Can one find a set of four positive integers in which the product of any two distinct integers is a perfect square plus one? This post is motivated by a famous IMO problem. It implies $\{2,5,13,x\}$ ($x$ denotes a positive integer) is not a set having this property. I am wondering if there really exists a set $\{a,b,c,d\}$ (each element is a positive integer) such that $ab-1$, $ac-1$, $ad-1$, $bc-1$, $bd-1$, $cd-1$ are all perfect squares. Could you help me?
 A: it seems the task may be impossible with four numbers. Compare the theorem that there cannot be four squares in arithmetic progression. Triples with largest number up to 100:
    2    5   13
    2   13   25
    5   10   29
    5   13   34
    2   25   41
   10   17   53
    5   29   58
    2   41   61
    5   34   65
   10   29   73
   13   25   74
   17   26   85
    2   61   85
   13   34   89
    5   58   97

A: Last I heard this is still an open problem, though it is known that there are at most finitely many such $4$-tuples (reducing the problem to an enormous but finite computation), and there are no $5$-tuples with this property.
Here is a recent paper on the topic: https://www.math.tugraz.at/~elsholtz/WWW/papers/elsholtz-filipin-fujitaNumDQIIIv19.pdf
A well-studied related problem is that of Diophantine tuples, where it's $ab+1$ rather than $ab-1$ that's a square.  In that case there is a simple construction providing infinitely many $4$-tuples but at most finitely many $5$-tuples with no known examples.
Update: it appears there is a paper that brings down the upper bound for Diophantine $5$-tuples far enough to finally establish nonexistence: https://arxiv.org/abs/1610.04020
A: It is better to use a more General approach. We write the system.
$$\left\{\begin{aligned}&xy+T=a^2\\&xz+T=b^2\\&xq+T=c^2\\&yz+T=d^2\\&yq+T=k^2\\&zq+T=n^2\end{aligned}\right.$$
If the number $T$, lay at the multipliers. We find then the desired settings.
$$T=3(p-t-s)(p+t-s)(p+s)^2$$
Then the solution can be written as.
$$x=t^2+2s^2+2ps-p^2$$
$$y=t^2-s^2+2ps+2p^2$$
$$z=4t^2-(p-s)^2$$
$$q=3(p+s)^2$$
$$a=p^2+ps+s^2-t^2$$
$$b=2t^2+s^2+ps-2p^2$$
$$c=3(p+s)s$$
$$d=2t^2-2s^2+ps+p^2$$
$$k=3(p+s)p$$
$$n=3(p+s)t$$
In the rational numbers there is a solution.
