Thursday, April 29, 2004


Factorizaton idea, revisited


(jk - Tk + T)(jk + Tk + T) = T^4

where T = M + 1, or T = M - 1, where M is some integer to be factored.

Solving for k gives

k = (-jT +/- T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2)

and the choice for T means that you can factor T^4 so that you have

jk - Tk + T = f_1 and

jk + Tk + T = f_2, where

f_1 f_2 = T^4,

and solve for j, and you'll get rational j's that are not integers for the interesting solutions.

Using those j's you should be able to factor M rather easily, as then you have

k = (-j(M+1) +/- (M+1)^2 sqrt(j^2 - M(M+2))/(j^2 - (M+1)^2)

using T = M+1.

I want readers to consider the insulting posts that came in response to my previous posting, as I want you to think about those people.

Friday, April 09, 2004


Factoring idea, off M

Now then, on to my latest idea, which is an attempt at finding a simple way to factor a positive integer M.

I use a tautological space.

It's such a simple way to go, but I've gotten a LOT of ridicule since December 1999 when I started working in tautological spaces (didn't call them that then).

Just a quick refresher, I have x+y+vz = x+y+vz, written as

x+y+vz = 0(mod x+y+vz)

where 0(mod x+y+vz) is what I call a tautological space, where anything that equals 0 in that space has to have x+y+vz as a factor *no matter what their value* is.

It's a four dimensional space because there are four variables.

To approach factoring, I use the five dimensional tautological space x+wy+vz, so I have

x+wy+vz = 0(mod x+wy+vz), so

x+wy = -vz(mod x+wy+vz), and squaring gives

x^2 + 2xyw + w^2 y^2 = v^2 z^2 (mod x+wy + vz).

Now I rather arbitrarily set a conditional statement, which is

x^2 + ax + by + y^2 = z^2,

which is basically a way for me to pick any x, y and z that I want, since there will always exists integers 'a' and 'b' that will allow that to work.

Then subtracting I have

(v^2 - 1) z^2 =- ax - by + 2xyw + (w^2 -1) y^2 (mod x+wy + vz), so

(v^2 - 1) z^2 - (w^2 - 1) y^2 - 2xyw + ax + by = 0(mod x+wy + vz)

and now I let

(v^2 - 1) z^2 - (w^2 - 1) y^2 - 2xyw +ax + by = M,

where M is the positive integer I want to factor, and because of the space I'm in, it necessarily has x+wy+vz as a factor.

Now I just need to find w and v, as I can pick x, y and z.

So now I'm going to creatively add in some factors to get a difference of squares,

((v^2 + 2vk + k^2) z^2 - (w^2 y^2 - 2wxy + x^2) + y^2 + x^2 - (1 + 2vk + k^2) z^2 - 4xyw +ax + by = M, so

(v + k)^2 z^2 - (w y - x)^2 -(1 + 2vk + k^2) z^2 - 4xyw = M - ax - by - y^2 - x^2,

which is

(v + k)^2 z^2 - (w y - x)^2 -(1 + 2vk + k^2) z^2 - 4xyw = M - z^2,

which just leaves adding in the additional constraint

(1 + 2vk + k^2)z^2 + 4xyw = 0

which now means that I can pick an integer z of my choosing, and factor M-z^2 to get two equations from the difference of squares. That is, if

M-z^2 = f_1 f_2, then I can have

(v+k)z - (wy - x) = f_1, and

(v+k)z + (wy - x) = f_2

and then I have my additional constraint

(1 + 2vk + k^2)z^2 + 4xyw = 0

to give me my third equation, so I have three equations so I can now solve for w, v or k.

My guess is that'd give a quadratic, BUT then I can choose any x and y that are integers, since 'a' and 'b' can always be adjusted&mbox;just keep z constant&mbox;to try and make k, v, and w integers.

If I succeed then I now have integer x+wy+vz which must be a factor of M.

Notice you don't even need to calculate 'a' and 'b', as it just matters that they exist and ax + by is an integer.

If you succeed then x+wy+vz must be a factor of M, but it could just equal M (as it must if M is prime) or even equal 1, -1, or -M.

But maybe, it's a non-trivial factor.

The attempt here is to shift factoring M into factoring some other number off M.

Does it work? I don't know. I toss ideas like this out to see if someone else will find out.

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