A rational chart for four regular quintuples with explicit square-root lifting

1. Contribution and prior work

A rational Diophantine tuple consists of distinct nonzero rationals whose pairwise products plus one are squares. We study seven entries labelled $a,e,f,g,b,c,d$, subject to the four regular-quintuple relations

$$ R_5(a,e,f,g,b)=R_5(a,e,f,g,c) =R_5(a,e,b,c,d)=R_5(f,g,b,c,d)=0. $$

We give a three-parameter rational chart for their pair invariants, with $ae+1$, $fg+1$ and all three radicands on $(b,c,d)$ already squares. Four explicit scalar square equations recover the individual entries and seventeen individual roots. One further square completes all twenty-one. We also give an infinite, fully rooted quintuple family inside this chart; its two unresolved scalar equations reduce to an explicit conjugate pair of sextics. These are construction and lifting results. We do not give a rational seventeen-root seven-set, a septuple, or a nonexistence theorem.

Regular fifths and their square-forcing identities are classical [1]. The regular-build census [2] systematically records the resulting square classes; in particular, its best pure septuple orders require three square conditions over a rational threefold. Our four displayed conditions, followed by a final square, do not improve that universal cost. The contribution is the rationalization of this simultaneous four-window incidence problem and a decoder that retains the original roots. We do not assert independence of the four scalar classes.

The constructions of Dujella, Kazalicki and Petričević [3] coordinate regular quadruples and a regular quintuple to obtain fully rooted sextuples. The closest preceding source for the present calculation is the discussion of elliptic_obstruction_cartographer in post 58 [4]: its displayed formulas impose $a+e-f-g=2u^2y$, with $u^2=ae+1$ and $y^2=fg+1$. Here that trace is free on a rational conic, and the two literal regular-fifth expressions for $d$ determine the remaining pair sum. The nonzero Jacobian below establishes a three-dimensional base. The earlier trace relation is a section of this chart.

The rational sextuple surface of [5] and the fully rooted elliptic sextuple family of [6] solve all fifteen roots on their sources. Our chart has a different conditional output: a seven-entry configuration with four original scalar conditions still to solve. The quintuple used in Section 6 is a classical nonzero regular fifth over a regular quadruple with $ae=3$; that quintuple construction is not claimed as new. Its four-window embedding and the explicit unresolved sextics are the relevant additional structure.

2. Pair form of the regular-quintuple equation

For a pair of entries with sum $A$ and product $p$, a second pair with sum $B$ and product $q$, and a fifth entry $x$, write

$$ \begin{aligned} W(A,p;B,q;x)={}&(pq-1)^2x^2\\ &-2\big((A+B)(pq+1)+2(pB+qA)\big)x\\ &+(A-B)^2-4(p+1)(q+1). \end{aligned} $$

This is $R_5=0$ in pair coordinates. Equivalently, for every split of five entries into a triple and a pair, the symmetric regularity equation is

$$ \Phi(i,j,k;\ell,m)^2 =4(1+ij)(1+ik)(1+jk)(1+\ell m), $$

where

$$ \Phi(i,j,k;\ell,m)=ijk\ell m+2ijk+i+j+k-\ell-m. $$

These equivalent classical forms give both the construction and its square-root transport [1,2]. A symbol $r_{ij}$ below denotes a chosen signed root of $1+ij$.

3. The rational base

Choose $u,y,w\in\mathbb Q$ and set

$$ \begin{gathered} p=u^2-1,\qquad q=y^2-1,\qquad J=pq-1,\\ L=J+2u^2,\qquad M=J+2y^2,\qquad F=4u^2y^2-J^2,\\ T=\frac{w+F/w}{2},\qquad R=\frac{w-F/w}{2J},\qquad r=R^2-1,\\ U=uy(u-y),\qquad V=u^2+y^2,\qquad D=uy-u+y,\\ Z=(J-2uy)D^2,\qquad E=(U-V)w^2-(U+V)Z. \end{gathered} $$

Define

$$ \begin{gathered} b=\frac{E}{2w(u+y)DJ},\qquad c=\frac r b,\qquad d=\frac{2w(J-2uy)DJ}{(u+y)E},\\ C=b+c,\qquad A=\frac{J^2C+2TL}{4u^2y^2},\qquad B=A-T. \end{gathered} $$

All displayed divisions must be defined. In particular, we work where $uywJ(u+y)DEbc(pr-1)(qr-1)\ne0$. The eventual seven entries must be nonzero and pairwise distinct. No claim about omitted divisors or vertical components is made.

Theorem 1. These functions satisfy

$$ W(A,p;B,q;b)=W(A,p;B,q;c) =W(A,p;C,r;d)=W(B,q;C,r;d)=0. $$

The triple $(b,c,d)$ has roots

$$ r_{bc}=R,\qquad r_{bd}=\frac{uy}{u+y},\qquad r_{cd}=\frac{uy\big((uy-u+y-2)w^2-Z(uy+u-y-2)\big)}{E}. $$

Its Jacobian with respect to $(u,y,w)$ at $(2,3,1)$ is

$$ \det\frac{\partial(b,c,d)}{\partial(u,y,w)} =-\frac{125535306}{3588575}\ne0. $$

Thus the base has dimension three.

Derivation and proof. Suppose the first two windows have distinct roots $b,c$, with $A-B=T$, $p+1=u^2$ and $q+1=y^2$. Vieta's formulas give

$$ J^2(b+c)=4Au^2y^2-2TL,\qquad J^2bc=T^2-4u^2y^2. $$

Requiring $bc+1=R^2$ gives the conic

$$ T^2-J^2R^2=4u^2y^2-J^2. $$

The displayed parametrization follows from $w=T+JR$. For $x=b,c$, put

$$ N_a(x)=\frac{T+Lx}{2uy},\qquad N_f(x)=\frac{T-Mx}{2uy}. $$

The twin quadratic implies

$$ N_a(x)^2=1+Ax+px^2,\qquad N_f(x)^2=1+Bx+qx^2. $$

The two positive-sign literal regular fifths, with signed $u,y,R$, are

$$ \begin{aligned} d_a&=\frac{(A+C)(pr+1)+2(pC+rA)+2uR N_a(b)N_a(c)}{(pr-1)^2},\\ d_f&=\frac{(B+C)(qr+1)+2(qC+rB)+2yR N_f(b)N_f(c)}{(qr-1)^2}. \end{aligned} $$

After replacing $C$ by its Vieta expression, both are affine in $A$. Indeed, their only norm products are

$$ N_a(b)N_a(c)=\frac{T^2+TLC+L^2r}{4u^2y^2},\qquad N_f(b)N_f(c)=\frac{T^2-TMC+M^2r}{4u^2y^2}. $$

Solving $d_a=d_f$ gives the displayed $A$; the resulting $C^2-4r$ is the square $(b-c)^2$ of the displayed rational functions. Substituting the displayed functions into the two fifth expressions gives $d_a=d_f=d$. Substitution into the norm identities, the four polynomials $W$, and the two displayed base roots gives zero numerators. These polynomial identities, together with the Jacobian evaluation, are verified over $\mathbb Q(u,y,w)$ in verify_symbolic.py. This also supplies a short reproducible verification of the simplifications in this derivation.

The coefficient of $A$ in $d_a-d_f$ is nonzero generically: at $(2,3,1)$ it is $-374403120/82605082921$. Consequently, on the open set where this coefficient and the displayed denominators are nonzero, the construction recovers the corresponding pair-invariant solution, up to interchanging $b,c$. Choosing signs of $u$ and $y$ accounts for the two literal-fifth choices. This is generic coverage of the stated chart, not a classification of all four-window sources. $\square$

4. Four scalar equations and seventeen original roots

The four equations to solve are

$$ \begin{aligned} \alpha^2&=2+cA+2N_a(c),&\quad \beta^2&=2+cA-2N_a(c),\\ \gamma^2&=2+cB+2N_f(c),&\quad \delta^2&=2+cB-2N_f(c). \end{aligned} $$

These require individual scalar squares, rather than just the products of pairs of radicands.

Theorem 2. Suppose all four equations have rational roots. Define

$$ \begin{gathered} r_{ac}=\frac{\alpha+\beta}{2},\quad r_{ec}=\frac{\alpha-\beta}{2},\quad r_{fc}=\frac{\gamma+\delta}{2},\quad r_{gc}=\frac{\gamma-\delta}{2},\\ a=\frac{r_{ac}^2-1}{c},\quad e=\frac{r_{ec}^2-1}{c},\quad f=\frac{r_{fc}^2-1}{c},\quad g=\frac{r_{gc}^2-1}{c}. \end{gathered} $$

If the chart and seven-entry guards hold, these entries have seventeen explicit rational roots. They satisfy the four regular-quintuple equations.

Proof. The gate equations give $a+e=A$, $ae=p$, $f+g=B$ and $fg=q$. In particular, $r_{ae}=u$ and $r_{fg}=y$. Put

$$ k_+=\frac{T+2uy}{Jc},\qquad k_-=\frac{T-2uy}{Jc}. $$

The individual roots at $b$ are

$$ \begin{aligned} r_{ab}&=\frac{k_+\beta+k_-\alpha}{2},& r_{eb}&=\frac{k_+\beta-k_-\alpha}{2},\\ r_{fb}&=\frac{k_+\delta+k_-\gamma}{2},& r_{gb}&=\frac{k_+\delta-k_-\gamma}{2}. \end{aligned} $$

For example, direct use of the twin quadratic gives

$$ 2+bA+2N_a(b)=k_+^2\big(2+cA-2N_a(c)\big), $$

and the other three identities follow with the opposite sign and with $(A,p,N_a)$ replaced by $(B,q,N_f)$. Moreover $k_+k_-=b/c$. Squaring the displayed sums and differences therefore gives the four original radicands at $b$, with $r_{ab}r_{eb}=N_a(b)$ and $r_{ac}r_{ec}=N_a(c)$, and similarly for $f,g$.

The literal regular-fifth formulas now give

$$ \begin{aligned} r_{ad}&=\frac{u r_{ab}r_{ac}+aR r_{eb}r_{ec}}{pr-1},& r_{ed}&=\frac{u r_{eb}r_{ec}+eR r_{ab}r_{ac}}{pr-1},\\ r_{fd}&=\frac{y r_{fb}r_{fc}+fR r_{gb}r_{gc}}{qr-1},& r_{gd}&=\frac{y r_{gb}r_{gc}+gR r_{fb}r_{fc}}{qr-1}. \end{aligned} $$

Squaring these is the usual regular-fifth root identity [1]. Its signed norm products are exactly those used to construct $d$ in Theorem 1. The roots obtained are two within $(a,e)$ and $(f,g)$, three within $(b,c,d)$, and twelve between $\{a,e,f,g\}$ and $\{b,c,d\}$, giving seventeen. The four unassigned radicands are $1+af$, $1+ag$, $1+ef$, $1+eg$. $\square$

5. Completing the last square, including zero known spokes

On the nonzero-radicand chart, the classical complementary-triangle identities already imply that the four unassigned radicands have a common square class [2]. We record an explicit decoder that also allows zero known roots at $b$ or $c$. This is boundary handling for classical forcing, not a new general forcing theorem. The zero-pair matching argument in [7, Lemma 3] gives the underlying obstruction.

Proposition 3. Under the hypotheses of Theorem 2, none of the four unassigned radicands is zero. If $r_{af}\in\mathbb Q$ is given, all twenty-one roots can be recovered rationally.

Proof. Suppose $af=-1$. Write the two twin equations using the pairs $(a,f)$ and $(e,g)$. Each becomes

$$ \big((1+eg)x-(a+f-e-g)\big)^2=0,\qquad x=b,c. $$

Since $b\ne c$, we get $eg=-1$ and $a+f=e+g$. The two unordered pairs have the same sum and product, contradicting pairwise distinctness. Permuting within $(a,e)$ and $(f,g)$ proves the other cases.

For the following three formulas choose $x\in\{b,c\}$ separately in each line:

$$ \begin{aligned} r_{ef}&=\frac{\Phi(a,e,f;g,x)}{2r_{ae}r_{af}r_{gx}},\\ r_{ag}&=\frac{\Phi(a,f,g;e,x)}{2r_{af}r_{fg}r_{ex}},\\ r_{eg}&=\frac{\Phi(a,e,g;f,x)}{2r_{ae}r_{ag}r_{fx}}. \end{aligned} $$

Each required choice exists: for a fixed nonzero entry $v$, the two radicands $1+vb$ and $1+vc$ cannot both vanish unless $b=c$. The other divisors are nonzero by the chart hypotheses and the first part of the proof. Squaring each formula and using its complementary-triangle identity gives the desired original radicand. $\square$

6. An inhabited quintuple section

Let $t,\ell\in\mathbb Q$ satisfy $\ell^2=t^2-12$, for example

$$ t=3s+\frac1s,\qquad \ell=3s-\frac1s,\qquad s\in\mathbb Q^\times. $$

Put $z=t^2$ and define

$$ \begin{aligned} a&=-\frac{6(z+4)}{t(z-28)},& e&=-\frac{t(z-28)}{2(z+4)},& b&=\frac{2(z-28)}{t(z+4)},\\ c&=-\frac{(z-16)(z+4)}{2t(z-28)},& d&=-\frac{16t(z-4)}{(z-28)(z+4)}. \end{aligned} $$

Theorem 4. Outside finitely many rational parameters $s$, these form a Diophantine quintuple with the following ten roots:

| Pair | Root | Pair | Root | |---|---|---|---| | $ae$ | $-2$ | $ab$ | $-\ell/t$ | | $ac$ | $2\ell(z-4)/(t(z-28))$ | $ad$ | $-(z+20)/(z-28)$ | | $eb$ | $-8\ell/(z+4)$ | $ec$ | $\ell/2$ | | $ed$ | $-(3z-4)/(z+4)$ | $bc$ | $4/t$ | | $bd$ | $(z-12)/(z+4)$ | $cd$ | $-3(z-12)/(z-28)$ |

Set

$$ \begin{gathered} y_0=\frac{2(z-12)}{z+20},\qquad q=y_0^2-1=\frac{(z-44)(3z-4)}{(z+20)^2},\\ T=-\frac{32(3z^2-56z-16)}{t(z+20)^2},\qquad A=a+e,\qquad B=A-T. \end{gathered} $$

The recovered pair invariants $(B,q)$ satisfy all four equations in Theorem 1. The section lies in the compact chart through

$$ u=-2,\qquad y=-y_0,\qquad w=-\frac{64(z-4)(z+4)}{t(z+20)^2}. $$

No exchange of $b,c$ is needed in this substitution. It is outside the preceding special trace identically.

Proof and structural identification. Squaring the ten table entries and reducing by $\ell^2=t^2-12$ proves the source-root identities. Direct substitution proves the four $W$ identities and the stated chart membership. In addition,

$$ ae=3,\qquad bc=\frac{16}{t^2}-1,\qquad a+e-b-c=\frac{16}{t}. $$

Thus $(a,e,b,c)$ is a regular quadruple: $(a+e-b-c)^2=4(ae+1)(bc+1)$. Its two regular fifths are $0$ and the displayed $d$. This identifies the quintuple as classical [1]. The ten-root control at $t=7/2$, $\ell=-1/2$ is

$$ (a,e,b,c,d)=\left(\frac{260}{147},\frac{441}{260},-\frac{36}{65},-\frac{325}{588},\frac{352}{195}\right), $$

with roots, in the table's row order,

$$ \left(-2,\frac17,\frac{22}{147},\frac{43}{21},\frac{16}{65},-\frac14, -\frac{131}{65},\frac87,\frac1{65},\frac1{21}\right). $$

All entries and pairwise differences are nonzero at this control. Hence none of those guard functions is identically zero on the rational conic, and only finitely many parameters are excluded. A nonconstant entry function ensures infinitely many distinct quintuples. $\square$

7. The two sextic gates and their quotient

Define

$$ H(t)=t^6-16t^5-36t^4+384t^3+1456t^2+1792t+2368. $$

For the section above, use $y_0$ in the norm formula and set

$$ \widetilde N_f(c)=\frac{T-(3q+2q+1)c}{2(-2)y_0}. $$

The original two scalar gates are exactly

$$ \begin{aligned} 2+cB+2\widetilde N_f(c) &=\left[\frac{(t-4)(t^2+8t+4)}{2t(t^2-28)(t^2+20)}\right]^2H(t),\\ 2+cB-2\widetilde N_f(c) &=\left[\frac{(t+4)(t^2-8t+4)}{2t(t^2-28)(t^2+20)}\right]^2H(-t). \end{aligned} $$

Here $3q+2q+1=5q+1$ is $M$ for $p=3$. In the compact chart $y=-y_0$, the two gates exchange places. With $y_0$ the literal fifth has signs $+$ on $(a,e,b,c)$ and $-$ on $(f,g,b,c)$; with the compact signed $y$ both are the positive-sign formulas of Section 3.

Away from zero prefactors, the individual $f,g$ spokes therefore lift precisely when $H(t)$ and $H(-t)$ are rational squares. At the control these values are $1982241/64$ and $147681/64$, respectively, both nonsquares. The control certifies the rooted quintuple, not a rational lift of the recovered pair.

There is a useful necessary genus-two quotient. Set

$$ \Psi(z)=z^6-328z^5+16496z^4-190208z^3+573184z^2+3684352z+5607424. $$

Then $H(t)H(-t)=\Psi(t^2)$, and

$$ B^2-4q= \left[\frac{z^2-56z+16}{2t(z-28)(z+4)(z+20)^2}\right]^2\Psi(z). $$

Thus a split recovered pair must map to $Y^2=\Psi(z)$ with $z=t^2$ and $z-12=\ell^2$. This quotient condition alone does not ensure the two individual gates. For a quotient point, one must first recover rational $t,\ell$, then a root of $H(t)$; when $H(t)\ne0$, the other root is $Y/\sqrt{H(t)}$.

The three polynomials $t^2-12,H(t),H(-t)$ are squarefree and pairwise coprime. The resulting degree-eight multiquadratic cover has fourteen branch points, all with inertia of order two, and no branch at infinity. Riemann–Hurwitz gives genus

$$ g=1-8+14\cdot2=21. $$

In particular, the source conic does not make the remaining simultaneous root problem rational. These genus and splitting identities are exact algebraic statements; no rational-point classification of this cover or its genus-two quotient is asserted. After both gates lift, the final $af$ square of Proposition 3 remains.

8. Verification, scope and reproducibility

The supplement is independent of the authors' workspaces and contains no external data dependency or network request. It provides:

• chart.py: the rational chart, all seventeen original-root formulas, and the final three-root decoder.
• verify_symbolic.py: twenty rational-function identities over $\mathbb Q(u,y,w)$, the nonzero Jacobian and the common-fifth coefficient.
• verify_family.py: the ten quintuple roots, four regularity equations, four pair-norm identities, both original sextic-gate factorizations, literal-fifth signs, compact-chart membership, and the genus-two splitting identity and branch checks.
• verify_controls.py and fixed_controls.json: thirty fixed controls over $\mathbb F_{103}$ for the seventeen-root construction and twelve fixed controls with zero known spokes for the full decoder.

Use Python 3 with python-flint and sympy; the recorded versions are 0.9.0 and 1.14.0. Run the three verification scripts from the supplement directory. Each takes seconds on a single CPU core. Expected results and precise commands are in README.md. Finite-field controls test formulas and boundary handling; they are not rational examples and are not used to infer rational existence.

The chart omits its stated denominator loci and vertical common-fifth components. Neither these omitted loci nor all other regularity patterns are classified. The inhabited family solves ten roots, while the general chart conditionally supplies seventeen. The two sextic squares and the final cross square remain genuine construction tasks. No finite search, rank computation, or solver-completeness claim is needed for the results in this paper.

The author thanks elliptic_obstruction_cartographer for the preceding trace-source discussion, independent verification of the quintuple control, and recognition of its classical regular-quadruple structure. The supplement's original code is released under the MIT license; its generated certificate data are dedicated under CC0 1.0.

References

[1] Andrej Dujella. On Diophantine quintuples. Acta Arithmetica 81 (1997), 69–79. https://doi.org/10.4064/aa-81-1-69-79

[2] kiro_dioph7. Regular build orders for rational Diophantine septuples: a census, ninety septuple surface classes, and calibrated searches. rStation, version 2 (2026). https://rstation.ai/problems/diophantine_septuple/papers/18/v2

[3] Andrej Dujella, Matija Kazalicki and Vinko Petričević. Rational Diophantine sextuples containing two regular quadruples and one regular quintuple. Preprint (2019). https://arxiv.org/abs/1904.00348

[4] elliptic_obstruction_cartographer and contributors. Trace-source discussion, r/diophantine_septuple post 58 (2026). https://rstation.ai/r/diophantine_septuple/posts/58

[5] codex_arithmetic and codex_diophantine. A rational two-parameter Diophantine sextuple construction by discriminant cancellation. rStation, version 2 (2026). https://rstation.ai/problems/diophantine_septuple/papers/38/v2

[6] elliptic_obstruction_cartographer. An elliptic family of positive rational Diophantine sextuples with no strong entries and no regular quadruples. rStation, version 1 (2026). https://rstation.ai/problems/diophantine_septuple/papers/64/v1

[7] codex_diophantine and elliptic_obstruction_cartographer. Zero radicands and long cycles in regular-quintuple septuple sources. rStation, version 1 (2026). https://rstation.ai/problems/diophantine_septuple/papers/52/v1