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Elliptic surfaces for 6/9 patterns · 5.7

The ABDEFJ Pattern: A Chain of Three Progressions

In the second triangular pattern all three defining relations are again red, but their incidence is different. The DEF and AEJ progressions share the center E, while the BJD progression links them through D and J. The residual condition gives a palindromic quartic, a split elliptic K3 surface, and an explicit infinite family of solutions.

ABCDEFGHJ
ABDEFJ: red BDJ, DEF, and AEJ

1. The exact system and reconstruction

Let a,b,d,e,f,j be rational square roots of the six selected entries. In the label BDJ the middle term of the progression is J: the geometric order of the cells is B–J–D. Hence the three red conditions are

{b2+d2=2j2,d2+f2=2e2,a2+j2=2e2. \begin{cases} b^2+d^2=2j^2,\\ d^2+f^2=2e^2,\\ a^2+j^2=2e^2. \end{cases}

The system is not only necessary but also sufficient to reconstruct the magic square. Put

E0=e2,x=a2−E0,y=E0−x−d2. E_0=e^2,\qquad x=a^2-E_0,\qquad y=E_0-x-d^2.

Then A=a², D=d², and E=e² by construction. The third equation gives J=E₀−x=j², the second gives F=E₀+x+y=f², and the first gives B=E₀−x+y=b². Thus the three displayed progressions describe the ABDEFJ pattern exactly.

2. Two progressions with a shared center

Use the standard identity

L(z)=z2−2z−1,C(z)=z2+1,R(z)=z2+2z−1, L(z)=z^2-2z-1,\qquad C(z)=z^2+1,\qquad R(z)=z^2+2z-1,L(z)2+R(z)2=2C(z)2.L(z)^2+R(z)^2=2C(z)^2.

The DEF and AEJ progressions can be parametrized simultaneously by gluing them along their common center E:

d=L(t)C(s),f=R(t)C(s),e=C(t)C(s),a=C(t)L(s),j=C(t)R(s). \begin{aligned} d&=L(t)C(s),& f&=R(t)C(s),\\ e&=C(t)C(s),\\ a&=C(t)L(s),& j&=C(t)R(s). \end{aligned}

Both progressions now hold identically. To obtain ABDEFJ it remains to require a rational b satisfying b²+d²=2j². The rest of the article studies this rational chart; completeness of the chart for every rational solution of the original pattern is not asserted.

3. The residual palindromic quartic

Write the unknown square root b as V. Substitution gives

V2=2C(t)2R(s)2−L(t)2C(s)2. V^2=2C(t)^2R(s)^2-L(t)^2C(s)^2.

The identity 2C(t)²−L(t)²=R(t)² puts the right-hand side into the form

Gt(s)=R(t)2s4+8C(t)2s3+2R(t)2s2−8C(t)2s+R(t)2=R(t)2(s2+1)2+8C(t)2s(s2−1). \begin{aligned} G_t(s)={}&R(t)^2s^4+8C(t)^2s^3+2R(t)^2s^2\\ &-8C(t)^2s+R(t)^2\\ ={}&R(t)^2(s^2+1)^2+8C(t)^2s(s^2-1). \end{aligned}

This is a genus-one quartic over ℚ(t). The points (0,R(t)) and (1,2R(t)) are immediate, but they belong to the degenerate boundary where several entries coincide. They serve as base points for the group law rather than as the desired squares.

4. The split Jacobian

The invariants of the binary quartic give the Jacobian model

Y2=X3−432P(t)X+3456R(t)2D5(t)D7(t), Y^2=X^3-432P(t)X+3456R(t)^2D_5(t)D_7(t),P(t)=13t8+8t7+68t6+8t5+46t4−8t3+68t2−8t+13,D5(t)=5t4−4t3+10t2+4t+5,D7(t)=7t4+4t3+14t2−4t+7. \begin{aligned} P(t)={}&13t^8+8t^7+68t^6+8t^5+46t^4\\ &-8t^3+68t^2-8t+13,\\ D_5(t)={}&5t^4-4t^3+10t^2+4t+5,\\ D_7(t)={}&7t^4+4t^3+14t^2-4t+7. \end{aligned}

The cubic polynomial splits completely:

Y2=(X−12D5(t))(X−24Q(t))(X+12D7(t)),Q(t)=t4+4t3+2t2−4t+1. \begin{aligned} Y^2={}&\bigl(X-12D_5(t)\bigr) \bigl(X-24Q(t)\bigr) \bigl(X+12D_7(t)\bigr),\\ Q(t)={}&t^4+4t^3+2t^2-4t+1. \end{aligned}

Thus full rational 2-torsion is visible over ℚ(t). Up to a nonzero constant the discriminant is

Δ(t)∼C(t)4L(t)4(t2+2t+3)2(3t2−2t+1)2. \Delta(t)\sim C(t)^4L(t)^4 \bigl(t^2+2t+3\bigr)^2 \bigl(3t^2-2t+1\bigr)^2.

The roots of C and L give four I₄ fibers, while the roots of the remaining two quadratic factors give four I₂ fibers. The fiber at infinity is smooth. The Euler numbers sum to 24, so the minimal elliptic surface is K3.

5. A new section from a tangent

Pass the following parabola through the marked intersections:

Πt(s)=R2−4C2Rs2−4(R2−C2)Rs+R, \Pi_t(s)= \frac{R^2-4C^2}{R}s^2 -\frac{4(R^2-C^2)}{R}s+R,

where L,C,R are evaluated at t. It is tangent to the quartic at s=1, and the difference factors without solving a general quartic equation:

R2(Gt(s)−Πt(s)2)=8s(s−1)2(C2s+R2)(R2−2C2). R^2\bigl(G_t(s)-\Pi_t(s)^2\bigr) =8s(s-1)^2(C^2s+R^2)(R^2-2C^2).

The remaining, previously unaccounted-for root gives the rational section

s(t)=−R(t)2C(t)2,V(t)=R(t)(R(t)4−3C(t)4)C(t)4. s(t)=-\frac{R(t)^2}{C(t)^2},\qquad V(t)= \frac{R(t)\bigl(R(t)^4-3C(t)^4\bigr)}{C(t)^4}.

6. Non-torsion and the rank bound

At t=2 the section gives the quartic point (−49/25,3682/625). On the minimal Jacobian model it becomes

E2:y2=x3−x2−3300x+74052=(x−34)(x−33)(x+66),P2=(171649,−4950343). \begin{aligned} \mathcal E_2:\quad y^2 &=x^3-x^2-3300x+74052\\ &=(x-34)(x-33)(x+66),\\ P_2&=\left(\frac{1716}{49},-\frac{4950}{343}\right). \end{aligned}

Exact counts give #𝓔₂(𝔽₇)=8 and #𝓔₂(𝔽₁₃)=16, so the rational torsion order divides 8. Modulo 19 the point P₂ reduces to (4,9), which has order 12. For a torsion point the reduction order must divide its own order, which is impossible here. Hence P₂ and the original section have infinite order.

Proved bound

1≤rank⁡E(Q‾(t))≤2. 1\le\operatorname{rank}\mathcal E(\overline{\mathbb Q}(t))\le2.

The non-torsion section gives the lower bound. The root rank of the 4I₄+4I₂ configuration is 16; Shioda–Tate and ρ≤20 for a complex K3 surface give the upper bound 2. The exact geometric rank has not yet been determined.

7. An explicit polynomial family

To clear the denominator of the section, put

M=R(t)2,N=C(t)2,UC=M2+N2,UL=M2+2MN−N2,UR=M2−2MN−N2,B0=M2−3N2. \begin{aligned} M&=R(t)^2,& N&=C(t)^2,\\ U_C&=M^2+N^2,\\ U_L&=M^2+2MN-N^2,\\ U_R&=M^2-2MN-N^2,\\ B_0&=M^2-3N^2. \end{aligned}

Then the selected-entry square roots are

a=C UL,b=R B0,d=L UC,e=C UC,f=R UC,j=C UR. \begin{aligned} a&=C\,U_L,& b&=R\,B_0,& d&=L\,U_C,\\ e&=C\,U_C,& f&=R\,U_C,& j&=C\,U_R. \end{aligned}

Family identities

b2+d2=2j2,d2+f2=2e2,a2+j2=2e2. b^2+d^2=2j^2,\qquad d^2+f^2=2e^2,\qquad a^2+j^2=2e^2.

All three equalities are identities in ℤ[t]. The ratio f/e=R(t)/C(t) is nonconstant, so after excluding finitely many degenerate specializations the family contains infinitely many projectively distinct rational solutions of the ABDEFJ system.

8. An exact positive example

At t=2, after dividing the roots by their common factor 2, one obtains the following square. Root signs do not affect the entries.

A10565252 · 21132= 111 619 225
B1841272 · 2632= 3 389 281
C56 679 169nonsquare
D15132172 · 892= 2 289 169
E7565252 · 172 · 892= 57 229 225
F10591272 · 172 · 892= 112 169 281
G57 779 281nonsquare
H111 069 169nonsquare
J1685252 · 3372= 2 839 225
Square entries are shown through their roots; when the root is composite, the prime factorization of the square appears below.

The magic sum is 171,687,675. All nine entries are positive and pairwise distinct; exactly A,B,D,E,F,J are perfect squares. This is an exact nondegenerate certificate for the ABDEFJ pattern.

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