TITLE: A question on elementary number theory
QUESTION [22 upvotes]: Just came across the following question:

Let $S=\{2,5,13\}$. Notice that $S$ satisfies the following property: for any $a,b \in S$ and $a \neq b$, $ab-1$ is a perfect square. Show that for any positive integer $d \not\in S$, $S \cup \{d\}$ does not satify the above property.

This question can be done by considering modulo 4.
Here comes my question:

What is the greatest value of $|A|$ if all elements in $A$ are different and for any $a,b \in A$ and $a \neq b$, $ab-1$ is a perfect square?

Remark: $A$ may not contain any of $\{2,5,13\}$, example: $\{17, 26, 85\}$
Edit: From the link here, there are infinite many 3-element sets statisfy the poperty. These sets are of the form $\{a, b, a+b+2r\}$ where $r^2 = ab-1$. Are we able to find a 4-element set that statisfies the property?

REPLY [7 votes]: By the paper A. Dujella and C. Fuchs, Complete solution of a problem of Diophantus and Euler, J. London Math. Soc. 71 (2005), 33-52. (see Theorem 1b), 
there does not exist 4-element set the considered property and with all elements greater than 1. 
For results on sets contaning 1, see e.g. N. C. Bonciocat, M. Cipu, M. Mignotte, On D(-1)-quadruples, Publ. Mat. 56 (2012), 279-304. In particular, if {1,b,c,d} has considered property, then b>10^{13}.