Monday, February 25, 2008


JSH: Here comes alpha

Given a target composite T that you wish to factor, it can be shown that if you have

z^2 = y^2 + nT

where n is a non-zero integer, then there exists an integer k such that

k = 2a*z/(1+2a^2)

where 'a' is alpha, though I just use 'a' for text postings, and it is a non-zero integer.

Further z = x+ak, and 2ax = k.

And finally,

k^2 = (1+a^2)^{-1}(nT) mod p

where p is an odd prime.

Also k will be near the maximum value of k such that

abs(nT - (1+a^2)k^2)

is a MINIMUM, which is the powerful bit of mathematics which makes this very likely to be a solution to the factoring problem. That is the most crucial finding.

So if you've noticed me posting a lot on this subject you may have seen postings where I said z should be divisible by 3, that is because if alpha is coprime to 3, then 1+2a^2 is divisible by 3, so z must be, if that value of alpha works.

And it is about finding an alpha value that works as some value WILL work, if you have non-zero integers z and y such that z^2 = y^2 + nT.

So the first and most likely factor of z is 3, the next is 9, and the next is 19 when a=3, so yes, you can have z coprime to 3, followed by 33 when a=4, and 51 when a=5, and 73, when a=6.

Those must all be factors of z, where again

z^2 = y^2 + nT.

So alpha=1 is the most likely, and then you have other possible values for alpha since z is set for each non-trivial factorization so the question is finding alpha and k.

Finding k is about looking near the maximum value of k such that

abs(nT - (1+a^2)k^2)

is a minimum, and the most likely alpha is 1, but it may be others.

And you use

k^2 = (1+a^2)^{-1}(nT) mod p

where p is an odd prime of your choice, to get the residue of k modulo p, where you pick an odd prime and go looking.

Most likely for your prime, alpha=1, but it can equal the other values though the probability is less.

Those forced factors of z, again are

3, when a=1, 9, when a=2, and the next is 19 when a=3, followed by 33 when a=4, and 51 when a=5, and 73, when a=6.

So, for instance, if for your picked prime a=5 works, then z does not have to be divisible by 3, and it will have 73 as a factor as

k = 2a*z/(1+2a^2).

I went to the factoring problem to end an impasse where modern mathematicians around the world as it has taken a good bit of effort by these people from what I've seen, have been successfully blocking knowledge of major mathematical finds that overturn results that have a lot to do with their careers.

And they're blocking knowledge of certain key things, like that Andrew Wiles did not prove Fermat's Last Theorem, as the research that I have that does, also shows that a crucial error slipped into the mathematical field around the time of Dedekind—over a hundred years ago.

Those of you who know a bit about math know that an error can allow people to make "proofs" which are in fact, not mathematical proofs as mathematics does not tolerate error, so this error allows these people to create an ever growing body of useless and wrong research, indefinitely.

But they have to stop real researchers, so they have turned to various tactics as they claim they are keeping the field "pure", including interestingly enough casting doubt on mathematical proof itself so that people talk of delicate proofs.

As this answer to the factoring problem shows you, real mathematical proofs are not delicate.

These people are con artists who got into an area where they could lie with apparent impunity until they made a mistake, which was the introduction of a technique for information security based on factoring supposedly being a hard problem.

They have nowhere to go from here so do not expect them to tell the truth.

For them, now it's just about waiting until the world catches up, and they face the consequences of their actions.

<< Home

This page is powered by Blogger. Isn't yours?