F9+: Quadratic Substitutions and the Cancellation Theorem

1. Searching for solutions by substitution

For a rational pair, put

f(a,b)=ab(a-b)(a+b)=ab(a^2-b^2).

One way to obtain two pairs with the same tf is to express a new pair (c,d) in terms of the original parameters (a,b) and require

\frac{f(c,d)}{f(a,b)}=y^2.

In the F1, F2, and F3 families, linear substitutions reduce the residual condition to rational conics and yield complete elementary parametrizations. Here we consider the next level: quadratic substitutions whose residual condition, in the nondegenerate case, defines a genus-one curve and, after choosing a rational point, an elliptic curve.

2. The basic F9 example

An empirical search leads to the substitution

c=a^2-b^2,\qquad d=b^2.

It does not make the two f-values identically equal. Instead, direct expansion gives the exact cancellation

\begin{aligned} f(a^2-b^2,b^2) &=f(a,b)\,ab(a^2-2b^2),\\ \frac{f(a^2-b^2,b^2)}{f(a,b)} &=ab(a^2-2b^2). \end{aligned}

Consequently, this substitution preserves tf precisely on the rational solutions of

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

For b≠0, the normalization x=a/b and y=s/b² turns the condition into the elliptic curve

y^2=x^3-2x.

This example suggests the general question: which homogeneous quadratic forms C and D allow f(C,D)/f(a,b) to cancel to a single polynomial residual?

3. The general quadratic substitution

Let C and D be arbitrary homogeneous quadratic forms over ℚ:

\begin{aligned} C(a,b)&=A a^2+B ab+C_0 b^2,\\ D(a,b)&=E a^2+F ab+G b^2. \end{aligned}

A common factor of all six coefficients is immaterial. Replacing (C,D) by (λC,λD) multiplies f(C,D) by λ⁴ and hence leaves its rational square class unchanged. The coefficients are therefore naturally projective.

The numerator splits into four quadratic factors:

f(C,D)=C\,D\,(C-D)(C+D).

The denominator to be cancelled consists of four distinct lines:

f(a,b)=a\,b\,(a-b)(a+b).

4. The cancellation theorem

F9+ theorem

The polynomial f(C,D) is divisible by f(a,b) if and only if, on each of the four lines a=0, b=0, a=b, and a=−b, at least one of C, D, C−D, and C+D vanishes identically.

Necessity

If, for example, a divides f(C,D), then the restriction of f(C,D) to a=0 is the zero polynomial in b. On that line each of C, D, C−D, and C+D becomes a constant times b². Since ℚ[b] has no zero divisors, their product is zero only when one factor vanishes identically. The same argument applies to the other three lines.

Sufficiency

If one factor vanishes on every base line, then f(C,D) vanishes on that line. Hence the corresponding linear form divides f(C,D). The forms a, b, a−b, and a+b are pairwise coprime, so their product also divides f(C,D).

5. The finite linear arrangement

For each of the four base lines there are four choices of a vanishing factor, giving 4⁴=256 assignments. Every assignment imposes four linear conditions on the six coefficients A, B, C₀, E, F, and G.

After identical linear systems are identified, 224 subspaces remain: 220 projective lines and four projective planes. The four planes are precisely the identically degenerate cases

C=0,\qquad D=0,\qquad C=D,\qquad C=-D.

A further 16 of the 220 lines lie entirely inside those planes. Removing them leaves 204 projective lines whose open parts have f(C,D) not identically zero. This is an exact cover of the polynomial layer before quotienting by the natural symmetries of the input and output pairs.

What the number 204 means

This is the number of linear branches in the chosen coordinates, not the number of essentially distinct families. Swaps, sign changes, and linear symmetries of f may carry branches into one another; no quotient by those symmetries is claimed here.

The basic F9 construction belongs to this arrangement as a point on the one-dimensional branch

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

At λ=0 this recovers the substitution of the preceding section. Thus the original F9 is a special case of the general F9+ layer, not a separate construction.

6. The residual quartic and the exact tf criterion

On every nondegenerate branch, the quotient

q(a,b)=\frac{f(C(a,b),D(a,b))}{f(a,b)}

is a homogeneous form of degree 4. At a rational point where both f-values are nonzero, one has the exact equivalence

\operatorname{tf}(C(a,b),D(a,b)) = \operatorname{tf}(a,b) \iff q(a,b)\in(\mathbb Q^\times)^2.

Indeed, two nonzero rational numbers have the same squarefree part exactly when their quotient is a positive rational square. If pair orientation is forgotten, changing the sign of one output parameter replaces q by −q.

Homogeneity makes the condition projective. For b≠0, set r=a/b and s=b²z. Then

s^2=q(a,b) \iff z^2=q(r,1).

7. Why genus-one curves appear

If the quartic q(r,1) is squarefree and has degree 3 or 4, the smooth projective model of z²=q(r,1) has genus one. Its factorization controls not the genus but the availability of a rational base point and the structure of 2-torsion.

Factorization of q	Geometric meaning
1+1+2	A genus-one curve with rational points above the linear roots; after choosing an origin, an elliptic curve with visible rational 2-torsion.
2+2	A genus-one curve, but a rational point is not guaranteed. In general it is a torsor under its Jacobian.
1+1+1+1	A genus-one curve with four rational branch points; after choosing an origin, its full rational 2-torsion subgroup is visible.
Repeated factor	The projective curve is singular; its normalization has genus zero or is degenerate.

8. Exact completeness boundary

What is classified completely

The theorem covers every pair of homogeneous quadratic forms C,D for which f(C,D)/f(a,b) is a polynomial. Within each such substitution, z²=q(r,1) is necessary and sufficient for equality of the oriented tf values.

This does not exhaust every quadratic way of producing equal tf values: a rational quotient without polynomial cancellation may still become a square on a curve. Nor does it classify all coincidences between arbitrary rational pairs; their complete coordinate descriptions are supplied by F4+ and F7+.

9. The role of F9+

F4+ and F7+ describe all solutions in two universal coordinate systems, but do not themselves enumerate rational points. F9+ serves a different purpose: it selects explicit one-parameter layers on which the search for pairs is transferred to concrete genus-one curves. Rational points on those curves directly produce pairs with equal tf.