The Early F1–F8 Families: Explicit tf Coincidences

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1. The equation being solved

Recall the notation from the preceding article:

f(a,b)=ab(a-b)(a+b),\qquad \operatorname{tf}(a,b)=t\!\left(f(a,b)\right).

A pair (a,b) is called nondegenerate when none of the four factors of f vanishes. For two nondegenerate integral pairs, equality of tf values is equivalent to

\operatorname{tf}(a,b)=\operatorname{tf}(c,d) \iff \frac{f(a,b)}{f(c,d)}\in(\mathbb Q^\times)^2.

All formulas below may be read over ℚ. Multiplying each pair by a common denominator makes it integral and multiplies f by a fourth power, so its square class and tf value are unchanged.

Status of the numbering

The index F records the historical shape of a substitution rather than a modern structural class. Different indices may therefore describe equivalent mechanisms, and the absence of an index does not imply the absence of solutions.

2. The linear families F1, F2, and F3

In the first three families, the second pair depends linearly on the first:

\begin{array}{c|c} \mathrm{F1} & (a,b)\longmapsto(2a,a+b)\\ \mathrm{F2} & (a,b)\longmapsto(2a,a-b)\\ \mathrm{F3} & (a,b)\longmapsto(a+b,b) \end{array}

Expanding f cancels three of its four linear factors:

\begin{aligned} f(2a,a+b) &=2a(a+b)(a-b)(3a+b),\\ f(2a,a-b) &=2a(a-b)(a+b)(3a-b),\\ f(a+b,b) &=(a+b)ba(a+2b). \end{aligned}

Hence equality of square classes reduces to

\begin{aligned} \mathrm{F1}:&\quad \frac{f(a,b)}{f(2a,a+b)} =\frac{b}{2(3a+b)}\in(\mathbb Q^\times)^2,\\ \mathrm{F2}:&\quad \frac{f(a,b)}{f(2a,a-b)} =\frac{b}{2(3a-b)}\in(\mathbb Q^\times)^2,\\ \mathrm{F3}:&\quad \frac{f(a,b)}{f(a+b,b)} =\frac{a-b}{a+2b}\in(\mathbb Q^\times)^2. \end{aligned}

In every case only a ratio of two linear forms remains. This is why all three equations reduce to conics and admit complete quadratic parametrizations.

3. Deriving the complete parametrizations of F1–F3

The coefficients in the parametrizations need not be guessed. In each of the three cases, the reduced ratio from the preceding section must be a nonzero rational square. Write that square as n²/m², where m,n∈ℚ×.

\begin{aligned} \mathrm{F1}:&\quad \frac{b}{2(3a+b)}=\frac{n^2}{m^2} \iff b(m^2-2n^2)=6an^2,\\ \mathrm{F2}:&\quad \frac{b}{2(3a-b)}=\frac{n^2}{m^2} \iff b(m^2+2n^2)=6an^2,\\ \mathrm{F3}:&\quad \frac{a-b}{a+2b}=\frac{n^2}{m^2} \iff a(m^2-n^2)=b(m^2+2n^2). \end{aligned}

Once m and n are fixed, each equality leaves only a common scale for the pair (a,b). Denoting it by λ∈ℚ× gives

\begin{aligned} \mathrm{F1}:&\quad \frac{a}{m^2-2n^2} =\frac{b}{6n^2} =\lambda,\\ \mathrm{F2}:&\quad \frac{a}{m^2+2n^2} =\frac{b}{6n^2} =\lambda,\\ \mathrm{F3}:&\quad \frac{a}{m^2+2n^2} =\frac{b}{m^2-n^2} =\lambda. \end{aligned}

It remains only to apply the linear substitutions defining F1, F2, and F3. This derives both pairs at once:

\begin{array}{c|c|c} & (a,b) & (c,d)\\ \hline \mathrm{F1} & \lambda(m^2-2n^2,\;6n^2) & \lambda(2m^2-4n^2,\;m^2+4n^2)\\[2pt] \mathrm{F2} & \lambda(m^2+2n^2,\;6n^2) & \lambda(2m^2+4n^2,\;m^2-4n^2)\\[2pt] \mathrm{F3} & \lambda(m^2+2n^2,\;m^2-n^2) & \lambda(2m^2+n^2,\;m^2-n^2) \end{array}

Every step was an equivalence. The derivation therefore proves both the validity of the formulas and the completeness of each parametrization within its linear ansatz: every nondegenerate rational solution determines a square n²/m² and, after representatives m,n have been chosen, the common scale λ.

The conditions λmn≠0 exclude zero scale and a zero ratio square. For F2 one must additionally require m²≠4n², and for F3, m²≠n²; precisely at those equalities one of the constructed pairs degenerates. In F1 no degeneration over ℚ is possible when λmn≠0.

4. F4: equality of the f values themselves

F4 starts from a stronger condition: the first entries of the two pairs agree, and the f values are not merely in the same square class but are equal:

f(a,b)=f(a,d).

For a≠0, expansion and cancellation of the common factor give

\begin{aligned} b(a^2-b^2)&=d(a^2-d^2),\\ a^2(b-d)&=b^3-d^3\\ &=(b-d)(b^2+bd+d^2). \end{aligned}

The branch b=d is trivial. On the nontrivial branch b≠d one obtains the conic

a^2=b^2+bd+d^2.

Complete parametrization of F4

All rational points on the nontrivial branch, apart from degenerate points, are given by

\begin{aligned} a&=\lambda(m^2+mn+n^2),\\ b&=\lambda(m^2-n^2),\\ d&=\lambda(2mn+n^2), \end{aligned} \qquad \lambda\in\mathbb Q^\times.

Nondegeneracy requires mn(m−n)(m+n)(2m+n)(m+2n)≠0.

Verification

After removing the common scale λ, denote the right-hand sides by A,B,D. Their linear factors are

\begin{aligned} B&=(m-n)(m+n),& A-B&=n(m+2n),& A+B&=m(2m+n),\\ D&=n(2m+n),& A-D&=m(m-n),& A+D&=(m+n)(m+2n). \end{aligned}

Thus both sides consist of exactly the same factors:

\begin{aligned} f(A,B) &=A\,mn(m-n)(m+n)(m+2n)(2m+n)\\ &=f(A,D). \end{aligned}

Why the parametrization is complete

For b≠0, divide the conic by b² and put x=d/b, y=a/b. This gives y²=1+x+x² with the rational point (0,1). The line y=1+sx meets the conic again at

x=\frac{1-2s}{s^2-1},\qquad y=\frac{s^2-s+1}{1-s^2}.

This is the standard parametrization of every rational point on the conic. Substituting s=(m+2n)/(2m+n) and restoring the common scale gives the formulas for a,b,d above. The value 2m+n=0 corresponds to the projective limiting point with d=0 and is already excluded by nondegeneracy.

5. F7: self-recurrence of a progression of squares

F7 assigns to the original pair (m,n) the middle term and the difference of the progression of squares it generates:

(m,n)\longmapsto(C,D),\qquad C=(m^2+n^2)^2,\qquad D=4f(m,n).

Put

S=m^2+n^2,\qquad R=-m^2+2mn+n^2,\qquad W=m^2+2mn-n^2.

The progression-of-squares identities give C−D=R² and C+D=W². Therefore

\begin{aligned} f(C,D) &=CD(C-D)(C+D)\\ &=S^2\cdot4f(m,n)\cdot R^2W^2\\ &=f(m,n)(2SRW)^2. \end{aligned}

The ratio of the two f values is a square, hence

\operatorname{tf}(m,n) = \operatorname{tf}\!\left((m^2+n^2)^2,\;4f(m,n)\right).

The formula holds for every nondegenerate rational pair. It constructs one new representation of the same tf value, but does not claim that every pair with that value is obtained by iterating F7. On the congruent-number elliptic curve, this operation is point doubling.

6. F8: a quartic lift

F8 uses the four expressions

\begin{aligned} A&=u^4+2v^4,&\qquad B&=2v^4,\\ C&=u^4+4v^4,&\qquad D&=4u^2v^2. \end{aligned}

The relevant factors split particularly simply:

A-B=u^4,\qquad A+B=C,\qquad C-D=(u^2-2v^2)^2,\qquad C+D=(u^2+2v^2)^2.

Hence

\frac{f(C,D)}{f(A,B)} = \frac{ 2(u^2-2v^2)^2(u^2+2v^2)^2 }{ u^2v^2(u^4+2v^4) }.

Every factor in this ratio except 2/(u⁴+2v⁴) is already a square. Thus for nondegenerate u,v, the equality tf(A,B)=tf(C,D) holds exactly when there exists w∈ℚ such that

w^2=\frac{u^4}{2}+v^4.

Under this condition the ratio becomes the explicit square

\frac{f(C,D)}{f(A,B)} = \left( \frac{(u^2-2v^2)(u^2+2v^2)}{uvw} \right)^2.

The elliptic curve of F8

The condition on w is a genus-one curve. The substitution

x=\frac{2u^2}{v^2},\qquad y=\frac{4uw}{v^3}

maps it to the elliptic curve

y^2=x^3+8x.

Indeed, substituting w²=u⁴/2+v⁴ gives

y^2 =\frac{16u^2}{v^6}\left(\frac{u^4}{2}+v^4\right) =\frac{8u^6}{v^6}+\frac{16u^2}{v^2} =x^3+8x.

Thus the F8 identity itself is elementary, but choosing admissible u,v is already an elliptic problem. F8 gives an exact family of solutions arising from rational points of the displayed quartic lift, not a classification of every tf coincidence of this kind.

7. What the old indices actually classify

Families	Proved scope	Modern interpretation
F1–F3	Complete coverage of each stated linear ansatz	Different representatives of one linear structural layer
F4	The complete nontrivial branch of f(a,b)=f(a,d)	The conic boundary layer of the later F4+ construction
F7	An identity for every nondegenerate starting pair	Point doubling on the curve of a fixed tf value
F8	An exact condition within the stated quartic ansatz	A quartic lift of a norm family to an elliptic curve

The early numbering is therefore useful as a dictionary of concrete formulas, but it cannot serve as a completeness proof for the general equation tf(a,b)=tf(c,d). A more general theory must classify mechanisms and geometric spaces of solutions rather than the sequence in which substitutions were historically found.

General theory of 4/9 and 5/9 patterns →Back to fmn and tfmn