## 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,    $b^2 \,/ \,a^2$

math grad - Interest: Number theory
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### 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.

• benvitalis says:

Good! B – A is a better choice

• benvitalis says:

I posted a general solution. Not all possible solutions are generated