Harristotelian Logic

Oh well, enough bravado on my part as I'm not certain this will work and waiting for Wednesday just seems silly. The Bulletin of the AMS did reject. Why would they change their minds between now and then?

The expert opinion is noted. Here is what my research says, which presumably then will not work, but I do not know why it would not.

Given a target composite T, from theory using x^2 = y^2 mod T and k = 2x mod T, it can be proven that

(x+k)^2 = y^2 + 2k^2 mod T

must be true for any solution of a difference of squares.

Explicitly to solve you need solutions for

(x+k)^2 = y^2 + 2k^2 + nT.

The algorithm picks x directly, choosing x = floor(sqrt(T)), so k = 2x, and then ranges for the n's from

n_max = floor(((x+k)^2 - 2k^2)/T)

and

n_min = floor((4(x+k-1) - 2k^2)/T)

which with my program has meant roughly 32 surrogates to factor.

By the theory, if you can fully factor all 32 surrogates for any target T, then you will non-trivially factor T.

If you cannot factor all 32 with the given x, you can increment it by 1 and try again, indefinitely.

Note that you can also use x = floor(sqrt(2T)) to have about 64 surrogates and much greater odds but I'm not clear how that works exactly and besides if you can factor 32 with the first one then you have the target in hand.

It is so weirdly simple and I think the theory is correct, but I guess I could be wrong.

I have tried to implement with my own programs but as I pointed out in a previous post, I use recursion and with big numbers fewer and fewer of the surrogates get factored, so it craps out.

I am not confident that I can work that problem out so what I said earlier was bravado on my part.

[A reply to someone who wrote that (x + k)² − 2k² − nT will not, in general, be a perfect square.]

Wo, hey, I didn't think of that until you mentioned it, but it is possible that my latest theory would REQUIRE that one of the n's from n_max to n_min give you a perfect square with that x and k, which would also be required to give a non-trivial factorization of T.

The governing limits are different as they are

-(k+2xT)/T < m_1 < -(k + 2xp_1)/T

where with m_1 all you really need to know is that its absolute value should be greater than or equal to 1, but you don't know p_1 of course as that is a prime factor of T, so you guess at it.

You DO know that one of the prime factors of T MUST be less than sqrt(T) which is why I picked

x = floor(sqrt(T))

to force x larger than the smallest prime factor with the limit:

-(k+2xp)/p < m < -(k + 2x)/p

assuming p is the smallest prime factor. But if you are within the previous limit with m_1 big enough then yeah, theory says you should non-trivially factor T by looping through possible n's with that k and x, looking for a perfect square solution to

(x+k)^2 - 2k^2 - nT

as that would be your y.

So then, with x = floor(sqrt(T)), k = 2x,

n_min = floor((4(x+k-1) - 2k^2)/T)

and

n_max = floor(((x+k)^2 - 2k^2)/T) ,

you could loop through y = sqrt((x+k)^2 - 2k^2 - nT), looking for a perfect square and the theory says there must be at least one, if for a prime factor p_1 of T with

-(k+2xT)/T < m_1 < -(k + 2xp_1)/T

the absolute value of m_1 is 1 or greater.

Oh, hey, and the theory says it MUST non-trivially factor T, so it is an absolute as in 100% probability of factoring T non-trivially.

And that is a definitive test of the theory!

Thanks Marcus!!!

[A reply to someone who wrote that surrogate factoring has the big advantage of getting James some sitting time with God and that he would wager that that is probably not something people get with Fermat's method.]

You people are so damn dumb.

Marcus pointed out a problem with my earlier algorithm as it turns out I have an assumption that

x^2>y^2

so x=floor(sqrt(T)) is too small, so add one.

Here's the corrected algorithm:

Kind of odd really, but fascinating to contemplate as by the theory the following algorithm should be applicable against an RSA sized number factoring it in a maximum of 32 trials:

Note, still with x^2 = y^2 mod T, and k = 2x mod T.

So then, with x = floor(sqrt(T))+1, k = 2x,

n_min = floor((4(x+k-1) - 2k^2)/T)

and

n_max = floor(((x+k)^2 - 2k^2)/T) ,

you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min to n_max, looking for a perfect square and the theory says there must be at least one, if for
a prime factor p_1 of T with

-(k+2xT)/T < m_1 < -(k + 2xp_1)/T

the absolute value of m_1 is 1 or greater.

Guess I should add that now x+y must have one prime factor of T, and x- y must have another, if you find an integer y, for at LEAST ONE of the n's so you take a gcd with T.

So the theory says it will give you a solution to a difference of squares, AND that for at least one of the n's that solution must non-trivially factor T.

That's the theory rejected by the Bulletin of the AMS.

That tests my theory here and intriguingly enough in case some peoplethink there are a lot of n's, there are 32.

32 n's.

So if that theory is correct then you could factor an RSA sized number within 32 trials.

Math people are not very bright. They are PRETEND bright.

Pretend smart, but not real smart.

---

Yuck. You're slimey. I'm sure you're smiling to yourself as you read this.

Maybe I will never crack the factoring problem.

Maybe you people will win with lies, but win what?

Satisfaction at convincing people too stupid to care about the truth?

So many others have done that and done it better.

But all you win is the death of the human species, and if that's what you're after, then fine.

I'm beaten. Let them die. I can do other things than continue to care.

Somehow I lost when I didn't think that possible which is why it happened. Maybe it's just a learning experience for me. Training for bigger battles down the line when it matters.

The worst can now begin. And the real pain for the planet can now start.

I have stalled it as long as I can and now I'm just tired.