## A Property

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• Here is just a comment. It is possible to compile a data image as a number, as in xxx.arXiv.org/physics/0510148. That number, when factored in primes, can be
Message 1 of 64 , Oct 20, 2005
Here is just a comment.

It is possible to compile a data image as a number, as in
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
• ... 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
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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