Re: JSH: Factoring and residues

Tim Peters wrote:
[Tim Peters]
...
And these things are generally false, although James wishes they were
true:

x = x_res modulo T
x^2 = y^2 modulo T
1 < gcd(x+y, T) < T
1 < gcd(x-y, T) < T


[jstevh@xxxxxxx]
The correct mathematical view is that given S and k, you have an
infinite range of possible x_res's and T's that will satisfy

S = 2x_res*k mod T

and

x^2 - y^2 = 0 mod T

so you can find an S and k that will work with a particular x_res you
choose, your desired residue for x, but there are an infinity of other
x_res's and T's possible for which S and k will also work.

It's as simple as that and there is no reason for anyone to act like
there's a big mathematical mystery here.

I agree that there isn't, but stop pretending that was your _original_ view
of this method. At its start, this was all delusional nonsense like:

I was wrong in thinking that picking x_res and getting k with a chosen
S would guarantee T.

However, while I have been reasonable, you have not, as you keep
fighting a political battle to convince other people that there is
nothing to these ideas.


[JSH]
Kind of one of those odd things, but this idea just did come to me
Saturday in that I just wrote down some equations and hey! I solved
the factoring problem.

Etc, etc.

You didn't solve it, and the primary purpose of my post was to show that,
and explain _why_ the method usually fails to factor the T you give it. A
secondary purpose was to help others understand what your method _is_, since
almost nobody else replying to you appeared to have any real idea what you
were trying to say (your own writeups were unclear on several key points).


But you haven't shown that the method usually fails.

You've stated it.

I've stepped in to show what's mathematically shown, having to revise
my own views as there were things I missed or was wrong about, as I
learn more, myself, about this approach.


Now you've backtracked to something that's true: while there's no reason to
hope this method can efficiently factor the T you give it, it does
efficiently factor /other/ integers, vaguely related to T via chains of
information-losing congruences. That's incredibly weaker than the original
claims, but at least it's a true claim.


I haven't back-tracked as I've learned more given more information, and
had some mistaken ideas cleared up.

It's basic research. Some expansion of knowledge along the way is
understandable.

For example, for those who didn't follow the full story on other newsgroups,
given T=91 x_res = S = 1 as inputs, it can end up deducing that 459 is
divisible by 17 and 27. The doesn't look useful to me, but if you can make
something useful of it, more power to you.

That opens the real question of how often when you find S and k for
your x_res, will you get some other answer coming in,

More often the larger T's smallest prime factor, and you're already very
unlikely to factor incoming T with 6 digits.


But you just make statements without proof as if the idea of
mathematics has escaped you.

Mathematics is not about just making statements, as people who are
mathematicians believe that you can actually prove things to be true,
or not true, using the tools of mathematics.

instead as the math shifts to some T other than the one you want, and
the good news is that other T's will tend to be integers of absolute
value bigger than yours,

Because the largest T' satisfying the congruences is |S - 2*x*k|, and when
you pick S=1 x_res=1 (as you've always done so far), k = (T+1)/2. Therefore
the largest T' satisfying the congruences is |1 - (T+1)*x|, which has a
_minimum_ value of T at x=1 (x=0 isn't useful).


Well, it seems to me that there must be some pressure against larger
T's, as there are an infinity of possible ones, correct?

But as you go up you get larger and larger numbers, so the likelihood
of getting them drops rapidly.

Sounds like a strategy to me for maximizing your chances of getting the
T you want could be lurking in there--or in plain sight.

which means there are probably strategies for limiting that happening.

See above. Pick larger S, and, more importantly, since

x = z-k = (f_1 + f_2)/2 - k

pick f_1 and f_2 to minimize the absolute value of that expression.


Which is not a mathematical statement.

Now I think you're playing politics in an effort to convince others to
ignore this research.

People who look over my posting of a roadmap of my research may wonder
how it's possible I could have so much research and it not be
recognized.

Well, people like Tim Peters are part of the problem.

I LIKE mathematics. And I LIKE talking about mathematics.

And there are people who LOVE to reply to me just to tell people that
I'm wrong or that what I found is useless.

I'm just one voice.

I'm tempted to delete off the rest of Tim Peters reply, but it might
help to leave it in, especially so that some of you can understand the
other weapon of people like him--tedium.

They make long posts which are a pain to reply to where they say
nothing of mathematical importance.

Eventually I get tired of bothering with them and abandon threads,
which they take as an opportunity to have the last word.

There are LOTS of them and one of me.

Now the mathematics here is simple and I think neat, but the politics
are high, so rather than admit that future development is possible, or
even bother giving mathematical reasons for how certain things can
happen Tim Peters is just pushing the position that it can't work well.

I explained earlier that, given any inputs and any specific way of factoring
S+k^2, your method ends up computing specific integers x and y. The set of
T for which the congruences hold is then the set of integer divisors of
x^2-y^2, no more, no less. Just like in the example above, given T=91 x_res
= S = 1, your method computes x=5 and y=22 (given one way of factoring
S+k^2), so that T then satisfies the congruences if and only if T divides
22^2-5^2 = 459.

There's no mystery about that either.


The only thing mathematically that is true is that two congruences
hold:

x^2 - y^2 = 0 mod T

and

S - 2*x_res*k = 0 mod T

where my ideas have you selecting S and k, and solving an equation that
follows using

x^2 - y^2 = S - 2*x*k

where you factor S+k^2 to get y, which means mathematically at that
point, the algebra JUST has S and k^2, so it can give you solutions
that work with x_res and T different from your choices, which fit with
the congruences.

That's what the mathematics says.

Well, maybe, maybe not, but why one way or the other?

See above. You have no way here to find useful solutions to:

x^2 = y^2 (modulo T)

for a /fixed/ T. Instead you're effectively picking x and y, then solving
for T. But that's trivial to solve -- none of the machinery in your method
is needed for that, and it's of no use for factoring a /fixed/ T regardless.


I've repeated what the mathematics actually says, while you've mainly
just STATED things without proof.

Your statements follow the trend of downplaying the method, which I say
is a political position.

Readers who think such politics is a minor thing need only look over
the roadmap of my current mathematical discoveries, to get some
comprehension of how powerful it is when people play politics with
mathematics.

Just take one thing--I have a short proof of Fermat's Last Theorem.

To shoot down consideration of it, people need only make fun of the
CLAIM.

That's why I hate how political this damn field has become as you run
into operatives like Peters who are fighting a war--what I call the
math wars--and the mathematics is just a pawn in a game about
convincing people one way or another.

I have no stake in whether you "solve the factoring problem" or not. It's
your ego that's hugely involved in that, not mine. When you post claims
that aren't true, sometimes I take time to refute them; when you're going
down a road that I clearly see is a dead end, sometimes I try to tell you
why. You haven't noticed that I haven't been wrong about one of those yet?
I have ;-)

The nature of mathematical discovery is that until you're right, you're
wrong.

I've had years of working to find mathematical answers and in every
case, guess what?

In every case, I was wrong, until I was right.

Past failures have no impact on whether or not the current ideas work
or not.

Each idea stands alone, and necessarily, guess what?

Ideas that are wrong, failed.

It just does not matter how many times I failed before, when current
ideas are right here and the question is about their validity.

The answer is objectivity: consider these ideas on their merits.

Now I have experience with my other mathematical results where people
kept claiming I was wrong when I was right, and there's a dead math
journal to tell you how powerful human denial can be.

That dead journal published a paper outlining the key controversial
mathematics that holds together my proof of Fermat's Last Theorem.

But it died quietly, with little fanfare or notice...social forces are
more powerful than most of you are willing to admit, which is why they
have been winning for so long.

People may be able to shoot down your research just by SAYING it's
wrong.

James Harris

.

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