A,B,C in geometrical progression; Each of B-A,C-B,C-A is a square

 

 
To find three numbers   (A, B, C)   in geometrical progression such that

each of   B – A,   C – B,   C – A   is a square

 

A general solution:

[It may not give all possible solutions, though]

 
 
Take   n \, a^4,     n \, a^2 \, b^2,     n \, b^4

n \; = \; a^2 \; - \; b^2

(a^2 - b^2) \, a^4 \; - \; (a^2 - b^2) \, (a \, b)^2 \; = \; (a^3-a \, b^2)^2
(a^2 - b^2) \, (a \, b)^2 \; - \; (a^2 - b^2) \, b^4 \; = \; (b^3-a^2 \, b)^2
(a^2 - b^2) \, a^4 \; - \; (a^2 - b^2) \, b^4 \; = \; (a^2-b^2)^2 \, (a^2+b^2)

If we set the condition,    a^2 + b^2 = c^2
this condition will be satisfied if

b \; = \; m^2 \; - \; n^2
a \; = \; 2 \, m \, n

the legs of a pythagorean triple.

(a^2 - b^2) \, a^4
(a^2 - b^2) \, (a \, b)^2
(a^2 - b^2) \, b^4

the common ratio is,    b^2 \,/ \,a^2

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

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

3 Responses to A,B,C in geometrical progression; Each of B-A,C-B,C-A is a square

  1. paul says:

    I’m choosing to use B – A for this instead of A – B

    We can have {a, b, c} = {567, 1008, 1792}
    1008 – 567 = 21^2
    1792 -1008 = 28^2
    1792 – 567 = 35^2

    Common ratio is 16/9

    Paul.

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