Re: What puzzles me, discovery ignored

On Sep 7, 11:58 am, Mark Murray <w.h.o...@xxxxxxxxxxx> wrote:
Enrico wrote:
I've got a couple of Excel speadsheets that will let you plug in
numbers and see what happens. You will need to have the Greatest
Common Divisor function GCD(X, Y) working - can't remember if GCD
came with the original Excel package or with one of those mysterious
"Add-Ins"

Yikes! :-) I don't do Excel.

It would be safer to receive a text - only explanation. I can pull
comments off the spreadsheets and organize them a bit to the point
where you should be able to try your own calculations. This will
take a few days - I'll have to remove variable name conflicts between
my stuff and JSH's

I can email the text-only stuff or post it here.
Let me know which one you want.

Text is vastly preferable. Please do both, if that is OK?

If your reply-to: is set, may I email you my real email address?

My intention is to code the work as (probably) C++, and then give
it a hammering (pun intended). C++ means I can then trivially modify
the code if I need to start using bignums by using operator
overloading.

There is another constraint that can be used to
limit the choices for x and y.
Ah, useful!

The constraint is: (2*x + 1) divides (2*y +1)^2 -1

Does James mention that anywhere?

Also - I'm more interested in the Diophantine solutions than in
factoring. Does James actually have a general solution, or is
your decryption of his work focussing on factorisation?

M
--
Mark Murray

===================================================================

If your reply-to: is set, may I email you my real email address?

I got one from James, so it should be working.

The constraint is: (2*x + 1) divides (2*y +1)^2 -1

Does James mention that anywhere?

No. James' factorization method relates to forms like:

P^2 - D*Q^2 = R^2

where D is an odd number to be factored.
My method involved the dot (scalar) product of a pair of vectors,
each of which contain 3 terms, any one of which can be calculated
from the other two. I found a way to convert them to the Pell - like
form above, but its not too obvious.


Also - I'm more interested in the Diophantine solutions than in
factoring. Does James actually have a general solution, or is
your decryption of his work focussing on factorisation?

I've been assuming that it's general because I can start with a
factorization, generate two equations of the form:

a x2 + b xy + c y2 + dx + ey + f = 0

and see that James' method correctly solves each for (x+y)

There are probably equations that I cannot generate from a
factorization, but have solutions. I haven't looked into this.
James mentions a solution existence check, so I have some reason
to think his solution is general. (I have not proved this!)

I'll write up a procedure to let you generate the pairs of
diophantine equations and see James' solution method work.
I expect to have it ready to post sometime tomorrow.


Enrico
.

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