Find these triples such that the product of any two increased by the third one gives a square.
(The three numbers are not equal to each other otherwise the result will be trivial.)
For example, he desired three rational numbers, the product of any two of which increased by the third shall be a square.
"History Of The Theory Of Numbers Vol-II" by Leonard Eugene Dickson
Preface, p-iii
https://archive.org/details/HistoryOfTheTheoryOfNumbersVolII/page/n3
\begin{align*} \left\{ \begin{split} xy+z&=a^2\\ xz+y&=b^2\\ yz+x&=c^2\\ xy+x+y&=d^2\\ xz+x+z&=e^2\\ yz+y+z&=f^2 \end{split} \right.\quad(1) \qquad\qquad\left\{ \begin{split} xy+z&=a^2\\ xz+y&=b^2\\ yz+x&=c^2\\ xy+x+y&=d^2\ne\square\\ xz+x+z&=e^2\ne\square\\ yz+y+z&=f^2\ne\square \end{split} \right.\quad(2) \end{align*}
I find that there are two situations. Here are some examples.
\begin{align*} \color{black}{ \begin{array}{|c|c|c|} \hline &\begin{split} \boxed{\left(d,e,f\right)\in\mathbb{Q}^3}\qquad\quad\,\\ \,\,\left(x,y,z\right)\quad\left(a,b,c\right)\quad\left(d,e,f\right)\, \end{split}& \begin{split} \boxed{\left(d,e,f\right)\notin\mathbb{Q}^3}\qquad\quad\,\\ \,\,\left(x,y,z\right)\quad\left(a,b,c\right)\quad\left(d,e,f\right)\, \end{split}\\ \hline \begin{split} \left(x,y,z\right)\in{\mathbb{Z}_-}\!\!^3 \end{split}& \begin{split} \\ \left(-3,-6,-14\right)\,\left(2,6,9\right)\,\left(3,5,8\right)\\ \\ \end{split} & \begin{split} \\ \left(-5,-21,-69\right)\,\left(6,18,38\right)\,\left(\sqrt{79},\sqrt{271},3\sqrt{151}\right)\\ \\ \end{split}\\ \hline \begin{split} \begin{split} \left(1,y,z\right)\in{\mathbb{Z}_+}\!\!^3 \end{split} \end{split}& \begin{split} \\ \left(1,4,12\right)\,\left(4,4,7\right)\,\left(3,5,8\right)\\ \\ \end{split} & \begin{split} \\ \left(1,16,33\right)\,\left(7,7,23\right)\,\left(\sqrt{33},\sqrt{67},3\sqrt{577}\right)\\ \\ \end{split} \\ \hline \begin{split} \begin{split} \left(x,y,z\right)\in{\mathbb{Z}_+}\!\!^3 \end{split} \end{split}& \begin{split} \\ \left(12,33,88\right)\,\left(22,33,54\right)\,\left(21,34,55\right)\\ \\ \end{split} & \begin{split} \\ \left(8,17,89\right)\,\left(15,27,39\right)\,\left(\sqrt{161},\sqrt{809},\sqrt{1619}\right)\\ \\ \end{split} \\ \hline \begin{split} \left(1,y,z\right)\quad\\ \in{\mathbb{Q}_+}\!\!^3\setminus{\mathbb{Z}_+}\!\!^3 \end{split}& \begin{split} \\ \left(1,\tfrac{15}{2},\tfrac{35}{2}\right)\,\left(5,5,\tfrac{23}{2}\right)\,\left(4,6,\tfrac{25}{2}\right)\\ \\ \end{split} & \begin{split} \\ \left(1,\tfrac{13}{5},\tfrac{32}{5}\right)\,\left(3,3,\tfrac{21}{5}\right)\,\left(\sqrt{\tfrac{31}{5}},\sqrt{\tfrac{69}{5}},\tfrac{\sqrt{641}}{5}\right)\\ \\ \end{split} \\ \hline \begin{split} \begin{split} \left(x,y,z\right)\quad\\ \in{\mathbb{Q}_+}\!\!^3\setminus{\mathbb{Z}_+}\!\!^3 \end{split} \end{split}& \begin{split} \\ \left(\tfrac{2}{3},\tfrac{23}{3},\tfrac{50}{3}\right)\,\left(\tfrac{13}{3},\tfrac{14}{3},\tfrac{34}{3}\right)\,\left(\tfrac{11}{3},\tfrac{16}{3},\tfrac{37}{3}\right)\\ \\ \end{split} & \begin{split} \\ \left(\tfrac{18}{7},\tfrac{67}{7},\tfrac{130}{7}\right)\,\left(\tfrac{46}{7},\tfrac{53}{7},\tfrac{94}{7}\right)\,\left(\tfrac{\sqrt{1801}}{7},\tfrac{4\sqrt{211}}{7},\tfrac{3\sqrt{1121}}{7}\right)\\ \\ \end{split} \\ \hline \begin{split} \begin{split} \left(\sqrt{x},\sqrt{y},\sqrt{z}\right)\\ \in{\mathbb{Q}_+}\!\!^3\setminus{\mathbb{Z}_+}\!\!^3\,\, \end{split} \end{split}& \begin{split} \\ \left(\tfrac{25}{9},\tfrac{64}{9},\tfrac{196}{9}\right)\,\left(\tfrac{58}{9},\tfrac{74}{9},\tfrac{113}{9}\right)\,\left(\tfrac{49}{9},\tfrac{83}{9},\tfrac{122}{9}\right)\\ \\ \end{split} & \begin{split} \\ \left(?,?,?\right)\,\left(?,?,?\right)\,\left(?,?,?\right)\\ \\ \end{split} \\ \hline \end{array} } \end{align*}
According to individ, the solution of Diophantine system can be given by it
\begin{align*} \left\{ \begin{split} x&=\left(\frac{1}{u}-\frac{u}{2}+v\right)^2-uv\\ y&=\frac{2}{u}\left(\frac{2}{u}-\frac{u}{2}\right)\\ z&=\left(\frac{1}{u}-\frac{u}{2}-v\right)^2+uv\\ a&=\frac{2}{u}\left(\frac{1}{u}-\frac{u}{2}-v\right)\\ b&=\left(\frac{1}{u}-\frac{u}{2}-v\right)^2+2uv+\frac{u^2}{2}\\ c&=\frac{2}{u}\left(\frac{1}{u}-\frac{u}{2}+v\right)\\ d&=\frac{2}{u}\left(\frac{1}{u}-v\right)\\ e&=\frac{u^2}{4}-\frac{1}{u^2}+v^2\\ f&=\frac{2}{u}\left(\frac{1}{u}+v\right) \end{split} \right.\qquad\quad(1) \end{align*}
But how to find out the result in another situation?
Square Eulerian Quadruples: https://arxiv.org/pdf/math/0609127.pdf
System of Diophantine equations (a): https://artofproblemsolving.com/community/c7h1682351p10728828