Re: Factoring idea, problems with math community

jstevh@xxxxxxx wrote:
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.


It was pointed out to me that these are also trivial, so I figured out
a way around that by turning the problem around a bit:

One approach is to find some quadratic residue r, where

(f_1 + f_2)^2 - 4*S = r + n*T

where n is a natural number, as then solving for f_1 gives

f_11 = (sqrt(4*S + r + n*T) +/- sqrt(r + (n-1)*T))/2

so you can arbitrarily pick some integer w, square it, and get the
quadratic residue modulo T, which is then your r, so now you have

w^2 = r + (n-1)*T

so you can easily solve for n, and then you pick S so that the second
square root is an integer.

So now you have

2*j - (f_1 + f_2) = w

is a solution.

Neat!!! I like solving problems!!!

Why don't more of you?

Now you can get k.

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.


Did any of you think of that next piece to the puzzle?

If not, why not?

Think about what's at stake here.

Isn't it a big enough deal to do your very best?


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: 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)
  • Reality of modern math world
    ... have that is not even in doubt--a new factoring method. ... mathematicians were not petty people willing to do just about anything ... They just lie about my results, lie about the very proofs, lying about ... The reality of the math world is that you can have mathematical proof ...
    (sci.math)