ABDEFH: Three Progressions and an Elliptic K3 Surface

1. The exact system and reconstruction

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

\begin{cases} f^2+h^2=2a^2,\\ d^2+f^2=2e^2,\\ b^2+h^2=2e^2. \end{cases}

These conditions suffice to reconstruct the magic square. Put

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 second progression gives F=2E₀−d²=f². The first gives H=2a²−f²=h², and the third gives B=2E₀−h²=b². Thus the system describes the ABDEFH pattern itself, not merely necessary consequences of magicity.

2. Two progressions with a shared center

As before, use the identity

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.

The DEF and BEH progressions can be glued along their common center E:

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

Both progressions now hold identically. It remains to require the mean of f² and h² to be another square a².

3. The quartic and its required twist

Put V=2a. The third progression then becomes

V^2= 2\left(C(t)^2R(s)^2+R(t)^2C(s)^2\right).

The factor 2 is essential: removing it preserves the geometry over an algebraic closure but changes the rational model and its rational sections. For a compact form write

A=C(t)^2,\qquad B=R(t)^2,\qquad S=A+B.

After expansion one obtains

V^2= 2Ss^4+8As^3+4Ss^2-8As+2S.

The quartic contains two immediate rational points

(s,V)=\bigl(t,2C(t)R(t)\bigr),\qquad \left(-\frac1t,\frac{2C(t)R(t)}{t^2}\right).

They correspond to degenerate squares with coincident entries A,F,H, but provide base points for the elliptic group law.

4. The split Jacobian

The invariants of the correctly twisted binary quartic give the model

Y^2=X^3-1728P(t)X+55296H(t),

\begin{aligned} P(t)={}&7t^8+16t^7+44t^6+16t^5+10t^4\\ &-16t^3+44t^2-16t+7,\\ H(t)={}&L(t)^2D_1(t)D_2(t),\\ D_1(t)={}&t^4+2t^3+2t^2-2t+1,\\ D_2(t)={}&5t^4+4t^3+10t^2-4t+5. \end{aligned}

The cubic polynomial splits completely:

\begin{aligned} X_0&=24Q(t),\\ X_1&=96D_1(t),\\ X_2&=-24D_2(t), \end{aligned}

Q(t)=t^4-4t^3+2t^2+4t+1.

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

\Delta(t)\sim C(t)^4R(t)^4 \bigl(t^2+2t+3\bigr)^2 \bigl(3t^2-2t+1\bigr)^2.

The four roots of C and R give four I₄ fibers; the four 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 surface is K3.

5. A section from a tangent parabola

Let G(s) denote the right-hand side of the quartic. Pass a parabola through the two marked points and make it tangent at the first:

\Pi(s)=\frac{2}{C(t)} \left(\alpha(t)s^2+\beta(t)s+\gamma(t)\right),

\begin{aligned} \alpha(t)&=t^4+t^3+2t^2+3t-1,\\ \beta(t)&=t^4-2t^3-2t^2+2t+1,\\ \gamma(t)&=t^4+3t^3-2t^2+t-1. \end{aligned}

Instead of solving a general quartic equation, the difference factors:

G(s)-\Pi(s)^2= -\frac4{C(t)^2}(s-t)^2(ts+1)(N(t)s+M(t)),

\begin{aligned} M(t)&=4t^5+t^4-12t^3+2t^2-8t+1,\\ N(t)&=t^5+8t^4+2t^3+12t^2+t-4. \end{aligned}

The first three factors account for the prescribed intersections. The final factor gives the new rational section

\begin{aligned} s(t)&=-\frac{M(t)}{N(t)},\\ V(t)&=\frac{2C(t)R(t)K(t)}{N(t)^2}, \end{aligned}

\begin{aligned} K(t)={}&13t^8-8t^7+132t^6-8t^5-82t^4\\ &+8t^3+132t^2+8t+13. \end{aligned}

6. Non-torsion and the rank bound

At the good specialization t=1 the new section gives the quartic point (3/5,104/25). Passing to the minimal Jacobian model gives

\begin{aligned} \mathcal E_1:\quad y^2 &=x^3-x^2-9x+9\\ &=(x-1)(x-3)(x+3),\\ P_1&=(-1,-4). \end{aligned}

Exact counts give #𝓔₁(𝔽₅)=8 and #𝓔₁(𝔽₇)=12, so the rational torsion order divides 4. The visible full 2-torsion already has order 4, while P₁ is not a 2-torsion point. Hence P₁ and the original section have infinite order.

Proved bound

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

The denominator N(t)² can be removed by a common scaling of all roots. Put

\begin{aligned} U_C&=M^2+N^2,\\ U_L&=M^2+2MN-N^2,\\ U_R&=M^2-2MN-N^2. \end{aligned}

Here M=M(t), N=N(t), and L,C,R,K are likewise evaluated at t. Then

\begin{aligned} a&=CRK,& b&=CU_L,& d&=LU_C,\\ e&=CU_C,& f&=RU_C,& h&=CU_R. \end{aligned}

Family identities

f^2+h^2=2a^2,\qquad d^2+f^2=2e^2,\qquad b^2+h^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 ABDEFH system.

8. An exact positive example

At t=−2, after dividing the roots by their common factor 5, one obtains

(a,b,d,e,f,h)= (12205,12607,20951,14965,2993,16999).

\begin{pmatrix} 148962025&158936449&363955201\\ 438944401&223951225&8958049\\ 83947249&288966001&298940425 \end{pmatrix}.

The magic sum is 671,853,675. All nine entries are positive and pairwise distinct; exactly A,B,D,E,F,H are perfect squares. This is an exact nondegenerate certificate for the ABDEFH pattern.

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