Monday, January 26, 2009
JSH: Probably right, but kind of depressing, you know?
Several months ago I found a general solution to binary quadratic Diophantine equations. In retrospect it kind of makes sense that with that first in human history, a solution to the factoring problem might kind of spontaneously emerge, but this thing I'm considering now is so freaking simple, and I kind of didn't quite figure it out myself anyway as others suggested simple things I just hadn't noticed, that it boggles the mind.
I hope is wrong, but it is, well, probably right.
Turns out you can solve for x^2 - Ty^2 = c^2, using some easy algebra. Trivial freaking algebra.
I posted it in one thread just a second ago, but I'll copy and paste it in here to explain some freaking things.
I'm going to switch variables here.
Given x^2 - Dy^2 = 1
you have solutions for an ellipse or Pythagorean triples with
(D-1)j^2 + (j+/-1)^2 = (x+y)^2
where j = ((x+Dy) -/+1)/D, and j can be a fraction.
That's the result I started trumpeting just a little bit ago, though I first figured out that you could connect Pell's Equation to Pythagorean triples—when D-1 is a square—or discrete ellipses back in September of last year.
I just kind of wandered off from the result at that time.
Recently I thought about letting D-1 be a target to be factored, but figured out that D-1 always factors out, so that was a no-go, to my relief. Then a poster pointed out that you might factor with x^2 - Dy^2 = 1, using
(x-1)(x+1) = Dy^2.
And here now is the answer I worked out for THAT approach.
Where you can easily replace D with T, and x is a fraction. Solving for the variables is done trivially:
(D-1)j^2 = (x+y)^2 - (j+/-1)^2 = (x+y + j +/- 1)(x + y - j -/+ 1)
and it's just a matter of choosing a factorization of the left side. One possibility is:
x+y + j +/- 1 = (D-1)j
x + y - j -/+ 1 = j
but notice you can also factor D-1, for instance, assuming D is an odd number, I know it is even, so I could factor as:
x+y + j +/- 1 = (D-1)/(2j)
x + y - j -/+ 1 = 2j
So there are as many ways to choose factorizations are there are factorizations of D-1.
That's key. As you are taking D-1 apart, so the question is, does the math put it back together on the other side to force a trivial factorization? Maybe. That's the open question. If it does then this approach doesn't work.
It's as simple as that.
So you solve for (x+y) and j, given the 2 equations, using whichever factorization you pick.
Once you have them you find x and y, using that first solution for (x+y) and j, with
j = ((x+Dy) -/+1)/D
where now you have two equations again. In one of you have x+y with a number, in the second it is x+Dy, with a number, so it's easy to solve for x and y exactly. Trivial algebra.
Now you can substitute into:
x^2 - Dy^2 = 1
But I keep talking about x^2 - Dy^2 = c^2, because the x and y you have are likely to be fractions, but it doesn't matter as you just multiply out the denominators.
What makes this thing so damn scary is that you have to solve for variables TWICE using simultaneous equations, which kind of scrambles things in a way that one would think would make it difficult for the algebra to de-scramble on the other side to force trivial factorizations.
If the algebra does not care, then factoring was never a hard problem.
People just didn't know how to do it, so they thought it was hard.
But hey, the mathematics may save your butts still. It may de-scramble everything on the other side and force a trivial factorization.
Otherwise, it is the end of RSA encryption. Overnight.
Damn thing is actually right. I don't know how it's possible, though there STILL is some minor possibility that it may not work well, but theoretically, it IS the solution to the factoring problem.
What amazes me though is that it can probably sit out indefinitely.
What other explanation for that except the human species is worthless?
So far I have none. If I get some other explanation I will consider something else to do.
But right now, a simple mathematical method for factoring numbers of arbitrary size—it should factor an RSA number faster than you can encrypt with the damn thing—seems to be sitting out in public view—with no impact.
No impact. No acknowledgment.
I feel like maybe I should hold my breath, except that would be useless I think, as it appears, there is no reason to think that will change!
Humanity is, well, proved to be stupid in aggregate. The full evidence is that despite billions of people on the planet, the human species is, dumb.
I live on the planet of dumb and dumber.
Was I cursed or something? Maybe in a former life?
I'm on the planet of the idiots.
[A reply to someone who told James that his method does not work and that if he wanted to know why then he should try to actually factor some numbers.]
No need. You still don't know my modus operandi? I rarely need to test things, as I brainstorm on newsgroups and then put them on my math blog.
I also use Google search rankings to see when something is wrong.
If, as you say, the idea fails, it will drop in Google search rankings.
Right now the primary thread connected to it is RISING, so the Google search engine indicator says you are wrong.
I do not rely on single human beings unless you give me facts. Simply stating something without support does nothing.
Being someone who moves Google search engine results, I can use the Internet as a kind of global computer, seeing a continual vote that pulls in data worldwide.
Based on stats from my math blog that pull is over 2500 cities in 100 plus countries.
You as a single human being are nothing in comparison.
I hope is wrong, but it is, well, probably right.
Turns out you can solve for x^2 - Ty^2 = c^2, using some easy algebra. Trivial freaking algebra.
I posted it in one thread just a second ago, but I'll copy and paste it in here to explain some freaking things.
I'm going to switch variables here.
Given x^2 - Dy^2 = 1
you have solutions for an ellipse or Pythagorean triples with
(D-1)j^2 + (j+/-1)^2 = (x+y)^2
where j = ((x+Dy) -/+1)/D, and j can be a fraction.
That's the result I started trumpeting just a little bit ago, though I first figured out that you could connect Pell's Equation to Pythagorean triples—when D-1 is a square—or discrete ellipses back in September of last year.
I just kind of wandered off from the result at that time.
Recently I thought about letting D-1 be a target to be factored, but figured out that D-1 always factors out, so that was a no-go, to my relief. Then a poster pointed out that you might factor with x^2 - Dy^2 = 1, using
(x-1)(x+1) = Dy^2.
And here now is the answer I worked out for THAT approach.
Where you can easily replace D with T, and x is a fraction. Solving for the variables is done trivially:
(D-1)j^2 = (x+y)^2 - (j+/-1)^2 = (x+y + j +/- 1)(x + y - j -/+ 1)
and it's just a matter of choosing a factorization of the left side. One possibility is:
x+y + j +/- 1 = (D-1)j
x + y - j -/+ 1 = j
but notice you can also factor D-1, for instance, assuming D is an odd number, I know it is even, so I could factor as:
x+y + j +/- 1 = (D-1)/(2j)
x + y - j -/+ 1 = 2j
So there are as many ways to choose factorizations are there are factorizations of D-1.
That's key. As you are taking D-1 apart, so the question is, does the math put it back together on the other side to force a trivial factorization? Maybe. That's the open question. If it does then this approach doesn't work.
It's as simple as that.
So you solve for (x+y) and j, given the 2 equations, using whichever factorization you pick.
Once you have them you find x and y, using that first solution for (x+y) and j, with
j = ((x+Dy) -/+1)/D
where now you have two equations again. In one of you have x+y with a number, in the second it is x+Dy, with a number, so it's easy to solve for x and y exactly. Trivial algebra.
Now you can substitute into:
x^2 - Dy^2 = 1
But I keep talking about x^2 - Dy^2 = c^2, because the x and y you have are likely to be fractions, but it doesn't matter as you just multiply out the denominators.
What makes this thing so damn scary is that you have to solve for variables TWICE using simultaneous equations, which kind of scrambles things in a way that one would think would make it difficult for the algebra to de-scramble on the other side to force trivial factorizations.
If the algebra does not care, then factoring was never a hard problem.
People just didn't know how to do it, so they thought it was hard.
But hey, the mathematics may save your butts still. It may de-scramble everything on the other side and force a trivial factorization.
Otherwise, it is the end of RSA encryption. Overnight.
Damn thing is actually right. I don't know how it's possible, though there STILL is some minor possibility that it may not work well, but theoretically, it IS the solution to the factoring problem.
What amazes me though is that it can probably sit out indefinitely.
What other explanation for that except the human species is worthless?
So far I have none. If I get some other explanation I will consider something else to do.
But right now, a simple mathematical method for factoring numbers of arbitrary size—it should factor an RSA number faster than you can encrypt with the damn thing—seems to be sitting out in public view—with no impact.
No impact. No acknowledgment.
I feel like maybe I should hold my breath, except that would be useless I think, as it appears, there is no reason to think that will change!
Humanity is, well, proved to be stupid in aggregate. The full evidence is that despite billions of people on the planet, the human species is, dumb.
I live on the planet of dumb and dumber.
Was I cursed or something? Maybe in a former life?
I'm on the planet of the idiots.
[A reply to someone who told James that his method does not work and that if he wanted to know why then he should try to actually factor some numbers.]
No need. You still don't know my modus operandi? I rarely need to test things, as I brainstorm on newsgroups and then put them on my math blog.
I also use Google search rankings to see when something is wrong.
If, as you say, the idea fails, it will drop in Google search rankings.
Right now the primary thread connected to it is RISING, so the Google search engine indicator says you are wrong.
I do not rely on single human beings unless you give me facts. Simply stating something without support does nothing.
Being someone who moves Google search engine results, I can use the Internet as a kind of global computer, seeing a continual vote that pulls in data worldwide.
Based on stats from my math blog that pull is over 2500 cities in 100 plus countries.
You as a single human being are nothing in comparison.
# posted by James Harris @ 22:17 
