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| Guest |
 Posted: Tue Jul 11, 2006 9:53 pm |
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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.
It's a quadratic residue solution, which may have been helped along by
Tim Peters noting when talking about a previous failed attempt that if
you could get multiple quadratic residues that were equal you could
factor.
He claimed it was a hard thing to do.
At the time I thought to myself that I didn't think it was, but I
didn't know how to do it.
That though may have lead my mind down the paths that eventually caused
the solution to just kind of pop out fully formed.
So one way to look at what I found is that I came up with a way to get
a lot of equal quadratic residues.
The proof is trivial.
The impact is not.
So why post it?
I've talked about that before, and maybe at this point the real answer
is that I've been wrong so many times that part of me still doesn't
believe it here, even though I think I have the proof as I've thought
that before and been wrong.
So I don't want to invest myself in the damn thing beyond the usual
ways--posts!!!
If it's right, well, well, I think it is right, but this situation is
just so wacky anyway.
I've been looking for federal people all day, expecting the FBI or even
maybe someone from the NSA to show up, but nothing...
But it's such a natural, simple solution that in retrospect is obvious.
It's also kind of clear why most people wouldn't go down those paths,
as I am a lateral thinker, trained in it very well.
I did something that I think most would consider unnatural, and that's
what it took.
So the short answer is that the question of how to get a bunch of
quadratic residues of equal value is done. That answer leads to a
factoring method. That factoring method should be a solution to the
factoring problem.
And it's very short, very straightforward, and so it should be
unbelievably fast.
Now what? I don't know. I kind of wish there had been another way to
go, but the fact that I posted this thing on Usenet yesterday, even
though it was on sci.skeptic, and nothing has happened up until today
shows how mucked up the math community is.
You people have to be banged on the head and forced to look at even
simple proofs of famous problems.
It's weird. It's like, you can't just look and get it. Someoe has to
push it on you, and hammer, hammer hammer to get you to pay attention
till MAYBE you'll think that possibly there's something there.
Which to me sounds like you're not really the mathematicians some of
you seem to think you are.
While I find these things in the wild.
James Harris |
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| Guest |
| Posted: Tue Jul 11, 2006 10:54 pm |
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Tim Peters wrote:
[quote:9c6d8b8116][jstevh@msn.com]
...
It's a quadratic residue solution, which may have been helped along by
Tim Peters noting when talking about a previous failed attempt that if
you could get multiple quadratic residues that were equal you could
factor.
_May_, not "could". If x^2 = y^2 modulo T, then gcd(x+y, T) and gcd(x-y, T)
have a chance of revealing a non-trivial factor of T. Saying more than that
requires knowing more than that. For example, if T is the product of two
distinct odd primes, T = p*q, and C is a quadratic residue of T coprime to
T, then:
x^2 = C modulo T
has 4 distinct solutions in [1, T). If r1 and r2 are two of them picked at
random (note that r1 = r2 is possible), then 1 < gcd(r1+r2, T) < T with
probability 1/2.
If r is one of them, then the 4 solutions are in fact:
+/- r
+/- u*r mod T
where:
u = (p^(q-1) - q^(p-1)) mod T
is a "non-obvious" square root of 1 modulo T (the "obvious" square roots of
1 modulo T are +1 and -1). Do you see why u computed that way _is_ a square
root of 1 modulo T? Pause to prove it (or to learn enough to prove it), or
remain forever lost ;-)
[/quote:9c6d8b8116]
Pedantic.
What I figured out is that I could just pick a residue modulo T for x.
No problems there as given
x^2 - y^2 = 0 mod T
there are non-trivial factorizations given for ANY residue of x, as you
have explicitly
x^2 - y^2 = k*T
where k can be any non-zero integer.
So you can just pick a residue and then find x and y as I've shown, and
importantly the modular inverse makes an appearance.
If you want to show real mathematical ability versus just playing at
it--as if all mathematics is, is acting--why don't you talk about why
it's key that my current research has the modular inverse?
Show what you know.
James Harris |
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| Tim Peters |
| Posted: Tue Jul 11, 2006 10:54 pm |
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Guest
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[jstevh@msn.com]
[quote:ca202eec40]...
It's a quadratic residue solution, which may have been helped along by
Tim Peters noting when talking about a previous failed attempt that if
you could get multiple quadratic residues that were equal you could
factor.
[/quote:ca202eec40]
_May_, not "could". If x^2 = y^2 modulo T, then gcd(x+y, T) and gcd(x-y, T)
have a chance of revealing a non-trivial factor of T. Saying more than that
requires knowing more than that. For example, if T is the product of two
distinct odd primes, T = p*q, and C is a quadratic residue of T coprime to
T, then:
x^2 = C modulo T
has 4 distinct solutions in [1, T). If r1 and r2 are two of them picked at
random (note that r1 = r2 is possible), then 1 < gcd(r1+r2, T) < T with
probability 1/2.
If r is one of them, then the 4 solutions are in fact:
+/- r
+/- u*r mod T
where:
u = (p^(q-1) - q^(p-1)) mod T
is a "non-obvious" square root of 1 modulo T (the "obvious" square roots of
1 modulo T are +1 and -1). Do you see why u computed that way _is_ a square
root of 1 modulo T? Pause to prove it (or to learn enough to prove it), or
remain forever lost ;-)
A less direct way to find the non-obvious square roots of 1 modulo T is to
use the Chinese Remainder Theorem to find the unique u in [1, T) such that:
u = +1 modulo p
u = -1 modulo q
or:
u = -1 modulo p
u = +1 modulo q
In either case, u^2 = 1 modulo both p and q, and so u^2 = 1 modulo lcm(p, q)
= p*q too.
For example, if p=7 & q=11, u=34 (=-1 mod 7 and =1 mod 11) or u=43 (=1 mod 7
and =-1 mod 11). Then from
1^2 = 34^2 modulo 77
you get at once that gcd(34+1, 77)=7 and gcd(34-1, 77)=11 are T's factors,
and similarly from:
1^2 = 43^2 modulo 77
OTOH, gcd(34 +/- 43, 77) is just as useless as gcd(1 +/- 76, 77).
[quote:ca202eec40]He claimed it was a hard thing to do.
[/quote:ca202eec40]
Yes, he did ;-)
[quote:ca202eec40]At the time I thought to myself that I didn't think it was, but I
didn't know how to do it.
[/quote:ca202eec40]
You still don't, at least not efficiently and without knowing T's factors
first. Try it. Start with a product of distinct odd primes, so you can
check where you get stuck against the theory above.
> ... |
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| Tim Peters |
| Posted: Tue Jul 11, 2006 10:54 pm |
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Guest
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[cut sci.crypt]
[jstevh@msn.com]
[quote:4cf06a9e63]...
If you want to show real mathematical ability versus just playing at
it--as if all mathematics is, is acting--why don't you talk about why
it's key that my current research has the modular inverse?
[/quote:4cf06a9e63]
It's not key, because this method isn't what you hope it is.
[quote:4cf06a9e63]Show what you know.
[/quote:4cf06a9e63]
Already did, and you rejected the opportunity to learn something real.
That's fine. No more from me on this one until you take the good advice
already given:
Try it. Start with a product of distinct odd primes, so you
can check where you get stuck against the theory above.
Still waiting. |
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| Guest |
| Posted: Wed Jul 12, 2006 2:42 am |
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Tim Peters wrote:
[quote:c4d5abc609][cut sci.crypt]
[jstevh@msn.com]
...
If you want to show real mathematical ability versus just playing at
it--as if all mathematics is, is acting--why don't you talk about why
it's key that my current research has the modular inverse?
It's not key, because this method isn't what you hope it is.
[/quote:c4d5abc609]
You didn't give anything before. Just babbled a bit about
trivialities.
[quote:c4d5abc609]Show what you know.
Already did, and you rejected the opportunity to learn something real.
That's fine. No more from me on this one until you take the good advice
already given:
Try it. Start with a product of distinct odd primes, so you
can check where you get stuck against the theory above.
Still waiting.
[/quote:c4d5abc609]
There wasn't anything of mathematical interest in what you posted.
And you didn't address anything from the method I stepped out.
You keep waiting then. I think you overvalue your own input.
I am only interested if you can actually point out failure. In the
past you have been useful in doing so.
Being recalcitrant here, you are of no use to me.
James Harris |
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| Tim Peters |
| Posted: Wed Jul 12, 2006 3:54 am |
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Guest
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[Tim Peters]
[quote:9f87555006]...
It's not key, because this method isn't what you hope it is.
[/quote:9f87555006]
[jstevh@msn.com]
[quote:9f87555006]You didn't give anything before. Just babbled a bit about
trivialities.
[/quote:9f87555006]
LOL -- a few days ago you didn't even know that quadratic congruences modulo
a composite generally have more than 2 roots. Admit it, bubba: you didn't
know any of what I said in the post you're talking about. While I tried not
to speak over your head, in grown-up words I showed you that the square
roots of 1 modulo p*q form a Klein group (which is why your previous method
_could_ get stuck in a useless subgroup of order 2), gave you explicit
formulas for its elements, and showed how the square roots of 1 can be used
to determine the full set of modular square roots for quadratic residues
other than 1. But no, you'd rather complain. Good luck with that ;-)
[quote:9f87555006]...
There wasn't anything of mathematical interest in what you posted.
[/quote:9f87555006]
As above, that says more about you than me.
[quote:9f87555006]And you didn't address anything from the method I stepped out.
[/quote:9f87555006]
Indeed -- it's not worth my bother. You'll discover for yourself why it
doesn't work as soon as you try it, and not a second before. You don't
listen, and are too lazy/frightened/whatever to try. Go ahead, factor 77
with it, and show your work. Oops -- not so hot after all, eh?
[quote:9f87555006]You keep waiting then. I think you overvalue your own input.
[/quote:9f87555006]
Not at all: input to _you_ is indeed of no value, because you don't listen.
I try to put in a bit of actual math that _others_ might find interesting.
The theory of modular square roots is really quite lovely, and I'm more than
satisfied if someone else reads a bit about that and smiles.
[quote:9f87555006]I am only interested if you can actually point out failure. In the
past you have been useful in doing so.
Being recalcitrant here, you are of no use to me.
[/quote:9f87555006]
If I were trying to be of use to you, that might be devastating  This
_would_ be of use to you if you weren't recalcitrant:
Try it. Start with a product of distinct odd primes, so you
can check where you get stuck against the theory above. |
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| Guest |
| Posted: Wed Jul 12, 2006 6:44 am |
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Tim Peters wrote:
[quote:d5336aa77c][Tim Peters]
...
It's not key, because this method isn't what you hope it is.
[jstevh@msn.com]
You didn't give anything before. Just babbled a bit about
trivialities.
LOL -- a few days ago you didn't even know that quadratic congruences modulo
a composite generally have more than 2 roots. Admit it, bubba: you didn't
know any of what I said in the post you're talking about. While I tried not
to speak over your head, in grown-up words I showed you that the square
roots of 1 modulo p*q form a Klein group (which is why your previous method
_could_ get stuck in a useless subgroup of order 2), gave you explicit
formulas for its elements, and showed how the square roots of 1 can be used
to determine the full set of modular square roots for quadratic residues
other than 1. But no, you'd rather complain. Good luck with that ;-)
...
There wasn't anything of mathematical interest in what you posted.
As above, that says more about you than me.
And you didn't address anything from the method I stepped out.
Indeed -- it's not worth my bother. You'll discover for yourself why it
doesn't work as soon as you try it, and not a second before. You don't
listen, and are too lazy/frightened/whatever to try. Go ahead, factor 77
with it, and show your work. Oops -- not so hot after all, eh?
You keep waiting then. I think you overvalue your own input.
Not at all: input to _you_ is indeed of no value, because you don't listen.
I try to put in a bit of actual math that _others_ might find interesting.
The theory of modular square roots is really quite lovely, and I'm more than
satisfied if someone else reads a bit about that and smiles.
I am only interested if you can actually point out failure. In the
past you have been useful in doing so.
Being recalcitrant here, you are of no use to me.
If I were trying to be of use to you, that might be devastating This
_would_ be of use to you if you weren't recalcitrant:
Try it. Start with a product of distinct odd primes, so you
can check where you get stuck against the theory above.
Indeed -- it's not worth my bother. You'll discover for yourself why it
doesn't work as soon as you try it, and not a second before. You don't
isten, and are too lazy/frightened/whatever to try. Go ahead, factor 77
with it, and show your work. Oops -- not so hot after all, eh?
[/quote:d5336aa77c]
Haha, *I* found that 77 is 7 times 11. Bet you thought that was
impossible. Well no, it just came to me, I thought, "hmm, 11, eh, yep,
multiply by 7, and wow, you get 77". You probably don't understand such
insight not being a real mathematician. Go on, test me again, I dare
you... (nothing bigger than 3 digits please). |
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| Chuck Grempu |
| Posted: Wed Jul 12, 2006 12:02 pm |
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Guest
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<gjedwards@gmail.com> wrote in message
news:1152697444.358568.226620@h48g2000cwc.googlegroups.com...
[quote:d0f12b7f5c]
Tim Peters wrote:
[Tim Peters]
...
It's not key, because this method isn't what you hope it is.
[jstevh@msn.com]
You didn't give anything before. Just babbled a bit about
trivialities.
LOL -- a few days ago you didn't even know that quadratic congruences
modulo
a composite generally have more than 2 roots. Admit it, bubba: you
didn't
know any of what I said in the post you're talking about. While I tried
not
to speak over your head, in grown-up words I showed you that the square
roots of 1 modulo p*q form a Klein group (which is why your previous
method
_could_ get stuck in a useless subgroup of order 2), gave you explicit
formulas for its elements, and showed how the square roots of 1 can be
used
to determine the full set of modular square roots for quadratic residues
other than 1. But no, you'd rather complain. Good luck with that ;-)
...
There wasn't anything of mathematical interest in what you posted.
As above, that says more about you than me.
And you didn't address anything from the method I stepped out.
Indeed -- it's not worth my bother. You'll discover for yourself why it
doesn't work as soon as you try it, and not a second before. You don't
listen, and are too lazy/frightened/whatever to try. Go ahead, factor 77
with it, and show your work. Oops -- not so hot after all, eh?
You keep waiting then. I think you overvalue your own input.
Not at all: input to _you_ is indeed of no value, because you don't
listen.
I try to put in a bit of actual math that _others_ might find
interesting.
The theory of modular square roots is really quite lovely, and I'm more
than
satisfied if someone else reads a bit about that and smiles.
I am only interested if you can actually point out failure. In the
past you have been useful in doing so.
Being recalcitrant here, you are of no use to me.
If I were trying to be of use to you, that might be devastating This
_would_ be of use to you if you weren't recalcitrant:
Try it. Start with a product of distinct odd primes, so you
can check where you get stuck against the theory above.
Indeed -- it's not worth my bother. You'll discover for yourself why it
doesn't work as soon as you try it, and not a second before. You don't
isten, and are too lazy/frightened/whatever to try. Go ahead, factor 77
with it, and show your work. Oops -- not so hot after all, eh?
Haha, *I* found that 77 is 7 times 11. Bet you thought that was
impossible. Well no, it just came to me, I thought, "hmm, 11, eh, yep,
multiply by 7, and wow, you get 77". You probably don't understand such
insight not being a real mathematician. Go on, test me again, I dare
you... (nothing bigger than 3 digits please).
[/quote:d0f12b7f5c]
I have made a HUGE discovery, I have been working on for years, my own prime
factoring theory.
Let me show you.
If a number is two digits, and both the digits are the same, then pick an S.
Then pick a T, but make sure it is 11.
Then divide T into the two digit number and the number gets factored!
This will make HUGE money for all the change machines around the world, the
FBI will want it. |
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| Tim Peters |
| Posted: Wed Jul 12, 2006 1:59 pm |
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Guest
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[gjedwards@gmail.com]
[quote:ef970a8048]...
Haha, *I* found that 77 is 7 times 11. Bet you thought that was
impossible.
[/quote:ef970a8048]
It is darned impressive. The way I sketched for James works like so:
[quote:ef970a8048]T = 77
u = pow(7, 10, T) - pow(11, 6, T) # a square root of 1 wrt 77
u
34
pow(u, 2, T)
1[/quote:ef970a8048]
Then take any integer i in [1, T) coprime to T, and then (i*u)^2 = i^2*u^2 =
i^2*1 = i^2 modulo T, and gcd(i*u +/- i, T) = gcd(i*(u +/- 1), T) = gcd(u
+/- 1, T) are non-trivial factors of T. Like i=13:
[quote:ef970a8048]i = 13
pow(i, 2, T)
15
pow(i*u, 2, T)
15
gcd(i*u + i, T)
7
gcd(i*u - i, T)
11[/quote:ef970a8048]
[quote:ef970a8048]Well no, it just came to me, I thought, "hmm, 11, eh, yep,
multiply by 7, and wow, you get 77". You probably don't understand such
insight not being a real mathematician.
[/quote:ef970a8048]
Ah, so _that's_ how mathematicians work! Cool. I'm starting to see now.
After I found the factors p and q, I plugged them into p^(q-1)-q^(p-1) to
get a square root of 1, and then used that to deduce p and q. But your way
stopped as soon as you guessed the factors to begin with! That's like magic
compared to my mess -- it must be real math.
[quote:ef970a8048]Go on, test me again, I dare you... (nothing bigger than 3 digits
please).
[/quote:ef970a8048]
OK, you asked for it: 111. It looks like it's divisible by 11: 11/11 is
1, and then tack that 1 on to the remaining 1 to get 11, and that's
divisible by 11 too. I'm not sure what the quotient works out to -- looks
like it should be 11, but 11 = 2 modulo 3, so 11*11 = 2*2 = 1 modulo 3, but
111 = 0 modulo 3, so there's some flaw in the integers blocking 11. I find
these things without even trying  |
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| Guest |
| Posted: Wed Jul 12, 2006 3:24 pm |
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Tim Peters wrote:
[quote:927272e6a8][gjedwards@gmail.com]
...
Haha, *I* found that 77 is 7 times 11. Bet you thought that was
impossible.
It is darned impressive. The way I sketched for James works like so:
T = 77
u = pow(7, 10, T) - pow(11, 6, T) # a square root of 1 wrt 77
u
34
pow(u, 2, T)
1
Then take any integer i in [1, T) coprime to T, and then (i*u)^2 = i^2*u^2 =
i^2*1 = i^2 modulo T, and gcd(i*u +/- i, T) = gcd(i*(u +/- 1), T) = gcd(u
+/- 1, T) are non-trivial factors of T. Like i=13:
i = 13
pow(i, 2, T)
15
pow(i*u, 2, T)
15
gcd(i*u + i, T)
7
gcd(i*u - i, T)
11
Well no, it just came to me, I thought, "hmm, 11, eh, yep,
multiply by 7, and wow, you get 77". You probably don't understand such
insight not being a real mathematician.
Ah, so _that's_ how mathematicians work! Cool. I'm starting to see now.
After I found the factors p and q, I plugged them into p^(q-1)-q^(p-1) to
get a square root of 1, and then used that to deduce p and q. But your way
stopped as soon as you guessed the factors to begin with! That's like magic
compared to my mess -- it must be real math.
Go on, test me again, I dare you... (nothing bigger than 3 digits
please).
OK, you asked for it: 111. It looks like it's divisible by 11: 11/11 is
1, and then tack that 1 on to the remaining 1 to get 11, and that's
divisible by 11 too. I'm not sure what the quotient works out to -- looks
like it should be 11, but 11 = 2 modulo 3, so 11*11 = 2*2 = 1 modulo 3, but
111 = 0 modulo 3, so there's some flaw in the integers blocking 11. I find
these things without even trying
[/quote:927272e6a8]
I admit it, I was wrong, my method was crap. But I've already got a new
one which will shut down your measly math departments. |
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| Chuck Grempu |
| Posted: Sun Jul 16, 2006 12:57 am |
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Guest
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<jstevh@msn.com> wrote in message
news:1152672539.475899.125310@p79g2000cwp.googlegroups.com...
[quote:52ed86d064]Tim Peters wrote:
[jstevh@msn.com]
...
It's a quadratic residue solution, which may have been helped along by
Tim Peters noting when talking about a previous failed attempt that if
you could get multiple quadratic residues that were equal you could
factor.
_May_, not "could". If x^2 = y^2 modulo T, then gcd(x+y, T) and gcd(x-y,
T)
have a chance of revealing a non-trivial factor of T. Saying more than
that
requires knowing more than that. For example, if T is the product of two
distinct odd primes, T = p*q, and C is a quadratic residue of T coprime
to
T, then:
x^2 = C modulo T
has 4 distinct solutions in [1, T). If r1 and r2 are two of them picked
at
random (note that r1 = r2 is possible), then 1 < gcd(r1+r2, T) < T with
probability 1/2.
If r is one of them, then the 4 solutions are in fact:
+/- r
+/- u*r mod T
where:
u = (p^(q-1) - q^(p-1)) mod T
is a "non-obvious" square root of 1 modulo T (the "obvious" square roots
of
1 modulo T are +1 and -1). Do you see why u computed that way _is_ a
square
root of 1 modulo T? Pause to prove it (or to learn enough to prove it),
or
remain forever lost ;-)
Pedantic.
What I figured out is that I could just pick a residue modulo T for x.
No problems there as given
x^2 - y^2 = 0 mod T
there are non-trivial factorizations given for ANY residue of x, as you
have explicitly
x^2 - y^2 = k*T
where k can be any non-zero integer.
So you can just pick a residue and then find x and y as I've shown, and
importantly the modular inverse makes an appearance.
[/quote:52ed86d064]
trivial
this is already known as a dead end.
Seems like you are stuck.
you will need to actually factor a number to ever be believeable.
we are waiting.................................... |
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| a.khanm@gmail.com |
| Posted: Sun Jul 16, 2006 8:52 am |
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Guest
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[quote:8f020ca978]I've been looking for federal people all day, expecting the FBI or even
maybe someone from the NSA to show up, but nothing...
[/quote:8f020ca978]
The only one to ever knock at your door would be the "The Department
for Mentally Disabled".
jstevh@msn.com wrote:
[quote:8f020ca978]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.
It's a quadratic residue solution, which may have been helped along by
Tim Peters noting when talking about a previous failed attempt that if
you could get multiple quadratic residues that were equal you could
factor.
He claimed it was a hard thing to do.
At the time I thought to myself that I didn't think it was, but I
didn't know how to do it.
That though may have lead my mind down the paths that eventually caused
the solution to just kind of pop out fully formed.
So one way to look at what I found is that I came up with a way to get
a lot of equal quadratic residues.
The proof is trivial.
The impact is not.
So why post it?
I've talked about that before, and maybe at this point the real answer
is that I've been wrong so many times that part of me still doesn't
believe it here, even though I think I have the proof as I've thought
that before and been wrong.
So I don't want to invest myself in the damn thing beyond the usual
ways--posts!!!
If it's right, well, well, I think it is right, but this situation is
just so wacky anyway.
I've been looking for federal people all day, expecting the FBI or even
maybe someone from the NSA to show up, but nothing...
But it's such a natural, simple solution that in retrospect is obvious.
It's also kind of clear why most people wouldn't go down those paths,
as I am a lateral thinker, trained in it very well.
I did something that I think most would consider unnatural, and that's
what it took.
So the short answer is that the question of how to get a bunch of
quadratic residues of equal value is done. That answer leads to a
factoring method. That factoring method should be a solution to the
factoring problem.
And it's very short, very straightforward, and so it should be
unbelievably fast.
Now what? I don't know. I kind of wish there had been another way to
go, but the fact that I posted this thing on Usenet yesterday, even
though it was on sci.skeptic, and nothing has happened up until today
shows how mucked up the math community is.
You people have to be banged on the head and forced to look at even
simple proofs of famous problems.
It's weird. It's like, you can't just look and get it. Someoe has to
push it on you, and hammer, hammer hammer to get you to pay attention
till MAYBE you'll think that possibly there's something there.
Which to me sounds like you're not really the mathematicians some of
you seem to think you are.
While I find these things in the wild.
James Harris[/quote:8f020ca978] |
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| Proginoskes |
| Posted: Sun Jul 16, 2006 2:16 pm |
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Guest
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a.khanm@gmail.com wrote:
[quote:bfed69bcf1]I've been looking for federal people all day, expecting the FBI or even
maybe someone from the NSA to show up, but nothing...
The only one to ever knock at your door would be the "The Department
for Mentally Disabled".
[/quote:bfed69bcf1]
Actually, THEY may be using James Harris to spread disinformation about
factoring techniques. The hope would be that the terrorists/bad guys
spend a lot of time on what is known to be a dead end.
--- Christopher Heckman
[quote:bfed69bcf1]jstevh@msn.com wrote:
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.
It's a quadratic residue solution, which may have been helped along by
Tim Peters noting when talking about a previous failed attempt that if
you could get multiple quadratic residues that were equal you could
factor.
He claimed it was a hard thing to do.
At the time I thought to myself that I didn't think it was, but I
didn't know how to do it.
That though may have lead my mind down the paths that eventually caused
the solution to just kind of pop out fully formed.
So one way to look at what I found is that I came up with a way to get
a lot of equal quadratic residues.
The proof is trivial.
The impact is not.
So why post it?
I've talked about that before, and maybe at this point the real answer
is that I've been wrong so many times that part of me still doesn't
believe it here, even though I think I have the proof as I've thought
that before and been wrong.
So I don't want to invest myself in the damn thing beyond the usual
ways--posts!!!
If it's right, well, well, I think it is right, but this situation is
just so wacky anyway.
I've been looking for federal people all day, expecting the FBI or even
maybe someone from the NSA to show up, but nothing...
But it's such a natural, simple solution that in retrospect is obvious.
It's also kind of clear why most people wouldn't go down those paths,
as I am a lateral thinker, trained in it very well.
I did something that I think most would consider unnatural, and that's
what it took.
So the short answer is that the question of how to get a bunch of
quadratic residues of equal value is done. That answer leads to a
factoring method. That factoring method should be a solution to the
factoring problem.
And it's very short, very straightforward, and so it should be
unbelievably fast.
Now what? I don't know. I kind of wish there had been another way to
go, but the fact that I posted this thing on Usenet yesterday, even
though it was on sci.skeptic, and nothing has happened up until today
shows how mucked up the math community is.
You people have to be banged on the head and forced to look at even
simple proofs of famous problems.
It's weird. It's like, you can't just look and get it. Someoe has to
push it on you, and hammer, hammer hammer to get you to pay attention
till MAYBE you'll think that possibly there's something there.
Which to me sounds like you're not really the mathematicians some of
you seem to think you are.
While I find these things in the wild.
James Harris[/quote:bfed69bcf1] |
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| Guest |
| Posted: Sun Jul 16, 2006 2:22 pm |
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I don't think *James* actually solved the factorisation problem but
maybe _someone_ did.
Proginoskes skrev:
[quote:00effc2e1f]a.khanm@gmail.com wrote:
I've been looking for federal people all day, expecting the FBI or even
maybe someone from the NSA to show up, but nothing...
The only one to ever knock at your door would be the "The Department
for Mentally Disabled".
Actually, THEY may be using James Harris to spread disinformation about
factoring techniques. The hope would be that the terrorists/bad guys
spend a lot of time on what is known to be a dead end.
--- Christopher Heckman
jstevh@msn.com wrote:
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.
It's a quadratic residue solution, which may have been helped along by
Tim Peters noting when talking about a previous failed attempt that if
you could get multiple quadratic residues that were equal you could
factor.
He claimed it was a hard thing to do.
At the time I thought to myself that I didn't think it was, but I
didn't know how to do it.
That though may have lead my mind down the paths that eventually caused
the solution to just kind of pop out fully formed.
So one way to look at what I found is that I came up with a way to get
a lot of equal quadratic residues.
The proof is trivial.
The impact is not.
So why post it?
I've talked about that before, and maybe at this point the real answer
is that I've been wrong so many times that part of me still doesn't
believe it here, even though I think I have the proof as I've thought
that before and been wrong.
So I don't want to invest myself in the damn thing beyond the usual
ways--posts!!!
If it's right, well, well, I think it is right, but this situation is
just so wacky anyway.
I've been looking for federal people all day, expecting the FBI or even
maybe someone from the NSA to show up, but nothing...
But it's such a natural, simple solution that in retrospect is obvious.
It's also kind of clear why most people wouldn't go down those paths,
as I am a lateral thinker, trained in it very well.
I did something that I think most would consider unnatural, and that's
what it took.
So the short answer is that the question of how to get a bunch of
quadratic residues of equal value is done. That answer leads to a
factoring method. That factoring method should be a solution to the
factoring problem.
And it's very short, very straightforward, and so it should be
unbelievably fast.
Now what? I don't know. I kind of wish there had been another way to
go, but the fact that I posted this thing on Usenet yesterday, even
though it was on sci.skeptic, and nothing has happened up until today
shows how mucked up the math community is.
You people have to be banged on the head and forced to look at even
simple proofs of famous problems.
It's weird. It's like, you can't just look and get it. Someoe has to
push it on you, and hammer, hammer hammer to get you to pay attention
till MAYBE you'll think that possibly there's something there.
Which to me sounds like you're not really the mathematicians some of
you seem to think you are.
While I find these things in the wild.
James Harris[/quote:00effc2e1f] |
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| Tim Peters |
| Posted: Sun Jul 16, 2006 4:22 pm |
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Guest
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[rossum]
[quote:f20d39906a]I seem to remember that he did factor 15 in one of his postings. Does
that count?
[/quote:f20d39906a]
Partially. He _started_ to factor 35 with the current method but didn't
finish it. I finished it for him, using his choices, and showed that they
didn't work. In a different post, I factored 35 using his method (by using
different choices). I don't believe any of that was posted to sci.crypt or
sci.math. |
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