ABCDEG: Two Gaussian Gluings and a 12I₂ Elliptic K3 Surface

1. The exact system

Let a,b,c,d,e,g be rational square roots of the selected entries. The ABCDEG pattern leaves three independent quadrics

\begin{cases} c^2+g^2=2e^2,\\ a^2+d^2=c^2+e^2,\\ a^2+b^2=e^2+g^2. \end{cases}

The first equation is the red CEG progression; the next two are yellow equalities of Gaussian norms on ACDE and ABEG. The system is necessary and sufficient for the six selected entries. Put

E_0=e^2,\qquad x=a^2-E_0,\qquad y=E_0-g^2.

Then the whole magic square is recovered in the standard form

\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. The central CEG progression

Introduce the standard forms

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

The identity L²+R²=2C² gives the complete projective parametrization of a nonconstant progression of squares. On the chosen affine chart put

c=L(t),\qquad e=C(t),\qquad g=R(t).

The difference of the endpoint entries is

\delta(t):=c^2-g^2=L(t)^2-R(t)^2=-8t(t^2-1).

Subtracting the two yellow equations now eliminates a and gives the particularly simple relation

d^2-b^2=c^2-g^2=\delta(t).

3. The residual even quartic

Put v=b+d. For v≠0 the preceding equality is inverted completely:

b=\frac12\left(v-\frac{\delta}{v}\right),\qquad d=\frac12\left(v+\frac{\delta}{v}\right).

It remains to require that the recovered value a² also be a square. After setting Y=2av, both yellow equations give the same equation

\mathcal C_t:\qquad Y^2=-v^4+8C(t)^2v^2-\delta(t)^2.

Exact lift back

Every rational point (v,Y) on this quartic with v≠0 lifts to an ABCDEG solution through

\begin{aligned} a&=\frac{Y}{2v},& b&=\frac12\left(v-\frac{\delta}{v}\right),& c&=L(t),\\ d&=\frac12\left(v+\frac{\delta}{v}\right),& e&=C(t),& g&=R(t). \end{aligned}

The quartic has rational points v=L+R, Y=2C(L+R) and v=L−R, Y=±2C(L−R). They correspond to sign lifts of the diagonal solution a²=e², b²=g², d²=c². Thus the generic smooth fiber is a pointed genus-one curve rather than a nontrivial torsor without a rational point.

4. The Jacobian and full 2-torsion

The classical invariants of the binary quartic are

\begin{aligned} I={}&64(t^8+16t^6-18t^4+16t^2+1),\\ J={}&-1024C(t)^2 (t^4-6t^3+2t^2+6t+1)\\ &\hspace{5.7em}\cdot(t^4+6t^3+2t^2-6t+1). \end{aligned}

Take the short Jacobian model

\mathcal E_t:\qquad \mathsf Y^2=\mathsf X^3-27I(t)\mathsf X-27J(t).

Its cubic splits completely over ℚ(t):

\begin{aligned} \mathsf Y^2={}& \bigl(\mathsf X-24(t^4+6t^3+2t^2-6t+1)\bigr)\\ &\cdot\bigl(\mathsf X-24(t^4-6t^3+2t^2+6t+1)\bigr)\\ &\cdot\bigl(\mathsf X+48C(t)^2\bigr). \end{aligned}

Hence the Jacobian has full rational 2-torsion. This is a property of the chosen fibration of the surface; it does not mean that all rational points have already been enumerated.

5. The K3 surface passport

Up to a nonzero constant, the discriminant is

\begin{aligned} \Delta(t)\sim{}& t^2(t-1)^2(t+1)^2\\ &\cdot(t^4-2t^3+2t^2+2t+1)^2\\ &\cdot(t^4+2t^3+2t^2-2t+1)^2. \end{aligned}

The eleven finite roots of the reduced discriminant support are simple, and c₄ does not vanish at them. They therefore give eleven I₂ fibers. After setting s=1/t and applying the minimal rescaling, the discriminant has order 2 at s=0 while c₄ remains nonzero: there is one further I₂ fiber at infinity.

\text{fiber configuration}=12I_2.

The Euler numbers sum to 24, so the minimal elliptic surface is K3.

6. A non-torsion section and the rank

Write

\begin{aligned} U={}&t^8-2t^6+18t^4-2t^2+1,\\ V_-={}&t^4-2t^3+2t^2+2t+1,\\ V_+={}&t^4+2t^3+2t^2-2t+1. \end{aligned}

The covariant map from the quartic to its Jacobian gives the section

P(t)=\left( \frac{96U(t)}{C(t)^2}, \frac{864L(t)R(t)V_-(t)V_+(t)}{C(t)^3} \right).

Non-torsion is certified by the exact specialization at t=2:

\mathcal E_2:\quad y^2=x^3-1826496x-463795200, \qquad P_2=\left(\frac{39264}{25},-\frac{2909088}{125}\right).

Good reductions satisfy #E₂(𝔽₇)=4 and #E₂(𝔽₂₃)=20, so the rational torsion order divides 4. Yet the image of P₂ modulo 11 has order 6. This is impossible for a torsion point, so P has infinite order.

Proved rank bound

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

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

7. An explicit polynomial family

To obtain a non-diagonal point on the quartic itself, take a parabola through

(L+R,\,2C(L+R)),\qquad (L-R,\,-2C(L-R))

and require tangency at the first point. The fourth intersection has coordinate

v_1=-\frac{4tQ_1(t)Q_2(t)}{P_-(t)P_+(t)},

\begin{aligned} P_\pm&=t^4\pm4t^3+10t^2\pm4t+1,\\ Q_1&=t^4-2t^2+5,\qquad Q_2=5t^4-2t^2+1,\\ H&=P_-P_+Q_1Q_2. \end{aligned}

Define two further polynomials

\begin{aligned} A_{16}={}&7t^{16}-168t^{14}+516t^{12}-2456t^{10}+5994t^8\\ &-2456t^6+516t^4-168t^2+7,\\ B_{16}={}&t^{16}+48t^{15}-88t^{14}-16t^{13}+92t^{12}+688t^{11}\\ &-872t^{10}+752t^9+1990t^8-752t^7-872t^6-688t^5\\ &+92t^4+16t^3-88t^2-48t+1. \end{aligned}

After lifting back and clearing the common denominator, the six square roots are

\begin{aligned} a&=C(t)A_{16}(t),& b&=-R(t)B_{16}(t),& c&=L(t)H(t),\\ d&=L(t)B_{16}(-t),& e&=C(t)H(t),& g&=R(t)H(t). \end{aligned}

Identity-level correctness

Substitution makes all three ABCDEG equations vanish in ℤ[t]. The ratio a/e is nonconstant, so outside a finite set of degenerate parameters the family contains infinitely many projectively distinct rational solutions. The Jacobian group law gives further families, but the displayed family is not claimed to enumerate every rational point.

8. An exact positive example

At t=4/5, after removing the common rational scale, one obtains the following magic square. Root signs do not affect the entries.

A19² · 41² · 1904865750013²

B31² · 39066764315519²

C5² · 7⁴ · 13² · 17² · 29² · 89² · 1601² · 8161²

D7⁴ · 11² · 129089² · 26757859²

E5² · 13² · 17² · 29² · 41² · 89² · 1601² · 8161²

F1202155296701953230994185478129

G5² · 13² · 17² · 29² · 31² · 89² · 1601² · 8161²

H3201706256042037743233998634129

J2466465282865062936601360603921

Square entries are written as prime factorizations.

The magic sum is 7,002,594,088,855,587,635,573,178,990,075. All nine entries are positive and pairwise distinct; exactly A,B,C,D,E,G are perfect squares.

9. Exact scope of the result

The quartic model describes the generic nondegenerate open part of the ABCDEG pattern. The excluded cases v=0, a constant red progression, and the parameter chart at infinity require adjacent projective charts, but they do not change the computed Jacobian of the generic fiber.

The existence of a non-torsion section, the geometric rank bound, and an explicit infinite family are proved. Neither the exact rank, completeness of one family, nor automatic positivity of every rational specialization is asserted.

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