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Elliptic surfaces for 6/9 patterns · 5.12

The ABCDFG Pattern: Red, Yellow, and Blue Quadrics

Another three-color type in the series combines the BFG progression, the Gaussian norm on BCDG, and an x²+2y² norm on ACFG. After parametrizing the red equation, the remaining two quadrics reduce to an even genus-one curve. Its Jacobian is a split 12I₂ elliptic K3 surface with a non-torsion section.

ABCDEFGHJ
ABCDFG: red BFG; yellow BCDG; blue ACFG

1. The exact three-color system

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

{b2+f2=2g2,b2+c2=d2+g2,2a2+g2=c2+2f2. \begin{cases} b^2+f^2=2g^2,\\ b^2+c^2=d^2+g^2,\\ 2a^2+g^2=c^2+2f^2. \end{cases}

The first equation is the red BFG progression; the second is the yellow equality of Gaussian norms on BCDG; the third is the blue equality of X²+2Y² norms on ACFG. Together they are necessary and sufficient for the six selected entries.

To recover the whole square, put

E0=c2+g22,x=a2−E0,y=c2−g22. E_0=\frac{c^2+g^2}{2},\qquad x=a^2-E_0,\qquad y=\frac{c^2-g^2}{2}.(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. The red BFG progression

Use 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 L²+R²=2C² parametrizes the first quadric. On the chosen affine chart take

b=L(t),g=C(t),f=R(t). b=L(t),\qquad g=C(t),\qquad f=R(t).

For the two remaining conditions, isolate the shifts

α(t):=L(t)2−C(t)2=−4t(t2−1),β(t):=2R(t)2−C(t)2=t4+8t3+2t2−8t+1. \begin{aligned} \alpha(t)&:=L(t)^2-C(t)^2=-4t(t^2-1),\\ \beta(t)&:=2R(t)^2-C(t)^2\\ &=t^4+8t^3+2t^2-8t+1. \end{aligned}

The yellow and blue equations now become

d2−c2=α(t),2a2−c2=β(t). d^2-c^2=\alpha(t),\qquad 2a^2-c^2=\beta(t).

3. The residual even quartic

Put v=c+d. For v≠0 the yellow condition is inverted completely:

c=12(v−αv),d=12(v+αv). c=\frac12\left(v-\frac{\alpha}{v}\right),\qquad d=\frac12\left(v+\frac{\alpha}{v}\right).

Substituting c into the blue quadric and setting Y=4av gives

Ct:Y2=2v4+(8β(t)−4α(t))v2+2α(t)2. \mathcal C_t:\qquad Y^2=2v^4+(8\beta(t)-4\alpha(t))v^2+2\alpha(t)^2.

Exact lift back

Every rational point (v,Y) with v≠0 lifts to an ABCDFG solution:

a=Y4v,b=L(t),c=12(v−αv),d=12(v+αv),f=R(t),g=C(t). \begin{aligned} a&=\frac{Y}{4v},& b&=L(t),& c&=\frac12\left(v-\frac{\alpha}{v}\right),\\ d&=\frac12\left(v+\frac{\alpha}{v}\right),& f&=R(t),& g&=C(t). \end{aligned}

The quartic already has the rational points (C+L,4R(C+L)) and (C−L,±4R(C−L)). They correspond to the diagonal lifts a²=f², c²=g², d²=b². Thus the generic smooth fiber is a pointed genus-one curve.

4. The Jacobian and full 2-torsion

The classical binary-quartic invariants are

I=64(t8+20t7+116t6+20t5−218t4−20t3+116t2−20t+1),J=−1024R(t)2(t4+10t3+2t2−10t+1)⋅(t4+16t3+2t2−16t+1). \begin{aligned} I={}&64(t^8+20t^7+116t^6+20t^5-218t^4\\ &\qquad-20t^3+116t^2-20t+1),\\ J={}&-1024R(t)^2 (t^4+10t^3+2t^2-10t+1)\\ &\qquad\cdot(t^4+16t^3+2t^2-16t+1). \end{aligned}

Take the short Jacobian model

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

Its cubic splits completely over ℚ(t):

Y2=(X−24(t4+16t3+2t2−16t+1))⋅(X−24(t4+4t3+2t2−4t+1))⋅(X+48(t4+10t3+2t2−10t+1)). \begin{aligned} \mathsf Y^2={}& \bigl(\mathsf X-24(t^4+16t^3+2t^2-16t+1)\bigr)\\ &\cdot\bigl(\mathsf X-24(t^4+4t^3+2t^2-4t+1)\bigr)\\ &\cdot\bigl(\mathsf X+48(t^4+10t^3+2t^2-10t+1)\bigr). \end{aligned}

Hence the Jacobian has full rational 2-torsion. The three colors of the original system do not prevent the surface from splitting as completely as in the preceding cases.

5. The K3 surface passport

Up to a nonzero constant, the discriminant is

Δ(t)∼t2(t−1)2(t+1)2⋅(t4+8t3+2t2−8t+1)2⋅(t4+12t3+2t2−12t+1)2. \begin{aligned} \Delta(t)\sim{}& t^2(t-1)^2(t+1)^2\\ &\cdot(t^4+8t^3+2t^2-8t+1)^2\\ &\cdot(t^4+12t^3+2t^2-12t+1)^2. \end{aligned}

The eleven finite roots of the reduced discriminant support are simple, and c₄ does not vanish there: these are eleven I₂ fibers. After the minimal transformation in the chart s=1/t, one further I₂ appears at infinity.

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

The total Euler number is 24, so the minimal elliptic surface is K3.

6. A non-torsion section and the rank

The covariant map from the pointed quartic to its Jacobian sends the diagonal point to the section

P(t)=(3U16(t)(LCR)2,−27V1(t)V2(t)V3(t)(LCR)3), P(t)=\left( \frac{3U_{16}(t)}{(LCR)^2}, -\frac{27V_1(t)V_2(t)V_3(t)}{(LCR)^3} \right),
U16=11t16+56t15+200t14+536t13+3604t12+248t11−11784t10−232t9+18754t8+232t7−11784t6−248t5+3604t4−536t3+200t2−56t+11,V1=t8−12t7−28t6−12t5+70t4+12t3−28t2+12t+1,V2=3t8+12t7−20t6+12t5+82t4−12t3−20t2−12t+3,V3=t8+4t7+36t6+4t5−58t4−4t3+36t2−4t+1. \begin{aligned} U_{16}={}&11t^{16}+56t^{15}+200t^{14}+536t^{13}+3604t^{12}\\ &+248t^{11}-11784t^{10}-232t^9+18754t^8+232t^7\\ &-11784t^6-248t^5+3604t^4-536t^3+200t^2-56t+11,\\ V_1={}&t^8-12t^7-28t^6-12t^5+70t^4+12t^3-28t^2+12t+1,\\ V_2={}&3t^8+12t^7-20t^6+12t^5+82t^4-12t^3-20t^2-12t+3,\\ V_3={}&t^8+4t^7+36t^6+4t^5-58t^4-4t^3+36t^2-4t+1. \end{aligned}

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

E2:y2=x3−13231296x+13933624320,P2=(522917611225,−37679625675942875). \mathcal E_2:\quad y^2=x^3-13231296x+13933624320, \quad P_2=\left(\frac{52291761}{1225}, -\frac{376796256759}{42875}\right).

Good reductions satisfy #E₂(𝔽₁₁)=12 and #E₂(𝔽₁₃)=16, so the rational torsion order divides 4. Yet the image of P₂ modulo 11 has order 3. Therefore P has 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 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. This is the rank of global sections, not a universal upper bound for the ranks of individual specializations.

7. An explicit polynomial family

Take a parabola through (C+L,4R(C+L)) and (C−L,−4R(C−L)), tangent at the first point. The fourth intersection has coordinates

v1=−2(t+1)Q8(t)D8(t),Y1=−8(t+1)R(t)M16(t)D8(t)2. v_1=-\frac{2(t+1)Q_8(t)}{D_8(t)},\qquad Y_1=-\frac{8(t+1)R(t)M_{16}(t)}{D_8(t)^2}.
Q8=7t8+20t7−44t6−28t5+82t4−52t3+4t2−4t−1,D8=t8−4t7−4t6−52t5−82t4−28t3+44t2+20t−7,M16=5t16+8t15−72t14+808t13+940t12−696t11−3704t10−1496t9+6942t8+1496t7−3704t6+696t5+940t4−808t3−72t2−8t+5. \begin{aligned} Q_8={}&7t^8+20t^7-44t^6-28t^5+82t^4-52t^3+4t^2-4t-1,\\ D_8={}&t^8-4t^7-4t^6-52t^5-82t^4-28t^3+44t^2+20t-7,\\ M_{16}={}&5t^{16}+8t^{15}-72t^{14}+808t^{13}+940t^{12}-696t^{11}\\ &-3704t^{10}-1496t^9+6942t^8+1496t^7-3704t^6+696t^5\\ &+940t^4-808t^3-72t^2-8t+5. \end{aligned}

For a compact formula, define also

S4=t4+12t3+2t2−12t+1,F12=t12+28t11+6t10−172t9+15t8+824t7+20t6−824t5+15t4+172t3+6t2−28t+1,D16=t16−56t15−424t14−1048t13+188t12+264t11−3608t10+1256t9+7942t8−1256t7−3608t6−264t5+188t4+1048t3−424t2+56t+1. \begin{aligned} S_4={}&t^4+12t^3+2t^2-12t+1,\\ F_{12}={}&t^{12}+28t^{11}+6t^{10}-172t^9+15t^8+824t^7+20t^6\\ &-824t^5+15t^4+172t^3+6t^2-28t+1,\\ D_{16}={}&t^{16}-56t^{15}-424t^{14}-1048t^{13}+188t^{12}+264t^{11}\\ &-3608t^{10}+1256t^9+7942t^8-1256t^7-3608t^6-264t^5\\ &+188t^4+1048t^3-424t^2+56t+1. \end{aligned}

After lifting back and clearing the common denominator, one obtains degree-18 roots:

a=−R M16,b=−L D8Q8,c=C S4F12,d=−L D16,f=−R D8Q8,g=−C D8Q8. \begin{aligned} a&=-R\,M_{16},& b&=-L\,D_8Q_8,& c&=C\,S_4F_{12},\\ d&=-L\,D_{16},& f&=-R\,D_8Q_8,& g&=-C\,D_8Q_8. \end{aligned}

Identity-level correctness

Substitution makes the red, yellow, and blue quadrics vanish in ℤ[t]. The ratio a/g is nonconstant, so outside a finite set of degenerate parameters the family contains infinitely many projectively distinct rational solutions.

8. An exact positive example

At t=−3, after removing the common factor of the roots, one obtains the following magic square. Root signs do not affect the entries.

A41369712
B72 · 9372 · 12172
C52 · 472 · 532 · 6432
D72 · 292 · 1032 · 4672
E48322934532625
F9372 · 12172
G52 · 9372 · 12172
H32928707881441
J79531340010409
Square entries are written as prime factorizations.

The magic sum is 144,968,803,597,875. All nine entries are positive and pairwise distinct; exactly A,B,C,D,F,G are perfect squares.

9. Exact scope of the result

The quartic model describes the generic nondegenerate open part of ABCDFG. The cases v=0, a constant red progression, denominator zeros, and the chart t=∞ require adjacent projective charts, but do not change the generic Jacobian.

The K3 passport, existence of a non-torsion section, geometric rank bound, and an explicit infinite family are proved. The exact rank, completeness of the displayed family, and positivity of every rational specialization are not asserted.

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