### Saturday, February 07, 2009

## JSH: Why factoring solution must work

Turns out you can prove through mathematical logic that given a solution like:

(r(v) - c(v))(r(v) + c(v)) = D(s(v))^2

where r(v), c(v) and s(v) are non-zero integer functions of v, if D is a composite, you MUST have a non-trivial factorization if

abs(r(v) - c(v)) < D and abs(r(v) + c(v)) < D.

So if you can FIND such function that smoothly traverse through all possible solutions in integers, then you know that factoring D is just a minima problem.

It goes without saying then, since minima problems are calculus problems and I think I would have heard if the factoring problem were considered to be an area of the calculus that NO ONE prior in all of human history has a known result doing the above.

But here is one now, and ironically it comes from Pell's Equation.

In rationals—I'll explain more on that later—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.

I use rationals so that it is all more compact.

The result above links solving the Pell's Equation to solving a discrete ellipse—a much easier problem!!!

One approach I've given is to let j = uv, introducing yet another degree of freedom, and use the discrete ellipse:

(D-1)j^2 = (x+y - (j+/-1))(x+y - (j+/-1))

so

(D-1)(uv)^2 = (x+y - (uv+/-1))(x+y - (uv+/-1))

and using f_1*f_2 = D-1 you can factor with:

(x+y - (uv+/-1)) = f_1*u

and

(x+y - (uv+/-1)) = f_2u*v^2

That's just one way to go, but notice I can now solve for u in terms of v (or vice versa but I think it's easier) and solve for x+y, and then solve for x and y directly as functions of v.

Which turns the factoring problem into a minima problem, which makes it work for the calculus.

I've already had some flak on newsgroups from posters who spent YEARS talking down my research where I noted often enough that they lie, but of course, with a lot of them saying that I was lying, while I was saying they were lying, things didn't really move much.

But now, if you believe them, you have to dismiss some easy mathematical logic, some easy algebra, and even then, just wait for other people to exploit the new mathematical find, and try to explain why you didn't do anything.

Kind of like sitting quietly while gasoline is poured on you and someone begins striking a match—or I am in error.

Break the logical chain in this post, prove me wrong, or wait for authorities to come and take some of you to jail when they find out I am correct and start looking at prosecutions.

You are not in a dream. What is happening is not a joke. What you do not do for some of you can get you prosecuted at this point.

And your life will be over. The world will go on. This result can have negatives but I am certain we will be ok.

But for some of you, the last days of your happy lives outside of prison are ticking away, and it may be for what you do NOT do.

Which may not seem fair, but, prove me wrong! The argument I have is in this post.

If no one can break the logic here then you DO SOMETHING RIGHT FOR ONCE, or wait. When I know you'll whine and cry as they lead you off to prison saying you didn't do anything—but that will be why you get prosecuted.

(r(v) - c(v))(r(v) + c(v)) = D(s(v))^2

where r(v), c(v) and s(v) are non-zero integer functions of v, if D is a composite, you MUST have a non-trivial factorization if

abs(r(v) - c(v)) < D and abs(r(v) + c(v)) < D.

So if you can FIND such function that smoothly traverse through all possible solutions in integers, then you know that factoring D is just a minima problem.

It goes without saying then, since minima problems are calculus problems and I think I would have heard if the factoring problem were considered to be an area of the calculus that NO ONE prior in all of human history has a known result doing the above.

But here is one now, and ironically it comes from Pell's Equation.

In rationals—I'll explain more on that later—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.

I use rationals so that it is all more compact.

The result above links solving the Pell's Equation to solving a discrete ellipse—a much easier problem!!!

One approach I've given is to let j = uv, introducing yet another degree of freedom, and use the discrete ellipse:

(D-1)j^2 = (x+y - (j+/-1))(x+y - (j+/-1))

so

(D-1)(uv)^2 = (x+y - (uv+/-1))(x+y - (uv+/-1))

and using f_1*f_2 = D-1 you can factor with:

(x+y - (uv+/-1)) = f_1*u

and

(x+y - (uv+/-1)) = f_2u*v^2

That's just one way to go, but notice I can now solve for u in terms of v (or vice versa but I think it's easier) and solve for x+y, and then solve for x and y directly as functions of v.

Which turns the factoring problem into a minima problem, which makes it work for the calculus.

I've already had some flak on newsgroups from posters who spent YEARS talking down my research where I noted often enough that they lie, but of course, with a lot of them saying that I was lying, while I was saying they were lying, things didn't really move much.

But now, if you believe them, you have to dismiss some easy mathematical logic, some easy algebra, and even then, just wait for other people to exploit the new mathematical find, and try to explain why you didn't do anything.

Kind of like sitting quietly while gasoline is poured on you and someone begins striking a match—or I am in error.

Break the logical chain in this post, prove me wrong, or wait for authorities to come and take some of you to jail when they find out I am correct and start looking at prosecutions.

You are not in a dream. What is happening is not a joke. What you do not do for some of you can get you prosecuted at this point.

And your life will be over. The world will go on. This result can have negatives but I am certain we will be ok.

But for some of you, the last days of your happy lives outside of prison are ticking away, and it may be for what you do NOT do.

Which may not seem fair, but, prove me wrong! The argument I have is in this post.

If no one can break the logic here then you DO SOMETHING RIGHT FOR ONCE, or wait. When I know you'll whine and cry as they lead you off to prison saying you didn't do anything—but that will be why you get prosecuted.