### Thursday, February 19, 2009

## JSH: Hammer falls, Pell's Equation used to solve factoring problem

Well, the factoring problem is solved easily enough with Pell's Equation by a direct solution with x and y functions of an independent variable:

x^2 - Dy^2 = 1

y = [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)

and

x = +/-(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v) - [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)

where f_1*f_2 = D-1, and the f's are non-zero integer factors, while v is a free variable.

What makes this solution so remarkable is that there is a factoring dependency in it, but it's for D-1, not D, but you can go after factoring D by considering x and y as ratios of integer functions r, s, and t:

x = r(v)/t(v), y = s(v)/t(v)

gives

(r(v) - t(v))*(r(v) + t(v)) = D*(s(v))^2

which means oddly enough that just playing around, like running through some values of v, you might accidentally factor D, but the mathematics also allows you to be very serious and just directly find minima.

So why minima?

Well, if abs(r(v) - t(v)) < D and abs(r(v) + t(v)) < D, then of course neither can have D itself as a factor!

So those conditions cannot even occur if D is prime, but MUST occur at some value for v if it's not, and then you have a non-trivial factorization of D from the gcd's.

Absolutely. With no mathematical doubt.

So the factoring problem becomes a calculus problem. Calculus. Yup. It becomes a freaking calculus problem.

Kind of like a homework one too, as the highest power is 2.

So finally you get to see the Hammer.

Your best course as a newsgroup at this point is to acknowledge the solution. Start accepting correct mathematics in general.

And apologize for all the years of abuse, lying about my research, and the math journal you killed.

Just start apologizing and wait for someone to tell you to stop.

x^2 - Dy^2 = 1

y = [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)

and

x = +/-(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v) - [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)

where f_1*f_2 = D-1, and the f's are non-zero integer factors, while v is a free variable.

What makes this solution so remarkable is that there is a factoring dependency in it, but it's for D-1, not D, but you can go after factoring D by considering x and y as ratios of integer functions r, s, and t:

x = r(v)/t(v), y = s(v)/t(v)

gives

(r(v) - t(v))*(r(v) + t(v)) = D*(s(v))^2

which means oddly enough that just playing around, like running through some values of v, you might accidentally factor D, but the mathematics also allows you to be very serious and just directly find minima.

So why minima?

Well, if abs(r(v) - t(v)) < D and abs(r(v) + t(v)) < D, then of course neither can have D itself as a factor!

So those conditions cannot even occur if D is prime, but MUST occur at some value for v if it's not, and then you have a non-trivial factorization of D from the gcd's.

Absolutely. With no mathematical doubt.

So the factoring problem becomes a calculus problem. Calculus. Yup. It becomes a freaking calculus problem.

Kind of like a homework one too, as the highest power is 2.

So finally you get to see the Hammer.

Your best course as a newsgroup at this point is to acknowledge the solution. Start accepting correct mathematics in general.

And apologize for all the years of abuse, lying about my research, and the math journal you killed.

Just start apologizing and wait for someone to tell you to stop.