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interests / rec.puzzles / Re: 3 numbers... A, B, C (in geometric progression)

SubjectAuthor
* Re: 3 numbers... A, B, C (in geometric progression)Edward Murphy
`* Re: 3 numbers... A, B, C (in geometric progression)henh...@gmail.com
 `* Re: 3 numbers... A, B, C (in geometric progression)Edward Murphy
  `- Re: 3 numbers... A, B, C (in geometric progression)Ilan Mayer

1
Re: 3 numbers... A, B, C (in geometric progression)

<t8ofvt$hoc$1@gioia.aioe.org>

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From: emurph...@zoho.com (Edward Murphy)
Newsgroups: rec.puzzles
Subject: Re: 3 numbers... A, B, C (in geometric progression)
Date: Sun, 19 Jun 2022 17:45:43 -0700
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 by: Edward Murphy - Mon, 20 Jun 2022 00:45 UTC

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.

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

<0238f8a2-8a55-4750-9773-fe336cfe7a66n@googlegroups.com>

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Subject: Re: 3 numbers... A, B, C (in geometric progression)
From: henha...@gmail.com (henh...@gmail.com)
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 by: henh...@gmail.com - Mon, 20 Jun 2022 00:54 UTC

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 ?

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

<t8ohb9$t4v$1@gioia.aioe.org>

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From: emurph...@zoho.com (Edward Murphy)
Newsgroups: rec.puzzles
Subject: Re: 3 numbers... A, B, C (in geometric progression)
Date: Sun, 19 Jun 2022 18:08:50 -0700
Organization: Aioe.org NNTP Server
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 by: Edward Murphy - Mon, 20 Jun 2022 01:08 UTC

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?

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

<b6353e5b-cae4-41b8-af9c-b917ba90a254n@googlegroups.com>

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Subject: Re: 3 numbers... A, B, C (in geometric progression)
From: ilan_no_...@hotmail.com (Ilan Mayer)
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Content-Type: text/plain; charset="UTF-8"
 by: Ilan Mayer - Mon, 20 Jun 2022 02:05 UTC

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).

Please reply to ilanlmayer at gmail dot com

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||

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