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.

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