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(This is a follow-up to a previous post.) A rational Diophantine $m$-tuple is a set of rationals {$a_1,a_2,\dots a_m$} such that (with $i\neq j$), all $a_i a_j+1$ is a square. Problem: Find a class of triples that can be extended to sextuples.

I. Elliptic curve

One solution is described in the paper by Dujella et al. Find rational $u,v$ such that,

$$(-27 - 9 u^2 - 27 u v + u^3 v - 54 v^2) (1 + v^2) (-1 + u v + 2 v^2)=y^2\tag1$$

This is a quartic in $u$ to be made a square. Since it has a rational point, it is birationally equivalent to an elliptic curve. Define,

$$\alpha_2 = \frac{-3-6uv-12v^2+u^2v^2}{4+4v^2}\tag2$$

Let $a,b,c$ be the roots of,

$$z^3-uz^2+\alpha_2 z-v =0\tag3$$

Then $a,b,c$ are special rational triples such that $ab+1,\;ac+1,\;bc+1$, and,

$$\begin{aligned} &a^2b^2c^2+1 = p^2\\ &2(a^2+b^2+c^2)-(a+b+c)^2-3 = q^2\\ &2(a^2+b^2+c^2+d^2)-(a+b+c+d)^2-3-6abcbd+(abcd)^2=(pd+q)^2 \end{aligned}\tag4$$

are all rational squares (for any $d$ for the last), implying they obey,

$$\big(2(a^2+b^2+c^2)-(a+b+c)^2-3\big)(1+a^2b^2c^2)=(a + b + c + 3 a b c)^2$$

A second elliptic curve can then yield infinitely many three rational $x_i$ such that $a,b,c,x_1,x_2,x_3$ is a sextuple, but we need not go to that step.

II. Solutions $a,b,c$

Using $v = \frac{t^2-1}{2t}$ and $u = \frac{1 + 130 t^2 - 390 t^4 + 130 t^6 + t^8}{-3 t + 105 t^3 - 105 t^5 + 3 t^7}$, Dujella et al found,

$$\begin{aligned} a &= \frac{18t\,(t^2-1)}{(t^2-6t+1)(t^2+6t+1)}\\ b &= \frac{(t - 1)(t^2 + 6t + 1)^2}{6t\,(t + 1)(t^2 - 6t + 1)}\\ c &= \frac{(t + 1)(t^2 - 6t + 1)^2}{6t\,(t - 1)(t^2 + 6t + 1)} \end{aligned}\tag5$$

Using $v = \frac{t^2-1}{2t}$ and $u = \frac{1 + 1996 t^2 + 102 t^4 + 1996 t^6 + t^8}{-8 t + 2584 t^3 - 2584 t^5 + 8 t^7}$, I found,

$$\begin{aligned} a &= \frac{128t\,(t^2+1)^2}{(t^2-1)(t^2-18t+1)(t^2+18t+1)}\\ b &= \frac{(t^2 - 1)(t^2 + 18 t + 1)^2}{16t\,(t^2 + 1)(t^2 - 18t + 1)}\\ c &= \frac{(t^2 - 1)(t^2 - 18 t + 1)^2}{16t\,(t^2 + 1)(t^2 + 18t + 1)} \end{aligned}\tag6$$

Notice that the triples $(5)$ and $(6)$ have an aesthetically pleasant similarity. And since $(1)$ is an elliptic curve, there should be infinitely many families.

Q: Using $(1)$, can you find another triple $a,b,c$ that has polynomial numerator/denominator of small degree?

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  • $\begingroup$ In terminology of mentioned paper formula (5) corresponds to the point 2R, while formula (6) corresponds to the points 3R. Other multiples of R produce less esthetical formulas. $\endgroup$ – duje Mar 13 '16 at 20:14
  • $\begingroup$ It might be interesting to mention that elliptic curves y^2=(ax+1)(bx+1)(cx+1) of this shape have torsion group Z/2Z x Z/6Z over Q. In fact, the most of known (and some new) record rank curves with this torsion (over Q and over Q(t)) can be obtained from the triple (5) by specializations of the parameter t (this is joint work in progress with J. P. Peral). $\endgroup$ – duje Mar 13 '16 at 21:05
  • $\begingroup$ @duje: The elliptic curve you cite below was specialized from $v = \frac{t^2-1}{2t}$. Do you think we can find another $v$ that may also yield simple families? $\endgroup$ – Tito Piezas III Mar 13 '16 at 21:07
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    $\begingroup$ @duje: Oh, I just recognized your name. :) By the way, in your paper, Lemma 1 can be aesthetically expressed as, $$\big(-3 - (a + b + c)^2 + 2 (a^2 + b^2 + c^2)\big)\,(1 + a^2 b^2 c^2) = (a + b + c + 3 a b c)^2$$ Since the second factor of the $LHS$ is a square, that implies the first one is as well and implies these special triples obey $(4)$ above. $\endgroup$ – Tito Piezas III Mar 13 '16 at 21:20
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    $\begingroup$ @duje: By the way, I've analyzed Gibbs' 45 sextuples and found 12 that belong to a family distinct from yours. (They do not contain triples that obey $(4)$.) I believe a different elliptic curve is involved. I'll post the family here when I've studied it more. $\endgroup$ – Tito Piezas III Mar 13 '16 at 21:34
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Similar formulas can be obtained by taking multiples $mR$ for the point $R=[0,(t^2+1)^3]$ on the elliptic curve $$ y^2=x^3+(3t^4-21t^2+3)x^2+(3t^8+12t^6+18t^4+12t^2+3)x+(t^2+1)^6. $$ The above formulas (5) and (6) correspond to $m=2$ and $m=3$. Other possible formulas appear if the rank of this curve is $\geq 2$. E.g. if $t=(T^2-6T+1)/(-1+T^2)$, then we get $$ a=\frac{(3T-1)(T-3)(T+1)}{2(T-1)(T^2-6T+1)}, $$ $$ b=\frac{8T(T-1)}{(T+1)(T^2-6T+1)}, $$ $$ c= -\frac{T^2-6T+1}{2(T-1)(T+1)}, $$ (note that in this case the point $P=[0,1]$ on $y^2=(ax+1)(bx+1)(cx+1)$ has finite order, so this triple does not extends to a sextuple by the method from mentioned paper), and $$ a=\frac{2T(T^2-5T+2)^2(T+1)^3}{(2T^2-T+1)(T^2-6T+1)(T^2-8T-1)(T-1)^2}, $$ $$ b=\frac{2(T-3)(T^2-6T+1)(2T^2-T+1)^2}{(T^2-5T+2)(T^2-8T-1)(T-1)^2(T+1)^2}, $$ $$ c=-\frac{(3T-1)(T^2-8T-1)^2(T-1)^3}{2(2T^2-T+1)(T^2-5T+2)(T^2-6T+1)(T+1)^2}. $$

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    $\begingroup$ Thanks! Your first family also has, $$(a+b-c)^2 = 4(ab+1)$$ and I forgot to specify that when this happens, then one element of the extended "sextuple" is zero, so effectively it is just a quintuple. $\endgroup$ – Tito Piezas III Mar 13 '16 at 20:58
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(Answer 1)

After considering this question for a few days, I found an answer. It uses a different family of elliptic curves from Dujella's and yields quadruples that can be extended to sextuples.

I. Theorems:

Theorem 1: "Given a quadruple $a,b,c,d$. Let $x_1,x_2$ be the two roots of,

$$(a b c d x + 2a b c + a + b + c - d - x)^2 = 4(a b + 1)(a c + 1)(b c + 1)(d x + 1)$$

If $x_1x_2+1$ is a square, then $a,b,c,d,x_1,x_2$ is a sextuple."

This is a known result. However, looking at Gibbs' list of $45$ sextuples, I noticed that $12$ of them were special cases.

Theorem 2: "If in Theorem 1, the quadruple $a,b,c,d$ also obeys,

$$(a + b - x_1 - x_2)^2 = 4(a b + 1)(x_1 x_2 + 1)$$

$$(c + d - x_1 - x_2)^2 = 4(c d + 1)(x_1 x_2 + 1)$$

which necessarily implies that $x_1 x_2+1$ is a square, then they have the simple relationship,

$$(a b c d - 3)^2 = 4(a b + c d + 3)$$

II. Elliptic curves:

A family of solutions to $(a b c d - 3)^2 = 4(a b + c d + 3)$ is,

$$a,b,c,d = \frac{2 (3 p^2 + 2p + 3) q}{(p + 1)^2 (q^2 - 1)} ,\;\frac{(p - 1)^2 (q^2 - 1)}{2 (p + 1)^2 q},\\ \frac{(p + 1)^2 (q^2 - 1)}{2 (p - 1)^2 q},\;\frac{2(3 p^2 - 2p + 3)q}{(p - 1)^2 (q^2 - 1)}$$

However, to fulfill Theorem 1 (hence Theorem 2), the $p,q$ must satisfy,

$$1 + p^2 = {z_1}^2\tag1$$

$$1 + \frac{2(17p^4 + 30p^2 + 17)}{(p^2 - 1)^2} q^2 + q^4 = {z_2}^2\tag2$$

The equation $(2)$ has the rational points $q = \frac{p+1}{p-1},\;q = \big(\frac{p+1}{p-1}\big)^2$, etc, thus is birationally equivalent to an elliptic curve so has infinitely many rational points.

III. Example:

Using $ = \frac{p+1}{p-1}$, then choosing rational $p$ to satisfy $(1)$ which is easily done, and applying it to Theorem 1, we get,

$$a= \frac{(n^2 - 2n - 1) (n^2 + 2n + 3) (3 n^2 - 2n + 1)}{4 n (n^2 - 1) (n^2 + 2n - 1)}$$

$$b= \frac{4 n(n^2 - 1) (n^2 - 2n - 1)}{(n^2 + 2n - 1)^3}$$

$$c= \frac{4 n(n^2 - 1) (n^2 + 2n - 1)}{(n^2 - 2n - 1)^3}$$

$$d= \frac{(n^2 + 2n - 1)(n^2 - 2n + 3)(3 n^2 + 2n + 1) }{4 n(n^2 - 1) (n^2 - 2n - 1)}$$

$$x_1 = \frac{-n^5 + 14 n^3 - n}{n^6 - 7n^4 + 7n^2 - 1}$$

$$x_2 =\frac{3 n^6 - 13n^4 + 13n^2 - 3}{4 n (n^4 - 6n^2 + 1)}$$

Ex. For rational $n = \frac{k}{2k+1}$, it seems all terms are positive if $k\geq3$. Thus for $n = \frac{3}{7}$,

$$a,b,c,d,x_1,x_2 = \frac{23001}{280},\;34440,\;\frac{840}{68921},\;\frac{1121}{11480},\;\frac{19383}{1640},\;\frac{2530}{287}$$

and so on for infinitely many sextuples.

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    $\begingroup$ Excellent finding! Congratulations! $\endgroup$ – duje Mar 18 '16 at 22:08
  • $\begingroup$ By taking $n=\pm\frac{P_k}{P_{k+1}}$, where $P_k$ is $k$-th Pell number, you obtain infinitely many sextuples with one non-zero integer element. $\endgroup$ – duje Mar 19 '16 at 8:47
  • $\begingroup$ You may take also $n=\pm \frac{P_k - P_{k-1}}{P_{k+1} - P_k}$ to get $b$ to be an integer. $\endgroup$ – duje Mar 19 '16 at 8:54
  • $\begingroup$ @duje: Ah, nice! Letting $n=p/q$, the denominator $D$ of $b$ or $c$ becomes, $$D=p^2\pm pq-q^2$$ and to solve $D=\pm1$ in integer $p,q$ would entail the Pell equation $x^2-2y^2=\pm1$. $\endgroup$ – Tito Piezas III Mar 20 '16 at 6:45
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    $\begingroup$ @JesperPetersen: Since $x_1x_2+1$ is a square, then $a,b,c,d$ obey, $$2(a^2+b^2+c^2+d^2)-(a+b+c+d)^2-3-6abcd+(abcd)^2 = y^2$$ as in Section III of this post, while $a,b,c,d,x_k$ satisfy, $$(\alpha_1-\alpha_5)^2=4(\alpha_2+\alpha_4+1)$$ where the $\alpha_i$ are in the elementary symmetric polynomials in this answer. However, there is no known general relation where all six variables appear at once. $\endgroup$ – Tito Piezas III Mar 21 '16 at 6:28
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(Answer 2)

I found another infinite family. It seems using the relation in Answer 1 is productive, namely to find a suitable triple $a,b,c$, and solve for $d$ in,

$$(abcd-3)^2=4(ab+cd+3)\tag0$$

Since $cd+1=\frac{ab+1}{(1\pm\sqrt{ab+1})^2}$, then one need only ensure that $ad+1,bd+1$ are squares so Theorems 1 & 2 of Answer 1 can apply.

I. Quadruple:

The new quadruple has one free parameter $n$, and two dependent variables $x,y$,

$$a,\,b = \frac{2 n (x^2 - 1) y}{(n - 1) x (n - y^2)},\;\frac{2 x (n - y^2)}{(n - 1)( x^2 - 1) y}$$

$$c,\,d = \frac{2 (n - 1) x y}{(x^2 - 1 )(n - y^2)},\;\frac{(n + 3) (x^2 - 1) (n - y^2)}{8 x y}$$

Then $x,y$ must satisfy,

$$F(x):=n-\frac{2(2+n+n^2)}{n+3}x^2+n\,x^4=\color{brown}m\,{t_1}^2\tag1$$

$$F(y):=n^2-\frac{2(2+n+n^2)}{n+3}y^2+y^4=\color{brown}m\,{t_2}^2\tag2$$

where $\color{brown}m=cd=\frac{1}{4}(n-1)(n+3)$.

II. Elliptic curves:

Given some constant $n$, then $(1),(2)$ are elliptic curves, with trivial points like $x_0=1$, and $y_0=1,n$, or those that result in either,

$$(a-b)(a-c)\dots(c-d) = 0\tag3$$

$$(a+b-c-d)^2=4(ab+1)(cd+1)\tag4$$

However, there are non-trivial ones like,

$$x_1 = \frac{n - 3}{n + 1},\;\;y_1 = \frac{3 n^2 - 2n + 3}{n^2 + 2n - 7}$$

We cannot pair $x_1,y_1$ together because it would satisfy $(4)$. But we can pair $x_1$ with subsequent rational points of $F(y)=z^2$ like,

$$y_2 = \frac{27 - 54 n + 277 n^2 - 180n^3 - 11 n^4 + 10 n^5 - 5 n^6}{-183 + 230 n - 65 n^2 - 12 n^3 - 41n^4 + 6 n^5 + n^6}$$

and the $y_k$'s numerator degrees are $2,6,12,20$, and so on. So this new family, like the previous one, has infinitely many polynomial solutions.

III. Example:

Let $x = \frac{n - 3}{n + 1}$. For convenience, let $n=\frac{3}{2},\;$ so $x =\tfrac{-3}{5}$ and,

$$F(x) = \big(\tfrac{88}{75}\big)^2 = {t_1}^2\tag5$$

$$F(y) = \tfrac{4}{81}(81 - 92 y^2 + 36 y^4) = {t_2}^2\tag6$$

The elliptic curve $(6)$ has infinitely many rational points and the small ones are trivial. The formula above yields $y_2 = \frac{533}{1247}$, but a smaller non-trivial one is $y = \frac{68}{55}$. Using the smaller, applying Theorem $1$, and getting $z_1,z_2$ of,

$$(a b c d z + 2 a b c + a + b + c - d - z)^2 = 4(a b + 1)(a c + 1)(b c + 1)(d z + 1)\tag7$$

then,

$$a,\,b,\,c,\,d,\,z_1,\,z_2 = \frac{47872}{173},\;\frac{519}{5984},\;\frac{14025}{346},\;\frac{519}{37400},\;\frac{13512232363}{161755000},\;\frac{2759132883}{647020000}$$

A feature of this family is that $ab$ and $cd$ are constants that depend only on $n$. Thus, this example belongs to infinitely many sextuples of form,

$$a,\;\frac{24}{a},\;c,\;\frac{9}{16c},\;z_1,\;z_2$$

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Alternative constructions of rational Diophantine sextuples can be found in

"Rational Diophantine sextuples containing two regular quadruples and one regular quintuple" https://arxiv.org/abs/1904.00348

"There are infinitely many rational Diophantine sextuples with square denominators" https://arxiv.org/abs/1903.02805

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    $\begingroup$ Thanks for the update $\endgroup$ – Tito Piezas III Aug 14 '19 at 15:52

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