Re: Factoring and rationals
From: Proginoskes (proginoskes_at_email.msn.com)
Date: 04/16/05
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Date: 15 Apr 2005 23:53:38 -0700
jst...@msn.com wrote:
> Proginoskes wrote:
> > jst...@msn.com wrote:
> > > [...]
> > > My point is not complicated. If you have j^2 T, and
> > > M^2 T, where j, T and M are integers, and consider
> > > factorizations in rationals, then you have two sets
> > > of rational factorizations.
> > >
> > > Like, with f_1, f_2, g_1, g_2 rational, where
> > >
> > > f_1 f_2 = j^2 T
> > >
> > > and
> > >
> > > g_1 g_2 = M^2 T
> > >
> > > you have an infinite number of values for f_1, f_2,
> > > g_1 and g_2.
> > >
> > > Right?
> > >
> > > Well, f_1 and f_2 while rationals can still be
> > > non-trivial in that their numerators give a single
> > > prime factor of j, and so can g_1 and g_2, which
> > > is the problem.
> > >
> > > That's because there is a one-to-one match, shown
> > > by the surrogate factoring theorem, between factors
> > > g_1 and g_2 and f_1 and f_2, in that you put in f_1
> > > and f_2 and get back g_1 and g_2, using the theorem.
> >
> > Correct as far as you go.
> >
> > My point is not complicated, either.
> >
> > You need to determine how to choose f_1 and f_2 so that
> > g_1 and g_2 are non-trivial. Since the SF Theorem
> > doesn't do this (as you've acknowledged), you would need
> > to use something else which does. Otherwise, you're not
> > getting anywhere with the factorization problem (find
> > the two primes that M factors into, given that it factors
> > into two primes).
> >
>
> If there is a mathematical reason to select trivial
> factors, then arbitrary choices of factors of j^2 T,
> will tend to give specialized choices of M^2 T.
What do you mean by "specialized choices"? Sure, if you choose a factor
of j^2 T, you get a factor of M^2 T. If you don't pay any attention to
what you're doing, you may get a trivial factor of M^2 T, or you may
get a non-trivial factor of M^2 T.
But if you plan ahead, then you can figure out which factors (f_1, f_2,
..., f_N) of M^2 T will give you non-trivial factors of M^2 T.
Then, to factor an integer, all you need to do is use your method with
ONE of f_1, f_2, .., or f_N. If you know that using these factors will
give you a non-trivial factor, then you solve the factorization problem
(the RSA version, where M = p * q) with 100% SUCCESS ON THE FIRST TRY.
This even beats the 50% "success rate" that you've been claiming all
along.
The mathematics behind a real SF Theorem would guarantee it. Period.
If you still haven't understood this vital point, consider solving the
following problem:
PROBLEM. Given r, a non-zero rational number, find an non-zero integer
N so that r * N is an integer.
What _you're_ saying is that it's okay to just try N's at random. Maybe
you solve the problem, maybe you don't. Does it really matter? The
mathematics doesn't mind, after all ... (This has been your whole
approach to the factoring problem.)
What _I'm_ saying is that if you choose your N correctly, with some
knowledge about r, you can solve the problem with ONLY ONE GUESS. In
short, I would say "Let N be the denominator of r." Then r * N is the
numerator of r, which is (by definition) an integer. Period. I'm done.
I get to go outside and play while you're stuck inside, choosing
numbers at random and hoping you get lucky.
It is a fact that an infinite number of N's that will NOT solve the
problem. So, if you're unlucky enough to choose N's from that set of
numbers, you'll never finish the problem.
With my theorem, I only have to test one; with your theorem, you may
have to test infinitely many. Whose theorem has more content?
> The problem with your claim is that you don't give a special
> reason for a special choice.
That's because I don't have a definite choice. But, conceivably, I
could discover a theorem which would tell me how to do it. Then I
_would_ have a reason for a special choice; the theorem says that it
works, that I get a non-trivial factor of M.
> The intuitive assumption is that human needs matter,
That's not the case. Proofs exist outside of the material world; they
still imply the same things.
> and because from a social perspective trivial versus
> non-trivial matters, that it has to matter mathematically.
Time to get The Shovel.
You made a claim earlier that mathematics doesn't care about factoring.
Well, there is a finite set of mathematical facts, all of which are
true, which do say that an integer has been factored:
(1) 20 = 2 * 2 * 5.
(2) For all positive integers m and n, (m+1)(n+1) is not 2.
(3) For all positive integers m and n, (m+1)(n+1) is not 5.
These are sufficient facts to prove the statement "20 can be factored
into a product of prime numbers." And they're all written in
Mathematics.
> But there has to be a reason.
>
> I see a lot of talk, and no reasons.
That's because I can't devote all of my time looking for one. I have to
eat, eject bodily waste, sleep, teach, grade papers, turn on my
computer to check the latest news, etc.
But the reason itself could exist out there, even if no one is looking
for it. And if someone finds it, then they have the reason why the
problem can be solved in one step.
But I may be wrong. Maybe there isn't a reason out there, and maybe
someone can find it why there isn't.
But in either case, the answer requires more research, which you are
unwilling to do (refusing to think that the KEY is out there, after
all), and in some sense unable to do (not having had the training to
know when you do have a real reason, which can be proven). And the idea
is powerful enough that someone _should_ search for it, just in case it
exists.
> > This would involve taking your equations and finding out
> > what values f_1, j, k_1 will make g_1 = p_1 and g_2 = p_2,
> > if M = p_1 p_2.
>
> There's a one-to-one correspondence between factors of j^2 T
> and factors of M^2 T,
Sure, it's the same set even.
But it's not really between factors of j^2 T and M^2 T, because you
also have k_1 and k_2 floating around in your equations (for Ax, in
particular). The factors you get for M^2 T depend on j, f_1, and k_1
(you say this in your SF Theorem, that all k_1 need to be considered).
These are all things which were not given to you. In the language of
mathematics, f_1 is a function of j, f_1, and k_1.
f_1 = F(j, f_1, k_1). (And technically, the function has to be 1-1 as
well. However, it still has to be a function.)
If you just know j and f_1, then there's no way to find Ax; the formula
for Ax uses k_1:
> and Ax is given by
>
> Ax = +/- (k_1 + k_2) + 2j^2
>
> where k_1 k_2 = -Tj^2
(This is from your Factoring Theorem post in the Surrogate Factoring
group, April 14.)
> so in the infinity of factors there have to be cases where
> you get both trivial and non-trivial factors.
Yes, and a theorem which had actual content would show how to choose
the factors of the first number wisely.
> So infinity is the answer.
What is the question, though? "How many factors are there?"
Or maybe a better question is "How many factors do you have to test, to
find one that gives you a non-trivial factor?"
Or maybe, related to that last question, there are an infinite number
of factors which give you trivial factors of M, so you could literally
spend forever checking factors and never finding any.
Is this the "great advance" you've been promising? At least someone
testing to see whether the integers between 2 and M-1 will finish the
task.
> My analysis is that many of you think that convincing others is key.
No, convincing YOU, James Steven Harris, the writer of the post I'm
responding to, is key. I've been talking to YOU this whole post.
Sometimes I talk to other people, but most of the time, I'm talking to
YOU.
I've been patient and explained, trying to give you analogies which
will help you despite your lack of formal training. But when you can't
(or won't) accept that, you turn to social rhetoric, while I have
remained on topic.
The simle fact is your theorem, as is, gives no help in solving the
factoring problem; a real theorem (provided it exists) would.
> So you post in reply to me in order to convince others,
> possibly believing that if no one believes the surrogate
> factoring theorem is important that nothing will happen.
The Surrogate Factoring Theorem isn't important.
Surrogate factoring may be important; there may be a way to salvage the
basic idea. (If you have a factor of T, you get a non-trivial factor of
M.)
But nothing is happening because you refuse to work with
mathematicians. When people say your proof doesn't say anything, or
your analysis is wrong, you don't look at what they're saying. You
become anti-social ...
> But you are not mathematicians. I am a mathematician.
... a case in point.
> I understand not only the mathematics, but enough of human
> nature to understand that just talking against it won't work.
You understand only the mathematics of your SF Theorem. If you knew the
complete mathematics of the factoring problem, you could solve the RSA
problems and prove that you're not a crank, a quack, a kook, or a
troll.
Using what I know about human behavior, I thus conclude that one of two
things must be true:
(1) You are unable to solve the factoring problem.
(2) You are able to solve the factoring problem, but won't, for some
deep psychological reason.
If you're not really a crank, then you want the Surrogate Factoring
idea to work. Why are you blocking the help that would make it work?
--- Christopher Heckman
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