The Elliptic Layers of F9+: From y²=x³−2x to Families of Curves

1. The basic F9 layer

Consider the quadratic substitution

C=a^2-b^2,\qquad D=b^2.

Its residual quotient follows by direct expansion:

\begin{aligned} f(C,D) &=(a^2-b^2)b^2\bigl((a^2-b^2)^2-b^4\bigr)\\ &=(a^2-b^2)b^2\bigl(a^4-2a^2b^2\bigr)\\ &=ab(a^2-b^2)\cdot ab(a^2-2b^2)\\ &=f(a,b)\cdot ab(a^2-2b^2). \end{aligned}

Hence, for nondegenerate pairs, equality of the oriented tf values is equivalent to the existence of s∈ℚ× such that

s^2=ab(a^2-2b^2).

For b≠0, put x=a/b and y=s/b². This gives the elliptic curve

E_9:\qquad y^2=x^3-2x.

2. Exact correspondence within the substitution

Basic-layer theorem

Up to a common scaling of each pair, nondegenerate solutions of the basic substitution are in bijection with rational points (x,y) on E₉ satisfying x≠0,±1 and x²≠2, modulo y↦−y.

(x,y) \longmapsto \bigl([x:1],\,[x^2-1:1]\bigr).

The forward direction is the identity for f proved above. Conversely, every projective input pair with b≠0 has a unique form [x:1]. If the substitution preserves tf, its residual x(x²−2) is a square, so a point (x,±y) exists on E₉. The two constructions are inverse.

3. Why there are infinitely many solutions

The point G=(2,2) lies on E₉. The doubling formula gives

2G=\left(\frac94,-\frac{21}{8}\right).

The curve y²=x³−2x is given by a minimal integral equation. If G were torsion, then 2G would also be torsion. By the Nagell–Lutz theorem its coordinates would have to be integral unless 2G were the identity. But 2G has nonintegral coordinates and is not 𝒪. Therefore G has infinite order.

The multiples nG give infinitely many distinct x-coordinates and hence infinitely many pairs

\bigl([x(nG):1],\,[x(nG)^2-1:1]\bigr)

with equal tf. For example, G gives [2:1] and [3:1]: f(2,1)=6 and f(3,1)=24=4·6.

4. A parametric narrow layer

The basic formula can be replaced by the three-parameter ansatz

C=b(a-nb),\qquad D=p a^2+mab-nb^2.

Before cancellation, the quotient is

\frac{f(C,D)}{f(a,b)} =- \frac{ (a-nb)\, (pa+(m-1)b)\, (pa^2+mab-nb^2)\, (pa^2+(m+1)ab-2nb^2) }{ (a-b)(a+b) }.

For polynomial cancellation, one of the four numerator factors must vanish at a=b and one at a=−b. These two choices impose linear conditions on p,m,n. After the identically zero branch is removed, 14 nonzero one-parameter branches remain; a typical residual is a binary quartic.

5. The parameter is not a scaling artifact

On the branch p=n, m=0, one has

\begin{aligned} C&=b(a-nb),\\ D&=n(a^2-b^2),\\ q_n(a,b) &=-n(a-nb)(na-b)(na^2+ab-2nb^2). \end{aligned}

This is a genuine family, not a reparametrization of one layer. At n=2 and n=3 one obtains smooth elliptic curves with distinct j-invariants:

j_{n=2}=\frac{2^2\,5^6}{3\cdot11}, \qquad j_{n=3}=\frac{2^6\,5^3\,11^3}{3^4\cdot73}.

If the parameter could be removed by a rational change of coordinates and scaling, the j-invariant would remain constant. The two different values rule this out.

6. Recorded elliptic layers

Two early lists record 16 concrete quadratic substitutions with residual factorization 1+1+2. Eight come from the narrow ansatz and eight from the full space of pairs of quadratic forms. The latter eight occupy only seven of the 204 branches of the general theorem; two formulas are different points on one branch.

Layer	Curve model
F9_A	$Y^2=-(6t+1)(12t^2+9t+1)$
F9_B	$Y^2=-4(3t+2)(6t^2+6t+1)$
F9_C	$Y^2=t^3-3t^2+t+1$
F9_D	$Y^2=(6t+1)(6t^2+6t+1)$
F9_E	$Y^2=-2(3t+2)(6t^2+9t+2)$
F9_F	$Y^2=-(2t+3)(2t^2-15t-9)$
F9_G	$Y^2=-3(8t+3)(24t^2+19t+3)$
F9_H	$Y^2=-3(2t+3)(6t^2+10t+3)$
F9__1	$Y^2=X^3+6X^2-8X$
F9__2	$Y^2=X^3-6X^2-8X$
F9__3	$Y^2=X^3-18X^2+64X$
F9__4	$Y^2=X^3-24X^2+256X$
F9__5	$Y^2=X^3-24X^2+162X$
F9__6	$Y^2=X^3+84X^2+1296X$
F9__7	$Y^2=X^3+24X^2+256X$
F9__8	$Y^2=X^3-84X^2+1296X$

This table is an inventory of constructed layers, not a classification of all elliptic branches. In particular, a 2+2 residual may also have rational points and yield an elliptic model even though no linear root is visible in advance.

7. How one point produces a tf pair

Suppose that on a chosen branch

f(C(a,b),D(a,b))=f(a,b)q(a,b)

and a rational point on the residual curve gives s²=q(a,b). Then

f(C(a,b),D(a,b))=s^2f(a,b),

so the input and output pairs have the same tf. If the residual curve has a point of infinite order and the map to r=a/b is nonconstant, its multiples give infinitely many such pairs; only the finite set of points where a factor of f vanishes is removed.

8. Embedding into F4+

Let b≠0, r=a/b, and write C(r)=C(r,1), D(r)=D(r,1), q(r)=q(r,1). Scaling the output pair by a/C produces a common first parameter:

(C,D)\sim\left(a,\frac{aD}{C}\right).

In F4+ coordinates, the corresponding curve is given by

y_F(r)=\frac{rD(r)}{C(r)}, \qquad \rho^2(r)=\frac{C(r)^4}{r^4q(r)}.

On the residual curve s²=q(r), the parameter ρ becomes the rational function

\rho(r,s)=\frac{C(r)^2}{r^2s}.

Thus every F9+ layer is an explicit curve or multisection inside the complete F4+ surface. This is a consequence of normalization, not the definition of F9+.

9. Passage to F7+ and the boundary of the result

Every F9+ point creates two pairs [a:b] and [C:D] in one square class T. Through the F7+ bijection they become two rational points on the same congruent-number curve E_T. After signs are chosen, the group law on E_T produces further representations of the same T.

What is proved and what remains open

The exact polynomial-cancellation layer and the residual-quartic criterion are completely classified. Each concrete layer still has its own arithmetic questions: rational solubility, Jacobian rank, independence of the induced points on E_T, and branch identifications after quotienting by symmetries.