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Elliptic surfaces for 6/9 patterns · 5.9

The ABCDEF Pattern: An Even Quartic and Twelve Singular Fibers

One progression of squares, a Gaussian norm, and an x²+2y² norm reduce to an even genus-one quartic. Its Jacobian is a split elliptic K3 surface with passport 12I₂; a non-torsion section generates an explicit polynomial family of ABCDEF squares.

ABCDEFGHJ
ABCDEF: red DEF, yellow ACDE, blue BCDE

1. The exact system

Let a,b,c,d,e,f be rational square roots of the selected entries. The three independent conditions for the ABCDEF pattern are

{d2+f2=2e2,a2+d2=c2+e2,b2+2c2=d2+2e2. \begin{cases} d^2+f^2=2e^2,\\ a^2+d^2=c^2+e^2,\\ b^2+2c^2=d^2+2e^2. \end{cases}

The first equation is the red progression DEF; the second is the yellow equality of two Gaussian norms; the third is the blue equality of x²+2y² norms. The system is sufficient to reconstruct the entire magic square. If

E0=e2,x=a2−E0,y=E0−c2, E_0=e^2,\qquad x=a^2-E_0,\qquad y=E_0-c^2,

then the remaining entries are recovered from the general form

(E0+xE0−x+yE0−yE0−x−yE0E0+x+yE0+yE0+x−yE0−x). \begin{pmatrix} E_0+x&E_0-x+y&E_0-y\\ E_0-x-y&E_0&E_0+x+y\\ E_0+y&E_0+x-y&E_0-x \end{pmatrix}.

2. Start with the DEF progression

Introduce the standard forms

L(t)=t2−2t−1,C(t)=t2+1,R(t)=t2+2t−1. L(t)=t^2-2t-1,\qquad C(t)=t^2+1,\qquad R(t)=t^2+2t-1.

The identity L(t)²+R(t)²=2C(t)² parametrizes the first equation. Up to a common scale, put

d=L(t),e=C(t),f=R(t). d=L(t),\qquad e=C(t),\qquad f=R(t).

The difference required by the yellow condition factors immediately as well:

K(t):=e2−d2=C(t)2−L(t)2=4t(t2−1). K(t):=e^2-d^2=C(t)^2-L(t)^2=4t(t^2-1).

Thus a²−c²=K(t). Rather than guessing a and c, factor this difference into two factors:

u=a+c,a−c=K(t)u, u=a+c,\qquad a-c=\frac{K(t)}u,
a=12(u+K(t)u),c=12(u−K(t)u). a=\frac12\left(u+\frac{K(t)}u\right),\qquad c=\frac12\left(u-\frac{K(t)}u\right).

On this affine chart, the red and yellow conditions now hold identically. Only the blue condition remains.

3. The residual even quartic

To avoid carrying the denominator u, put B=2ub. Substituting the expression for c into the third equation and multiplying by 4u² gives

Ct:B2=−2u4+12C(t)2u2−2K(t)2. \mathcal C_t:\quad B^2=-2u^4+12C(t)^2u^2-2K(t)^2.

Exact lift back

Every rational point (u,B) on this quartic with u≠0 returns a solution of the original system through

a=12(u+Ku),b=B2u,c=12(u−Ku),d=L(t),e=C(t),f=R(t). \begin{aligned} a&=\frac12\left(u+\frac Ku\right),& b&=\frac{B}{2u},& c&=\frac12\left(u-\frac Ku\right),\\ d&=L(t),& e&=C(t),& f&=R(t). \end{aligned}

The quartic has the marked rational point

u0=2t(t−1),B0=4t(t−1)R(t). u_0=2t(t-1),\qquad B_0=4t(t-1)R(t).

It corresponds to the diagonal degeneration a=e, b=f, c=d. Hence the smooth generic fiber is a pointed genus-one curve, not an unresolved torsor.

4. The Jacobian and full 2-torsion

For the binary quartic, compute the classical invariants

I=48(K(t)2+3C(t)4),J=3456C(t)2(K(t)2−C(t)4). \begin{aligned} I&=48\bigl(K(t)^2+3C(t)^4\bigr),\\ J&=3456C(t)^2\bigl(K(t)^2-C(t)^4\bigr). \end{aligned}

The short Jacobian model is

Et:Y2=X3−27I(t)X−27J(t). \mathcal E_t:\quad Y^2=X^3-27I(t)X-27J(t).

Its cubic splits completely in a particularly transparent way:

Y2=(X−36R(t)2)(X−36L(t)2)(X+72C(t)2). Y^2= \bigl(X-36R(t)^2\bigr) \bigl(X-36L(t)^2\bigr) \bigl(X+72C(t)^2\bigr).

Thus all three nonzero points of order 2 are rational over ℚ(t). The complete splitting explains the sign symmetries of the original quartic and makes the singular fibers directly computable.

5. The K3 surface passport

Up to a nonzero constant, the discriminant is

Δ(t)∼t2(t−1)2(t+1)2⋅(t2−2t+3)2(t2+2t+3)2⋅(3t2−2t+1)2(3t2+2t+1)2. \begin{aligned} \Delta(t)\sim{}& t^2(t-1)^2(t+1)^2 \cdot(t^2-2t+3)^2(t^2+2t+3)^2\\ &\cdot(3t^2-2t+1)^2(3t^2+2t+1)^2. \end{aligned}

The eleven finite roots are simple in the reduced discriminant support, while c₄ does not vanish there: these are eleven I₂ fibers. After setting s=1/t and applying the minimal rescaling X=s⁻⁴Xₛ, Y=s⁻⁶Yₛ, the discriminant has order 2 at s=0 while c₄ remains nonzero. There is one further I₂ fiber at infinity.

fiber configuration=12I2.\text{fiber configuration}=12I_2.

The Euler numbers sum to 12·2=24, so the minimal elliptic surface is K3.

6. A non-torsion section and the rank

Choose the marked quartic point as the origin of the group law. Another sign lift of the same diagonal square maps to the section

PX(t)=−36L(t)2,PY(t)=−432C(t)(t4−6t2+1). \begin{aligned} P_X(t)&=-36L(t)^2,\\ P_Y(t)&=-432C(t)(t^4-6t^2+1). \end{aligned}

Non-torsion is proved by exact specialization. At t=2 one obtains

E2:y2=x3−3176496x+114307200,P2=(−36,15120). \mathcal E_2:\quad y^2=x^3-3176496x+114307200,\qquad P_2=(-36,15120).

Good reductions give #E₂(𝔽₅)=8 and #E₂(𝔽₁₃)=16, so the rational torsion order divides 8. Yet the reduction of P₂ modulo 19 has order 7. This is impossible for a torsion point; therefore P and all its nonzero multiples have infinite order.

Proved rank bound

1≤rank⁡E(Q‾(t))≤6. 1\le\operatorname{rank}\mathcal E(\overline{\mathbb Q}(t))\le6.

The section P gives the lower bound. The configuration 12I₂ has root rank 12; Shioda–Tate and ρ≤20 for a complex K3 surface give rank≤20−2−12=6. The exact geometric rank is not claimed here.

7. A polynomial family from 2P

The section P itself encodes a degenerate diagonal square, but doubling leaves the diagonal. Mapping 2P back birationally and clearing the common denominator gives a compact expression in five auxiliary polynomials.

H1=t8−4t7+6t6+8t5+28t4+12t3−6t2+3,H2=3t8−6t6−12t5+28t4−8t3+6t2+4t+1,H=H1H2. \begin{aligned} H_1={}&t^8-4t^7+6t^6+8t^5+28t^4+12t^3-6t^2+3,\\ H_2={}&3t^8-6t^6-12t^5+28t^4-8t^3+6t^2+4t+1,\\ H={}&H_1H_2. \end{aligned}
A0=t16+44t14−96t13−92t12+96t11+788t10+192t9−1226t8−192t7+788t6−96t5−92t4+96t3+44t2+1,B0=5t16−28t15+52t14−108t13+100t12−284t11+524t10−204t9−82t8+204t7+524t6+284t5+100t4+108t3+52t2+28t+5,C0=t16−16t15−4t14−48t13+68t12−176t11−188t10−144t9+502t8+144t7−188t6+176t5+68t4+48t3−4t2+16t+1. \begin{aligned} A_0={}&t^{16}+44t^{14}-96t^{13}-92t^{12}+96t^{11} +788t^{10}+192t^9\\ &-1226t^8-192t^7+788t^6-96t^5-92t^4+96t^3+44t^2+1,\\ B_0={}&5t^{16}-28t^{15}+52t^{14}-108t^{13}+100t^{12}-284t^{11} +524t^{10}\\ &-204t^9-82t^8+204t^7+524t^6+284t^5+100t^4+108t^3+52t^2+28t+5,\\ C_0={}&t^{16}-16t^{15}-4t^{14}-48t^{13}+68t^{12}-176t^{11} -188t^{10}\\ &-144t^9+502t^8+144t^7-188t^6+176t^5+68t^4+48t^3-4t^2+16t+1. \end{aligned}

Then the square roots of the six selected entries are

a=C(t)A0(t),b=R(t)B0(t),c=L(t)C0(t),d=L(t)H(t),e=C(t)H(t),f=R(t)H(t). \begin{aligned} a&=C(t)A_0(t),& b&=R(t)B_0(t),& c&=L(t)C_0(t),\\ d&=L(t)H(t),& e&=C(t)H(t),& f&=R(t)H(t). \end{aligned}

Identity-level correctness

Substituting these six polynomials makes all three ABCDEF equations vanish in ℤ[t]. The ratio a/e is nonconstant, so the family contains infinitely many projectively distinct solutions outside a finite set of degenerate specializations. The multiples 4P,6P,… give further rational families on the same surface.

8. An exact positive example

At t=2 the family gives the following magic square. The signs of the roots do not affect the entries.

A52 · 4333932
B72 · 672 · 46572
C11236312
D32 · 1392 · 9072
E32 · 52 · 1392 · 9072
F32 · 72 · 1392 · 9072
G5889933973889
H2382043636361
J2456743286825
Square entries are written as prime factorizations.

The magic sum is 10,728,720,897,075. All nine entries are positive and pairwise distinct; exactly A,B,C,D,E,F are perfect squares. This is an exact nondegenerate certificate for the ABCDEF pattern.

9. Scope of the result

The result provides an explicit elliptic surface, a proved infinite subgroup of its sections, and a polynomial family of solutions. This is substantially stronger than an isolated fitted example.

It is not claimed that the family from 2P enumerates every rational ABCDEF solution, and the exact geometric rank of the K3 surface has not yet been determined.

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