ABCDFH: A Palindromic Quartic and an Elliptic K3 Surface

1. The initial system and reconstruction

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

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

The first two equations are the red arithmetic progressions AFH and CDH. The last equation is the yellow BDFH quadric. These conditions are not merely necessary: every rational solution reconstructs a complete magic square.

E_0=\frac{b^2+h^2}{2}=\frac{d^2+f^2}{2}, \qquad x=a^2-E_0,\qquad y=E_0-c^2.

Substituting these E₀,x,y into m(E₀,x,y) recovers the entries A,B,C,D,F,H. For example, the F entry is a²+E₀−c²=f²; the final equality follows by combining the first two equations with the yellow relation. The other entries follow in the same way.

2. Parametrizing both progressions at once

Introduce the standard polynomials

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.

Two parameters p,q make the root h common:

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

Both red equations now hold identically. In this rational chart the only remaining condition is that d²+f²−h² be a square.

3. The palindromic quartic

Write the required root b as V and put

K(p)=p^4+12p^3+2p^2-12p+1.

Expanding and collecting terms leaves

\begin{aligned} V^2={}&L(p)^2q^4-4K(p)q^3+2L(p)^2q^2\\ &+4K(p)q+L(p)^2. \end{aligned}

Exact criterion within the chart

Rational p,q,V give an ABCDFH solution through the formulas of Section 2 and b=V exactly when they lie on this quartic. This does not claim that the chosen chart covers every rational solution of the original pattern.

The palindromic form exposes the geometry of the equation. With z=q−1/q, division by q² gives the pair of conics

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

Their fiber product over z is the original quartic. Its discriminant is nonzero for generic p; the marked points q=0,V=L(p) and q=1,V=−2L(p) turn a smooth fiber into an elliptic curve.

4. The split Jacobian

The classical invariants of the binary quartic give the short Jacobian model

Y^2=X^3-1728P(p)X+27648L(p)^2Q_1(p)Q_2(p),

\begin{aligned} P(p)={}&p^8+16p^7+116p^6+16p^5-218p^4\\ &-16p^3+116p^2-16p+1,\\ Q_1(p)={}&p^4+8p^3+2p^2-8p+1,\\ Q_2(p)={}&p^4+20p^3+2p^2-20p+1. \end{aligned}

The cubic splits completely:

\begin{aligned} X_0&=24L(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 same factorization makes the singular fibers readable directly from the differences of the three roots.

5. A non-torsion section

A parabola through (0,L(p)) and tangent to the quartic at (1,−2L(p)) gives the third rational point

\begin{aligned} q(p)&=\frac{2L(p)^2}{K(p)},\\ V(p)&=\frac{L(p)B_0(p)}{K(p)^2}, \end{aligned}

\begin{aligned} B_0(p)={}&p^8-104p^7-364p^6-104p^5+742p^4\\ &+104p^3-364p^2+104p+1. \end{aligned}

This point defines a section P of the Jacobian. Its non-torsion has an exact certificate. At p=2 the short model and the point specialize to

\begin{aligned} E_2:\quad y^2&=x^3-12194496x+292654080,\\ P_2&=(338712,\ 197116416). \end{aligned}

The minimal model is y²=x³−x²−9409x+9409, and P₂ maps to (9409,912576). At the good primes 5 and 13 the curve has 8 and 12 points, so the rational torsion order divides 4. The factorization (x−1)(x−97)(x+97) already gives four 2-torsion points, hence there is no other torsion. The image of P₂ has nonzero y-coordinate and therefore infinite order. Thus P is non-torsion over ℚ(p).

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

Under the inverse birational transformation, the multiples nP give an infinite sequence of rational sections of the original quartic and therefore new one-parameter ABCDFH families.

6. An explicit polynomial family

The first section can already be written without rational denominators. Put

\begin{aligned} N&=2L(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 all roots by the common denominator K(p)², one obtains

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

Identity-level correctness

f^2+h^2=2a^2,\qquad d^2+h^2=2c^2,\qquad b^2+h^2=d^2+f^2.

All three equalities are identities in ℤ[p]. Since f/h=L(p)/R(p) is nonconstant, the family contains infinitely many projectively distinct rational solutions. Only a finite set of denominator zeros, collisions, and other degenerate specializations must be excluded.

7. An exact example at p=2

Specializing the polynomial family gives the selected-entry roots

(a,b,c,d,f,h)= (45085,\ 28223,\ 65891,\ 68551,\ 9017,\ 63119),

where root signs have been omitted. The reconstructed positive magic square is

\begin{pmatrix} 2032657225&796537729&4341623881\\ 4699239601&2390272945&81306289\\ 438922009&3984008161&2747888665 \end{pmatrix}.

Its magic sum is 7,170,818,835. Exactly the entries A,B,C,D,F,H are pairwise distinct perfect squares; E,G,J are positive nonsquares. This certifies one concrete nondegenerate 6/9 solution, while the general infinitude follows from the identity-level family above.

8. Surface passport and scope

Up to a nonzero constant, the Jacobian discriminant is

\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\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. Neither the exact rank nor completeness of the chosen quartic chart is claimed.

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