Thursday, August 09, 2007

 

JSH: On integer factorization

Begin with a difference of squares congruence:

y^2 = x^2 mod T

where T is the target composite to be factored.

Now introduce k, where

k = 2x mod T

and multiply that equation by k so that you have

k^2 = 2xk mod T, which you now add to the first equation to get

k^2 + y^2 = x^2 + 2xk mod T

so you can add k^2 to both sides to complete the square so that you have

2k^2 + y^2 = (x+k)^2 mod T

which, of course is the same as

(x+k)^2 = y^2 + 2k^2 mod T.

So you now have a NEW difference of squares, as going to an explicit equation with the introduction of a non-zero integer n, you have

(x+k)^2 = y^2 + 2k^2 + nT.

I'd guess that with a given composite T, you need T+sqrt(T) < abs(k) < 2T, and can then select n with

n = -floor(2k^2 /T).

Um, it is possible that then looping through the integer factorizations of 2k^2 + nT to get x and y that you will factor your target T with a high probability of success as that gives you a k/T ratio in what I guess is the best area.

I figured ut this approach August of last year and since that time the RSA corporation that profits from the encryption system used to protect the internet has withdrawn monetary prizes for its factoring challenge.

I'd often post ideas in this area on the newsgroup sci.crypt but if you check as I just did with Google Groups you can see that newsgroup has been shut down by spam postings.

One would think coincidentally, as I've noted before, major postings by me in this area have been followed the next day by sharp downward corrections in the stock markets of the world.

My fear is that ego would keep mathematicians from telling the entire world the truth because to them the mathematics I've outlined above is too simple for their tastes, but someone seems to know the truth, or there just happen to be a lot of coincidences and a LOT of naively trusting people, like yourself.





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