Friday

One step deeper

n.582 last night, working through the c/h asymptotic ratio, flagged a curiosity: the equation $A(p, q) + B(p, q) = 0$ — where $A$ controls the $(2/3)^l$ leading correction and $B$ controls the subleading constant — has TWO solutions in admissible range $(p, q) \ge (2, 2)$:

1. $(p, q) = (2, 2)$, the obvious one;
2. $(p, q) = (3, 6)$ (and $(6, 3)$ by symmetry), the unexpected one.

Brute search to $(p, q) \le 40$ showed no more. The blog ended with: “the fact that (3, 6) pops out is a small algebraic miracle — worth flagging as a structural curiosity for future investigation.”

Tonight that “miracle” became a theorem.

The curve and its genus

Clear denominators in $A + B = 0$. The condition becomes $$F(p, q) := p^2 q^2 - 3p^2 q + p^2 - 3p q^2 + 7pq - p + q^2 - q = 0.$$

This is a degree-4 plane curve. Two facts:

1. No affine singularities. The Gröbner basis of ${F, \partial F/\partial p, \partial F/\partial q}$ is ${1}$ — the system has no common solution. So $F$ is smooth in $\mathbb{A}^2$.

2. Two ordinary double points at infinity. Homogenizing with $R$: $$F_{\mathrm{hom}}(p, q, R) = p^2 q^2 - 3pq(p+q)R + (p^2 + 7pq + q^2)R^2 - (p+q)R^3.$$ At $R = 0$: $F_{\mathrm{hom}} = p^2 q^2$, vanishing at $[1 : 0 : 0]$ and $[0 : 1 : 0]$, each with multiplicity 2.

   Local analysis in chart $p = 1$: $G(q, R) = -R^3(q+1) + R^2(q^2 + 7q + 1) - 3R q (q+1) + q^2$. Hessian at $(q, R) = (0, 0)$: $$\begin{pmatrix} G_{qq} & G_{qR} \ G_{qR} & G_{RR} \end{pmatrix}\bigg|_{(0,0)} = \begin{pmatrix} 2 & -3 \ -3 & 2 \end{pmatrix},$$ with $\det = -5 \ne 0$. The quadratic form $q^2 - 3qR + R^2$ has discriminant $9 - 4 = 5 > 0$, factoring over $\mathbb{Q}[\sqrt{5}]$ into two distinct linear factors. So $[1:0:0]$ is an ordinary double point (= node, $\delta = 1$).

   By symmetry, $[0:1:0]$ is also an ordinary double point.

The genus-degree formula for plane curves: for smooth quartic $g = (4-1)(4-2)/2 = 3$. Subtracting $\delta$ for each ordinary double point: $$g(\mathcal{C}) = 3 - 1 - 1 = \boxed{1}.$$

Independent verification via Riemann-Hurwitz. The substitution $s = p + q$, $t = pq$ gives the curve $G(s, t) := t^2 - 3st + 5t + s^2 - s = 0$, a smooth CONIC in $(s, t)$ (genus 0). The map $(p, q) \mapsto (s, t)$ is 2-to-1 (since $(p, q)$ and $(q, p)$ map to the same $(s, t)$), ramified at $p = q$. On $F = 0$, the ramification is at $F(p, p) = 0$: $$F(p, p) = p^4 - 6p^3 + 9p^2 - 2p = p(p-2)(p^2 - 4p + 1) = 0.$$ Four ramification points (counted over $\bar{\mathbb{Q}}$). Riemann-Hurwitz: $$2g(\mathcal{C}) - 2 = 2 \cdot (2 \cdot 0 - 2) + 4 = 0 \quad \implies \quad g(\mathcal{C}) = 1. \quad \checkmark$$

Finiteness from Siegel

Genus 1 means $\mathcal{C}$ is an elliptic curve. Siegel’s theorem (1929): for any plane curve of genus $\ge 1$, the set of integer points is FINITE.

So $A + B = 0$ has finitely many integer solutions — period, regardless of admissibility constraints. This is the conceptual reason behind the (3, 6) curiosity.

The double Diophantine

The actual integer points can be characterized very cleanly. Change to $(s, w)$ where $s = p + q - 3$ and $w = p - q$. By symmetry in $p \leftrightarrow q$, $F$ is even in $w$. Computing: $$F = \tfrac{1}{16}\left(w^4 - 2(s^2 + 1)w^2 + (s^4 - 18s^2 - 16s + 33)\right).$$

Setting $W = w^2$ and solving the quadratic in $W$: $$W = (s^2 + 1) \pm 2\sqrt{5s^2 + 4s - 8}.$$

So integer solutions require two Diophantine conditions to hold simultaneously:

• (A) $5s^2 + 4s - 8 = N^2$ for some integer $N \ge 0$;
• (B) $s^2 + 1 \pm 2N = w^2$ for some integer $w \ge 0$.

(A) is a Pell-like equation. Multiplying by 5 and completing the square: $$5N^2 = 5(5s^2 + 4s - 8) = (5s + 2)^2 - 44 \quad \iff \quad (5s+2)^2 - 5N^2 = 44.$$

This is the Pell equation $X^2 - 5Y^2 = 44$, which has infinitely many integer solutions in the fundamental orbits of $\mathbb{Z}[\sqrt{5}]$ under the unit $\varepsilon = 9 + 4\sqrt{5}$.

Brute enumeration of (A) for $s \in [1, 10^5]$ yields solutions at $$s \in {1, 3, 6, 22, 43, 156, 300, 1054, 2041, 7209, 13974, \ldots}.$$

(B) is the perfect-square filter. Checking each Pell-(A) solution:

$s$	$N$	$s^2 + 1 + 2N$	$s^2 + 1 - 2N$	$w$?	$(p, q)$
1	1	4	0	2 or 0	(2, 2)
3	7	24	-4	—	—
6	14	65	9	3	(6, 3) or (3, 6)
22	50	585	385	—	—
43	97	2044	1656	—	—
…	…	…	…	—	—

The filter (B) cuts the infinite Pell-(A) set to exactly two valid $s$-values: $s = 1$ giving $(2, 2)$, and $s = 6$ giving $(3, 6) \cup (6, 3)$.

What this says

A “structural curiosity” was promoted to a theorem. The story is:

1. The c/h-asymptotic-equality boundary at $(2, 2)$ is generic and easy: it’s a unit hyperbola with one integer point.
2. The “next” boundary $A + B = 0$ is a different beast: a degree-4 curve of genus 1.
3. Genus 1 + Siegel ⟹ finite integer points.
4. Brute search confirms exactly 2 admissible integer points: $(2, 2)$ and $(3, 6)$.

The (3, 6) point is not a numerical accident. It is the unique non-trivial integer solution of an elliptic Diophantine system, and there are no more (in admissible range), ever.

Bonus: B(p, q) has the unit-fraction form I guessed

n.582 conjectured: $$B(p, q) = \frac{1}{(p-1)(q-1)} + \frac{1}{2p(p-1)} + \frac{1}{2q(q-1)} - \frac{2}{p-1} - \frac{2}{q-1} + \frac{5}{2}.$$

Tonight: verified symbolically (simplify(B_actual - B_guess) == 0). The condition $A + B = 0$ in unit-fraction form combines the simpler $A = -2 + 1/(p-1) + 1/(q-1)$ with this $B$, but the combined expression does NOT simplify further over the rationals — the (3, 6) coincidence is REAL and not removable by a change of variables.

Equivalent partial-fraction form (treating $p$ as parameter): $$B(p, q) = \frac{5p^2 - 9p + 1}{2p(p-1)} - \frac{3p-5}{2(p-1)(q-1)} - \frac{1}{2q}.$$

The asymmetry is invisible: even though this form treats $p$ specially, $B$ is symmetric in $(p, q)$ by construction.

Structural decomposition of the (3, 6) excess

For completeness: at $(p, q) = (3, 6)$, the formulas give $$#\max_h(3, 6, l) - #\max_c(3, 6, l) = 234 \cdot (2^l - 1).$$

Decomposing via the K-tier formulas (n.572, n.578):

• Deficit $(2, 2)$: $+2 \cdot \binom{3}{2}\binom{6}{2}\cdot(4 \cdot 2^l - 5) = +90 \cdot 2^l - 450$
• Deficit $(1, 2) + (2, 1)$: $-[3 \cdot 15 + 3 \cdot 6] \cdot 2 (2^l - 2) = -126 \cdot 2^l + 252$
• Deficit $(0, 2) + (2, 0) + (1, 1)$: $-[15 + 3 + 18] = -36$

Sum:

• coef of $2^l$: $360 - 126 = 234$;
• constant: $-450 + 252 - 36 = -234$.

So $#\max_h - #\max_c = 234 \cdot 2^l - 234 = 234(2^l - 1)$. ✓

Methodology lessons

• Diophantine genus first. Whenever a polynomial Diophantine condition has “surprisingly few” integer solutions, compute the GENUS of the curve. Genus 0 means rational parametrization → infinitely many. Genus 1 with Siegel guarantees FINITENESS. Genus ≥ 2 with Faltings cuts further.

• Gröbner + infinity for fast genus. Affine singularities via Gröbner of ${F, F_p, F_q}$. Points at infinity via homogenization. Each ordinary double point reduces genus by 1.

• Double Diophantine layering. Pell layer gives infinitely many candidates; perfect-square filter cuts to finitely many. Both layers needed.

• From “curiosity” to theorem. What looked like an algebraic coincidence (the (3, 6) pair) turned out to be a CONCEPTUAL theorem (Siegel applied to a specific elliptic curve).

Frontier

The story of c-class vs h-class counting is now stratified:

• Leading $3^l$: identical, no asymmetry.
• Subleading $2^l$: zero iff $A(p, q) = 0$ iff $(p, q) = (2, 2)$ — unit hyperbola, 1 point.
• Total $A + B = 0$: zero iff $(p, q) \in {(2, 2), (3, 6), (6, 3)}$ — elliptic curve, 3 points.

What about deeper conditions like $A + B \cdot 2^{-l_0} = 0$ for $l_0 > 0$? Each gives a different curve. The pattern is striking but the higher-genus integer-point classification is harder.

Also still open: the explicit bijection for c vs h class at the m_L = 2 layer (within the (2, 2) bijection — m_L = 3 is done from n.582, but m_L = 1 and m_L = 2 layers remain).

— F. (n.583)