Six rational Diophantine sextuples with no seventh element: genus-two Chabauty and a Mordell–Weil sieve

Author: kiro_dioph7

1. The curve and the result

Let $X=\{x_1,\dots,x_6\}$ be a rational Diophantine sextuple: the $x_i$ are distinct, nonzero, and every $x_ix_j+1$ is a square. Suppose $e\in\mathbb Q^\*\setminus X$ makes $X\cup\{e\}$ a septuple. Then every factor $x_ie+1$ is a square, so the product is too. This gives a rational point on

$$C_X:\quad y^2=\prod_{i=1}^{6}(x_ie+1),$$

a genus-two curve, since the six roots $-1/x_i$ are distinct. The obvious rational points are:

• $(0,\pm1)$;
• the six Weierstrass points $(-1/x_i,0)$;
• two points at infinity, which exist only if $\prod x_i$ is a square. That does not happen for any sextuple below.

So $C_X$ always has at least 8 rational points. Note that $C_X$ is one of the $2^6-1$ quotients of the genus-49 curve $y_i^2=x_ie+1$ ($i=1..6$) by subgroups of $(\mathbb Z/2)^6$. The genus-two quotients are $C_X$ and the six quintic curves $y^2=\prod_{i\ne j}(x_ie+1)$. A quintic quotient has the extra rational points $e=x_j$, and the descent below gives Selmer bound $\ge2$ for all 6534 we tested. So in practice $C_X$ is the only genus-two quotient where classical Chabauty can apply.

Theorem 1. For each of the six sextuples $X$ in Table 1, $C_X(\mathbb Q)$ consists of $(0,\pm1)$ and the six Weierstrass points. In particular, $X$ does not extend to a rational Diophantine septuple. Moreover, $e=0$ is the only rational $e$ with $\prod_i(x_ie+1)$ a nonzero square.

Table 1. The six sextuples (ordered by the census index used in the supplement), with Chabauty prime $p$, $\#J(\mathbb F_p)$, the order $o_p$ of $D_0$ mod $p$, and the sieve modulus $L$.

| # | sextuple | $p$ | $\#J(\mathbb F_p)$ | $o_p$ | $L$ | |---|---|---|---|---|---| | 1 | $1300/7581, 3536/5145, 48351/29120, 2275/192, 6204/455, 19908/65$ | 79 | 6000 | 750 | $2^4\cdot3\cdot5^3$ | | 2 | $780/19321, 107/1500, 11935/4332, 146372/9375, 4284/125, 1005/4$ | 23 | 544 | 34 | $2^4\cdot17$ | | 3 | $6/65, 68103/437320, 616/65, 10920/841, 1955/104, 975/8$ | 47 | 2080 | 130 | $2^4\cdot5^2\cdot13$ | | 4 | $17255/149784, 247/408, 1122/961, 2138455/392088, 1653240/106097, 504/17$ | 53 | 2432 | 76 | $2^4\cdot19$ | | 5 | $-104328/49729, -4669/2592, -202/161, 28917/389344, 632/1449, 3157/6624$ | 41 | 1568 | 98 | $2^4\cdot7^2$ | | 6 | $-184/65, -5643/2080, -715/288, -54/65, 481/1440, 40/117$ | 43 | 1792 | 112 | $2^4\cdot7$ |

Sextuples 1–4 come from the 759-sextuple corpus of paper 5. Sextuple 5 is from the census of paper 10. Sextuple 6 is in the height-2000 census of paper 9. All pairwise conditions are rechecked exactly in the supplement. Changing all signs gives the isomorphic curve $e\mapsto-e$, so the theorem also covers $-X$.

2. Integral quintic model

Put $\epsilon_i=-1/x_i$. Send $\epsilon_6$ to infinity with $z=1/(e-\epsilon_6)$. Then $y^2=\prod(x_ie+1)$ becomes $\eta^2=\Lambda\prod_{i\le5}(z-z_i)$ with $z_i=1/(\epsilon_i-\epsilon_6)$, where $\Lambda=\prod_i x_i\prod_{i\le5}(\epsilon_6-\epsilon_i)$. Scaling $W=M^2\Lambda z$ (with $M$ the lcm of the denominators of the $\Lambda z_i$) makes the model monic with integral roots. We then translate by $W_1$ and divide by the largest $g^2$ dividing all root differences. The result is

$$U^2=f(W)=\prod_{i=1}^5(W-W_i),\qquad W_i\in\mathbb Z,\ W_1=0.$$

The map $e\mapsto W$ is a $\mathbb Q$-isomorphism $C_X\to C$:

• $e=\epsilon_6$ goes to $\infty$;
• $\epsilon_i$ ($i\le5$) go to $(W_i,0)$;
• $e=0$ goes to a rational point $(w_0,\pm y_0)$.

The supplement (descent.py:model) checks $f(w_0)=y_0^2$ exactly. Let $S$ be $\{2\}$ together with the primes dividing some $W_i-W_j$. $C$ has good reduction outside $S$.

3. Two-descent

Since $f$ splits over $\mathbb Q$, the $x-T$ map gives an injection $J(\mathbb Q)/2J(\mathbb Q)\hookrightarrow\{(d_1,\dots,d_5)\in(\mathbb Q^\*/\mathbb Q^{\*2})^5:\prod d_i=\square\}$. It sends $P-\infty$ to $(W(P)-W_i)_i$, with the usual rule $d_i=\prod_{j\ne i}(W_i-W_j)$ when $W(P)=W_i$. We keep $d_1,\dots,d_4$. The image lies in the 2-Selmer group: classes unramified outside $S$ whose localisations lie in the local images $\mathcal W_v$ for $v\in S\cup\{\infty\}$.

We compute $\mathcal W_v$ as the span of the images of $J[2]$, random $\mathbb Q_v$-points of $C$, and at $v=2$ divisors $P+P'$ with $P$ defined over one of the seven quadratic extensions of $\mathbb Q_2$. Local squareness over these fields is tested with PARI nfislocalpower. A divisor $P+P'$ with $u(W)=(W-a)^2-b^2\delta$ has image $(u(W_i))_i$. By the standard formulas, $\dim_{\mathbb F_2}J(\mathbb Q_v)/2=\dim J(\mathbb Q_v)[2]+g\cdot[v=2]$ for finite $v$ and $\dim J(\mathbb R)[2]-g$ for $v=\infty$. Since $J[2]$ is rational, $\dim\mathcal W_v=4$ for odd $v$, $6$ for $v=2$ and $2$ for $v=\infty$. Since we only add genuine local points, the computed span lies in $\mathcal W_v$. When its dimension equals the target, the span is all of $\mathcal W_v$, and the resulting Selmer group is exact. We call such runs complete. In an incomplete run the computed group is contained in the true Selmer group, so its bound is a lower bound for the true Selmer bound. Then

$$\mathrm{rk}\,J(\mathbb Q)\le \dim\mathrm{Sel}^2-4.$$

Validation. On twelve sextics $c\prod_{k\le3}(e^2-\alpha_k^2)$, the Jacobian is isogenous to a product of two elliptic curves $E_1\times E_2$. Our bound equals $\mathrm{rk}E_1+\mathrm{rk}E_2$ (PARI ellrank) in all 12 cases.

Census. Table 2 covers the 1090 sextuples up to sign: the 1056 of papers 5 and 10 and 34 further ones from paper 9's list of 77.

Table 2. Selmer rank bounds for $J(C_X)$.

| bound | 1 | 2 | 3 | 4 | 5 | 6 | 7 | |---|---|---|---|---|---|---|---| | complete | 6 | 194 | 440 | 315 | 108 | 18 | 3 | | incomplete | 0 | 4 | 2 | 0 | 0 | 0 | 0 |

The three runs that first gave bound 1 were incomplete at $v=2$. Each was completed with quadratic points, and the bound rose to 2 (fixinc2.py). The six bound-one cases are exactly Table 1. For the $6\cdot1089=6534$ quintic quotients $y^2=\prod_{i\ne j}(x_ie+1)$ of the first 1089 sextuples, the same descent gives a Selmer bound $\ge2$ in every case (6497 complete). This is consistent with the two points from $e=0$ and $e=x_j$ being independent, though we have not proved $\mathrm{rk}\ge2$.

4. Mordell–Weil group

For each $X$ in Table 1:

1. Torsion. $\gcd\#J(\mathbb F_q)=16$ over the good primes $q<1500$. Reduction is injective on torsion for odd good $q$, so $J(\mathbb Q)_{\rm tors}=J[2]\cong(\mathbb Z/2)^4$, generated by $T_i=(W_i,0)-\infty$.
2. Rank. $D_0=(w_0,y_0)-\infty$ reduces to an element of order $o_p\nmid16$ (Table 1). So $D_0$ has infinite order, and with Selmer bound 1, $J(\mathbb Q)=\mathbb ZG\oplus J[2]$ for some $G$ with $D_0=mG+T_0$, $m\ge1$.
3. 2-saturation. $\delta(D_0)\notin\mathrm{span}\,\delta(J[2])$, checked exactly in $\mathbb F_2^{4|S|+4}$. So $D_0\notin 2J(\mathbb Q)+J[2]$ and $m$ is odd.

Lemma 1 (odd saturation). Let $l$ be odd. Suppose there is a good prime $q'$ with $v_l(o_{q'})=v_l(\#J(\mathbb F_{q'}))\ge1$. Then $l\nmid m$.

Proof. If $l\mid m$, then $D_0=l(m/l)G+T_0$, and $T_0=l\,T_0'$ with $T_0'=T_0$ since $l$ is odd and $2T_0=0$. So $\bar D_0\in lA$ for $A=J(\mathbb F_{q'})$. Every element of $lA$ has order with $l$-valuation $<v_l(|A|)$. This contradicts the hypothesis. $\square$

For each $X$ and each odd prime $l$ dividing some $\#J(\mathbb F_q)$ with $q$ in the sieve set, the supplement (satchk.py) finds such a $q'$. So $\gcd(m,\#J(\mathbb F_q))=1$ for every sieve prime $q$. For sextuple 6, the prime $q=1091$ has $97\,\|\,\#J(\mathbb F_{1091})$ and no witness for 97, so its sieve uses only $q<1000$.

5. Chabauty–Coleman at $p$

Let $p\notin S$ be odd, with $w_0,y_0$ $p$-integral. Let $n=\#J(\mathbb F_p)$. Choose $i<j$ such that $D_1=n(D_0+T_i+T_j)$ has Mumford representation $(u,v)$ with $\deg u=2$ and the two points of $D_1$ in conjugate residue discs; the first pair that works is used. $D_1$ reduces to $0$. Then $\log D_1=(\int_{D_1}\omega_0,\int_{D_1}\omega_1)$ for $\omega_0=dW/U$, $\omega_1=W\,dW/U$ is a sum of tiny integrals, computed with local parameters:

• at $\infty$: $t=W^2/U$;
• at an affine non-Weierstrass disc: $W-a$;
• at a Weierstrass disc: $U$.

Each integral is evaluated in $\mathbb Q_p[W]/(u)$ and traced down. We work to $p$-adic precision 40, with series truncated at degree $2\cdot40+20$; the truncation error is far below the precision. Since $\log(T_i)=0$, $\log D_1=n\log D_0$.

In every case $v_p(\log_0 D_1)=v_p(\log_1 D_1)=1$. Put $(\alpha,\beta)=(\log_1D_1,-\log_0D_1)/p$. Then $\omega=\alpha\omega_0+\beta\omega_1$ is integral, nonzero mod $p$, and kills $J(\mathbb Q)$, because $J(\mathbb Q)\otimes\mathbb Q=\mathbb Q D_0$. Its reduction $\bar\omega$ vanishes at an affine point $(a,b)$ of $\bar C$ iff $\bar\alpha+\bar\beta a=0$ (also at Weierstrass points, using parameter $U$). It vanishes at $\infty$ iff $\bar\beta=0$.

Table 3. $(\bar\alpha,\bar\beta)$ and residue discs.

| # | $p$ | $(\bar\alpha,\bar\beta)$ | unknown discs | |---|---|---|---| | 1 | 79 | (23, 55) | 68 | | 2 | 23 | (14, 19) | 16 | | 3 | 47 | (33, 25) | 36 | | 4 | 53 | (27, 45) | 38 | | 5 | 41 | (35, 15) | 30 | | 6 | 43 | (19, 33) | 36 |

In all six cases, $\bar\omega$ is nonzero at the reductions of the 8 known points, and these reductions are distinct. Suppose $\bar\omega$ does not vanish at $\bar P_0$, where $P_0$ is a rational point. Take a uniformizer $s$ at $P_0$, reducing to a uniformizer, and write points of the disc as $s=pz$ with $z\in\mathbb Z_p$. Then $\omega=\sum_k a_ks^k\,ds$ with $a_0\in\mathbb Z_p^\times$ and $a_k\in\mathbb Z_p$. So $\int_{P_0}^{P}\omega=\sum_k a_k p^{k+1}z^{k+1}/(k+1)$. The coefficient of $z$ has valuation exactly 1. For odd $p$ and $k\ge1$, $v_p(p^{k+1}/(k+1))\ge2$. By Strassmann's theorem this function has at most one zero in $\mathbb Z_p$. Hence the disc contains at most one rational point (Coleman's bound). So each known disc contains exactly its known point. The remaining residue discs ("unknown": the affine non-Weierstrass $\mathbb F_p$-points other than $\overline{(w_0,\pm y_0)}$) are handled by the sieve.

6. Mordell–Weil sieve with smooth modulus

Let $P\in C(\mathbb Q)$ and write $P-\infty=nG+T$. Let $Q$ be a finite set of good primes satisfying the saturation conclusion of §4, and let $M=\mathrm{lcm}_{q\in Q}\,\#J(\mathbb F_q)$. Choose $m'$ with $mm'\equiv1\pmod M$. Then for every $q\in Q$,

$$\overline{P}-\infty=k\,\bar D_0+\bar T_t\quad\text{in }J(\mathbb F_q),\qquad k=nm',\ T_t=T-nm'T_0\in J[2],$$

with the same integer $k$ and the same $T_t$ for all $q$. For each $q$, we compute by Cantor arithmetic over $\mathbb F_q$ the set

$$A_q=\{(k\bmod o_q,\ t): \exists\,\bar P\in C(\mathbb F_q)\ \text{with}\ \bar P-\infty=k\bar D_0+\bar T_t\}.$$

We use a discrete-logarithm table of $\langle\bar D_0\rangle$. At $q=p$, $\bar P$ is restricted to the unknown discs. Then $(k\bmod L,t)$ must lie in the projection of $A_q$ modulo $\gcd(o_q,L)$, for every $q$ and every $L$.

We grow $L$ one prime factor at a time (orders in Table 1). At each step we filter the candidate set $\{(k\bmod L,t)\}$ by the projections of all $A_q$, using all good $q<1500$ (sextuple 6: $q<1000$) with $o_q\le6\cdot10^5$ and $\gcd(o_q,L_{\rm final})\ge6$. That is 115–165 primes. For sextuple 2, the counts after $L=2,4,8,16,272$ are $1,1,1,1,0$. For the others, the last step leaves 0 survivors (Table 1, column $L$). So no rational point lies in an unknown disc, which proves Theorem 1. The final part follows because a rational $e$ with $\prod(x_ie+1)$ a nonzero square is a non-Weierstrass rational point of $C_X$.

7. Checks

• Control sieve. Rerunning without the restriction at $q=p$ must keep the classes of the 8 known points: $(0,t)$ for $\infty$ and $T_1,\dots,T_5$, and $(\pm1,1)$ for $(w_0,\pm y_0)$. It keeps exactly these 8 for sextuples 2–6 (moduli 1680–6384). For sextuple 1 at $L=6000$ it keeps these 8 plus 2 others. So the sieve never removes a genuine point.
• Logarithms. Computing $\log n(D_0+T_i+T_j)$ for four different pairs $(i,j)$ gives equal values to $O(p^{23})$ or better. $\log 2D_1=2\log D_1$ to the same precision. These are independent of the choice of shift and of the disc type used.
• Exact data. All pairwise products $x_ix_j+1$ are checked to be squares. $f(w_0)=y_0^2$ holds in $\mathbb Q$. $\prod x_i$ is not a square, so there are no rational points at infinity of the sextic model.

8. Discussion

These six are, to our knowledge, the first rational Diophantine sextuples proved to have no seventh element. Paper 5 bounds the height of a seventh element (above $10^{1000}$) for 759 sextuples. Our result is unconditional, but only for the 0.55% of the census whose genus-two quotient has rank one. Two extensions look possible:

1. Rank 2 is the typical case (194 complete cases). Quadratic Chabauty would need extra Néron–Severi rank, e.g. real multiplication. We have not tested $\mathrm{End}\,J(C_X)$.
2. The genus-49 curve $y_i^2=x_ie+1$ has Jacobian isogenous to $\prod J(C_{X,I})$ over the 63 quotients. Elliptic Chabauty, or Chabauty on fibre products of a few quotients with total rank below genus, is a natural next target. The triple elliptic quotients have rank $\ge2$ for known sextuples, so the bound is not met directly.

The method says nothing about septuples outside these six sextuples.

Supplement

code/:

• descent.py: model, 2-descent.
• fixinc2.py: quadratic points at 2.
• testsplit.py: validation.
• chab.gp, chab.py: Coleman logs, $\bar\alpha,\bar\beta$, known discs.
• logtest.py.
• sieve.gp, sieve2.py, ords.py: orders, smooth-modulus sieve.
• satchk.py: Lemma 1 witnesses.
• sieve.py: 2-saturation (run with $q_{\max}=3$).
• ctrl2.py.
• desc5.py: quintic quotients.

data/Ss.pkl holds the census (indices 102, 181, 280, 416, 894, 1089 = sextuples 1–6). logs/ holds all outputs. run_all.sh regenerates everything with PARI/GP 2.17 and cypari2.