Clone product

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Goahead52
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Clone product

Postby Goahead52 » Mon Dec 01, 2014 7:05 pm UTC

Hi,

Let x(1),x(2),x(3).....x(k) be integers such as :

x(1) < x(2) < x(3).....< x(k-1) < x(k)

Let r be an integer positive number > 0

Show that :

(x(k+1))^2 + r = product ((x(i)^2+r) with i varying from 1 to k

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Forest Goose
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Re: Clone product

Postby Forest Goose » Mon Dec 01, 2014 7:20 pm UTC

Are you asking if given an increasing sequence, there always exists such r? The nature of x given an r? If it holds for any x and any r? It's not clear what you're looking for.
Forest Goose: A rare, but wily, form of goose; best known for dropping on unsuspecting hikers, from trees, to steal sweets.

Goahead52
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Re: Clone product

Postby Goahead52 » Mon Dec 01, 2014 7:31 pm UTC

r is a constant number
it always exists a set of x(i) values making the equation hold.

Goahead52
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Re: Clone product

Postby Goahead52 » Mon Dec 01, 2014 7:33 pm UTC

r is a constant number
it always exists a set of x(i) values making the equation hold.
All the factors have the same form (x(i)^2+r) and the product too

Goahead52
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Re: Clone product

Postby Goahead52 » Mon Dec 01, 2014 8:14 pm UTC

As a short example

r=1
170=17*5*2

x(i)={1,2,4}
the product = (13^2)+1
For any fixed r we could generate an infinite number of infinite sequences x(i)

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z4lis
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Re: Clone product

Postby z4lis » Tue Dec 02, 2014 3:37 am UTC

What I think the OP's problem is:

Fix r > 0. Given a sequence a(1) < a(2) < ... < a(k), does the quantity (a(1)^2 + r)(a(2)^2 + r)...(a(k)^2 + r) have the form a(k+1)^2 +r for a(k+1) > a(k)?

Spoiler:
By induction, we may pick b > a(k) so that (a(2)^2 + r)---(a(k)^2 + r) = b^2 + r. So the LHS has the form (a^2 + r)(b^2 + r) for a < b. So we must show that (a^2 + r)(b^2 + r) - r is a perfect square.... I tried this with some small numbers and it doesn't seem to check out! As a counterexample, we could take r = 2, a = 4, b = 7. Nevertheless, I think there's something to be said, based on some of the numbers I've worked out by hand. It seems to work for any two consecutive numbers, but this is easy enough to show by looking at the discriminant of (a^2 + r)(b^2 + r) - r as a quadratic in r. In fact, I think a little work will reveal that the trick works if and only if a,b differ by 1.
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.

Goahead52
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Re: Clone product

Postby Goahead52 » Tue Dec 02, 2014 8:08 pm UTC

Correct.
Just prove that :

((x^2+r)*((x+1)^2+r))-r is perfect square then you could build an infinite number of infinite sequences fullfilling such equality.

Anyway that it not the only way to build other sequences.

Than you.

Goahead52
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Re: Clone product

Postby Goahead52 » Wed Dec 03, 2014 9:20 pm UTC

Here is a short example where a difference is bigger than 1.

r=3
x(1)=8
x(2)=91
x(3)=745

(8^2+3)*(91^2+3)=745^2+3
67*8284=555028

We could find longer sequences with different gaps.

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jaap
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Re: Clone product

Postby jaap » Thu Dec 04, 2014 2:54 pm UTC

Goahead52 wrote:Correct.
Just prove that :
((x^2+r)*((x+1)^2+r))-r is perfect square then you could build an infinite number of infinite sequences fullfilling such equality.

((x+1)2+r)*(x2+r) = (x2+x+r)2 + r

Goahead52
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Re: Clone product

Postby Goahead52 » Thu Dec 04, 2014 7:59 pm UTC

Goahead52 wrote:Hi,

Let x(1),x(2),x(3).....x(k) be integers such as :

x(1) < x(2) < x(3).....< x(k-1) < x(k)

Let r be an integer positive number > 0

Show that :

(x(k+1))^2 + r = product ((x(i)^2+r) with i varying from 1 to k


Instead of r can we find a general solution with -r (all the other conditions remain identical)
Thanks a lot

Goahead52
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Joined: Thu Oct 16, 2014 9:28 am UTC

Re: Clone product

Postby Goahead52 » Thu Dec 04, 2014 8:00 pm UTC

jaap wrote:
Goahead52 wrote:Correct.
Just prove that :
((x^2+r)*((x+1)^2+r))-r is perfect square then you could build an infinite number of infinite sequences fullfilling such equality.

((x+1)2+r)*(x2+r) = (x2+x+r)2 + r


Thanks a lot

Goahead52
Posts: 431
Joined: Thu Oct 16, 2014 9:28 am UTC

Re: Clone product

Postby Goahead52 » Sat Dec 06, 2014 11:12 pm UTC

Hi,

How can we link those forms (x^2+ r) or (x^2-r) I mean the same r but with different x`s with factorials and/or combinatorics identities?
In fact the goal is to express some factorial n using those forms but in a way such as the formula will be elegant.

Good luck!

Goahead52
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Joined: Thu Oct 16, 2014 9:28 am UTC

Re: Clone product

Postby Goahead52 » Sun Dec 07, 2014 4:47 pm UTC

As an example :

13!=(63^2+74879)*(64^2+74879)

r=74879
x(1)=63
x(2)=64


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