ABCDEH: An Elliptic K3 Surface for a 6/9 Pattern

1. The initial system

Write the rational square roots of the selected entries in lowercase. The general form of a magic square leaves three independent conditions for the ABCDEH pattern:

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

The first two equations are arithmetic progressions of squares with the common endpoint h². The third equation couples them and is the only remaining obstruction.

2. Parametrizing both progressions at once

Put

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

Direct expansion gives the identity

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

To make the two progressions share the same root h, take two parameters p,q and multiply the corresponding factors:

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

The first two equations now hold identically. No numerical coefficients have been guessed: each row is a copy of L²+R²=2C² multiplied by a common square.

3. The residual quartic

Substitute the expressions for b,c,d,e,h into the third condition and write a=V. Collecting terms gives the single equation

V^2=C(p)^2(q^2+1)^2+8K(p)q(q^2-1),

K(p)=p^4+2p^3+2p^2-2p+1.

Exact construction criterion

For any rational p,q,V satisfying this quartic, the six numbers

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

give the square entries ABCDEH of a rational magic square. Its coordinates are recovered as E=e², x=a²−e², and y=e²−c².

For fixed p, the right-hand side is a quartic polynomial in q. The point q=1, V=±2C(p) is rational, so the smooth generic fiber has genus one and a marked rational point.

4. The Jacobian and the split cubic

Computing the classical invariants I and J of the binary quartic sends the pointed genus-one curve to the short Weierstrass form

Y^2=X^3-432P(p)X+3456Q(p),

\begin{aligned} P(p)={}&13p^8+48p^7+100p^6+48p^5-18p^4\\ &-48p^3+100p^2-48p+13,\\ Q(p)={}&(p^2+1)^2 \cdot(5p^4+12p^3+10p^2-12p+5)\\ &\cdot(7p^4+12p^3+14p^2-12p+7). \end{aligned}

The cubic splits completely over ℚ(p):

\begin{aligned} X_0&=24(p^2+1)^2,\\ X_1&=12(5p^4+12p^3+10p^2-12p+5),\\ X_2&=-12(7p^4+12p^3+14p^2-12p+7). \end{aligned}

Thus the Jacobian has full rational 2-torsion over ℚ(p). This is useful both for computing the singular fibers and for an exact independence test for the sections found below.

5. Two independent sections and an infinite subgroup

Besides the base point q=1, the quartic has the symmetric pair of points

q=-\frac1p,\qquad V=\pm\frac{(p^2+2p-1)^2}{p^2}.

The standard pointed-quartic transformation maps the two sign choices to two sections of the Weierstrass model. Their independence is not inferred from numerical sampling: at p=3 their images together with two 2-torsion classes have rank 4 in the exact Kummer map. Hence the two sections are independent over ℚ(p).

Denote one of these sections by P. It is non-torsion, hence

\langle 2P\rangle =\{\,2nP:n\in\mathbb Z\,\} \cong\mathbb Z.

Every multiple 2nP returns under the inverse birational map to a rational quartic point (qₙ(p),Vₙ(p)), and then, through the formulas of Section 3, to an ABCDEH solution. Thus one non-torsion section produces an infinite sequence of rational parametrizations, not a single numerical example.

6. The K3 surface passport

Up to a nonzero constant, the Jacobian discriminant is

\begin{aligned} \Delta(p)\sim{}& (p^2+2p-1)^4 \cdot(p^2+2p+3)^2\\ &\cdot(3p^2-2p+1)^2 \cdot(p^4+2p^3+2p^2-2p+1)^2. \end{aligned}

The roots of the first factor give two I₄ fibers; the roots of the remaining factors give eight I₂ fibers. The fiber at infinity is smooth. The Euler numbers sum to 24, so the minimal elliptic surface is a K3 surface.

Proved rank bound

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

The two independent sections give the lower bound. For the upper bound, the configuration 2I₄+8I₂ has root rank 14; Shioda–Tate and ρ≤20 for a complex K3 surface give rank≤20−2−14=4. Exact rank 2 is not claimed.

7. An explicit parametrization from the section 2P

The first nontrivial even multiple already gives a one-parameter family. Here p remains free, while q and V are no longer chosen independently:

\begin{aligned} q(p)&=-\frac{p^4+p^3-p+1}{p(p^2-2p-1)},\\ V(p)&=-\frac{(p^2+1)A_0(p)} {p^2(p^2-2p-1)^2}, \end{aligned}

where, for compactness, put

\begin{aligned} A_0(p)={}&p^8-2p^7-10p^6-10p^5+10p^4\\ &+10p^3-10p^2+2p+1,\\ U(p)={}&p^8+2p^6+8p^5+2p^4-8p^3+2p^2+1,\\ C_0(p)={}&p^8+2p^7+2p^6-6p^5+2p^4\\ &+6p^3+2p^2-2p+1,\\ D_0(p)={}&p^8+4p^7-2p^6-4p^5-6p^4\\ &+4p^3-2p^2-4p+1. \end{aligned}

After multiplying all roots by the common denominator p²L(p)², one obtains the completely polynomial parametrization

\begin{aligned} a&=-C(p)A_0(p),& b&=L(p)U(p),\\ c&=R(p)C_0(p),& d&=R(p)D_0(p),\\ e&=C(p)U(p),& h&=R(p)U(p). \end{aligned}

Identity-level correctness

These six polynomials identically satisfy all three ABCDEH equations. Outside a finite set of degenerate p-values they give nondegenerate rational squares. Further families arise in the same way from 4P,6P,… via the Jacobian group law.

8. Scope of the result

The result is an explicit generating surface, not a collection of isolated examples. Rational points on its fibers give families of ABCDEH squares, and two independent sections guarantee a nontrivial supply of such points.

What is not proved here is that the chosen pair of parameters p,q enumerates every rational solution of the original pattern, nor is the exact geometric rank of the surface determined. Both statements remain outside the proved result.

Next: the ABCDEJ surface →The F7+, F4+, and F9+ series Theory contents