There exist infinitely many sets of three positive integers $(a,b,c)$ where $a\not=b,b\not=c$ and $c\not=a$ such that each of $$ab-1,\ bc-1,\ ca-1,\ ab-a-b+c,\ bc-b-c+a,\ ca-c-a+b$$ is a perfect square (note that $0=0^2$ is a perfect square). For example, for $$(a,b,c)=(F_{2n-1},F_{2n+1},F_{2n+3})$$ where $F_k$ is the $k$-th Fibonacci number defined by $F_0=0,F_1=1,F_{k}=F_{k-2}+F_{k-1}\ (k\ge 2)$, we have $${ab-1}={{F_{2n}}^2},\ {bc-1}={F_{2n+2}}^2,\ {ca-1}={F_{2n+1}}^2$$ $${ab-a-b+c}={(F_{2n}+1)^2},{bc-b-c+a}={(F_{2n+2}-1)^2},ca-c-a+b={(F_{2n+1}-1)^2}.$$
By the way, I noticed that for every such set $(a,b,c)$ I found, each of $$ab+a+b-c,\ bc+b+c-a,\ ca+c+a-b$$ is also a perfect square.
For example, for $(a,b,c)=(F_{2n-1},F_{2n+1},F_{2n+3})$, we have $${ab+a+b-c}={(F_{2n}-1)^2},{bc+b+c-a}={(F_{2n+2}+1)^2},{ca+c+a-b}={(F_{2n+1}+1)^2}.$$
So, I asked if proposition 1 is true on math.SE and knew that it is false.
Proposition 1 (False) : For positive integers $(a,b,c)$ such that $a\not=b,b\not=c$ and $c\not=a$, if each of $$ab-1,\ bc-1,\ ca-1,\ ab-a-b+c,\ bc-b-c+a,\ ca-c-a+b$$ is a perfect square, then each of $$ab+a+b-c,\ bc+b+c-a,\ ca+c+a-b$$ is a perfect square.
Every counterexample for proposition 1 found on math.SE satisfies $$a=1\ \ \text{and}\ \ ab+a+b-c\lt 0$$ such as $$(a,b,c)=(1,5,65),(1,2,145),(1,10,325).$$
Then, I asked the following question on math.SE, but I have not received any complete answers.
Question : Are the following propositions true?
Proposition 2 (True) : For positive integers $(a,b,c)$ such that $\color{red}{1\le} a\lt b\lt c$ and $ab+a+b-c\ge 0$, if each of $$ab-1,\ bc-1,\ ca-1,\ ab-a-b+c,\ bc-b-c+a,\ ca-c-a+b$$ is a perfect square, then each of $$ab+a+b-c,\ bc+b+c-a,\ ca+c+a-b$$ is a perfect square.
Proposition 3 : For positive integers $(a,b,c)$ such that $\color{red}{2\le} a\lt b\lt c$, if each of $$ab-1,\ bc-1,\ ca-1,\ ab-a-b+c,\ bc-b-c+a,\ ca-c-a+b$$ is a perfect square, then each of $$ab+a+b-c,\ bc+b+c-a,\ ca+c+a-b$$ is a perfect square.
Proposition 4 (True) : For positive integers $(a,b,c)$ such that $\color{red}{1\le} a\lt b\lt c$, if each of $$ab-1,\ bc-1,\ ca-1,\ ab\mp (a+ b-c),\ bc\mp (b+ c- a),\ ca\mp (c+ a-b)$$ is a perfect square, then $c=a+b+2\sqrt{ab-1}$ holds.
I would like to know if these propositions are true and any relevant references.
Additional information : On math.SE, a user Tito Piezas III found that if $$ab-1 = d^2,\ \ c = a + b - 2 d,$$ then $$\begin{aligned} a b-1\; &= d^2\\ a c-1\; &= (a - d)^2\\ b c-1 \; &= (b - d)^2\\ a b - a - b + c \; &= (1 - d)^2\\ b c + a - b - c \; &= (1 - b + d)^2\\ a c - a + b - c \; &= (1 - a + d)^2\\ a b + a + b - c \; &= (1 + d)^2\\ b c - a + b + c \; &= (1 + b - d)^2\\ a c + a - b + c \; &= (1 + a - d)^2 .\end{aligned}$$
The proposition 4 comes from Tito Piezas III's findings.
Added : From duje's comment, we can say
If each of $$ab-1,\ \ bc-1,\ \ ca-1$$ is a perfect square and $0\lt a\lt b\lt c\lt 3ab\in\mathbb Z$, then $c=a+b+2\sqrt{ab-1}$ holds.
By adding Tito Piezas III's findings to this result, we can say
If each of $$ab-1,\ \ bc-1,\ \ ca-1$$ is a perfect square and $0\lt a\lt b\lt c\lt 3ab\in\mathbb Z$, then each of $$ab\mp (a+ b-c),\ bc\mp (b+ c- a),\ ca\mp (c+ a-b)$$ is a perfect square.
Added : A user duje pointed that since $1\le a\lt b\lt c,ab+a+b-c\ge 0\Rightarrow c\lt 3ab,$ we can say that proposition 2 and 4 are true.