Diophantine Quintuple?

Question

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.

Answer 1

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

Answer 2

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).