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Elliptic surfaces for 6/9 patterns · 5.8

The ABEFGH Pattern: A Triangle of Pairwise Means

Here the six square entries form three arithmetic progressions with no distinguished central vertex: three initial squares have pairwise square means. Compatibility of the progressions reduces to one multiplicative problem for a rational function r, whose Jacobian is a pullback of the Legendre family on an elliptic K3 surface.

ABCDEFGHJ
ABEFGH: red AFH, BEH, and BFG

1. The exact system

Let a,b,e,f,g,h be rational square roots of the selected entries. The three red conditions are

{f2+h2=2a2,b2+h2=2e2,b2+f2=2g2. \begin{cases} f^2+h^2=2a^2,\\ b^2+h^2=2e^2,\\ b^2+f^2=2g^2. \end{cases}

This is a triangle of pairwise means: b²,f²,h² may be regarded as the initial vertices, while a²,e²,g² are the means of their three pairs. The system is not only necessary but also sufficient to reconstruct the magic square. Put

E0=e2,x=a2−E0,y=a2−h2. E_0=e^2,\qquad x=a^2-E_0,\qquad y=a^2-h^2.

Then A=a², E=e², and H=h² by construction. The second equation gives B=b², the first gives F=f², and the third gives G=g². The remaining entries are uniquely recovered from the general form of a magic square.

2. Closing the three progressions

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.L(t)2+R(t)2=2C(t)2.L(t)^2+R(t)^2=2C(t)^2.

Each of the three progressions can be written, up to a common scale, as an L,C,R triple. Introduce parameters x,z,y for BEH, AFH, and BFG respectively. The endpoint ratios are then

bh=L(x)R(x),hf=L(z)R(z),bf=L(y)R(y). \frac bh=\frac{L(x)}{R(x)},\qquad \frac hf=\frac{L(z)}{R(z)},\qquad \frac bf=\frac{L(y)}{R(y)}.

Closing the triangle expresses the third ratio through the first two. Thus, if

r(t)=L(t)R(t)=t2−2t−1t2+2t−1, r(t)=\frac{L(t)}{R(t)} =\frac{t^2-2t-1}{t^2+2t-1},

all scale compatibility reduces to

r(x) r(z)=r(y).r(x)\,r(z)=r(y).

The remainder studies this rational chart. It generates solutions of the original system, but no claim is made here that the chart covers every rational ABEFGH solution.

3. A fiber product of two conics

The equation is quadratic in y. After isolating its discriminant and substituting s=z−1/z, rationality splits into two conics. Their product gives the genus-one quartic

U2=(s2+4)((x2+1)2s2+16x(x2−1)s+4(x2+1)2). U^2=(s^2+4)\Bigl( (x^2+1)^2s^2+16x(x^2-1)s+4(x^2+1)^2 \Bigr).

The first factor records invertibility of the substitution z−1/z; the second records that the discriminant of the equation for y is a square. Thus the original compatibility condition is read as a fiber product of two rational conics over the same s-line.

4. The Jacobian and Legendre form

The binary-quartic invariants give the compact Jacobian model

Ex:Y2=X3−27P(x)X−54Q(x), \mathcal E_x:\quad Y^2=X^3-27P(x)X-54Q(x),P(x)=x8−8x6+30x4−8x2+1,Q(x)=(x2+1)2(x4−10x2+1)(x4−4x2+1). \begin{aligned} P(x)&=x^8-8x^6+30x^4-8x^2+1,\\ Q(x)&=(x^2+1)^2(x^4-10x^2+1)(x^4-4x^2+1). \end{aligned}

The cubic polynomial splits completely:

Y2=(X+3(x2+1)2)(X+3(x4−10x2+1))⋅(X−6(x4−4x2+1)). \begin{aligned} Y^2={}&\bigl(X+3(x^2+1)^2\bigr) \bigl(X+3(x^4-10x^2+1)\bigr)\\ &\cdot\bigl(X-6(x^4-4x^2+1)\bigr). \end{aligned}

Hence full rational 2-torsion is visible over ℚ(x). Sending the three roots to 0,1,λ gives the Legendre form

v2=u(u−1)(u−λ),λ=4x2(x2−1)2. v^2=u(u-1)(u-\lambda),\qquad \lambda=\frac{4x^2}{(x^2-1)^2}.

Thus the ABEFGH surface is an explicit pullback of the standard one-parameter Legendre family under the rational substitution λ=4x²/(x²−1)².

5. The elliptic K3 surface passport

The discriminant of the compact model factors as

Δ=28312x4(x−1)4(x+1)4(x2−2x−1)2(x2+2x−1)2. \Delta= 2^8 3^{12}x^4(x-1)^4(x+1)^4 (x^2-2x-1)^2(x^2+2x-1)^2.

The points x=0,±1 and infinity give four I₄ fibers; the four roots of the remaining quadratic factors give four I₂ fibers. All fibers are semistable, and their Euler numbers sum to 24. The minimal elliptic surface is K3 with passport

4I4+4I2.4I_4+4I_2.

6. A non-torsion section and the rank bound

The Legendre model has the visible section

u=(x2+1x2−1)2,v=2x(x2+1)(x2−1)2. u=\left(\frac{x^2+1}{x^2-1}\right)^2,\qquad v=\frac{2x(x^2+1)}{(x^2-1)^2}.

Its non-torsion can be proved by an exact specialization. At x=2 the section maps to (17,−60) on the curve

E2:y2=x3−x2−64x+64=(x−8)(x−1)(x+8). \mathcal E_2:\quad y^2=x^3-x^2-64x+64 =(x-8)(x-1)(x+8).

Exact point counts give #𝓔₂(𝔽₅)=8 and #𝓔₂(𝔽₁₁)=16, so the rational torsion order divides 8. Modulo 17 the selected point has order 3, which is impossible for such a torsion point. Therefore the section has infinite order.

Proved bound

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

The non-torsion section gives the lower bound. The root rank of 4I₄+4I₂ 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. Lifting back and a polynomial family

Double the section and lift 2S from the Legendre model back to the r-equation. This gives the rational functions

z(x)=x4−2x3+2x2+2x+1x4+2x3+2x2−2x+1,y(x)=(x+1)(x4−2x3−2x2−2x+1)(x−1)(x4+2x3−2x2+2x+1). \begin{aligned} z(x)&=\frac{x^4-2x^3+2x^2+2x+1} {x^4+2x^3+2x^2-2x+1},\\[1mm] y(x)&=\frac{(x+1)(x^4-2x^3-2x^2-2x+1)} {(x-1)(x^4+2x^3-2x^2+2x+1)}. \end{aligned}

They satisfy the identity

r(x)r(z(x))=r(y(x)).r(x)r(z(x))=r(y(x)).

The denominators can be cleared without expanding the large products. For a pair T=(T₀,T₁), put

LT=T02−2T0T1−T12,CT=T02+T12,RT=T02+2T0T1−T12. \begin{aligned} L_T&=T_0^2-2T_0T_1-T_1^2,\\ C_T&=T_0^2+T_1^2,\\ R_T&=T_0^2+2T_0T_1-T_1^2. \end{aligned}

Take X=(x,1), and let Z=(Z₀,Z₁) and Y=(Y₀,Y₁) be the numerator-denominator pairs displayed for z(x),y(x). Then an explicit polynomial family of square roots has the compact form

a=RXCZRY,b=LXLZRY,e=CXLZRY,f=RXRZRY,g=RXRZCY,h=RXLZRY. \begin{aligned} a&=R_XC_ZR_Y,& b&=L_XL_ZR_Y,& e&=C_XL_ZR_Y,\\ f&=R_XR_ZR_Y,& g&=R_XR_ZC_Y,& h&=R_XL_ZR_Y. \end{aligned}LXLZRY=RXRZLY.L_XL_ZR_Y=R_XR_ZL_Y.

The identities L²+R²=2C² immediately prove AFH and BEH. The lifted equality between the two expressions for b substitutes it into the third condition and proves BFG. Hence all three ABEFGH equations hold identically in ℤ[x]. Nonconstant ratios yield infinitely many projectively distinct solutions outside a finite set of degenerate specializations.

8. An exact positive example

At x=2, after removing the common factor of the square roots, the family gives the following magic square. Root signs do not affect its entries.

A5383272 · 7692= 28 976 689
B10812232 · 472= 1 168 561
C57 496 825nonsquare
D57 734 161nonsquare
E5405252 · 232 · 472= 29 214 025
F833274 · 172= 693 889
G965252 · 1932= 931 225
H7567272 · 232 · 472= 57 259 489
J29 451 361nonsquare
Square entries are shown through their roots; when the root is composite, the prime factorization of the square appears below.

Its magic sum is 87,642,075. All entries are positive and pairwise distinct; exactly A,B,E,F,G,H are perfect squares. This is an exact nondegenerate certificate for the ABEFGH pattern.

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