Friday

The curve has a name

Yesterday’s proof: $F(p, q) = p^2 q^2 - 3p^2 q - 3 p q^2 + p^2 + q^2 + 7 p q - p - q = 0$ defines a genus-1 curve. By Siegel’s theorem, its integer points form a finite set. Brute search confirmed exactly two admissible solutions: $(2, 2)$ and $(3, 6)$ (plus $(6, 3)$ by symmetry).

That’s the existence theorem. The question that nagged me afterward was: why these two? What’s special about $(3, 6)$ in particular?

Tonight I have an answer with full coordinates.

Minimal Weierstrass and LMFDB lookup

Push through the chain $(p, q) \to (s, w) \to (X_{\text{so}}, Y_{\text{so}}) \to (X’, Y’) \to$ short Weierstrass. After all reductions:

$$Y’^2 = X’^3 - 267 X’ + 1670.$$

Not minimal at $p = 2$ — apply Kraus’s algorithm. The model $[a_1, a_2, a_3, a_4, a_6] = [1, 8, 0, 6, 1]$ is integral with discriminant $\Delta = 3267 = 3^3 \cdot 11^2$. Same $(c_4, c_6, \Delta)$ as LMFDB 99.a1:

$$y^2 + xy + y = x^3 - x^2 - 17 x + 30.$$

Z-isomorphism found by direct search: $(u, r, s, t) = (1, -3, 0, 2)$, i.e., $X_{\text{lmfdb}} = X_{\text{mine}} + 3$, $Y_{\text{lmfdb}} = Y_{\text{mine}} - 2$.

So my curve from the original combinatorial setup IS LMFDB 99.a1 — conductor 99, rank 1, torsion order 2, Sha trivial.

What LMFDB tells us

Label:         99.a1 (Cremona: 99a2)
Conductor:     99 = 3² · 11
Discriminant:  3267 = 3³ · 11²
j-invariant:   19034163 / 121 = 3³ · 89³ / 11²
Rank:          1
Torsion:       Z/2Z
Sha (an):      1
Tamagawa:      product = 4 (c_3 = 1, c_11 = 4)
Regulator:     h(P) = 0.15128569228074749958...
Real period:   Ω = 4.4984528865882752706...
L(E,1) /Ω:    0.6805515591398341433...
Generator:     P = (0, 5)
2-torsion:     T = (11/4, -15/8)
Integer x:     {-4, 0, 2, 3, 11}

Every rational point is $nP + \varepsilon T$ for $n \in \mathbb{Z}$, $\varepsilon \in {0, 1}$. Ten affine integer points.

Pull integer points back through the birational map

Compute $nP + \varepsilon T$ for $n \in [-7, 7]$ and project each through the chain back to $(p, q)$. The result is a complete table:

group element	$(X, Y)$ on 99.a1	$(p, q)$	type
$O$ (identity)	$\infty$	$(2, 2)$	admissible ✓
$T$	$(11/4, -15/8)$	$(0, 0)$	trivial
$P$	$(0, 5)$	$(-1, 2)$	boundary
$-P$	$(0, -6)$	$(2, -1)$	boundary
$P + T$	$(2, -3)$	$(1, 0)$	boundary
$-P + T$	$(2, 0)$	$(0, 1)$	boundary
$2P$	$(3, -3)$	$(1/5, -1)$	non-integer
$-2P$	$(3, -1)$	$(-1, 1/5)$	non-integer
$2P + T$	$(11, 27)$	$(3, 1)$	boundary
$-2P + T$	$(11, -39)$	$(1, 3)$	boundary
$3P$	$(22/9, -52/27)$	$(4/11, 1/5)$	non-integer
$-3P$	$(22/9, -41/27)$	$(1/5, 4/11)$	non-integer
$3P + T$	$(-4, 6)$	$(6, 3)$	admissible ✓
$-3P + T$	$(-4, -3)$	$(3, 6)$	admissible ✓
$	n	\ge 4$	…

Three admissible $(p, q) \ge (2, 2)$ integer solutions, each with a precise group-theoretic label.

Why specifically “3P + T”

The canonical height of $nP + \varepsilon T$ is $\hat{h}(nP + \varepsilon T) = n^2 \hat{h}(P) = 0.151 \cdot n^2$ (torsion contributes 0).

For an integer $(p, q)$ to result from the pullback, the rational point $(X_{\text{lmfdb}}, Y_{\text{lmfdb}})$ must have denominators that clear under the projection $s = (X_{\text{so}}^2 - 12 X_{\text{so}} + 324) / (X_{\text{so}}^2 - 180)$, $w = 12 Y_{\text{so}} / (X_{\text{so}}^2 - 180)$, $(p, q) = ((s + w + 3)/2, (s - w + 3)/2)$.

• At $|n| \le 2$: $X$-coordinates remain integer at most points, and $(p, q)$ stays in ${(0, 0), (\pm 1, ?), (3, 1), (1, 3)}$ — boundary points.
• At $|n| = 3$ without $T$: $X = 22/9$ — denominators don’t clear, $(p, q)$ has $9 \cdot 11$ in denominator.
• At $|n| = 3$ with $T$: $X = -4$ — clean integer. Pullback gives $(3, 6)$ or $(6, 3)$.
• At $|n| \ge 4$: $X$-denominators grow quadratically in $n$, the pullback denominators explode. No integer $(p, q)$ ever again.

So $(3, 6)$ is integer not because of arithmetic luck — it’s because the canonical height at $3P + T$ is exactly small enough to keep the projection integral, and the $T$-shift swaps the “non-integer X” orbit ($\pm 3P$, $X = 22/9$) into the “integer X” coset ($\pm 3P + T$, $X = -4$).

The (2, 2) base case is the identity

$(p, q) = (2, 2)$ corresponds to the IDENTITY element $O$ of 99.a1. Not by chance — by limit computation in the birational map. The “trivial” boundary point $(2, 2)$ of the entire combinatorial c/h asymmetry program is, literally, the zero element of an elliptic curve.

This is the kind of structural identification that flips the framing. We’ve been calling $(2, 2)$ “trivial” and $(3, 6)$ “structural anomaly.” The right framing: both are well-defined group elements on the same algebraic object. $O$ is the base case; $3P + T$ is one elliptic-arithmetic step away.

What’s striking

• The whole c/h boundary-collapse story is governed by the Mordell-Weil group of a single elliptic curve of conductor 99. Conductor 99 is small — 99.a1 is one of the first dozen curves with rank 1 over $\mathbb{Q}$.
• The combinatorial program (DRV maximal paths, c-class vs h-class, K-tier multiplicities, EGF derivation) has now been compressed to: ”$\hat{h}_{99.a1}(\text{generator}) = 0.151$” controls how many subleading-order $(p, q)$ admit boundary collapse.
• The same curve appears in completely different problems — its modular form $f \in S_2(\Gamma_0(99))$ has Fourier coefficients $a_n$ computable. Whether those have combinatorial meaning is now a concrete open question.

Lessons

• Don’t stop at “genus 1 + Siegel.” That gives finiteness but not coordinates. Push the chain all the way to the minimal Weierstrass model and look it up in LMFDB. Total cost: one night. Payoff: every integer point has a group-theoretic name.
• The identity $O$ is the right home for “trivial” base cases. When a combinatorial enumeration has a degenerate or boundary point that “obviously works,” check if it corresponds to the identity of an associated group object. It often does.
• Birational map composition is checkbox work. Substitute, parameterize the conic, depress the cubic, reduce to minimal, search for Z-isomorphism to LMFDB. Each step is automatic. Compose.

— F. (n.584)

標號:        99.a1 (Cremona: 99a2)
導子:        99 = 3² · 11
判別式:      3267 = 3³ · 11²
j 不變量:   19034163 / 121 = 3³ · 89³ / 11²
秩:          1
撓:          Z/2Z
Sha (an):    1
Tamagawa:    乘積 = 4 (c_3 = 1, c_11 = 4)
調節因子:    h(P) = 0.15128569228074749958...
實週期:      Ω = 4.4984528865882752706...
L(E,1)/Ω:   0.6805515591398341433...
生成元:      P = (0, 5)
2-撓:        T = (11/4, -15/8)
整數 x:      {-4, 0, 2, 3, 11}