the invariants are

\begin{aligned} \Delta&=-432\,\tau^4(\tau-1)^2(\tau+1)^2,\\ c_4&=144\,\tau^2,\\ j&=-\frac{6912\,\tau^2}{(\tau-1)^2(\tau+1)^2}. \end{aligned}

Thus the smooth elliptic fibers degenerate at τ=0,±1 and at infinity. At τ=1, the original cubic splits into exactly two components:

R_1(x,y) =(1-y)(x^2-y^2-y-1).

The component y=1 means d=b and gives the trivial pairing of a pair with itself. The nontrivial component is the conic

x^2=y^2+y+1, \qquad\text{то есть}\qquad a^2=b^2+bd+d^2.

This is precisely the layer previously used to construct families of congruent-number curves of rank at least 2. At τ=−1, similarly,

R_{-1}(x,y) =(1+y)(x^2-y^2+y-1).

The fiber τ=−1 is needed for the geometric passport of the surface, although it does not arise from rational ρ in τ=ρ².

6. Visible sections and explicit families

The surface has two immediately visible rational sections

P(\tau)=\bigl(\tau,\tau(\tau-1)\bigr), \qquad Q(\tau)=\bigl(\tau^2,\tau(\tau^2-1)\bigr).

The sections P and Q themselves lie on the degenerate boundary: the inverse map gives (x,y)=(1,−1) and (1,0), respectively. The same holds for P+Q, which gives (−1,1). Their role is not in these three trivial solutions but in the Mordell–Weil group: other section combinations already recover nondegenerate pairs.

Section	$X$	$Y$	$x=a/b$	$y=d/b$
$P-Q$	$3\tau+4$	$(\tau+1)(\tau+8)$	$\dfrac{\tau+8}{3\tau}$	$\dfrac{4-\tau}{3\tau}$
$2P$	$-2\tau$	$-\tau(\tau-1)$	$\dfrac{\tau-1}{2\tau+1}$	$\dfrac{\tau+2}{2\tau+1}$
$P+2Q$	$4-3\tau$	$(\tau-1)(\tau-8)$	$\dfrac{8-\tau}{3\tau}$	$\dfrac{\tau+4}{3\tau}$

Each row is an identically valid rational family of solutions to x²−1=τy(x²−y²). A finite table of combinations does not replace the set 𝓔_τ(ℚ), however: for fixed τ, determining all rational points remains a separate arithmetic problem.

7. Passport of the rational elliptic surface

Regard τ as a coordinate on the projective line. The minimal elliptic surface 𝓔→ℙ¹_τ has four singular fibers:

Base point	Kodaira type	Root lattice	Euler number
$\tau=0$	IV	A₂	4
$\tau=1$	I₂	A₁	2
$\tau=-1$	I₂	A₁	2
$\tau=\infty$	IV	A₂	4

The Euler numbers sum to 12, so this is a rational elliptic surface. Over an algebraic closure its Picard number is 10. The total root-lattice rank of the singular fibers is 2+1+1+2=6. The Shioda–Tate formula gives

\operatorname{rank} \mathcal E_\tau(\overline{\mathbb Q}(\tau)) =10-2-6=2.

The sections P and Q are independent. Indeed, any integral relation between them would specialize on every good fiber where both sections are defined. At τ=9 one obtains

Y^2=X^3-243X+6642, \qquad P=(9,72),\quad Q=(81,720).

An exact computation of the rational-point group of this curve gives rank 2 and P,Q as independent generators. Hence P,Q give the lower bound 2 already over ℚ(τ), while Shioda–Tate gives the matching geometric upper bound:

\operatorname{rank}\mathcal E_\tau(\mathbb Q(\tau)) = \operatorname{rank}\mathcal E_\tau(\overline{\mathbb Q}(\tau)) =2.

The good specialization at τ=2 has trivial torsion, so specialization of torsion also rules out nontrivial torsion over ℚ(τ). Thus P and Q exhaust the free rank up to finite index. This is a statement over the function field; it does not say that every rational specialization has rank exactly 2.

8. Quadratic base change and the K3 surface

For equality of square classes, the parameter τ must have the form τ=ρ². This base change gives the original family

\mathcal E_\rho:\qquad Y^2=X^3-3\rho^4X+\rho^4(\rho^4+1).

The ramified pullback of the two type-IV fibers gives type-IV* fibers at ρ=0 and ρ=∞. The fibers τ=1 and τ=−1 lift to four type-I₂ fibers at ρ=±1,±i. Their Euler numbers sum to

8+8+2+2+2+2=24,

so the minimal surface is an elliptic K3 surface. The total rank of its root lattices is

6+6+1+1+1+1=16.

For a complex K3 surface the Picard number is at most 20. By Shioda–Tate,

\operatorname{rank} \mathcal E_\rho(\overline{\mathbb Q}(\rho)) \le 20-2-16=2.

The pulled-back sections P,Q remain independent, so equality holds. Thus the quadratic base change τ=ρ² creates no new free sections over an algebraic closure:

\operatorname{rank} \mathcal E_\rho(\overline{\mathbb Q}(\rho))=2.

9. A universal congruent-number curve of rank at least 2

F4+ has two distinct elliptic interpretations. The surface 𝓔_ρ parametrizes the triples themselves. Every such triple, in turn, gives two points on one congruent-number curve. In the normalization a=1, write

A=x(x^2-1),\qquad \mathcal C_A:\quad v^2=u^3-A^2u.

On the surface

x^2-1=\rho^2y(x^2-y^2)

the two Pythagorean triangles give the following rational points:

R_x= \left(x^2(x^2-1),\,x^2(x^2-1)^2\right), \qquad R_y= \left(\rho^2x^2(x^2-y^2),\, \rho^3x^2(x^2-y^2)^2\right).

Theorem on two independent points

Let K be the function field over ℚ of the irreducible surface x²−1=ρ²y(x²−y²). Then Rₓ and Rᵧ lie in 𝓒_A(K) and are independent modulo torsion. In particular,

\operatorname{rank}\mathcal C_A(K)\ge 2.

Proof

First note that the defining polynomial of the surface is irreducible: as a quadratic polynomial in ρ, it could factor over ℚ(x,y) only if (x²−1)/(y(x²−y²)) were a square, but its order at y=0 is −1.

Substitution of Rₓ into 𝓒_A is the standard passage from a Pythagorean triangle to the curve of its area. For Rᵧ, the same calculation, after removing common factors, reduces exactly to the defining equation of F4+:

\begin{aligned} v(R_x)^2-u(R_x)^3+A^2u(R_x)&=0,\\ v(R_y)^2-u(R_y)^3+A^2u(R_y) &=\rho^2x^4(x^2-y^2) \Bigl((x^2-1)^2-\rho^4y^2(x^2-y^2)^2\Bigr)\\ &=\rho^2x^4(x^2-y^2) \bigl(x^2-1-\rho^2y(x^2-y^2)\bigr) \bigl(x^2-1+\rho^2y(x^2-y^2)\bigr)=0. \end{aligned}

so the last expression vanishes on the surface. A single good specialization is enough to prove independence. Take

\rho=1,\qquad x=\frac73,\qquad y=\frac53, \qquad 7^2=3^2+3\cdot5+5^2.

After scaling all sides by 9/2, this gives the right triangles (20,21,29) and (12,35,37), both of area 210. Hence the specialized universal curve and sections are

E_{210}:\quad v^2=u^3-210^2u, \qquad R_x=(490,9800),\quad R_y=(294,3528).

Use the standard Kummer map for a curve with full rational 2-torsion, with its usual limiting values at the 2-torsion points:

\delta:E_{210}(\mathbb Q)/2E_{210}(\mathbb Q) \longrightarrow (\mathbb Q^\times/\mathbb Q^{\times2})^2, \qquad (u,v)\longmapsto(u,u-210).

For the two independent 2-torsion points T₀=(0,0), T₊=(210,0) and the two points under consideration, the square classes are

\delta(T_0)=(-1,-210),\quad \delta(T_+)=(210,2),\quad \delta(R_x)=(10,70),\quad \delta(R_y)=(6,21).

In the square-class basis (−1,2,3,5,7), these four elements give the matrix

\left( \begin{array}{ccccc|ccccc} 1&0&0&0&0&1&1&1&1&1\\ 0&1&1&1&1&0&1&0&0&0\\ 0&1&0&1&0&0&1&0&1&1\\ 0&1&1&0&0&0&0&1&0&1 \end{array} \right),

It has rank 4 over 𝔽₂. Moreover, direct counting gives #E₂₁₀(𝔽₁₁)=12 and #E₂₁₀(𝔽₁₃)=20. Both reductions are good, so the order of the rational torsion divides 4; the visible points (0,0), (±210,0) already form a group of order 4. Thus the torsion is exactly E₂₁₀[2]. In any relation aRₓ+bRᵧ=T with T∈E₂₁₀[2], the matrix rank forces a and b to be even and T to vanish. Divide the resulting relation by 2 and repeat: a and b are divisible by every power of 2, hence zero. The points Rₓ and Rᵧ are independent.

Any integral relation between the universal sections would specialize to the same relation on E₂₁₀, which is impossible. Therefore the sections are independent over K.

Exact scope of the result

F4+ gives a complete coordinate model for every equality of square classes between two nondegenerate Pythagorean parametrizations.
The two natural points are independent over the function field of the full F4+ surface; this is a separate theorem, not a consequence of the rank of the auxiliary surface 𝓔_ρ.
The theorem does not say that independence survives every rational specialization, nor does it by itself prove that dependent specializations have density zero.

10. Relation to previous work

What is known independently of F4+

Malvina Baica studied constructions of several rational Pythagorean triangles with one common area. 1988 paper.
Lorenz Halbeisen and Norbert Hungerbühler used the conic a²=b²+bd+d²—the ρ=1 layer of our model—to construct an infinite family of congruent-number curves of rank at least 2. 2019 paper.
Raiza Corpuz gave a 2-descent proof of the classical relation between nm²=uv(u²−v²) and y²=x³−n²x, and constructed families of rank at least 2 and 3. 2020 preprint.
Rational elliptic surfaces with four singular fibers were classified by Herfurtner; the configuration IV²I₂² occurs in the Persson–Miranda list. The abstract type of the auxiliary surface is therefore not claimed as new. Herfurtner · Miranda.

What F4+ adds

F4+ unifies these themes in a complete model for an arbitrary square ratio between two Pythagorean areas: it proves normalization to a common first parameter, gives explicit birational coordinates, determines the passport of the rational surface and its K3 base change, and proves independence of the two natural points on the universal congruent-number curve. We did not find this complete construction or the final theorem in the sources listed above; that statement is relative to the literature reviewed here and may be refined if an earlier source is identified.

The label F4+ arose historically within the study of magic squares: equality of square classes appeared when matching the differences of two arithmetic progressions of squares. This motivation is not used in the proof of the surface, but it explains the name.

Generating pairs with F4+ and F7+ →The congruent-curve bijection Theory contents