### Sunday, February 26, 2006

## SF: I hope I am wrong

At this point as I make progress explaining my non-polynomial factorization research and ending some specious objections by posters by going to the complex plane with a rather basic result, I am concerned that, well, maybe I did solve the factoring problem, but hopefully those who disagree with me, are right.

The equations with T the target composite:

T = (x+y+vz)(vz-x)

where x, y and z are given by

x^2 + xy + k_1 y^2 = k_2 z^2

and

(2(v^2 - k_2)z + vy)^2 = ((1-4k_1)y^2+4T)v^2 + 4k_2(k_1y^2 - T)

where you pick y, k_1 and k_2 to get what I call the surrogate, which is

k_2(k_1y^2 - T)

and by factoring it you can solve v, z and then x, are from research which I easily show overturns over a century of mathematics, as also from those same equations, I can prove that the quadratic

a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0

for integer x and f, where f is not 1 or -1 and is coprime to x that its roots must have f as a factor, where there's an odd quirk of the ring of algebraic integers where if the roots are non-rational then provably neither of them can have f as a factor IN THAT RING.

(An analogy is with evens where with just evens 2 is not a factor of 6 because 3 is not an even, so my result is that the ring of algebraic integers is incomplete.)

The proper interpretation of the result is that the ring of algebraic integers is flawed in that it is incomplete, showing over a century of mathematics to be false.

Those who just like to play with math programs can have their software find integer solutions and see the result directly.

So, if the factoring piece doesn't work to solve the factoring problem, whew!

But if it does then reasonable people may conclude that you people betrayed the world to protect yourselves from both results, trying to protect your careers.

So, your careers are probably the first thing that would be taken away.

There's a lot riding on this research now. I proved a key piece which posters have been claiming is false actually depends on the distributive property, thus proving they have been arguing against the distributive property.

The proof is easy.

In the complex plane, given

7C(x) = (A(x) + 7)(B(x) + 1)

true for all x, where A(0) = B(0) = 0

let

C(x) = (A'(x) + 1)(B'(x) + 1)

where A'(0) = B'(0) = 0

and making that substitution, gives

7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1)

and by the distributive property

A(x) = 7A'(x) and B'(x) = B(x).

That result valid over the complex plane allows me to cement the case for the full argument proving I have been right with my research where you can directly SEE the result with integer roots of

a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0

and if this research, which is so huge that it overturns over a century of mathematics does not solve the factoring problem as well, then we can all breathe a sigh of relief.

But if it does, you people are betraying a lot, and I think you're naively doing it to protect your careers in a short-sighted way.

If you're wrong, you're not the only ones who lose a lot, as you're costing people who don't know a lot about mathematics or who don't know anything about it, who just trust their society to protect them.

Too bad they're not like us, eh? We know not to trust anyone.

Not in this world, as when you trust, someone is going to make you pay.

And this time, it'll be the mathematical community betraying the entire world.

The equations with T the target composite:

T = (x+y+vz)(vz-x)

where x, y and z are given by

x^2 + xy + k_1 y^2 = k_2 z^2

and

(2(v^2 - k_2)z + vy)^2 = ((1-4k_1)y^2+4T)v^2 + 4k_2(k_1y^2 - T)

where you pick y, k_1 and k_2 to get what I call the surrogate, which is

k_2(k_1y^2 - T)

and by factoring it you can solve v, z and then x, are from research which I easily show overturns over a century of mathematics, as also from those same equations, I can prove that the quadratic

a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0

for integer x and f, where f is not 1 or -1 and is coprime to x that its roots must have f as a factor, where there's an odd quirk of the ring of algebraic integers where if the roots are non-rational then provably neither of them can have f as a factor IN THAT RING.

(An analogy is with evens where with just evens 2 is not a factor of 6 because 3 is not an even, so my result is that the ring of algebraic integers is incomplete.)

The proper interpretation of the result is that the ring of algebraic integers is flawed in that it is incomplete, showing over a century of mathematics to be false.

Those who just like to play with math programs can have their software find integer solutions and see the result directly.

So, if the factoring piece doesn't work to solve the factoring problem, whew!

But if it does then reasonable people may conclude that you people betrayed the world to protect yourselves from both results, trying to protect your careers.

So, your careers are probably the first thing that would be taken away.

There's a lot riding on this research now. I proved a key piece which posters have been claiming is false actually depends on the distributive property, thus proving they have been arguing against the distributive property.

The proof is easy.

In the complex plane, given

7C(x) = (A(x) + 7)(B(x) + 1)

true for all x, where A(0) = B(0) = 0

let

C(x) = (A'(x) + 1)(B'(x) + 1)

where A'(0) = B'(0) = 0

and making that substitution, gives

7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1)

and by the distributive property

A(x) = 7A'(x) and B'(x) = B(x).

That result valid over the complex plane allows me to cement the case for the full argument proving I have been right with my research where you can directly SEE the result with integer roots of

a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0

and if this research, which is so huge that it overturns over a century of mathematics does not solve the factoring problem as well, then we can all breathe a sigh of relief.

But if it does, you people are betraying a lot, and I think you're naively doing it to protect your careers in a short-sighted way.

If you're wrong, you're not the only ones who lose a lot, as you're costing people who don't know a lot about mathematics or who don't know anything about it, who just trust their society to protect them.

Too bad they're not like us, eh? We know not to trust anyone.

Not in this world, as when you trust, someone is going to make you pay.

And this time, it'll be the mathematical community betraying the entire world.