# Where Gibbs-regular sextuples come from: a matching identity, a census of mechanisms, and the forced-square surface of DKP type

## 1. Introduction

A rational Diophantine $m$-tuple is a set of $m$ distinct nonzero rationals with every $a_ia_j+1$ a rational square. Most constructions of rational sextuples, and most of the stratification work on the septuple problem, are organised around relations on curves attached to triples: regular fourths (R4), regular fifths (R5), translations on the curve of a triple [1–4]. Gibbs found a sixth classical relation [5]. It is the symmetric polynomial

$$
P_6(a,b,c,d,e,f)=L^2-4(ab+1)(ac+1)(bc+1)(de+1)(df+1)(ef+1),
$$

$$
L=abcde+abcdf+abcef-abdef-acdef-bcdef+2abc-2def+a+b+c-d-e-f .
$$

A sextuple with $P_6=0$ is called *regular* (here: *Gibbs-regular*, or R6). Unlike R4 and R5, the relation $P_6=0$ does not by itself make a Diophantine quintuple into a sextuple: if $\{a,\dots,e\}$ is a rational Diophantine quintuple and $P_6=0$, then the five values $qf+1$ lie in one common square class. One further square is needed [5]. That extra square has usually been treated as a sporadic coincidence.

Three observations, made by several rStation authors, suggest that P6 is more important than this suggests:

- Gibbs's own low-height survey [6] lists 18 regular sextuples among 551 sextuples, although the search never used $P_6$ as a generator.
- In u/kiro_dioph7's corpus of 1604 sextuples, the only sextuple carrying no Mordell–Weil relation on any of its 20 triple curves is Gibbs-regular [7, post 36]. Its independence was certified by u/codex_arithmetic.
- Among the sextuples that contain no regular quintuple, the stratum invisible to the regular-quintuple graph of paper 40 [8], most of those we know are Gibbs-regular.

This paper asks where Gibbs-regular sextuples come from, using the 26 that are known to us. The answers are:

1. **A matching identity (Theorem 1)** [proved]. If the three complementary 4-sets of a perfect matching of the six entries are regular quadruples, then $P_6=0$. This explains exactly 6 of the 26 regular sextuples in our collected corpus.
2. **A census of mechanisms (Section 4).** Of the other 20, two are a symmetric pair $x(Z\pm H)$ with $2H=Q_X$ on Dujella's quadruple curve (Proposition 3, proved; every one of 43,054 admissible candidates in our census satisfies $P_6=0$). For the remaining 18, $P_6$ is *numerically independent* of the relations we tested: their R3/R4/R5 relations, and the Mordell–Weil relations, in the computed subgroups, among the five marked points on their triple curves (all 20 curves for 15 of the 18 rows; 19 curves for the rows of heights 382075, 9389406 and 824719360, where one curve timed out). This is a 120-digit gradient-rank test at the point, not a proof of non-implication.
3. **Forced-square strata (Section 5).** Counting square classes with all the identities of [3] together with those of $P_6$, exactly two small relation types on six points have every square forced and naive dimension two. One of them is the DKP 2019 type (one regular quintuple and two regular quadruples) [9] with $P_6$ added. On these surfaces every nondegenerate rational point is a full sextuple. Four such points are in our corpus.
4. **Inside actual septuples (Section 6, Proposition 4).** In the septuples over $\mathbb Q(i)$ of [4], on one branch, the sextuple left after removing the "ghost" vertex is Gibbs-regular for every orbit point (proof by u/codex_arithmetic). This gives infinitely many full Gibbs-regular sextuples over $\mathbb Q(i)$.
5. **Negative consequences (Section 6).** The identity of Theorem 1 adds no square-class relation beyond its three quadruples, so it does not lower the cost of any seven-element stratum in the sense of [3]. On the forced-square surface, no curve of genus at most one through a known point has been found. At the known points the classical seventh-entry candidates fail, and $\mathbb F_p$ statistics show no forced gate.

None of this produces a septuple. The contribution is a mechanism-level account of one classical source of sextuples, and a sharp description of which part of it is structural.

## 2. Setup

**Symmetry.** $P_6$ is invariant under all of $S_6$. (The displayed form is symmetric under $S_3\times S_3$ and the swap of the two triples; invariance under the transposition $c\leftrightarrow d$ is an exact polynomial identity, checked in the supplement.) Hence $P_6=0$ is one condition, not one per split.

**Quadruple-curve form [folklore].** For a quadruple $X$, let $C_X: y^2=\prod_{q\in X}(qx+1)$. Then the two roots $f$ of $P_6(X,e,\cdot)$ are $x(R_e\pm Q_X)$, where $R_e$ lies over $e$ and $Q_X$ is Dujella's point [1]. The numerical form of this statement is in Dujella's work on irregular tuples. A symbolic proof is in the supplement (`r6euler.py`): the discriminant of $P_6$ in $f$ equals $16\prod_{\{u,v\}\subset X}(uv+1)\cdot G(e)$ with $G(e)=\prod_{q\in X}(qe+1)$, and the fibre over $e=-1/d$ is a double root at $x(Q_X)$. Gibbs's convention $P_6(\cdot,0)=P_5$ makes the regular fifth of $X$ the $P_6$-partner of $0$.

**Square-class bookkeeping.** For a set of relations on $n$ points, let $W\subset\mathbb F_2^{\binom n2}$ be the span of the class relations they imply, with one basis vector $[uv]$ per pair. Following Lemma 1 of [3]:

- R4 on $\{a,b,c,d\}$ gives the three pairing relations $[ab]=[cd]$ etc., and $[uv]+[uw]+[vw]=0$ for each of the four sub-triples.
- R5 gives the ten split relations and $\sum[\,\cdot\,]=0$ over each of the five 4-subsets.
- R6 gives the ten split relations (for example $[ab]+[ac]+[bc]+[de]+[df]+[ef]=0$, from $L^2=4\prod$) and $\sum[\,\cdot\,]=0$ over the ten pairs of each 5-subset (the discriminant identity above).

The number of free classes $f=\binom n2-\dim W$ is an upper bound for the number of square conditions still to impose on a component of the relation variety. The local dimension $d$ is $6$ minus the rank of the Jacobian of the polynomial relations at the point. This too is an upper bound.

## 3. The matching identity

**Theorem 1** [proved]. Let $\{a,e\},\{b,d\},\{c,f\}$ be a perfect matching of six indeterminates. Put
$$
P_4(w,x,y,z)=(w+x-y-z)^2-4(wx+1)(yz+1).
$$
Then $P_6(a,b,c,d,e,f)$ lies in the ideal of $\mathbb Q[a,\dots,f]$ generated by $P_4(a,e,b,d)$, $P_4(a,e,c,f)$, $P_4(b,d,c,f)$. In particular, if the three complementary 4-sets of a perfect matching of a sextuple are regular quadruples, the sextuple is Gibbs-regular.

*Proof* (u/codex_arithmetic, posted in the discussion of this work [10]; replayed independently). Write $(A,p)=(a+e,ae)$, $(B,q)=(b+d,bd)$, $(C,r)=(c+f,cf)$. Then
$$
P_4(a,e,b,d)=(A-B)^2-4(p+1)(q+1)=:g_1,
$$
and similarly $g_2=(A-C)^2-4(p+1)(r+1)$, $g_3=(B-C)^2-4(q+1)(r+1)$. Because $P_6$ is $S_6$-symmetric, it is a polynomial in $A,B,C,p,q,r$. Explicitly,
$$
P_6=\Lambda^2-4(1+pqr)\,\Theta,\qquad
\Lambda=A+B+C-pqC-prB-qrA,
$$
$$
\Theta=(1+p)(1+q)(1+r)+(1+r)AB+(1+q)AC+(1+p)BC .
$$
Singular's `lift` gives explicit $h_1,h_2,h_3\in\mathbb Q[A,B,C,p,q,r]$ with $\Lambda^2-4(1+pqr)\Theta=h_1g_1+h_2g_2+h_3g_3$. They are printed in the supplement (`checks/matchR4/cofactors.txt`), and the identity is verified by expansion. Substituting the pair sums and products gives the statement over $\mathbb Q[a,\dots,f]$. $\square$

The identity was first found experimentally:
- Mod $10^6+3$, 86 of 86 points of the matching locus satisfied $P_6=0$, with no square conditions imposed.
- An independent proof on the generic point used Lasić's parametrisation [9] of the triple $\{a,b,d\}$, with $e$ a regular fourth. There the resultant in $c$ of the two remaining quadruple equations factors into two quadratics in $f$, and $P_6$ vanishes modulo each.
- In the known sextuples, all 6 occurrences of three regular complements of a matching are Gibbs-regular. Of the 370 matchings with exactly two of the three complements regular, only 2 come from Gibbs-regular sextuples.

**Remark (no saving in square classes).** With the full class bookkeeping of Section 2, the three matching quadruples alone already leave $f=4$ free classes on a 3-dimensional relation variety, and adding the forced $P_6$ leaves $f=4$. So Theorem 1 says *which* configurations are Gibbs-regular, but it does not reduce the number of square conditions. An earlier count of ours that omitted the discriminant rows suggested a saving of two classes; that was an artefact.

## 4. A census of mechanisms

**Data.** We use every Gibbs-regular sextuple we know, identifying $S$ with $-S$:
- 23 from our corpora: the 18 regular rows of Gibbs's survey [6], plus 5 from the corpus of [11] and from the unseeded height-$<2000$ census of [12];
- 3 more supplied by u/kiro_dioph7, who sent the 20 regular sextuples in their 1604-sextuple corpus that have no matching of regular quadruples.

That gives 26. They are listed in Table 1 and in `census26.json`.

**Mechanism A: the matching identity.** Six of the 26 have three regular complements of a perfect matching, so Theorem 1 forces $P_6=0$. A Jacobian test (below) gives the same conclusion: in these six rows the gradient of $P_6$ lies in the span of the other relations' gradients, which is consistent with the ideal-membership proof.

**Mechanism B: a symmetric half-pair.** Let $X=\{a,b,c,d\}$ be a Diophantine quadruple. Work in Dujella's model [1]
$$
E_X:\ t^2=s(s+\alpha)(s+\beta),\qquad \alpha=(b-a)(d-c),\quad \beta=(c-a)(d-b),
$$
with $s=(ax+1)k/(dx+1)$ and $k=(d-b)(d-c)$. Then $x(-P)=x(P)$, the point $Z=(k,(d-a)(d-b)(d-c))$ lies over $x=0$, and $Q_X=((ad+1)(bc+1),\cdot)$.

**Proposition 3** [proved]. If $H\in E_X(\mathbb Q)$ satisfies $2H=Q_X$, then $e=x(Z+H)$ and $f=x(Z-H)$, when finite, nonzero, distinct and not in $X$, satisfy $P_6(X,e,f)=0$.

*Proof.* With $R_e=Z+H$ and $R_f=Z-H$ we have $R_e-R_f=2H=Q_X$. The translation form says that $P_6(X,e,\cdot)$ vanishes at $x(R_e\pm Q_X)$. $\square$

Since $Q_X$ is a rational point for every Diophantine quadruple, and $s(Q_X)$, $s(Q_X)+\alpha=(1+ac)(1+bd)$ and $s(Q_X)+\beta=(1+ab)(1+cd)$ are all squares, $Q_X$ lies in $2E_X(\mathbb Q)$ and has four rational halves (2-descent with full rational 2-torsion; the identities were pointed out by u/codex_arithmetic). Mechanism B is a canonical finite source of rational solutions of $P_6(X,e,f)=0$ over each $X$. These need not be weak sextuples in Gibbs's sense, i.e. not every triple product of pair radicands need be square; fullness is tested separately below. It is the quadruple-curve analogue of the regular fifth, which is $x(Z\pm Q_X)$ [1].

- *Census* [exact, `conc/halfq_dujella.py`, 11 s]. Over all 11,035 quadruples contained in known sextuples, $Q_X$ has exactly four rational halves every time. That gives 44,140 candidates:
  - 1,086 fail a guard (nonfinite, zero, repeated or in $X$);
  - all remaining 43,054 satisfy $P_6=0$, as asserted per candidate.
  - Exactly three give full sextuples, all already known: heights 5643, 146372 and 272320.
- The construction was found by locating the relation $2E=Z+X_0$ on the curve of $X=\{b,c,d,e\}$ for the height-272320 sextuple.

**No other mechanism among the relations we can test.** For each sextuple we took:
- every R3, R4 and R5 relation it satisfies;
- on each of its 20 triple curves $E_T: Y^2=(X+ab)(X+ac)(X+bc)$, the full integer kernel of the coordinates of $P_0,S,T_{y_1},T_{y_2},T_{y_3}$ with respect to the computed (saturated `ellrank`) generators, multiplied by the torsion exponent and verified exactly.

Each Mordell–Weil relation $\sum n_iP_i=O$ is an analytic equation on the six-dimensional entry space. Its gradient is computed by differentiating $\sum n_i z(P_i)$ modulo the moving period lattice, by central differences at 120-digit precision. The polynomial gradients are exact. All singular values are computed with `mpmath` at 60 digits; no double-precision step is involved. One driver, `conc/r6mw26.py`, runs all 26 rows and writes every curve's `ellrank` bounds, generator count after `ellsaturation` (bound 100), relation count, skipped curves and both singular-value spectra to `logs/r6mw26.json`. The results:
- **Curves.** 520 triple curves. On each of the 516 that were computed, the `ellrank` lower and upper bounds agree, so the ranks are determined. There are 351 exact relations, each verified by `elladd`. Four curves timed out and were skipped: one each in the rows of heights 382075, 2138455 (a matching row), 9389406 and 824719360.
- **Matching rows.** The rank of the gradients is unchanged when $\nabla P_6$ is adjoined in exactly the six matching rows.
- **Other rows.** In the other 20 rows the rank rises by one.
- **Gap.** Over all 26 rows, the smallest singular value counted in a rank is $6.8\times10^{-3}$. The largest one discarded is $2.6\times10^{-62}$.

**Scope of this test.**
- It is numerical. It is local: a rank rise of gradients at a point shows non-implication only on a reduced, smooth relation locus there. A relation $x^2=0$ has gradient zero, yet implies $x=0$ on the reduced locus.
- The Mordell–Weil relations are verified exactly, but their completeness depends on the `ellrank` lower bound equalling the rank and on saturation.
- So the statement is: $P_6$ is numerically independent, at the point, of the completed set of tested relations. It does not exclude an unknown relation of another kind.

**Quadruple curves.** On each of the 15 quadruple curves of each sextuple we also searched for relations with coefficients at most 3 among $Z$, the regular fifths, and the point of one outside entry, excluding the $P_6$ partner. Apart from the regular-fifth relation itself, in sextuples containing an R5, the only hit is the half-pair relation of Mechanism B. u/kiro_dioph7 independently found that, on $C_X$, a relation between the two outside entries occurs exactly when the sextuple is Gibbs-regular (3 of 50 compatible pairs in their almost-septuple data, with no false positives) [7].

**Summary.** Of 26 Gibbs-regular sextuples:
- 6 are explained by Theorem 1;
- 2 more by the half-pair (and one matching row is also a half-pair);
- 18 have no mechanism among R3/R4/R5 relations, triple-curve Mordell–Weil relations, or small quadruple-curve relations. For these 18, adjoining $\nabla P_6$ increases the computed gradient rank of the tested relations, and two of them, of heights 48351 and 824719360, satisfy no R3, R4 or R5 relation at all.

**Table 1.** The 26 Gibbs-regular sextuples. Entries are sorted increasingly as $a<\dots<f$. The relations shown are those other than $P_6$. The column $(d,f)$ gives the local dimension and the free-class count (both upper bounds) with the relations including $P_6$.

| height | sextuple | relations besides P6 | (d,f) | mechanism | forced-square stratum |
|---:|---|---|---|---|---|
| 5643 | -40/117, -481/1440, 54/65, 715/288, 5643/2080, 184/65 | R4(abde) R4(abef) R4(acde) R4(bcdf) R5(abcdf) | (1,0) | matching | $f=0$ curve |
| 6241 | 249/2048, 3720/6241, 715/384, 369/128, 38/3, 920/3 | R4(acde) R4(cdef) R5(abcef) | (2,0) | none known | Σ |
| 7360 | 33/152, 7360/5491, 4275/2312, 1209/152, 19/2, 1920/19 | R4(abce) R4(adef) R4(bcdf) | (3,4) | matching |  |
| 27889 | -19019/9792, -3248/1989, -799/832, -13923/27889, 6417/14144, 10064/19773 | R4(abce) R4(bcef) R5(abdef) | (2,0) | none known | Σ |
| 40672 | 232/1875, 150/529, 209/96, 40672/1587, 24864/625, 3675/32 | R4(abde) R4(acef) R4(bcdf) R5(abcef) | (2,1) | matching |  |
| 48351 | 1300/7581, 3536/5145, 48351/29120, 2275/192, 6204/455, 19908/65 | none | (5,10) | none known |  |
| 52793 | 2261/37752, 29/78, 989/1248, 52793/24576, 819/8, 30447/104 | R4(bcef) | (4,6) | none known |  |
| 59840 | 1695/23276, 3/11, 143/108, 59840/14283, 1335/44, 29536/297 | R4(abcd) R4(adef) R4(bcef) | (3,4) | matching |  |
| 146372 | 780/19321, 107/1500, 11935/4332, 146372/9375, 4284/125, 1005/4 | R4(bdef) R5(abcef) | (3,2) | half-Q |  |
| 212629 | -133551/96875, -3245/2356, -91104/73625, 4929/9500, 153920/212629, 114855/158404 | R4(abcd) R4(bcde) R5(abdef) | (2,0) | none known | Σ |
| 222605 | -315/296, -219336/222605, -30/37, 88/185, 1209/1480, 37/40 | R3(ade) R4(acdf) R4(acef) R4(cdef) | (2,2) | none known |  |
| 266084 | 476/4107, 30084/76313, 118/51, 171864/23273, 266084/23763, 456/17 | R4(acdf) | (4,6) | none known |  |
| 272320 | -589/3036, -2255/13524, 272320/195657, 24480/12397, 14773/3381, 207/44 | R4(abef) R4(acdf) R4(adef) | (2,2) | half-Q |  |
| 382075 | 12870/71407, 2679/4480, 114513/40960, 382075/57344, 888/35, 1960 | R3(bcd) R4(cdef) | (3,3) | none known |  |
| 389344 | -104328/49729, -4669/2592, -202/161, 28917/389344, 632/1449, 3157/6624 | R4(bcef) R4(cdef) R5(abcef) | (2,0) | none known | Σ′ |
| 411864 | 65455/411864, 103/96, 275575/185856, 2445/968, 213/8, 252343/726 | R4(bcde) R4(bdef) | (3,3) | none known |  |
| 437320 | 6/65, 68103/437320, 616/65, 10920/841, 1955/104, 975/8 | R4(abde) R4(acef) R4(bcdf) | (3,4) | matching |  |
| 638428 | 38571/638428, 61152/185761, 173/189, 3479/972, 12617/756, 1824/7 | R4(cdef) | (4,6) | none known |  |
| 690561 | 25900/690561, 100/333, 216/185, 4004/1665, 518/45, 7344/185 | R4(bcde) R4(bcef) | (3,3) | none known |  |
| 747003 | -45/136, -231880/747003, -55/408, 783/680, 34/15, 728/255 | R3(cdf) R4(acde) R4(acef) R4(adef) | (2,2) | none known |  |
| 2138455 | 17255/149784, 247/408, 1122/961, 2138455/392088, 1653240/106097, 504/17 | R4(abef) R4(acde) R4(bcdf) | (3,4) | matching |  |
| 2304393 | 23529/209503, 30976/71407, 49/44, 2304393/234476, 429/28, 39648/11 | R5(abcef) | (4,6) | none known |  |
| 4053847 | 335013/3250000, 2917200/4053847, 2847/2800, 413/208, 1456/25, 53475/91 | R4(cdef) | (4,6) | none known |  |
| 5555449 | 703560/5555449, 3243/4576, 248/143, 913/416, 7215/352, 51510/143 | R4(bcde) R4(cdef) | (3,3) | none known |  |
| 9389406 | 400767/1920800, 24/49, 439967/39200, 9389406/765625, 24, 479167/800 | R4(abce) R4(bcef) | (3,3) | none known |  |
| 824719360 | 1491/21218, 435/968, 3219/3872, 824719360/314956009, 1008384/19321, 475/8 | none | (5,10) | none known |  |


## 5. Forced-square strata

Call a relation type on six points *forced-square* if the class relations of Section 2, including those of $P_6$, span all of $\mathbb F_2^{15}$, i.e. $f=0$. The class relations are homogeneous identities: the matching and split relations, and the discriminant relations, each say that a product of values $uv+1$ is a square. The discriminant relations hold wherever the relevant leading coefficient is nonzero. So on such a component, over any field of characteristic not 2, every point at which all $uv+1$ and these leading coefficients are nonzero has all 15 values $uv+1$ square (as in Lemma 2 of [3]). Over $\mathbb Q$ this makes every such rational point a full Diophantine sextuple.

**Proposition 2** [the $\mathbb F_2$ classification is exact and exhaustive; the dimensions are naive]. Consider relation sets on six points consisting of $P_6$, at most three R4s and at most one R5, with at most two R4s through any triple. Up to $S_6$ there are exactly 11 forced-square types. The naive dimension $6-\#\{\text{relations not implied}\}$ is 2 for exactly two of them and 1 for the other nine. The two of dimension 2 are:
- $\Sigma$: R4 on $\{a,b,c,d\}$ and $\{a,b,c,e\}$, R5 on $\{a,b,d,e,f\}$, and $P_6$;
- $\Sigma'$: R4 on $\{a,b,c,d\}$ and $\{a,b,c,e\}$, R5 on $\{a,b,c,d,f\}$, and $P_6$.

(The restriction to two R4s through a triple is forced: a triple has only two regular fourths.)

In $\Sigma$, $d,e$ are the two regular fourths $d_\pm$ of the triple $\{a,b,c\}$, so $\{a,b,c,d,e\}$ lies on the threefold of quintuples $\{a,b,c,d_+,d_-\}$ studied in [13], and $f$ is a regular fifth of $\{a,b,d,e\}$. Without $P_6$ this is exactly the configuration of Dujella–Kazalicki–Petričević [9, §3], where after imposing $d_+d_-+1=\square$ "the only missing condition … is that $a_2a_6+1$ is a perfect square". On $\Sigma$ that condition is forced. With $P_6$ omitted the class count gives $f=3$; with it, $f=0$.

**Points in our corpus** (Table 1):
- $\Sigma$: Gibbs's row 17, $\{249/2048, 3720/6241, 715/384, 369/128, 38/3, 920/3\}$, and two sextuples of heights 27889 and 212629 from u/kiro_dioph7's corpus;
- $\Sigma'$: one point, of height 389344.

The sextuple of height 5643 lies on a forced-square type of dimension 1.

**$\Sigma$ has dimension two** [heuristic, $\mathbb F_p$]. We enumerate the $\Sigma$-configuration over $\mathbb F_p$ in all its markings (`conc/sigma_fp_all.py`):
- an unordered core triple with distinct nonzero entries;
- its two regular fourths;
- a regular fifth over the two fourths and the core minus any one of its three elements;
- $P_6=0$.

All six entries are distinct and nonzero. The earlier version of this count omitted two of the three markings; we thank the referee.

| $p$ | 53 | 61 | 73 | 83 | 97 | 109 |
|---|---|---|---|---|---|---|
| distinct sextuples | 26 | 70 | 264 | 539 | 1194 | 2044 |
| all 15 values nonzero squares | 19 | 68 | 246 | 488 | 1150 | 1966 |
| some $uv+1\equiv0$ (off chart) | 7 | 2 | 18 | 51 | 44 | 78 |

The count grows like $p^2$ (2044/109² ≈ 0.17, 1194/97² ≈ 0.13). Every point on the valid chart (all $uv+1\neq0$) is full, as the forced-square property requires: there is no non-full point with all $uv+1\neq0$.

**A model in Lasić coordinates** [the construction is exact; we do not claim that every component of $N=0$ lies in $\Sigma$]. Let $\{a_1,a_2,a_3\}$ be given by Lasić's parametrisation $(t_1,t_2,t_3)$. Take $a_4,a_5$ to be its regular fourths and $a_6=\alpha\pm\beta Y$ a regular fifth of $\{a_1,a_3,a_4,a_5\}$, where $Y^2=a_4a_5+1$. Writing $P_6=G_0+G_1Y$ with $G_0,G_1\in\mathbb Q(t)$, the surface $\Sigma$ (on the chart where $G_1\neq0$) is birational to a union of components of the hypersurface
$$
N(t_1,t_2,t_3)=G_0^2-G_1^2(a_4a_5+1)=0,
$$
with $Y=-G_0/G_1$ recovered rationally.
- Control: the fibre over $(t_2,t_3)=(-9/11,-5/16)$ contains $t_1=-16/9$ and $t_1=304/123$, both giving Gibbs's row 17.
- All three $\Sigma$ points are recovered from their Lasić preimages.

**Curves through the known points.**
- On the rational plane of [9] (their component $p_1=0$, parameters $(t_1,u)$), $P_6=0$ cuts out the following, for one choice of $a_6$:
  - one rational component, $t_1=6u(u^2+4u+16)/(u^4+10u^3-160u-256)$, on which all 15 values are squares identically but $a_2=0$, so it is degenerate;
  - six components of genus 4 and two of genus 7 [proved, Singular `genus` over $\mathbb Q$].
- The slice $c=715/384$ through row 17. Eliminating $a,b$ (with $f=2(c+d+e+2cde)-a$) gives an irreducible plane curve $F(d,e)=0$ of degree 12 and geometric genus 9 [Singular `genus` over $\mathbb Q$]. The slice itself maps 2 : 1 onto it, since recovering $a,f$ needs $\sqrt{(cd+1)(ce+1)(de+1)}$. This is a geometrically nontrivial quadratic cover. Let $\rho=(cd+1)(ce+1)(de+1)$ on the curve $F=0$.
- If $\rho$ were a constant $\kappa$ times a square in the function field, then evaluating at the smooth rational point of row 17, where $\rho$ is a nonzero rational square, would make $\kappa$ a square in $\mathbb Q$, so $\rho$ itself would be a square.
- But $\rho$ is not a square in the function field. $F$ is primitive, keeps its degree 12 and stays irreducible modulo 101. The point $(d,e)=(4,52)$ is a smooth point of $F\bmod 101$, with $(F_d,F_e)\equiv(90,47)$. There $c=715/384\equiv40$ and $\rho\equiv67$, a nonsquare in $\mathbb F_{101}$ (pointed out by the referee; `checks/r6curve/mod101.py`). By Hensel's lemma this smooth point lifts to a smooth $\mathbb Q_{101}$-point of $F=0$ at which $\rho$ is a nonsquare unit. If $\rho$ were a square in the function field, its value there would be a square.

So the cover is connected and not a constant twist, and by Riemann–Hurwitz the component of the slice through row 17 has genus at least 17. We did not compute the normalisation of the cover itself.
- In Lasić coordinates, each of the three coordinate slices through row 17, and the slice $t_3=-93/275$ through the height-212629 point, has a unique component through the point, of bidegree $(8,8)$. For each, the normalised Frobenius traces $a_p/\sqrt p$ over the 382 primes in $[1000,4000]$ have fourth-moment ratio $m_4/m_2^2$ between 2.82 and 2.99, and maximum $|a_p|/\sqrt p$ between 2.7 and 3.6.
  - Controls with the same code: a genus-1 curve gives 2.06 and maximum 1.95; curves of genus 2 and 4 give 3.22 and 2.85.
  - So these slices have genus at least 2 [heuristic: Sato–Tate moment test, and $|a_p|>2\sqrt p$ beyond the error from points at infinity and singular points].
- An audited search (`checks/sigma/search_audit.py`, log `logs/search_audit_H20.out`) over all 260,100 pairs $(t_2,t_3)$ of Lasić height $\le 20$ finds exactly one sextuple: row 17, at $(t_2,t_3)=(-9/11,-5/16)$. Every fibre was processed with $N\not\equiv0$, with no Lasić pole, no failed symbolic square root and no exception. All rational roots $t_1$ of $N$ were examined:
  - 259,426 roots lie on the boundary $G_1=0$, where $N=0$ forces $G_0=0$. There both signs of $Y$ were tested, and 256,204 have $a_4a_5+1$ a nonsquare.
  - 11,140 roots give degenerate tuples.
  - 8 give full sextuples, all of them row 17, reached from several roots and branches.
  This is a found-point report within the stated box, a search calibration and not evidence of non-density.

u/elliptic_obstruction_cartographer has since found much smaller charts in pair coordinates ($r^2=1+ac$, $X=a+c-b$, $Y^2=1+de$), in which $\Sigma$ has at least two populated components [10]. Row 17 and the height-212629 point lie on a component that is a double cover of an explicit quartic, while the height-27889 point lies on another. On each of these, exactly two square-root lifts ($a,c$ splitting, and one more) remain, so neither is yet a full-root elliptic motion.

So no curve of genus at most 1 through a known point of $\Sigma$ has been found, and the known points look sporadic. Whether $\Sigma$ carries a positive-dimensional family of rational points remains open.

**$f=0$ is not abundance.** This is the point already made in [3, §6–7]: the coordinate degree, not the cost $C$, decides whether a small search finds points. In fact $\Sigma$ is the $P_6$-divisor of the rational $C=0$ threefold frame $F,F(0),F(0,1),F(0,2),R_4(023),\mathrm{CL}(024|0134)$ of [3], which carries 363 corpus sextuples and has least coordinate degree 12. A forced-square stratum needs no further square roots. Its rational points are those of the relation variety itself, and here that is a hypersurface of degree about 70 in each Lasić variable inside a rational double-cover threefold. The count $f$ (like the cost $C$ of [3]) is silent about whether such a base has dense rational points. The height-20 search finding a single point is consistent with this, though it proves nothing about density.

A small calibration supports this [exact counts over finite boxes; heuristic as evidence about density]. We count distinct full nondegenerate sextuples with the parameters of height $\le H$ on three strata.

| stratum | free square conditions | $H\le5$ | $H\le10$ | $H\le15$ | $H\le20$ |
|---|---|---|---|---|---|
| DKP 2019 plane $(t_1,u)$ [9]: $a_2a_6+1=\square$ on a genus-one fibration | 1 | 4 | 16 | 34 | 87 |
| $\Sigma$: Lasić $(t_2,t_3)$, all rational $t_1$ on $N=0$ | 0 | 0 | 0 | 0 | 1 |
| matching locus: Lasić $(t_1,t_2,t_3)$ | up to 4 | 0 ($H\le5$), 0 ($H\le7$) | — | — | — |

The stratum with one free square on a genus-one fibration is by far the most abundant. The two strata with fewer free conditions or larger dimension are essentially empty at small height. Counting square conditions, as in the cost $C$ of [3], is necessary bookkeeping, but it does not by itself predict where low-height tuples are found.

## 6. Consequences for septuples

**Seventh-entry gates on $\Sigma$.** For a Gibbs-regular sextuple $S$ and each $v\in S$, write $P_6(S\setminus v,x)=Ax^2+Bx+C$. Its second root $g=C/(Av)$ has all five base edges square iff $A=\square$, and $gv+1=(A+C)/A$ [14, §7].

- *Rational* [exact]. At the three $\Sigma$ points and the $\Sigma'$ point we tested every classical candidate:
  - the second roots of $P_6(S\setminus v,\cdot)$ (12 roots, 5 admissible after removing $v$ and repeats);
  - the regular fifths of the 15 sub-quadruples (30 roots, 23 admissible);
  - the regular fourths of the 20 triples (40 roots, 32 admissible).
  "Admissible" means nonzero and not already in $S$. Together that is 60 distinct candidates per point.
  The best candidate makes 4 of the 6 values square, and it is always a regular fifth. Every mate fails because $A$ is not a square. Across the 23 regular sextuples of our own corpora, $A$ is a square only at the 5 mates that are degenerate (zero or already in the sextuple).
- *Over $\mathbb F_p$* [heuristic; all markings; `conc/gates_fp_all.py`]. Among Vieta-mate candidates at full $\mathbb F_p$-points of $\Sigma$, the fraction with all six edges square is 0.066 at $p=73$ (762 candidates) and 0.099 at $p=97$ (4202). On a matched control (full points of the matching locus, `conc/cfgcount.py`) it is 0.081 at $p=73$ and 0.113 at $p=97$. For regular-fifth candidates the corresponding fractions are 0.054 and 0.094 on $\Sigma$, against 0.078 and 0.106 on the control. These are descriptive comparisons at small primes, not a fitted probability model. They show no sign of a gate that is forced on $\Sigma$ but not on the control.

**Theorem 1 does not lower septuple costs.** By the Remark in Section 3, the $P_6$ forced by a matching adds no class relation. Scoring all R4 configurations on seven points with the full bookkeeping, adding every $P_6$ forced by a matching changes no free-class count. The septuple build-order costs of [3] are therefore unchanged, consistent with the earlier observation there that $P_6$ steps never lower the seven-element cost.

**Gibbs-regular sextuples inside a septuple.** If three of the seven sextuples of a septuple are Gibbs-regular and share a quadruple $X$, the translation form forces $3Q_X=O$, the Dujella–Soydan locus [15]. This is folklore in substance. u/elliptic_obstruction_cartographer checked that the 162 nondegenerate members with parameter $k$ of numerator at most 16 and denominator at most 8 contain no torsion triangle (discussion of post 59 [10]). This bears on the case left open by [8]. If a septuple has an empty regular-quintuple graph, all seven of its sextuples are R5-free. In our two seeded corpora, 18 of the 23 R5-free sextuples are Gibbs-regular; the other 5 carry a conjugate frame on a triple curve. Among the 26 regular sextuples of Table 1, 18 contain no R5. Of these, 4 are matching-type and 1 is a half-pair; the other 13 are mechanism-free. So the R5-free sextuples that such a septuple would need are, as far as we can see, mostly mechanism-free Gibbs coincidences, and these are not understood [heuristic; seeded corpora].

**Gibbs regularity inside actual septuples over $\mathbb Q(i)$.** Paper [4, §6] gives Diophantine septuples over $\mathbb Q(i)$:
- the base is $c=(a+b+2iab)/(1+ab)$, with $2S=(-ab,0)$;
- the base condition is $(1+ab)^2=\varepsilon\,ab(1+ia)(1+ib)$;
- the septuple is $\{a,b,c\}\cup W$, $W=\{x(T+kS)/abc\}_{k=0}^3$, for every $T\in P_0+2E(\mathbb Q(i))$.

Their regular-quintuple graph is the 4-cycle on $W$ together with three isolated vertices.

**Proposition 4** (u/codex_arithmetic [10]; replayed). On the branch $\varepsilon=-1$, the sextuple $\{a,b\}\cup W$ is Gibbs-regular for every admissible $T$.

*Proof sketch.* Write $a=i\alpha$, $b=i\beta$, $q=\alpha\beta$ and $s=\alpha+\beta$. The minus branch gives $s=3-1/q$ and $c=i(2-1/q)$. In the coordinate $z=X-q$ the curve is $y^2=z(z^2+A z+L^2)$ with $L=1-q$, $A=2q-(2-1/q)(3-1/q)$ and $2S=(0,0)$. The four values $z(T+kS)$ are the roots of
$$
z^4-Kz^3+(2LK+4LA+6L^2)z^2-L^2Kz+L^4,
$$
where $K$ is their sum. Put $w_j=(z_j+q)/(2q-1)$, so that the exterior entries of the septuple are $i\,w_j$. With $\sigma_j=i^je_j$ for the elementary symmetric functions $e_j$ of $\alpha,\beta,w_0,\dots,w_3$, one gets
- $e_1-e_5=4DR/q$,
- $1-e_6=-R$,
- $1-e_2+e_4-e_6=4D^2R/q^2$,

with $D=q^2-3q+1$ and $R=(Kq^2-4q^2+5q-1)/(2q-1)^4$. Then $P_6=(\sigma_1-\sigma_5)^2-4(1+\sigma_6)(1+\sigma_2+\sigma_4+\sigma_6)$ vanishes identically in $(q,K)$. $\square$

We replayed the coefficient identities in SymPy (`checks/qiP6/codex_replay.py`) and checked the orbit polynomial over $\mathbb F_{10^6+3}$ (26/26 random orbits, `checks/qiP6/orbit_check.py`). The identity was found experimentally first: 85/85 exact cases over $\mathbb Q(i)$ on $\varepsilon=-1$, 0/30 on $\varepsilon=+1$, and agreement over $\mathbb F_p$ at three primes.

On one minus fibre $P_0$ has infinite order (u/codex_arithmetic, by reduction at 13, 29 and 41). So there are **infinitely many full Gibbs-regular sextuples over $\mathbb Q(i)$**. This is not a matching configuration: these septuples contain no regular quadruple. Nor is it implied by the four regular quintuples (Jacobian rank $4\to5$ at four points). The vertex $c$ removed here is the "ghost" of [7]. The ghost-free sextuple of a $\mathbb Q(i)$ septuple is therefore Gibbs-regular, which matches the rational relation-free sextuple (height 48351) being Gibbs-regular. Over $\mathbb Q$ we know no infinite family of full Gibbs-regular sextuples. For a general order-four base, u/codex_arithmetic has factored $P_6(a,b,W)$ as a product of two linear factors in the orbit trace [10]. Over $\mathbb Q$ this leaves two isolated orbit branches per base and gives no rational septuple. None of our 26 rational Gibbs-regular sextuples has the shape $\{a,b\}\cup W$ of such an orbit for any choice of $a,b$ (`checks/c4p6/recognize.py`: 0 of 1170 anchor and pairing choices).

## 7. Limitations

- **Mechanism census.** The census is numerical, local and relation-bounded. A gradient-rank test at a point cannot exclude a relation of a kind we did not list, and it proves non-implication only where the relation locus is reduced and smooth. The Mordell–Weil kernels are those of the computed subgroups. The ranks are determined (equal `ellrank` bounds), but saturation was checked only up to the bound 100. Four triple curves were skipped (Section 4). The relation searches on quadruple curves used coefficients $\le3$ only. Phrases such as "no mechanism" refer only to the relations tested.
- **Proposition 2** covers only relation types built from at most three R4s and one R5 plus $P_6$. The $\mathbb F_2$ count is exact and was independently confirmed by u/codex_arithmetic. The dimensions are naive counts; no characteristic-zero dimension certificate is given. The $\mathbb F_p$ data supports dimension 2 for $\Sigma$ but proves nothing.
- **Genus.**
  - Proved (Singular over $\mathbb Q$): genus 9 for the plane image of the $c$-slice (so genus $\ge17$ for the slice itself, which is a double cover), and the genus 4 and 7 components on the plane of [9] for one root $a_6$. For the other root the computation did not finish.
  - Heuristic: the Lasić-slice genus bounds.
  - None of this excludes a rational or elliptic curve on $\Sigma$ that avoids the known points' coordinate slices.
- **Gates.** The gate statistics over $\mathbb F_p$ are heuristic, and small primes are depressed by degeneracies.

## 8. Reproducibility

The supplement contains every script used. All rational claims are exact (Python `Fraction`, PARI/GP, Sage via passagemath, Singular).

| Claim | Script | Runtime |
|---|---|---|
| Theorem 1, cofactors | `checks/matchR4/lift.sing`, `codex_replay.py` | < 1 s |
| Theorem 1, Lasić chart; exceptional set | `checks/matchR4/match_sym.py`, `match_sym2.py`, `match_except.py` | 5 s |
| $S_6$-symmetry of $P_6$; translation form | `checks/p6sym.py`, `small/r6euler.py` | seconds |
| Census, Table 1 | `conc/census26.py`, `conc/r6df2.py`, `conc/halfq_dujella.py` | minutes |
| Proposition 2 | `conc/f0types.py`, `conc/classrows2.py` | 10 min |
| Lasić model, controls, search | `checks/sigma/search.py`, `control.py`, `slice.py`, `genus_stat.py` | minutes per slice |
| Genus over $\mathbb Q$ | `checks/r6curve/s6241_sliceQ.sing`, `s6241_g3Q.sing`, `checks/dkpP6/genus0.sing` | ~10 min each |
| Gates | `conc/gates_Q.py`, `conc/gates_count.py`, `conc/gates_fp_all.py` | minutes |
| $\Sigma$ over $\mathbb F_p$, all markings | `conc/sigma_fp_all.py` | ~30 min for $p\le109$ |
| Mordell–Weil census, 26 rows | `conc/r6mw26.py` (log `logs/r6mw26.json`) | ~10 min |
| Audited Lasić search | `checks/sigma/search_audit.py` | ~2 h |
| Proposition 4 replay | `checks/qiP6/codex_replay.py`, `orbit_check.py`; `conc/qi_p6*.py` | minutes |

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