Factoring idea, problems with math community

A little while back I posted what I thought was this brilliant
factoring idea here because I have this problem where mathematicians
lie about my research which is a statement I know won't go over well,
but it's true. In any event, that idea was I later found out crap, but
here's another as my quest to find a solution in this area is to prove
that mathematicians routinely lie--even about VERY important things.
And this time, the equations are correct and the simplicity should jump
out at you.

Desperate to find some way to break through major lying about my
research by the mathematical community, I was doodling, playing around
with some simple equations and noticed that with

x^2 - a^2 = S + T

and

x^2 - b^2 = S - k*T

I could subtract the second from the first to get

b^2 - a^2 = (k+1)*T

which is, of course, a factorization of (k+1)*T:

(b - a)*(b+a) = (k+1)*T

with integers for S and T, where T is the target composite to factor,
so you have to pick this other integer S, and factor S+T.

Really simple.

But how do you find all the variables?

Well, if you pick S, and have a T you want to factor, then using

f_1*f_2 = S+T

it must be true that

a = (f_1 - f_2)/2

And

x=(f_1 + f_2)/2

so, you need the sum of factors of (S-k*T)/4 to equal the sum of the
factors of (S+T)/4, so I introduce j, where

S - k*T = (f_1 + f_2 - j)*j

and now you solve for k, to get

k = (S - (f_1 + f_2 - j)*j)/T

so you also have

S - (f_1 + f_2 - j)*j = 0 mod T

so

j^2 - (f_1 + f_2)*j + S = 0 mod T

and completing the square gives

j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T

so

(2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T

so you have the quadratic residue of ((f_1 + f_2)^2 - 4*S) modulo T, to
find j, which is kind of neat, while it's also set what the quadratic
residue is, so there's no search involved.

The main residue is a trivial result that gives k=-1, but you have an
infinity of others found by adding or subtracting T.

And then you can find b, from

b^2 = x^2 - S + kT

and you have the factorization:

(b-a)*(b+a) = (k-1)*T.

It is possible to generalize further using

j = z/y

and then the congruence equation becomes

(2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T.

If you're skeptical you may consider the question of finding k when you
already have the factorization of T.

And just like that I may have succeeded at showing the problem with
modern mathematicians, as, these people lie about so much mathematics
you'd be shocked.

Their bold lying has forced me to turn to a practical problem to prove
it, so I work on the factoring problem not because I'm really
interested in it, but because I'm desperate.

And here they seem to have missed a trivial solution in the area of
factoring.

But how? I don't know. It's a mystery to me. But just look over the
equations. Simple stuff.

James Harris

.

Relevant Pages

  • JSH: Proper evaluation of surrogate factoring
    ... To use surrogate factoring you pick k, and then everything you do is ... I am being very serious here, modern mathematicians lie a lot. ...
    (sci.math)
  • Re: JSH: Contradictory behavior, issue of math fraud
    ... is not a hard problem after all, then how can mathematicians who not ... and that math journals routinely publish false papers!!! ... fraction of them were by the editor himself. ... implement your many variants of surrogate factoring. ...
    (sci.math)
  • JSH: Lets recrap
    ... Surrogate factoring preferentially yanks out small prime factors. ... but that was without modern computing technology, ... There is no way that I can see that mathematicians ignoring this ...
    (sci.math)
  • Re: Factoring idea, problems with math community
    ... factoring idea here because I have this problem where mathematicians ... lie about my research which is a statement I know won't go over well, ... that mathematicians routinely lie--even about VERY important things. ... quadratic residue modulo T, which is then your r, so now you have ...
    (sci.physics)
  • Re: Surrogate factoring and the k/T ratio
    ... actually it is harder to lie about mathematics than other areas ... since mathematicians give proofs of their theorems that can be checked ... Like with surrogate factoring, if you knew that for a given k, ... or you brilliant avoid proof that you are lying about it. ...
    (sci.crypt)