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 ?
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.
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:
[spoiler space]
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 ?
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 ?
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:
[spoiler space]
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 ?
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, 1225Well, 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?
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