• #### 3 numbers... A, B, C (in geometric progression)

From henhanna@gmail.com@21:1/5 to All on Sun Jun 19 10:13:30 2022
3 numbers... A, B, C (in geometric progression)

that is... A:B = B:C

such that the 3 differences are all squares.

Are there 5 such triples ?

Are there MANY such triples ?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Edward Murphy@21:1/5 to henh...@gmail.com on Sun Jun 19 17:45:43 2022
On 6/19/2022 10:13 AM, henh...@gmail.com wrote:

3 numbers... A, B, C (in geometric progression)

that is... A:B = B:C

such that the 3 differences are all squares.

Are there 5 such triples ?

Are there MANY such triples ?

[spoiler space]

The numbers are (for some integer x)
a
b = ax
c = ax^2
so the differences are
b - a = a(x-1)
c - a = a(x^2-1) = a(x+1)(x-1)
c - b = ax(x-1)

Divide all three by a(x-1), which must be a square, and note that one
square divided by another square must be a square:
1
x+1
x
The only integer x where x and x+1 are both squares is 0, which would
lead to (a, 0, 0), and the ratio 0/0 is indeterminate. So there are
no such triples.

Maybe "x is an integer" is an incorrect assumption; consider x = y/z
and a = w*z^2 (y, z, w) are all integers. But I'll leave that for
someone else to finish working out.

--- SoupGate-Win32 v1.05
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• From henhanna@gmail.com@21:1/5 to Edward Murphy on Sun Jun 19 17:54:14 2022
On Sunday, June 19, 2022 at 5:45:51 PM UTC-7, Edward Murphy wrote:
On 6/19/2022 10:13 AM, henh...@gmail.com wrote:

3 numbers... A, B, C (in geometric progression)

that is... A:B = B:C

such that the 3 differences are all squares.

Are there 5 such triples ?

Are there MANY such triples ?
[spoiler space]

The numbers are (for some integer x)
a
b = ax
c = ax^2
so the differences are
b - a = a(x-1)
c - a = a(x^2-1) = a(x+1)(x-1)
c - b = ax(x-1)

Divide all three by a(x-1), which must be a square, and note that one
square divided by another square must be a square:
1
x+1
x
The only integer x where x and x+1 are both squares is 0, which would
lead to (a, 0, 0), and the ratio 0/0 is indeterminate. So there are
no such triples.

Maybe "x is an integer" is an incorrect assumption; consider x = y/z
and a = w*z^2 (y, z, w) are all integers. But I'll leave that for
someone else to finish working out.

( 567, 1008, 1792 )

the 3 diff.'s are : 441, 784, 1225

-------- Are there 5 such triples ?

------ Are there MANY such triples ?

--- SoupGate-Win32 v1.05
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• From Edward Murphy@21:1/5 to henh...@gmail.com on Sun Jun 19 18:08:50 2022
On 6/19/2022 5:54 PM, henh...@gmail.com wrote:

On Sunday, June 19, 2022 at 5:45:51 PM UTC-7, Edward Murphy wrote:
On 6/19/2022 10:13 AM, henh...@gmail.com wrote:

3 numbers... A, B, C (in geometric progression)

that is... A:B = B:C

such that the 3 differences are all squares.

Are there 5 such triples ?

Are there MANY such triples ?
[spoiler space]

The numbers are (for some integer x)
a
b = ax
c = ax^2
so the differences are
b - a = a(x-1)
c - a = a(x^2-1) = a(x+1)(x-1)
c - b = ax(x-1)

Divide all three by a(x-1), which must be a square, and note that one
square divided by another square must be a square:
1
x+1
x
The only integer x where x and x+1 are both squares is 0, which would
lead to (a, 0, 0), and the ratio 0/0 is indeterminate. So there are
no such triples.

Maybe "x is an integer" is an incorrect assumption; consider x = y/z
and a = w*z^2 (y, z, w) are all integers. But I'll leave that for
someone else to finish working out.

( 567, 1008, 1792 )

the 3 diff.'s are : 441, 784, 1225

Well, that's that confirmed, then. Put another way, the numbers are
2^0 * 3^4 * 7^1
2^4 * 3^2 * 7^1
2^8 * 3^0 * 7^1
and the ratio of successive numbers is 16/9, and the differences are
3^2 * 7^2
4^2 * 7^2
5^2 * 7^2

-------- Are there 5 such triples ?

------ Are there MANY such triples ?

Any triple of the form (81x, 144x, 256x) (x is a positive integer)
probably works. The above example fits this form for x = 7.

Followup questions:

* Are there such triples not fitting this form?

* Are there such triples not fitting the form (p^4 * x, p^2 * q^2 * x,
q^4 * x) where p and q and x are all positive integers?

--- SoupGate-Win32 v1.05
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• From Ilan Mayer@21:1/5 to Edward Murphy on Sun Jun 19 19:05:13 2022
On Sunday, June 19, 2022 at 9:08:59 PM UTC-4, Edward Murphy wrote:
On 6/19/2022 5:54 PM, henh...@gmail.com wrote:

On Sunday, June 19, 2022 at 5:45:51 PM UTC-7, Edward Murphy wrote:
On 6/19/2022 10:13 AM, henh...@gmail.com wrote:

3 numbers... A, B, C (in geometric progression)

that is... A:B = B:C

such that the 3 differences are all squares.

Are there 5 such triples ?

Are there MANY such triples ?
[spoiler space]

The numbers are (for some integer x)
a
b = ax
c = ax^2
so the differences are
b - a = a(x-1)
c - a = a(x^2-1) = a(x+1)(x-1)
c - b = ax(x-1)

Divide all three by a(x-1), which must be a square, and note that one
square divided by another square must be a square:
1
x+1
x
The only integer x where x and x+1 are both squares is 0, which would
lead to (a, 0, 0), and the ratio 0/0 is indeterminate. So there are
no such triples.

Maybe "x is an integer" is an incorrect assumption; consider x = y/z
and a = w*z^2 (y, z, w) are all integers. But I'll leave that for
someone else to finish working out.

( 567, 1008, 1792 )

the 3 diff.'s are : 441, 784, 1225
Well, that's that confirmed, then. Put another way, the numbers are
2^0 * 3^4 * 7^1
2^4 * 3^2 * 7^1
2^8 * 3^0 * 7^1
and the ratio of successive numbers is 16/9, and the differences are
3^2 * 7^2
4^2 * 7^2
5^2 * 7^2
-------- Are there 5 such triples ?

------ Are there MANY such triples ?
Any triple of the form (81x, 144x, 256x) (x is a positive integer)
probably works. The above example fits this form for x = 7.

Followup questions:

* Are there such triples not fitting this form?

* Are there such triples not fitting the form (p^4 * x, p^2 * q^2 * x,
q^4 * x) where p and q and x are all positive integers?

As noted above, x and x+1 are squares; x is a rational number.
x can be written as a^2/b^2, and then (a^2+b^2)/b^2 is also a square, and this is true if a and b are the first two numbers of a Pythagorean triplet (a^2+b^2=c^2). There is an infinite number of those.
Each triplet yields a solution of the form
(b^2-a^2)*a^4, (b^2-a^2)*a^2*b^2, (b^2-a^2)*b^4
The case a=3, b=4 yields the solution 567, 1008, 1792 (given above)
The case a=5, y=12 yields the solution 74375, 428400, 2467584
etc.
So there are triplets other than (81*x, 144*x, 256*x), but they all have the form (p^4*x,p^2*q^2*x, q^4*x).