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Elliptic surfaces for 6/9 patterns · 5.4

The ABCDHJ Pattern: Two Progressions Sharing the Entry D

The BDJ and CDH progressions meet at the endpoint D. The yellow ABHJ relation leaves a palindromic genus-one quartic whose Jacobian is a split elliptic K3 surface with an explicit non-torsion section.

ABCDEFGHJ
ABCDHJ: red BDJ, CDH; yellow ABHJ

1. The initial system and reconstruction

Let a,b,c,d,h,j be the rational square roots of the six selected entries. The general form of a magic square gives three independent relations:

{b2+d2=2j2,d2+h2=2c2,a2+j2=b2+h2. \begin{cases} b^2+d^2=2j^2,\\ d^2+h^2=2c^2,\\ a^2+j^2=b^2+h^2. \end{cases}

The first two equations are the red progressions BDJ and CDH with the common endpoint D. The last equation is the yellow ABHJ quadric. Together these conditions suffice to reconstruct the complete magic square.

E0=a2+j22=b2+h22,x=a2−E0,y=E0−c2. E_0=\frac{a^2+j^2}{2}=\frac{b^2+h^2}{2}, \qquad x=a^2-E_0,\qquad y=E_0-c^2.D=c2+j2−E0=d2,B=j2+E0−c2=b2,H=E0−j2+c2=h2. \begin{aligned} D&=c^2+j^2-E_0=d^2,\\ B&=j^2+E_0-c^2=b^2,\\ H&=E_0-j^2+c^2=h^2. \end{aligned}

The entries A,J,C immediately become a²,j²,c², while the three displayed simplifications follow from the initial system. Thus the conditions describe the ABCDHJ pattern itself, not merely a collection of necessary consequences.

2. Gluing the two progressions at D

Use the standard polynomials

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 common root d is obtained by multiplying the right endpoints of the two parametrizations:

b=L(p)R(q),j=C(p)R(q),d=R(p)R(q),h=L(q)R(p),c=C(q)R(p). \begin{aligned} b&=L(p)R(q),& j&=C(p)R(q),\\ d&=R(p)R(q),\\ h&=L(q)R(p),& c&=C(q)R(p). \end{aligned}

Both red equations now hold identically. The only remaining condition is that b²+h²−j² be a square a².

3. The palindromic quartic and two conics

Write the required root a as V and put

K(p)=p4+8p3+2p2−8p+1. K(p)=p^4+8p^3+2p^2-8p+1.

Expanding and collecting terms gives

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

Exact criterion within the chart

Rational p,q,V give an ABCDHJ solution through the formulas of Section 2 and a=V exactly when they lie on this quartic. Completeness of the chosen chart for all rational solutions of the original pattern is not claimed.

The palindromic form permits the substitution z=q−1/q. After division by q², the original curve becomes the fiber product of two conics:

(Vq)2=C(p)2(z2+4)−4K(p)z,(q+1q)2=z2+4. \left(\frac Vq\right)^2=C(p)^2(z^2+4)-4K(p)z, \qquad \left(q+\frac1q\right)^2=z^2+4.

For generic p the quartic is smooth and contains the rational points (0,C(p)) and (1,−2C(p)); choosing either point turns it into an elliptic curve.

4. The split Jacobian

The binary-quartic invariants give the short Jacobian model

Y2=X3−1728P(p)X+27648C(p)2Q1(p)Q2(p), Y^2=X^3-1728P(p)X+27648C(p)^2Q_1(p)Q_2(p),P(p)=p8+12p7+52p6+12p5−90p4−12p3+52p2−12p+1,Q1(p)=p4+6p3+2p2−6p+1,Q2(p)=p4+12p3+2p2−12p+1. \begin{aligned} P(p)={}&p^8+12p^7+52p^6+12p^5-90p^4\\ &-12p^3+52p^2-12p+1,\\ Q_1(p)={}&p^4+6p^3+2p^2-6p+1,\\ Q_2(p)={}&p^4+12p^3+2p^2-12p+1. \end{aligned}

The cubic polynomial splits completely:

X0=24C(p)2,X1=24Q2(p),X2=−48Q1(p). \begin{aligned} X_0&=24C(p)^2,\\ X_1&=24Q_2(p),\\ X_2&=-48Q_1(p). \end{aligned}

Thus full rational 2-torsion is visible over ℚ(p). The differences of the three roots simultaneously determine the discriminant and the singular-fiber configuration.

5. How the non-torsion section arises

The section can be derived from the marked points rather than guessed. Let F(q) denote the right-hand side of the quartic and consider the parabola

Π(q)=C2+2KCq2−22C2+KCq+C. \Pi(q)= \frac{C^2+2K}{C}q^2 -2\frac{2C^2+K}{C}q+C.

It passes through (0,C) and is tangent to the branch V=−2C at q=1. The exact factorization of the difference is

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

The remaining intersection gives the rational point

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

This point defines a section P of the Jacobian. Under the good specialization p=−2 it maps to the minimal curve

E−2:y2=x3−x2−737x+7905,P−2=(32925,2208125). \begin{aligned} E_{-2}:\quad y^2&=x^3-x^2-737x+7905,\\ P_{-2}&=\left(\frac{329}{25},\frac{2208}{125}\right). \end{aligned}

The right-hand side is (x−17)(x−15)(x+31), so the curve already displays four rational 2-torsion points. Exact counts give #E(𝔽₅)=4 and #E(𝔽₇)=12; hence the full rational torsion order divides 4 and equals the 2-torsion subgroup. Since P₋₂ has nonzero y-coordinate, it has infinite order. Therefore the original section P is non-torsion over ℚ(p).

⟨P⟩={ nP:n∈Z }≅Z. \langle P\rangle=\{\,nP:n\in\mathbb Z\,\}\cong\mathbb Z.

The inverse birational transformation sends the multiples nP to new rational sections of the original quartic and hence to an infinite sequence of one-parameter ABCDHJ families.

6. An explicit polynomial family

The first section already gives a family without rational denominators. Put

N=2C(p)2,UR=N2+2NK−K2,UL=N2−2NK−K2,UC=N2+K2. \begin{aligned} N&=2C(p)^2,\\ U_R&=N^2+2NK-K^2,\\ U_L&=N^2-2NK-K^2,\\ U_C&=N^2+K^2. \end{aligned}

After multiplying the roots by the common denominator K(p)², one obtains

a=C(p)B0(p),b=L(p)UR,c=R(p)UC,d=R(p)UR,h=R(p)UL,j=C(p)UR. \begin{aligned} a&=C(p)B_0(p),& b&=L(p)U_R,\\ c&=R(p)U_C,& d&=R(p)U_R,\\ h&=R(p)U_L,& j&=C(p)U_R. \end{aligned}

Identity-level correctness

b2+d2=2j2,d2+h2=2c2,a2+j2=b2+h2. b^2+d^2=2j^2,\qquad d^2+h^2=2c^2,\qquad a^2+j^2=b^2+h^2.

All three equalities are identities in ℤ[p]. The ratio b/d=L(p)/R(p) is nonconstant, so the family contains infinitely many projectively distinct rational solutions. A finite set of degenerate specializations must be excluded.

7. An exact positive example

At p=−2 the selected-entry roots, with signs omitted, are

(a,b,c,d,h,j)=(4565, 2303, 3029, 329, 4271, 1645). (a,b,c,d,h,j)= (4565,\ 2303,\ 3029,\ 329,\ 4271,\ 1645).

The reconstructed magic square is

(2083922553038099174841108241117726252343700914370409182414412706025). \begin{pmatrix} 20839225&5303809&9174841\\ 108241&11772625&23437009\\ 14370409&18241441&2706025 \end{pmatrix}.

Its magic sum is 35,317,875. Exactly the entries A,B,C,D,H,J are pairwise distinct perfect squares; E,F,G are positive nonsquares. This is an exact certificate for one nondegenerate 6/9 solution.

8. Surface passport and scope

Up to a nonzero constant, the Jacobian discriminant is

Δ(p)∼p2(p−1)2(p+1)2R(p)4K(p)2. \Delta(p)\sim p^2(p-1)^2(p+1)^2R(p)^4K(p)^2.

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

Proved rank bound

1≤rank⁡E(Q‾(p))≤4. 1\le\operatorname{rank}E(\overline{\mathbb Q}(p))\le4.

The non-torsion section gives the lower bound 1. 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 rank and completeness of the chosen quartic chart remain undetermined.

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