Is there a positive integer $d$ in existence$?$

Question

Suppose there are three positive integers $a,b,c$ such that the product of any two is one less than an integer squared, i.e., $$\begin{align} ab+1=x^2\\ ac+1=y^2\\ bc+1=z^2 \end{align}$$ Then is it true that there exists a fourth positive integer $d$ such that its product with any of the first three is also one less than a square, i.e. $$\begin{align} ad+1=w^2\\ bd+1=u^2\\ cd+1=v^2 \end{align}$$

I tried factorizing the expressions but it did not prove to be of much help. Like we can say that $ab=(x+1)(x-1)$ and similarly for all expressions. Any help is greatly appreciated.

EDIT

A triplet satisfying $(a,b,c)$ is $(1,3,8)$ and the value of $d$ in this case is $120$. As far as I know, this is an old and very interesting problem and many famous mathematicians had attacked this problem but I guess none of them gave a detailed proof or maybe I don't know about it.

Answer 1

It turns out that this is a classical topic in number theory. In the terminology you're seeking, you're asking whether every "Diophantine triple" of positive integers can be extended to a "Diophantine quadruple" of positive integers. And the answer is yes: Arkin, Hoggatt, and Strauss proved in 1979 that (in your notation) you can choose $$ d = a + b + c + 2abc + 2xyz. $$ (It was a well-known conjecture in this field, for example, that there do not exist any Diophantine quintuples of positive integers; this was only proved recently, by He, Togbé, and Ziegler in 2019.)