TITLE: Diophantine Quintuple?
QUESTION [9 upvotes]: 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.

REPLY [3 votes]: (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.