# A conditional seven-quintuple bound from finite regularity graphs

## 1. Statement and contribution

Let $A=\{x_0,\ldots,x_6\}$ be seven distinct nonzero rational numbers
such that $1+x_ix_j$ is a rational square for every $i<j$. For five
entries with elementary symmetric functions $e_i$, put
$$
R_5=(e_5-e_1)^2-4(1+e_2+e_4).
$$
The regular-quintuple graph $G(A)$ has an edge $ij$ when
$R_5(A\setminus\{x_i,x_j\})=0$, following [1]. Its number of edges is
the number of regular five-element subsets of $A$. Zero is allowed as
a square throughout.

**Theorem 1 (computational, with an explicit completeness assumption).**
Assume the completeness of the characteristic-zero polynomial-system
classifications specified in Sections 4–5, including the imported
$K_{2,3}$ and three eight-edge exclusions of [2,3]. Then a rational
Diophantine septuple has at most seven regular quintuples.

The completeness assumption is the one supplied by msolve's
multimodular stopping criterion. The reported finite points, their
substitution into the input equations, and the ensuing arithmetic
obstructions have separate exact checks. Those checks do not prove
that the solver omitted no component. We do not assert an independent
deterministic degree or ideal-membership certificate for the new finite
classifications.

The contribution is a global numerical restriction, obtained from six
eight-edge graphs rather than a classification of every smaller graph.
Three exclusions are from [3]. We compute the other three, using the
four-cycle equations for one and small symmetric pair quotients for
two. One quotient forces a coordinate square to equal $21$; another
forces a strictly positive quartic in an actual pair product. A short
graph argument proves that these six cases exhaust the obstruction to
the bound. This argument does not require a heptagon degree hypothesis
or the unfinished classification of connected seven-edge graphs.

We also combine the known zero-pair matching lemma of [4] with the
four-cycle involution of [3]. Their combination shows that the
involution's nonvanishing hypotheses are automatic on the distinct
nonzero locus. Thus the generic pivot systems already cover every
zero-radicand case; separate boundary charts are unnecessary for this
application. Finally, an explicit full lift of one finite quotient
provides a septuple over a totally real field of degree eight, as an
existence control for the computation. It is not a rational septuple.

The rational septuple problem remains open. The theorem allows sparse
regularity graphs, including the empty graph.

## 2. Prior results and what is added here

The graph and its translation interpretation are due to [1]. The
$K_{2,3}$ obstruction and public atlas of triangle-free seven-vertex
graphs are due to [2]. The latter locates the finite strata treated
here, but finite-field points or their absence alone do not classify
their rational points. Paper [3] gives the explicit four-cycle pivot
equations and excludes three of our six eight-edge cases. We import
those exclusions with their actual solver qualification.

The zero-pair matching lemma is already [4, Lemma 3]; we do not claim it
as new. Section 3 records its immediate consequence for [3]. The
degree-four reduction and census in [5], and the five-edge theorem for
degree-four graphs in [6], remove all high-degree counterexamples.
The latter uses individual rational-square conditions in some source
components, so Theorem 1 is a statement about actual rational
Diophantine septuples, not arbitrary rational solutions of window
equations.

The present new arithmetic is the finite classification of T7092,
T42232 and T7864, coupled to a six-core proof of the global bound.
These three exclusions require only rational solutions of the edge
windows with distinct nonzero entries; they do not assume individual
rational square roots. The polynomial reductions hold over every
characteristic-zero field, but the resulting arithmetic exclusions
are field-specific. In particular, Section 6 exhibits a number-field
solution of T42232.
The pair quotients retain all distinct nonzero source configurations,
including zero radicands, and produce short arithmetic obstructions.
The broader geometric framework rests on classical regular extensions;
see Dujella [7] and the elliptic constructions of rational sextuples
in [8]. We do not claim novelty for the regular-fifth polynomial,
its discriminant, or the order-three star relation.

## 3. Local polynomial facts and automatic pivot coverage

### 3.1. Two distinct fifths

For a quartet with elementary symmetric functions $E_1,\ldots,E_4$,
the regular-fifth polynomial is
$$
Q(X)=(E_1+X(1-E_4))^2
      -4(1+E_2+E_4+X(E_1+E_3)).
$$
It cannot vanish identically for a distinct quartet: its coefficients
would imply $E_4=1$, $E_3=-E_1$, $E_2=E_1^2/4-2$, and hence the
quartet root polynomial would be
$$
(T^2-E_1T/2-1)^2.
$$
Thus a quartet has at most two distinct regular fifths. This proves
triangle-freeness of $G$: a triangle would give three distinct fifths
of one quartet. If two distinct fifths do exist, then $Q$ is quadratic
and has nonzero discriminant
$$
\operatorname{disc}(Q)=16\prod_{i<j\text{ in the quartet}}(1+x_ix_j).
\tag{1}
$$
In particular, all six quartet radicands are nonzero. The leading
coefficient also gives $E_4\ne1$. These facts are used in [1,6].

### 3.2. The known matching lemma and its four-cycle consequence

**Lemma 2 ([4, Lemma 3]).** If $1+x_ix_j=0$, then the graph induced
on the other five vertices has maximum degree at most one.

An alternative proof is immediate from (1). A wedge $za,zb$ in that
induced graph would give two distinct regular fifths $a,b$ of the
quartet $A\setminus\{x_z,x_a,x_b\}$. That quartet contains the zero
pair $i,j$, contradicting its nonzero discriminant. This argument
works over any field of characteristic zero with distinct entries;
no other square conditions are used.

**Corollary 3 (global coverage of the four-cycle pivot systems).**
Let $W$ be the vertices of a four-cycle in $G$, and $Z=A\setminus W$.
Every zero pair is wholly contained in $W$. Consequently all radicands
within $Z$ and between $Z$ and $W$ are nonzero.

Indeed, deleting a pair not wholly in $W$ leaves either the four-cycle
or a three-vertex path, both contradicting Lemma 2. Therefore the
hypotheses of [3, Proposition 1] hold automatically. For some pivot
$c\in Z$, writing $Z=\{c,p,q\}$, we have
$$
(1+pq)^2=pq(c-p)(c-q),
\tag{2}
$$
and for each opposite pair $a,a'$ of the cycle,
$$
pq(ca+1)(ca'+1)=(c-p)(c-q).
\tag{3}
$$
The proof and elliptic interpretation of (2)–(3) are in [3]. Their
three pivot choices cover the entire distinct nonzero locus. For
several four-cycles, all zero pairs must lie in the intersection of
their vertex sets. The systems used below invert entries and
differences only, so they retain any permitted zero pair within $W$.

The supplement checks (1) and the repeated-quartet identity symbolically,
and independently compares wedge enumeration with the induced-graph
degree test on all 107 public atlas rows. This audit is a check of
the graph bookkeeping, not a premise of the proof.

### 3.3. A degree-three star in symmetric coordinates

Suppose a vertex has three chosen neighbors $d,e,g$ and complementary
triple $a,b,c$. Set
$$
p=abc,\quad s=a+b+c,\quad q=ab+ac+bc,
\qquad L=d+e+g,\quad M=de+dg+eg,\quad N=deg.
$$
The three edge windows imply
$$
\begin{aligned}
G_1&=p^2M-2pL-2ps-4q-3=0,\\
G_2&=p^2N-L+4p+2s=0,\\
D_3&=p^2s^2-4p^2q-12p^2-6ps-4q-3=0.
\end{aligned}
\tag{4}
$$
For completeness, with $B=2p+s$, $C=-2ps-4q-2$ and
$D=s^2-4q-4$, the fixed-triple window is
$$
p^2d^2e^2-2p(d^2e+de^2)+d^2+e^2+Cde-2B(d+e)+D.
$$
Subtracting two windows with a common argument and dividing by the
nonzero neighbor differences gives the cubic having $d,e,g$ as
roots. Vieta gives $M=(2pL+1-C)/p^2$ and $N=(L-2B)/p^2$.
Substitution in the sum of the windows gives $D_3=0$, as in
[6, Section 2.2]. Geometrically, this is the $k=3$ case of the cycle
lemma in [1], forcing $3S=O$ on the complementary triple curve.
Only complementary entries and neighbor differences
are divided out. Thus (4) includes zero radicands. The supplement
also verifies the universal polynomial identities directly.

## 4. Why six eight-edge graphs suffice

First, the maximum degree is at most four. If a vertex had five
neighbors $c_1,\ldots,c_5$, let $f$ be the seventh vertex. Choosing
four neighbors and applying (4) to their three-element substars shows
that $c_1,\ldots,c_4$ are the four roots of $D_3(c_5,f,T)$.
Its leading coefficient is $c_5^2f^2$ and its root sum is $2(c_5+f)$.
Interchanging $c_4,c_5$ forces $3(c_4-c_5)=0$, a collision. The same
argument uses any five neighbors of a degree-six vertex. This is the
root-sum degree lemma of [5].

If a rational septuple graph has a degree-four vertex, [6, Theorem 1]
gives at most five edges. Any counterexample to Theorem 1 therefore
has maximum degree three. It is triangle-free by Section 3 and
$K_{2,3}$-free by [2], with the imported completeness qualification.

We next rule out a disconnected counterexample. A triangle-free,
$K_{2,3}$-free graph on six vertices has at most seven edges. If it
is nonbipartite, it contains a five-cycle; there are no chords, and
the sixth vertex has at most two neighbors on that cycle. If it is
bipartite, partitions $1+5$ and $2+4$ give at most five and six
edges respectively. In the $3+3$ partition, eight edges give
$K_{3,3}$ with at most one missing edge, which contains $K_{2,3}$.
The corresponding bounds for one through five vertices are
$0,1,2,4,5$. Summing over any proper partition of seven gives at
most seven edges. These elementary bounds are independently checked
by a small exact graph enumeration in the supplement.

A connected graph with at least eight edges has a connected spanning
subgraph $H$ with exactly eight edges: take a spanning tree and two
additional edges. It remains triangle-free, $K_{2,3}$-free and
subcubic. Its cycle rank is two. Remove pendant trees. Two cycles
meeting only at one vertex require degree four; disjoint cycles
require at least eight vertices. Therefore the core is a theta,
three internally disjoint paths joining two vertices.

Let their lengths be $a\le b\le c$. At most one is one,
all pair sums are at least four, and $a+b+c\le8$. Type $(2,2,2)$
is $K_{2,3}$ and is excluded. Exactly the following remain:

| Core path lengths | Extra vertex | Attachment types | Atlas names |
|---|---|---:|---|
| $(1,3,3)$ | one pendant | 1 | T7092 |
| $(2,2,3)$ | one pendant | 2 | T7090, T42218 |
| $(1,3,4)$ | none | 1 | T44340 |
| $(2,2,4)$ | none | 1 | T7864 |
| $(2,3,3)$ | none | 1 | T42232 |

![The six theta graphs: T7092 is theta(1,3,3) with a pendant; T7090
and T42218 are the two pendant attachments to theta(2,2,3); T44340,
T7864 and T42232 are theta(1,3,4), theta(2,2,4) and theta(2,3,3).
Dark vertices are the two hubs and orange vertices are pendants.](supplement/six_theta_cores.png)

The pendant must meet a degree-two core vertex. The four available
vertices of $(1,3,3)$ are equivalent; for $(2,2,3)$ the two choices
are a short-path midpoint and a vertex of the long path. The script
`eight_edge_core.py` constructs these cores and attachments first,
identifies isomorphism classes independently, and uses the public
atlas only to attach names.

Paper [3] excludes T7090, T42218 and T44340 over characteristic zero,
with its solver qualification. Section 5 supplies the other three
cases. Every extra edge merely adds an equation, so an exclusion of
these subgraphs excludes their supergraphs. This proves Theorem 1
under the stated completeness assumption.

## 5. The three remaining finite systems

### 5.1. T7092: nine globally covering pivot choices

The edges are
$$
03,05,06,13,14,15,23,24.
$$
The four-cycles have
$$
W_1=\{0,1,3,5\},\quad Z_1=\{2,4,6\},\qquad
W_2=\{1,2,3,4\},\quad Z_2=\{0,5,6\}.
$$
Apply (2)–(3) to both, together with all eight original windows and
$$
1-y\prod_i x_i\prod_{i<j}(x_i-x_j)=0.
\tag{5}
$$
There are nine assignments in $Z_1\times Z_2$. Corollary 3 proves
global coverage; no additional radicand guard is present. The
characteristic-zero calculations reported by kiro_dioph7 give:

| Pivot pair | Output |
|---|---|
| $(2,0)$ | degree-eight eliminant, one octic factor |
| $(4,5)$ | degree-fourteen eliminant, factors of degrees six and eight |
| other seven pairs | empty |

The separating coordinate for the nonempty leaves is $x_0$.
All three factors have root-free reduction modulo five, with leading
coefficients nonzero modulo five. Therefore there is no rational
point in the reported finite loci. This argument does not require an
irreducibility assertion: a checked factor product and root-free
reductions suffice. A rational root has denominator prime to five
when the leading coefficient is prime to five, and would reduce to
a forbidden root.

Explicitly the three factors, in order, are
$$
\begin{aligned}
h_{20}(X)&=88209X^8+40986X^6+1090593X^4-768768X^2+1982464,\\
h_{45,6}(X)&=1452X^6-7056X^4+11667X^2-9604,\\
h_{45,8}(X)&=5184X^8+31536X^6+175180X^4+40113X^2+331776.
\end{aligned}
$$
Their value lists at $0,1,2,3,4$ modulo five are respectively
$(4,4,3,3,4)$, $(1,4,1,1,4)$ and $(1,4,1,1,4)$; their leading
coefficients are $4,2,4$ modulo five. The independent exact verifier
checks all fifteen system equations in each field factor and checks
that the RUR denominators are coprime to the eliminants.

`quotient_T7092.py` independently reconstructs the systems from the
edge list and the symmetric definition of $R_5$. Six published
finite-field controls satisfy exactly one pivot assignment each.
The complete two rational univariate outputs and their exact replay
certificates are included. The seven empty cases are recorded in the
original computation manifest with input hashes and output values;
the mathematical inputs regenerate from the independent script.
The generator emits fully expanded polynomials. An independent run
of its leaf $(2,5)$ reproduces the empty output in approximately
31 seconds using two threads.

### 5.2. T42232: a four-point pair quotient

The edges are $04,05,06,12,13,16,25,34$. Put
$$
A=x_1,\quad B=x_0,\quad f=x_6,\qquad
S=x_2+x_3,\ P=x_2x_3,\ T=x_4+x_5,\ Q=x_4x_5.
$$
Let $r=x_2-x_3$, $t=x_4-x_5$, $k=rt$ and
$D_S=S^2-4P$, $D_T=T^2-4Q$. The two degree-three stars give six
equations (4). The first has complementary data
$(p,s,q)=(AP,A+S,AS+P)$ and neighbor data
$(L,M,N)=(T+f,Q+fT,Qf)$; exchange $(A,S,P)$ and $(B,T,Q)$ for
the other.

For the two remaining windows, put
$$
\begin{aligned}
L_0&=(S+T+r-t)/2,\\
N_0&=(ST-rt+Tr-St)/4,\\
p_0&=ABf,\quad s_0=A+B+f,\quad q_0=AB+Af+Bf,\\
F(r,t)&=(p_0N_0+2p_0+s_0-L_0)^2
       -4(p_0^2+p_0s_0+q_0+1)(N_0+1).
\end{aligned}
$$
The windows are $F(r,t)=F(-r,-t)=0$. Reduce by
$r^2=D_S,t^2=D_T$. Their sum is a polynomial in $k$; their
difference has form $\alpha r+\beta t$. Use the sum,
$\alpha D_S+\beta k=0$, and $k^2=D_SD_T$.
Every original point maps to these equations. Conversely, with
$D_S\ne0$, choosing $r^2=D_S$ and $t=k/r$ recovers both residual
windows. For exclusion only the forward implication is needed.

Adjoin $1-yH=0$, where
$$
H=ABPQfD_SD_T(A-B)(A^2-AS+P)(B^2-BT+Q).
$$
Every factor is nonzero for distinct nonzero source entries. No
radicand is inverted; the possible zero pair $01$ remains covered.

The quotient returned over $\mathbb Q$ has degree four. Its actual
coordinate $B=x_0$ has eliminant
$$
25B^4-666B^2+81=(5B^2-24B-9)(5B^2+24B-9).
\tag{6}
$$
Both quadratic discriminants are $756=36\cdot21$. An exact check
in the returned algebra verifies all ten input equations and the
identities
$$
\begin{gathered}
f^2=21,\quad AB=9/5,\quad 6f=5(A+B),\\
S=(5A+11B)/18,\quad T=(11A+5B)/18,\\
P=-S^2/16,\quad Q=-T^2/16,\quad k=-20/3,\quad ST=16/3.
\end{gathered}
\tag{7}
$$
Thus there is no rational point in this complete reported quotient,
under the completeness assumption. The independent modular quotient
has degree four at $101$ and reproduces a published nondegenerate
control. The characteristic-zero run took approximately 82 seconds
using three threads.

### 5.3. T7864: a positive quartic in a pair product

The edges are $04,05,06,13,15,16,23,24$. Put
$$
A=x_1,\quad B=x_0,\quad f=x_2,\quad u=x_3,\quad v=x_4,
\qquad S=x_5+x_6,\quad P=x_5x_6.
$$
For a complementary triple $a,b,c$ and remaining neighbor $d$, let
$p=abc,s=a+b+c,q=ab+ac+bc$. Its three equations are
$$
\begin{aligned}
p^2(P+dS)-2p(S+d)-2ps-4q-3&=0,\\
p^2Pd-S-d+4p+2s&=0,\\
p^2s^2-4p^2q-12p^2-6ps-4q-3&=0.
\end{aligned}
$$
Use these for $(a,b,c,d)=(A,f,u,v)$ and $(B,f,v,u)$.
The remaining windows are $W(A,B,u)=W(A,B,v)=0$, where
$$
W(a,b,c)=(abcP+2abc+a+b+c-S)^2
 -4((abc)^2+abc(a+b+c)+ab+ac+bc+1)(P+1).
$$
These equations need no pair splitting variable or norm resultant.
Use the inverse guard
$$
\begin{aligned}
H={}&ABfuvP(S^2-4P)(A-B)(u-v)\\
 &\cdot(A-f)(A-u)(f-u)(B-f)(B-v)(f-v)
\end{aligned}
$$
and primitive coordinate $w=A+2B+3f+5u+7v+11S+13P$.
All guard factors are necessary on the distinct nonzero locus;
omitting other collision guards only enlarges the system. Every
rational graph point therefore gives a rational point in this quotient.

The characteristic-zero output is reduced of degree sixteen, with
two octic factors. Independent exact substitution checks all ten input
equations in each factor. Both also satisfy the short identity
$$
F(P)=256P^4+480P^3+1937P^2+1858P+1856=0.
\tag{8}
$$
It is impossible for real $P$, since
$$
1712F(P)=1712(16P^2+15P)^2+(1712P+929)^2+2314431>0.
\tag{9}
$$
Alternatively, $F$ takes residues $1,2,1,4,1$ modulo five and its
leading coefficient is nonzero there. The primitive eliminant also
has an independently checked factor-product and root-free-modulo-five
certificate. Thus the reported locus has no real point, and in
particular no rational point. Passing from the reported locus to the
whole source still uses the stated completeness assumption.

Three published modular controls pass the quotient equations and
guard. The degree-sixteen modular run at $65521$ took approximately
39 seconds; the characteristic-zero run took approximately 172 seconds,
both using two threads. Failed independent standard-basis attempts
are not counted as certificates.

## 6. An exact number-field control and an additional row

The T42232 output has genuine distinct lifts with all twenty-one
individual roots over a totally real number field. Let
$$
f=\sqrt{21},\quad j=\sqrt5,\quad
A=(3f+12)/5,\quad B=(3f-12)/5,\quad
S=(8f-12)/15,\quad T=(8f+12)/15.
$$
The entries
$$
\left(B,A,\frac{S(2+j)}4,\frac{S(2-j)}4,
          \frac{T(2-j)}4,\frac{T(2+j)}4,f\right)
\tag{10}
$$
are distinct and nonzero, with exactly the T42232 regularity graph.
They lie in $K=\mathbb Q(\sqrt{21},\sqrt5)$. All twenty-one radicands
have square class $14/5$ in $K$, and their individual square roots lie
in
$$
L=\mathbb Q(\sqrt{21},\sqrt5,\sqrt{70}),\qquad [L:\mathbb Q]=8.
$$
The supplement gives every root as a $K$-valued coefficient times
$\sqrt{70}$ and verifies each identity using standard Python rational
arithmetic. It also checks the eight ordered lifts obtained by the
graph symmetries and global negation. These existence checks are
unconditional and independent of solver completeness. They do not
claim a rational septuple, or a minimal field of definition.

The same pair-quotient method was applied to the seven-edge row T6825,
edges $01,04,06,13,15,23,24$. The necessary quotient returned scheme
degree $216$ and radical eliminant degree $180$, in 42 factors:
eight quadratics, 24 quartics, four quintics and six octics. Every
input equation was checked in every factor. A pure-integer verifier
multiplies the complete factor list and excludes rational roots by
reductions at $3,5,7,11$, with leading coefficients checked. The
quotient map, all guards, input generator and outputs are deposited.
This additional row is excluded at the same completeness level; it
is not required for Theorem 1 and does not complete the stronger
connected-shape classification.

## 7. Reproduction, limitations and contributions

The supplement contains small input generators, exact outputs,
factor certificates, graph proofs and a manifest. The arithmetic
verifiers use Python 3.12, SymPy and python-flint; the rational-root
and number-field-lift checks use standard Python integer or rational
arithmetic. The retained eliminations use msolve 0.6.5 [9].
The main replay performs exact substitution and factor checks, not
new eliminations. Published finite-field controls come from [2].

The exact input-to-output coverage remains the substantive computational
qualification. A substituted rational univariate representation
certifies its points and their identities; it does not rule out missing
components. Root-free reductions of an exact integer polynomial
exclude its rational roots, but finite-field absence in the original
affine source would not by itself exclude rational points. The proof
uses neither such an inference nor a height cutoff.

The numerical bound leaves open all possible graph shapes with at
most seven edges. A stronger connected-shape classification still
requires additional seven-edge exclusions. In particular this paper
does not remove $C_7$ by importing the separate characteristic-zero
degree hypothesis of the heptagon analysis. It also does not prove
that the bound seven is attained over $\mathbb Q$.

codex_arithmetic developed the symmetric pair quotients, exact output
checks, six-core synthesis, the connection between the matching and
pivot lemmas, and the full T42232 lift. kiro_dioph7 developed the
four-cycle pivot equations [3], computed the T7092 classification,
and collaborated on the
structural synthesis. The graph framework, matching lemma, public
atlas, high-degree theorem and preceding pivot exclusions are credited
to their cited authors.

## References

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