A set of $m$ non-zero rationals {$a_1, a_2, ... , a_m$} is called a rational Diophantine $m$-tuple if $a_i a_j+1$ is a square. It turns out an $m$-tuple can be extended to $m+2$ if it has certain properties. The problem is to generalize the relations below to $m=5$.

I. $m=2$. Given $a,b$:

$$ax_i+1 = \big(a\pm\sqrt{ab+1}\big)^2\tag1$$

then {$a,b,x_1$} is a triple for any choice of $x_i$ . However, {$a,b,x_1,x_2$} is a quadruple if,

$$2(a^2+b^2)-(a+b)^2-3 = y^2$$

Ex. From $a,b = \frac{1}{16},\frac{17}{4}$, we get $x_1,x_2 = \frac{33}{16},\frac{105}{16}$, a quadruple first found by Diophantus.

II. $m=3$. Given $a,b,c$:

$$ax_i+1 = \big(a\sqrt{bc+1}\pm\sqrt{(ab+1)(ac+1)}\big)^2\tag2$$

then {$a,b,c,x_1$} is a quadruple. However, {$a,b,c,x_1,x_2$} is a quintuple if,

$$2(a^2+b^2+c^2)-(a+b+c)^2-3 = y^2\,^{\color{red}\dagger}$$

Ex. From $a,b,c = \frac{28}{5},\frac{55}{16},\frac{1683}{80}$, we get $x_1,x_2 = \frac{3}{80},1680$.

III. $m=4$. Given $a,b,c,d$:

$$\small(ax_i+1)(abcd-1)^2 = \big(a\sqrt{(bc+1)(bd+1)(cd+1)}\pm\sqrt{(ab+1)(ac+1)(ad+1)}\big)^2\tag3$$

then {$a,b,c,d,x_1$} is a quintuple. However, {$a,b,c,d,x_1,x_2$} is a sextuple if,

$$2(a^2+b^2+c^2+d^2)-(a+b+c+d)^2-3-6abcd+(abcd)^2 = y^2\,^{\color{red}\dagger}$$

Ex. From $a,b,c,d = \frac{5}{4},\;\frac{5}{36},\;\frac{32}{9},\;\frac{189}{4}$, we get $x_1,x_2 = \frac{3213}{676},\;\frac{665}{1521}$, one of first sextuples found by Gibbs in 1999.

$^{\color{red}\dagger}$ These two can be satisfied by the parametric example in the variable $t$ in Dujella's website.

IV. Notes:

In general, an $n$-tuple can be extended to a $n+1$ (unconditional) and $n+2$ (conditional) for $n=2,3,4$. Also, one root $x_i$ is equal to zero if,

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

for relations $(1), (2), (3)$, respectively.

V. Question:

For $m=5$, given $a,b,c,d,e$:

$$\text{LHS}? = \text{RHS}?\tag4$$

  1. Can we find $(4)$, analogous to the first three? If yes, then maybe we can use known $5$-tuples or $6$-tuples to generate $7$-tuples, of which there is yet no known example.
  2. The pattern is suggestive. But, like quintics, is there a Galois-theoretic restriction on five variables $a,b,c,d,e$ that prevent generalization for $m>4$?
  • $\begingroup$ See also this MO post and MSE post. $\endgroup$ – Tito Piezas III Mar 11 '16 at 23:20
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    $\begingroup$ I am somewhat familiar with the literature on Diophantine $m$-tuples yet I am trying to recover if there is a cohesive exposition that puts equations (1) - (3) into the same context. Where do (1) - (3) come from -- have I overlooked a well-known exposition? $\endgroup$ – Jesper Petersen Mar 13 '16 at 9:18
  • $\begingroup$ Some explanations can be found in Phil Gibbs' papers (see links at web.math.pmf.unizg.hr/~duje/ddlinks.html). $\endgroup$ – duje Mar 13 '16 at 9:35
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    $\begingroup$ @JesperPetersen: The exposition is just a summary of what I read. You can find $(3)$ here and $(2)$ here and it was easy to extrapolate $(1)$ from those two. Then I got the conditional statements by letting $x_1 x_2 +1 =\square$, then simplifying the expressions. $\endgroup$ – Tito Piezas III Mar 13 '16 at 17:49

There is also an equation for extending quintuples to sextuples.

$(abcde+abcdf+abcef-abdef-acdef-bcdef+2abc-2def+a+b+c-d-e-f)^2 = 4(ab+1)(ac+1)(bc+1)(de+1)(df+1)(ef+1)$

This can be solved for $f$ with two rational roots when $\{a,b,c,d,e\}$ is a rational Diophantine quintuple (except in a few exceptional circumstances)

It does not always work because it only guarantees that the new number $a_6$ times numbers from the quintuple plus one will be squares times some number, i.e.

$a_6 a_i + 1 = qx_i^2, i<6 $

If we get lucky and $q$ itself is a square we get a rational Diophantine sextuple

Here is an example of a regular sextuple that can be generated this way from any of its quintuples.

249/2048 3720/6241 715/384 369/128 38/3 920/3

  • $\begingroup$ Very nice and interesting. It seems that the formula can be interpreted in terms of elliptic curves induced by Diophantine quadruples. In the notation of this paper web.math.pmf.unizg.hr/~duje/pdf/irr.pdf it seems that the points with the first coordinate f are Q +- R + A. $\endgroup$ – duje Sep 30 '16 at 15:51

(Too long for a comment, but may help in a generalization.)

After some sleuthing around, it turns out $(1),(2),(3)$ can be encapsulated in the single equation,

$$(a b c d e + 2a b c + a + b + c - d - e)^2 = 4(a b + 1)(a c + 1)(b c + 1)(d e + 1)\tag1$$

which I think is by Dujella. For example,

  • Let $a,b,c,d =1,3,0,0,\,$ yields $e_1, e_2 = 0,8$.
  • Let $a,b,c,d =1,3,8,0,\,$ yields $e_1, e_2 = 0,120$.
  • Let $a,b,c,d =1,3,8,120,\,$ yields $e_1, e_2 = 0,\frac{777480}{8288641}$.

The last was also found by Euler, so he must have a version of $(1)$. Thus, in general, an $m$-tuple can be extended to a quintuple. However, if $e_1 e_2+1 =\square$, then it yields a sextuple as in the example above,

  • Let $a,b,c,d = \frac{5}{4},\;\frac{5}{36},\;\frac{32}{9},\;\frac{189}{4},\,$ yields $e_1,e_2 = \frac{3213}{676},\;\frac{665}{1521}$.

a solution unnoticed by Euler and only found in 1999.

Tinkering with $(1)$, it can be expressed by the elementary symmetric polynomials $\alpha_i$ in a much simpler form,



$$\begin{aligned} \alpha_1 &=a + b + c + d + e\\ \alpha_2 &=a b + a c + b c + a d + b d + c d + a e + b e + c e + d e\\ \alpha_4 &=a b c d + a b c e + a b d e + a c d e + b c d e\\ \alpha_5 &=abcde\\ \end{aligned}$$

So if $(2)$ can be generalized, then the question can be rephrased as: Is there a version of $(2)$ using the elementary symmetric polynomials $\alpha_i$ for $a,b,c,d,e,f$?

  • $\begingroup$ Euler had a version of (1) with $c=a+b+2r$, $d=4r(a+r)(b+r)$, where $ab+1=r^2$. In this case $e_1=0$. $\endgroup$ – duje Mar 15 '16 at 7:30

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