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• At my former employer, a French software outfit, we already had that idea. In a modified way, we incorporated it into a piece of software that compresses
Message 1 of 64 , Oct 21, 2005
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At my former employer, a French software outfit, we already had that idea.
In a modified way, we incorporated it into
a piece of software that compresses images obtained by LAR ( Large Aperture
Radar ) reconnaissance flights from the
French Air Force.
Any fool can have that idea. The proof: I had it in 2000, and the French
Air Force uses it now *sardonic grin*
Jan van Oort

On 10/21/05, gordon_as_number <gordon_as_number@...> wrote:
>
>
>
> Here is just a comment.
>
> It is possible to compile a data image as a number, as in
> xxx.arXiv.org/physics/0510148 <http://xxx.arxiv.org/physics/0510148>. That
> number, when factored
> in primes, can be stored in the latter to data compress that
> image by a factor of an approximate 10^12. As the labeling
> of primes is about a digit or two below the number in accordance
> with numbers up to 10^17 (the known primes), with density of
> an approximate x/ln(x) 10^17.
>
> The point is, with the prime factorization of a number up to
> 10^15 or so, and each prime in accordance with integers, a
> digit is lost with each prime factor. And assuming that there
> a maximum ln(N) prime factors, or rather less, most data
> compression schemes can be achieved with a factor of 10^12,
> which is actually quite amazing. (Images could be written as
> a number with approximately 10^260 digits.)
>
> So, depending on how fast you can factor a number with 10^100
> digits or so, compressions of 10^10 are achievable.
>
> Gordon
>
>
>
>
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
>
>
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--
Non sunt multiplicanda entia praeter necessitatem necnon voluptatem
( Ockam's razor, hack # 1, beta 0.1 )

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• ... Thank you David. I had been confused by consideration of the following: if x y = z and y x, then if we set t = (y+x)/2 and s = (y-x)/2 then z = t^2 - s^2
Message 64 of 64 , Oct 20, 2012
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On 10/20/2012 10:07 AM, primenumbers@yahoogroups.com wrote:
> 2.2. Re: Question
> Date: Fri Oct 19, 2012 9:02 am ((PDT))
>
>
>
> Kermit Rose<kermit@...> wrote:
>
>> >Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
>> >
>> >for what values of x would
>> >x^2 + 5 x + 6 be a perfect square.
> Set A = 1, B = 0, C = -1, D = 5, E = 0, F = 6.
>
> and you will get the obvious answer from Dario:
>
> x = -2
> y = 0
> and also:
> x = -3
> y = 0
> Calculation time: 0h 0m 0s
>
> David

Thank you David.

I had been confused by consideration of the following:

if x y = z and y > x,
then if we set t = (y+x)/2 and s = (y-x)/2

then z = t^2 - s^2
and x = y - 2 t.

So z = x y = (y - 2t) y = y^2 - 2 t y

y^2 - 2 t y - z = 0

Transforming w = y - k yields

(w + k)^2 - 2 t (w + k) - z = 0

w^2 + (2 k - 2 t) w + (k^2 - 2 t k - z) = 0

which has integral solution if and only if

(2k - 2 t)^2 + (z + 2 t k - k^2) is an integral square

In all that derivation, I forgot that I still did not know what value t was.

I confused myself too easily.

Kermit

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