### Tuesday, June 06, 2006

## SF: Easy math, theory behind the solution

Easier than giving the surrogate factoring equations that must work is explaing why they must.

Consider that given

w + x - 2z = f_1

w + 3x + 2y + 2z = f_2

w + x + y + z = g_1

3w + x - y - 3z = g_2

it's trivially true that you can solve for w, x, y and z in terms of the f's and g's as you have 4 unknowns with 4 linear equations.

If you get that, you are most of the way there.

What I did next was rather clever. I considered

(w + x - 2z)(w + 3x + 2y + 2z) = k + (w + x + y + z)(3w + x - y - 3z)

and if you multiply out, and solve for k, you get

k = 2x^2 + 2xy + y^2 - 2w^2 - z^2 - 2xz

and now I can substitute out for two of the variables using just two of the linear equations.

And next find rational solutions for the remaining variables by completing the square twice.

With these particular equations it turns out that if you solve for z using the 4 linear equations you get

z = (g_2 - g_1 + f_2 - 3f_1)/4

and if g_1 g_2 = T, your target, then it is obvious enough that part of the solution for z is a difference of factors of the target, and in fact, you can find a solution where you end up with a factorization dependent on T.

But that only works for z with these equations.

If you solve for y, you don't get something so simple as a solution with a difference of factors of T, so it's impossible for the mathematics to use T itself for the difference.

So the algebra uses a surrogate.

It factors the surrogate, so that you get a difference of the surrogate's factors.

So if h_1 h_2 = S, where S is the surrogate, you will get a solution where you have, say

h_1 - h_2

but the math can do that because h_1 h_2 is coprime to T.

So, surrogate factoring MUST work, as the algebra has no other choice, no other way to balance everything out.

If you believe that, then you may begin feverishly working to get the equations to try and get a piece of the $300k+ that RSA was offering for solutions to their factoring challenge.

But you are probably already too late.

That's a lot of money to some people. More than enough reason for them to shut-up and work, while the rest of you are adrift without a clue, unable to see the simple mathematics until someone besides me shows it to you.

So why do I make a post like this?

To make sure you feel betrayed.

[A reply to someone who mentioned that James doesn't seem interested in the RSA factoring challenge.]

I'm not interested in the RSA factoring challenge except as a means to an end.

It is the carrot that pulls others into figuring out the answer.

You people have won for years when you just faced me.

But all that money means that soon enough you will be facing more people, and then you will fall.

[A reply to someone who said that James should spend his time working, earning money, getting an education, etc.]

This is my hobby. I've got a degree from Vanderbilt University. I've done all kinds of things, traveled all over the United States.

I have fun.

You are just someone who tries to FEED off the misery of others so you don't want the truth. You want justification for your parasitic behavior. You need people, yes, but to try and cause them pain.

You are a parasitic personality.

For those who want to understand why this current method MUST be a surrogate factoring solution, consider what I pointed out in my original post. It's neat!!!

Given

w + x - 2z = f_1

w + 3x + 2y + 2z = f_2

w + x + y + z = g_1

3w + x - y - 3z = g_2

you have 4 unknowns with 4 linear equations so you get one solution for w, x, y and z.

That's basic. It's not a debatable point.

But importantly, for y, you get

y = (7g_1 - 3g_2 + 5f_1 - 3f_2)/4

which is the explicit solution for y that follows from the 4 linear equations.

It's easy to think that the algebra can handle 7 g_1 - 3 g_2, just by multiplying 21 times T and some posters have claimed that is the case, but there's a BIG problem with that claim as it forces another solution for y:

y = (5f_1 - 3f_2 + 21g_1 - g_2)/4

But you have 4 linear equations with 4 unknowns giving ONE solution, and centuries of mathematics saying that one is all you get.

Um, I think it's easier to believe two sci.math regulars are lying than centuries of mathematics is wrong.

So what must the algebra do to avoid the contradiction?

It uses a surrogate. It uses some number that it can factor such that you get this other weird relation that solves for y in terms of the surrogate factorization.

It has no choice.

If you try to solve the factors of the surrogate in terms of the factors of the target, you get a hyperbola!!!

Neat. I know that as it is what must follow logically, I think.

Trouble is, I can make mistakes, and have done so many times in the past. Maybe I'm missing something?

[A reply to someone who said that James' method works

He's lying.

Look at the stock market.

The world is treading water waiting on the result of this debate.

The answer is, I found an easy solution to the factoring problem.

Math people lie all the time. They're lying about my solution.

RSA is broken.

Consider that given

w + x - 2z = f_1

w + 3x + 2y + 2z = f_2

w + x + y + z = g_1

3w + x - y - 3z = g_2

it's trivially true that you can solve for w, x, y and z in terms of the f's and g's as you have 4 unknowns with 4 linear equations.

If you get that, you are most of the way there.

What I did next was rather clever. I considered

(w + x - 2z)(w + 3x + 2y + 2z) = k + (w + x + y + z)(3w + x - y - 3z)

and if you multiply out, and solve for k, you get

k = 2x^2 + 2xy + y^2 - 2w^2 - z^2 - 2xz

and now I can substitute out for two of the variables using just two of the linear equations.

And next find rational solutions for the remaining variables by completing the square twice.

With these particular equations it turns out that if you solve for z using the 4 linear equations you get

z = (g_2 - g_1 + f_2 - 3f_1)/4

and if g_1 g_2 = T, your target, then it is obvious enough that part of the solution for z is a difference of factors of the target, and in fact, you can find a solution where you end up with a factorization dependent on T.

But that only works for z with these equations.

If you solve for y, you don't get something so simple as a solution with a difference of factors of T, so it's impossible for the mathematics to use T itself for the difference.

So the algebra uses a surrogate.

It factors the surrogate, so that you get a difference of the surrogate's factors.

So if h_1 h_2 = S, where S is the surrogate, you will get a solution where you have, say

h_1 - h_2

but the math can do that because h_1 h_2 is coprime to T.

So, surrogate factoring MUST work, as the algebra has no other choice, no other way to balance everything out.

If you believe that, then you may begin feverishly working to get the equations to try and get a piece of the $300k+ that RSA was offering for solutions to their factoring challenge.

But you are probably already too late.

That's a lot of money to some people. More than enough reason for them to shut-up and work, while the rest of you are adrift without a clue, unable to see the simple mathematics until someone besides me shows it to you.

So why do I make a post like this?

To make sure you feel betrayed.

[A reply to someone who mentioned that James doesn't seem interested in the RSA factoring challenge.]

I'm not interested in the RSA factoring challenge except as a means to an end.

It is the carrot that pulls others into figuring out the answer.

You people have won for years when you just faced me.

But all that money means that soon enough you will be facing more people, and then you will fall.

[A reply to someone who said that James should spend his time working, earning money, getting an education, etc.]

This is my hobby. I've got a degree from Vanderbilt University. I've done all kinds of things, traveled all over the United States.

I have fun.

You are just someone who tries to FEED off the misery of others so you don't want the truth. You want justification for your parasitic behavior. You need people, yes, but to try and cause them pain.

You are a parasitic personality.

For those who want to understand why this current method MUST be a surrogate factoring solution, consider what I pointed out in my original post. It's neat!!!

Given

w + x - 2z = f_1

w + 3x + 2y + 2z = f_2

w + x + y + z = g_1

3w + x - y - 3z = g_2

you have 4 unknowns with 4 linear equations so you get one solution for w, x, y and z.

That's basic. It's not a debatable point.

But importantly, for y, you get

y = (7g_1 - 3g_2 + 5f_1 - 3f_2)/4

which is the explicit solution for y that follows from the 4 linear equations.

It's easy to think that the algebra can handle 7 g_1 - 3 g_2, just by multiplying 21 times T and some posters have claimed that is the case, but there's a BIG problem with that claim as it forces another solution for y:

y = (5f_1 - 3f_2 + 21g_1 - g_2)/4

But you have 4 linear equations with 4 unknowns giving ONE solution, and centuries of mathematics saying that one is all you get.

Um, I think it's easier to believe two sci.math regulars are lying than centuries of mathematics is wrong.

So what must the algebra do to avoid the contradiction?

It uses a surrogate. It uses some number that it can factor such that you get this other weird relation that solves for y in terms of the surrogate factorization.

It has no choice.

If you try to solve the factors of the surrogate in terms of the factors of the target, you get a hyperbola!!!

Neat. I know that as it is what must follow logically, I think.

Trouble is, I can make mistakes, and have done so many times in the past. Maybe I'm missing something?

[A reply to someone who said that James' method works

**if**one already has the factors.]He's lying.

Look at the stock market.

The world is treading water waiting on the result of this debate.

The answer is, I found an easy solution to the factoring problem.

Math people lie all the time. They're lying about my solution.

RSA is broken.