Diophantine Quintuple? I have come across the following set of numbers: $\{1, 3, 8, 120\}$
These are positive integers where the product of any two of the numbers equal to a number that is one less than a square number. (e.g. $1 \times 3$ = $3$, which is $1$ less than $4$, $2^2$; and $3 \times 8$ = $24$, which is $1$ less than $25$, $5^2$)
Is it possible to add another number to the list so that the product of any two numbers in the set will still equal to one less than a perfect square?
I have tried to prime factorise the already existing numbers in the list, but that got me nowhere (at least for now.)
Adding 0 to the list works, but it is a really trivial answer and I was wondering if there are any more numbers that can be added to the list.
A proof would be nice, thanks.
 A: Andrej Dujella is the world expert on diophantine quintuples. Start reading here. The conjecture is that there aren't any (not just that there's no 5th number after 1, 3,  8, 120, but that there's no set of 5 distinct positive integers with each product of two being one less than a square). 
A: In OEIS A192629 it is stated that four integers is the maximum, but you can add $\frac {777480}{8288641}$.  You can multiply it by any number in the set, add one, and get a square rational number.  No further extension is known
A: (This is an old question, but updates were found for the rational case by Dujella et al in $2015$.)

I. Sextuple

Given a quadruple {$a,b,c,d$}, if they satisfy an additional constraint, then it can be extended to a rational sextuple {$a,b,c,d,e,f$}. Let,
$$ab+1,\;ac+1,\;ad+1 = r_1^2,\; r_2^2,\; r_3^2\tag1$$
$$bc+1,\;bd+1,\;cd+1 = r_4^2,\; r_5^2,\; r_6^2\tag2$$
If also,
$$(abcd)^2-6abcd+2(a^2+b^2+c^2+d^2)-(a+b+c+d)^2-3=r_7^2\tag3$$
then one can find two more rationals {$e,f$},
$$e=\frac{1}{a}\Big(\frac{(r_1 r_2 r_3+a\, r_4 r_5 r_6)^2}{(abcd-1)^2}-1\Big)$$
$$f=\frac{1}{a}\Big(\frac{(r_1 r_2 r_3-a\, r_4 r_5 r_6)^2}{(abcd-1)^2}-1\Big)$$
such that the product of any two members of the sextuple {$a,b,c,d,e,f$} plus one is a square. 

II. Example 

$$a,b,c,d = \frac{5}{4},\;\frac{5}{36},\;\frac{32}{9},\;\frac{189}{4}$$ 
Using the formulas, we get,
$$e,f = \frac{3213}{676},\;\frac{665}{1521}$$

III. Remarks 



*

*The eq $(3)$ in fact has a parametric solution discussed in Dujella's website,


$$a = \frac{18t(t^2-1)}{(t^2-6t+1)(t^2+6t+1)}$$
$$b = \frac{(t - 1)(t^2 + 6t + 1)^2}{6t(t + 1)(t^2 - 6t + 1)}$$
$$c = \frac{(t + 1)(t^2 - 6t + 1)^2}{6t(t - 1)(t^2 + 6t + 1)}$$
$$d = \frac{r_3^2-1}{a}$$
Using an elliptic curve, appropriate values of $r_3$ will make all of $(2)$ also as squares. This implies there is an infinite number of sextuples which contain fixed $a,b,c$.


*However, if $(a + b - c - d)^2 = 4(a b + 1)(c d + 1)$, then one of $e,f$ will be zero, hence will only yield a quintuple. 

