Magic Squaresproof atlas
← Back to the 6/9 surfaces series

Elliptic surfaces for 6/9 patterns · 5.5

The ABCEGH and ABCEGJ Patterns: One Surface, Two Readings

In both patterns two progressions of squares share the center E. After a simultaneous parametrization, the two different yellow relations lead to the same genus-one quartic: a split elliptic K3 surface with two provably independent sections.

1. Two exact systems

Let lowercase letters denote rational square roots of the selected entries. For ABCEGH the independent conditions are

ABCEGH:{c2+g2=2e2,b2+h2=2e2,a2+c2=e2+h2. \mathrm{ABCEGH}:\qquad \begin{cases} c^2+g^2=2e^2,\\ b^2+h^2=2e^2,\\ a^2+c^2=e^2+h^2. \end{cases}

For ABCEGJ the second progression and the yellow relation change:

ABCEGJ:{c2+g2=2e2,a2+j2=2e2,b2+e2=g2+j2. \mathrm{ABCEGJ}:\qquad \begin{cases} c^2+g^2=2e^2,\\ a^2+j^2=2e^2,\\ b^2+e^2=g^2+j^2. \end{cases}

These are not merely necessary consequences of magicity. In either case put

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

The general form of a magic square then reconstructs C=c² and E=e². The first equation gives G=g². For ABCEGH the third equation gives H=h² and the second then gives B=b². For ABCEGJ the second equation gives J=j² and the third gives B=b². Hence each system is sufficient for its pattern.

2. The shared center E

Use the standard triple

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.

Parametrize the common progression CEG and the shared center of the two second progressions by

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

For ABCEGH attach the progression BEH:

b=C(p)L(q),h=C(p)R(q),a=V. b=C(p)L(q),\qquad h=C(p)R(q),\qquad a=V.

For ABCEGJ the same two endpoint roots occupy A and J, while B becomes the unknown entry:

a=C(p)L(q),j=C(p)R(q),b=V. a=C(p)L(q),\qquad j=C(p)R(q),\qquad b=V.

All four red relations now hold identically. In each reading only the corresponding yellow relation remains to be imposed.

3. Why the quartic is identical

For ABCEGH the unknown square is e²+h²−c²; for ABCEGJ it is g²+j²−e². Their equality follows from the elementary identity

C(p)2−L(p)2=R(p)2−C(p)2=4p(p2−1). C(p)^2-L(p)^2=R(p)^2-C(p)^2=4p(p^2-1).V2=C(p)2R(q)2+4p(p2−1)C(q)2=R(p)2q4+4C(p)2q3+2R(p)2q2−4C(p)2q+R(p)2. \begin{aligned} V^2 &=C(p)^2R(q)^2+4p(p^2-1)C(q)^2\\ &=R(p)^2q^4+4C(p)^2q^3+2R(p)^2q^2\\ &\qquad-4C(p)^2q+R(p)^2. \end{aligned}

One curve, two maps

A rational point (p,q,V) on this quartic simultaneously gives a solution to both systems: V occupies A in ABCEGH and B in ABCEGJ. Completeness of this chart for all rational solutions of either pattern is not claimed.

4. The split Jacobian

The binary-quartic invariants give the short model

Y2=X3−1728P(p)X+27648R(p)2Q1(p)Q2(p), Y^2=X^3-1728P(p)X+27648R(p)^2Q_1(p)Q_2(p),P(p)=p8+2p7+8p6+2p5−2p4−2p3+8p2−2p+1,Q1(p)=p4−2p3+2p2+2p+1,Q2(p)=p4+p3+2p2−p+1. \begin{aligned} P(p)={}&p^8+2p^7+8p^6+2p^5-2p^4\\ &-2p^3+8p^2-2p+1,\\ Q_1(p)={}&p^4-2p^3+2p^2+2p+1,\\ Q_2(p)={}&p^4+p^3+2p^2-p+1. \end{aligned}

The cubic polynomial splits completely:

(X−24R(p)2)(X−24Q1(p))(X+48Q2(p)). (X-24R(p)^2)(X-24Q_1(p))(X+48Q_2(p)).

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

Δ(p)∼p2(p−1)2(p+1)2C(p)4D(p)2, \Delta(p)\sim p^2(p-1)^2(p+1)^2C(p)^4D(p)^2,D(p)=p4+2p3+2p2−2p+1. D(p)=p^4+2p^3+2p^2-2p+1.

The two roots of C give I₄ fibers. The points p=0,±1, the four roots of D, and infinity give eight I₂ fibers. Their Euler numbers sum to 24, so the minimal elliptic surface is K3.

5. Two rational sections

The diagonal section

The substitution q=−p immediately gives the point

q=−p,V=C(p)2.q=-p,\qquad V=C(p)^2.

Some selected entries coincide in the corresponding squares, so this family is degenerate as a 6/9 family. On the elliptic surface, however, it defines a genuine section.

The tangent-parabola section

Let F(q) be the right-hand side of the quartic and write C=C(p), R=R(p). The parabola

Π(q)=R2−2C2Rq2−22R2−C2Rq+R \Pi(q)= \frac{R^2-2C^2}{R}q^2 -2\frac{2R^2-C^2}{R}q+R

passes through (0,R) and is tangent to the branch V=−2R at q=1. The difference factors exactly:

R2(F(q)−Π(q)2)=4q(q−1)2(C2q+2R2)(R2−C2). R^2\bigl(F(q)-\Pi(q)^2\bigr)= 4q(q-1)^2(C^2q+2R^2)(R^2-C^2).

The remaining intersection gives the second section

q=−2R(p)2C(p)2,V=R(p)B0(p)C(p)4,B0(p)=4R(p)4−3C(p)4. \begin{aligned} q&=-\frac{2R(p)^2}{C(p)^2},\\ V&=\frac{R(p)B_0(p)}{C(p)^4},\\ B_0(p)&=4R(p)^4-3C(p)^4. \end{aligned}

6. Independence and the rank bound

The sections are proved independent by an exact specialization at p=2, not by a numerical search. A birational transformation gives the curve

E2:y2=x3−x2−1425x+12177=(x−33)(x−9)(x+41), E_2:\quad y^2=x^3-x^2-1425x+12177 =(x-33)(x-9)(x+41),

and the two sections specialize to

P=(214,5558),Q=(−15949,44400343). P=\left(\frac{21}{4},\frac{555}{8}\right), \qquad Q=\left(-\frac{159}{49},\frac{44400}{343}\right).

The exact counts #E₂(𝔽₇)=8 and #E₂(𝔽₁₃)=20 bound rational torsion by order 4; the visible full 2-torsion already has this order. In the 2-descent, the two torsion classes have rank 2 over 𝔽₂, adjoining P raises the rank to 3, and adjoining Q raises it to 4. Therefore P and Q are independent modulo torsion.

Proved bound

2≤rank⁡E(Q‾(p))≤4. 2\le\operatorname{rank}E(\overline{\mathbb Q}(p))\le4.

The two independent sections give the lower bound. The root rank of the 2I₄+8I₂ configuration is 14; Shioda–Tate and ρ≤20 for a complex K3 surface give the upper bound 4. The exact geometric rank has not yet been determined.

7. A shared polynomial family

The tangent-parabola section can be cleared of denominators for both patterns at once. Put

N=2R(p)2,UR=N2−2NC(p)2−C(p)4,UL=N2+2NC(p)2−C(p)4,UC=N2+C(p)4. \begin{aligned} N&=2R(p)^2,\\ U_R&=N^2-2NC(p)^2-C(p)^4,\\ U_L&=N^2+2NC(p)^2-C(p)^4,\\ U_C&=N^2+C(p)^4. \end{aligned}

Then the ABCEGH family is given by the roots

a=RB0,b=CUL,c=LUC,e=CUC,g=RUC,h=CUR, \begin{aligned} a&=R B_0,& b&=C U_L,& c&=L U_C,\\ e&=C U_C,& g&=R U_C,& h&=C U_R, \end{aligned}

while the same section in the ABCEGJ reading gives

a=CUL,b=RB0,c=LUC,e=CUC,g=RUC,j=CUR. \begin{aligned} a&=C U_L,& b&=R B_0,& c&=L U_C,\\ e&=C U_C,& g&=R U_C,& j&=C U_R. \end{aligned}

Here L,C,R, and B₀ are evaluated at the same p. All six defining equations of the two patterns are identities in ℤ[p]. The ratio c/g=L(p)/R(p) is nonconstant, so after excluding finitely many degeneracies the construction yields infinitely many projectively distinct rational solutions.

8. A paired positive example

At p=−7 the shared section gives two positive magic squares at once. For ABCEGH the selected-entry roots are

(a,b,c,e,g,h)=(14242447,16648975,22465979,18117725,12320053,19476025), (a,b,c,e,g,h)= (14242447,16648975,22465979,18117725,12320053,19476025),
(20284729654780927718836855062550472021242844163012487505625732825195917562526379043294993151783705922809379315549800625453656621803441). \begin{pmatrix} 202847296547809&277188368550625&504720212428441\\ 630124875056257&328251959175625&26379043294993\\ 151783705922809&379315549800625&453656621803441 \end{pmatrix}.

In the ABCEGJ reading the roots in A and B exchange roles, and the final root occupies J:

(a,b,c,e,g,j)=(16648975,14242447,22465979,18117725,12320053,19476025), (a,b,c,e,g,j)= (16648975,14242447,22465979,18117725,12320053,19476025),
(277188368550625202847296547809504720212428441555783803053441328251959175625100720115297809151783705922809453656621803441379315549800625). \begin{pmatrix} 277188368550625&202847296547809&504720212428441\\ 555783803053441&328251959175625&100720115297809\\ 151783705922809&453656621803441&379315549800625 \end{pmatrix}.

Both squares have magic sum 984,755,877,526,875, nine distinct positive entries, and exactly the six square entries specified by their respective patterns.

Open the complete 6/9 atlas →Previous surface: ABCDHJNext surface: ABDEFHTheory contents