### Saturday, November 22, 2008

## Integer factorization equations

Given a target composite T to be factored remarkably you can simply solve for factors f_1 and f_2, of

f_1*f_2 = nT

where n is defined by

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

with

f_1 = ak mod p

and

f_2 = a^{-1}(1 + a^2)k mod p.

Here 'a' and k are chosen at will, but if they can be found such that n=1, for a prime p greater than sqrt(T), you have a factorization of T. Of course n=2, or n=3 or n equal some other known prime will work just as well for you.

But you may then have a trivial factorization.

However, it can be further shown that with all positive integers, for a given 'a' the minimum value for k which I call k_0 which will give a non-trivial factorization is given by finding k such that

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

is a minimum, which is just a rather cool and remarkable result, and also amazing to me in that it is the only situation I know of where the algebra cares about trivial versus non-trivial factorizations.

That result is one of my most beautiful research finds. It was sobering when the math community rejected it like so many others and in the months since I first made that discovery I've considered how real mathematicians could ignore a stunningly beautiful result.

I've concluded they could not. Given the reality of the result then, the people who the world sees as its mathematicians, are fakes.

And those equations define integer factorization. They also show what I call deep variables and a helper variable which is the prime number p.

f_1*f_2 = nT

where n is defined by

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

with

f_1 = ak mod p

and

f_2 = a^{-1}(1 + a^2)k mod p.

Here 'a' and k are chosen at will, but if they can be found such that n=1, for a prime p greater than sqrt(T), you have a factorization of T. Of course n=2, or n=3 or n equal some other known prime will work just as well for you.

But you may then have a trivial factorization.

However, it can be further shown that with all positive integers, for a given 'a' the minimum value for k which I call k_0 which will give a non-trivial factorization is given by finding k such that

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

is a minimum, which is just a rather cool and remarkable result, and also amazing to me in that it is the only situation I know of where the algebra cares about trivial versus non-trivial factorizations.

That result is one of my most beautiful research finds. It was sobering when the math community rejected it like so many others and in the months since I first made that discovery I've considered how real mathematicians could ignore a stunningly beautiful result.

I've concluded they could not. Given the reality of the result then, the people who the world sees as its mathematicians, are fakes.

And those equations define integer factorization. They also show what I call deep variables and a helper variable which is the prime number p.