Re: Factoring problem, solved

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Date: 22 Jan 2005 09:01:54 -0800

jstevh@msn.com wrote:
> I am increasingly certain that I've solved the factoring problem.
>
> That is a bold claim to add to previous bold claims of mine, but here
> in THIS post I'll give the underlying theory as it's very basic, and
> direct you to a site where you can download a rough, and somewhat
> flawed, but still good enough to show the idea, prototype factoring
> program, which implements some of the theory.
>
> I'll explain why I say some of the theory in just a bit.
>
> My work boils down to the analysis of a system of equations:
>
> a_1 x + b_1 = f_1
> a_2 x + b_2 = f_2
>
> a_1 b_2 + a_2 b_2 = A
>
> a_1 a_2 x^2 + Ax + b_1 b_2 = f_1 f_2
>

I thought I'd bring this thread back to the subject at hand, as some
posters have pushed it off into a different subject area.

Remember the problem I'm talking about is the factoring problem.

More than likely you used some software today that depends on the
supposed difficulty of factoring very large numbers.

I am telling you that security probably is a mirage based on my
analysis of the system of equations above.

I wrote a paper, and then wrote a program, as a proof of concept.

Whether you accept it or not, the theory I have now is rather well
tested and getting better and better tested every day, and it keeps
saying the same thing: the factoring problem is solved.

Now you may see me talking about 50% of the time factoring, or may have
noticed my own prototype program began failing rather massively with
big numbers.

That's a quirk of how it's implemented, and if you understand how my
program works then those results are terrifying.

My program calls the algorithm in order to try and use the algorithm to
factor a number, which means that even factoring a relatively small
number, it may call that algorithm dozens or even hundreds of times.

Each time the algorithm fails there is a greater likelihood that the
program will fail to factor, so if I were right and the algorithm
worked 50% of the time, then the program probably would have crapped
out sooner than it did.

More than likely the correct number is greater than 90% factoring, and
the heavy recursion of my program hides how well the algorithm works.

You may think, ok, well, 90% still isn't perfection.

Yes, you are right--though I think the mathematics can be pushed to
give 100% factoring--but say you take ANY large number that can be
processed in a reasonable amount of time with my algorithm, which means
you get numbers far bigger than currently even envisioned for Internet
encryption, and the algorithm probably will factor them.

So how long would it take?

For an RSA challenge sized number it'd probably take it a few
milleseconds on a typical pc to iterate through all the combinations
once T is factored.

So what is T? T is the surrogate that you have to factor to try and
get the factorization of your target number.

Ok, so you may be relieved again, as, what's the point of a factoring
method that needs to factor some other number, which it turns out is
bigger than your target?

Well, its *easier* to factor than your target!!!

T can be trivially factored by any number of current factoring
techniques from elliptic curve to the number field sieve for just about
any RSA challenge number you can think of, so my claims can in fact, be
checked immediately, by, well by a lot of people reading these posts.

> where to match the variables in my initial paper
>
> y = a_1 a_2, T = f_1 f_2, and j^2 = - b_1 b_2.
>
> (Here b_1 b_2 is a negative number so j is still an integer.)
>
> I have a rather complex looking solution in the paper, but remarkably
> you can encompass it by solving the first three equations for key
> variables:
>
> a_1 = A(b_1 - f_1)/(b_1 f_2 + b_2 f_1 - 2b_1 b_2)
>
> and
>
> x = (b_1 f_2 + b_2 f_1 - 2b_1 b_2)/A
>
> which shows that you can actually factor simply by cycling through
the
> factors of j and T, while my focus in the program is on cycling
> through the factors of T.
>
> All together you then get the factors of your target M, as
>
> M^2 = j^2 + T, so
>
> a_1 a_2 x^2 + Ax + b_1 b_2 = f_1 f_2
>
> is
>
> a_1 a_2 x^2 + Ax - M^2 = 0
>
> so x is a factor of M.
>
> The idea may seem pathetically simple, but that is because you are
> programmed to believe that the factoring problem is hard. It is not.
>
> I have just shown you in a few lines how to factor an arbitrary
> non-zero positive integer M in polynomial time.
>

The only way algorithms based on the method cannot work is if there
does not exist a *rational* x that will give a non-trivial factor of M.

That's the only way.

There are mathematical reasons as to when it will or will not work,
which I'm working on figuring out.

But with mathematics this simple at its base, there's not a question of
whether or not it's right or wrong, which is why I declare the
factoring problem solved.

Do the math.

James Harris

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