When (X,Y,Z) in geometrical progression {(X-Y), (X-Z), (Y-Z)} squares

To find three rational numbers in geometrical progression, the difference of any two of which is a square number.

Here’s one possible solution:

The integers,

$b \; = \; m^2 \; - \; n^2$
$a \; = \; 2 \; m \, n$
$c \; = \; m^2 \; + \; n^2$

form a Pythagorean triple.

and the following integers,

$X \; = \; (a^2 \; - \; b^2) \,(a^4)$
$Y \; = \; (a^2 \; - \; b^2) \,(a^2) \,(b^2)$
$Z \; = \; (a^2 \; - \; b^2) \,(b^4)$

are in geometrical progression

The common ratio is :        $a^2 \,/ \,b^2$

$X \; - \; Y$
$= \; (a^2 \; - \; b^2) \,(a^4) \; - \; (a^2 \; - \; b^2) \,(a^2) \,(b^2)$
$= \; (a^3 \; - \; a \, b^2)^2$

$X \; - \; Z$
$= \; (a^2 \; - \; b^2) \,(a^4) \; - \; (a^2 \; - \; b^2) \,(b^4)$
$= \; (a^2 \; - \; b^2)^2 \, (a^2 \; + \; b^2)$
$= \; (a^2 \; - \; b^2)^2 \, (c^2)$

$Y \; - \; Z$
$= \; (a^2 \; - \; b^2) \,(a^2) \,(b^2) \; - \; (a^2 \; - \; b^2) \,(b^4)$
$= \; (b^3 \; - \; a^2 \, b)^2$