3
$\begingroup$

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 $m>4$?
$\endgroup$
  • $\begingroup$ See also this MO post and MSE post. $\endgroup$ – Tito Piezas III Mar 11 '16 at 23:20
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

(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,

$$(\alpha_1-\alpha_5)^2=4(\alpha_2+\alpha_4+1)\tag2$$

where,

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

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 $a,b,c,d,e,f$?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.