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

 
 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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