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From: Enrico on 8 Nov 2009 12:08
On Nov 7, 10:55 pm, JSH <jst...(a)gmail.com> wrote:
> On Nov 7, 9:11 pm, Enrico <ungerne...(a)aol.com> wrote:
>
> > On Nov 7, 9:19 pm, JSH <jst...(a)gmail.com> wrote:
>
> <deleted>
>
>
>
>
>
> > > > > Mathematically that runs into the reality that if a trivial
> > > > > factorization will work, the very first one you check will work.
>
> > > > This is not true.
>
> > > Yes, it is.  if 2q - f_1^2 is NOT a quadratic residue modulo N, then
> > > you won't get the correct answer for k.
>
> > > > A trivial factorization will always work on any odd composite N
>
> > > No.  It will not.
>
> > > My guess is you're confused on the details.
>
> > > If you're looking for q, such that q is a quadratic residue modulo N,
> > > then you find f_1 and f_2 such that f_1*f_2 = 2q + jN, where j is a
> > > nonzero integer.
>
> > > With k^2 = q mod N, you will get k, from k = 3^{-1}(f_1 + f_2) iff 2q
> > > - f_1^2 is a quadratic residue modulo N.
>
> > > If you look only at f_1*f_2 where you have a trivial factorization,
> > > then f_1 will not change, so if it doesn't work once, it will NEVER
> > > work.  (f_1 and f_2 are interchangeable)
>
> > > > coprime to 3.  *** eventually ***, NEVER on the first one I check..
>
> > > I have given an example where the trivial factorization DID work on
> > > the first one I checked.
>
> > > > How do _you_ select "the very first one you check" ?
>
> > > I use 2q + N.  j=1.
>
> > > I GAVE an example.
>
> > > > I start at x=1 and increment by 1 for each iteration.
> > > > I find R = X^2 mod N
> > > > 2 * R + N factors as 1 and 2 * R + N, so
> > > > f1 + f2 = 1 + 2 * R + N
> > > > Multiply by 3^-1 mod N
> > > > (f1 + f2) * [3^-1 mod N]
> > > > Y = ((f1 + f2) * [3^-1 mod N]) mod N
> > > > then check Y^2 mod N to see if it equals X ^2 mod N
>
> > > > If they are equal, then the factors of N will split between
> > > > (X+Y) mod N and (X-Y) mod N
>
> > > >                                           Enrico
>
> > > I've explained the mathematics to you.  It is a mathematical absolute--
> > > unless I'm wrong--that 2q - f_1^2 MUST be a quadratic residue modulo N
> > > for you to get the correct, such that k^2 = q mod N.
>
> > > If that's wrong then my research is in serious trouble!  If a
> > > counterexample exists it should be easy for you to show one.
>
> > > Oh, sorry but I have to add that I have seen cases where when faced
> > > with absolute mathematics, people just kind of, make things up.  It's
> > > another way you can have mathematical proof and not get acceptance as
> > > people claim problems that do not exist.
>
> > > It is ANOTHER reason I went to the factoring problem, as with my other
> > > research posters clearly understood that making things up could work
> > > to prevent widespread knowledge of and acceptance of the research!
>
> > > But it won't here.  Yes, posters can deny the mathematics if they
> > > wish, but it will not change anything.
>
> > > It is your choice.  If you all as a group wish to leave this
> > > mathematics freely floating out there on the hopes no one will notice
> > > then you are allowed to so act.
>
> > > You've done so for a year.  Maybe you wish another?
>
> > > Remember your brain is what controls you.  Reality is so much about
> > > perception.  You will tell yourselves whatever you need to hear, but
> > > it does not mean that reality will not hold you accountable!
>
> > > Fantasy cannot save you.
>
> > > You are not royalty.  You never were.  You are not inherently better
> > > than anyone else.
>
> > > You are not noble blood.  Let go of the fantasy!!!  Let go.
>
> > > James Harris- Hide quoted text -
>
> > > - Show quoted text -
>
> > ================================================
>
> > >> How do _you_ select "the very first one you check" ?
> > >I use 2q + N.  j=1.
>
> > Terrific.
>
> > How do you select the first q without using
> > the factors of N ?
>
> That's trivial.  Just pick a quadratic residue.
>
> I like to use the first non-perfect square quadratic residue, but
> check to be sure that it doesn't share prime factors with N.
>
> Like with my example I used 8^2 = 29 mod 35 because while 7^2 = 14 mod
> 35, is a non-perfect square 14 is not coprime to 35.
>
> > (What you call q is what I call X^2 mod N)
>
> Yeah the quadratic residue.  I got that part.
>
> > Remember - I've got this whole thing on a speadsheet.
>
> Ok.
>
> > Conditional color coding to make the solutions
>
> Sounds really funky, and the significance escapes me.
>
> > really obvious. I can type in any N and see the results
> > instantly. Most importantly - it duplicates your results
> > on the few examples you provided.
>
> Including k=9, with N=35, gives q = 11, as 81 = 11 mod 35?  But the
> trivial factorization didn't work, so I used the non-trivial
> factorization.
>
> From what I saw you're only using the trivial factorization, so how
> would you replicate an example using a non-trivial one?
>
> > Experimental results trump theory unless the experiment
>
> But not mathematical proof.
>
> The concept of mathematical proof seems to be escaping you.
>
> It's not like just having a theory.  A mathematical proof trumps all
> else.
>
> If an argument fails it was not a proof!!!
>
> > can be shown to be not a test of the theory. You'll want
> > that second option, which is why I gave you my algorithm.
>
> >                                                     Enrico
>
> I like getting notice of errors.  If I yell "mathematical proof" and
> you can show a counterexample then I don't have a proof.  And in
> finding out why, I learn something.
>
> Here it's an easy process and you keep doing weird stuff, like only
> doing trivial factorizations.
>
> My take on what you think the argument is, is that you will keep
> shifting q, the quadratic residue, and you're trying to say you always
> can get to one that will work, with the trivial factorization.
>
> So, you're trying to test this with only the trivial factorization,
> which should only work with diminishing probability as the number of
> quadratic residues rises, which would confirm a deep internal need you
> may have for this to work less the larger N is.
>
> So you gamed it.  You put in your own arbitrary thing--not what I said
> to do--which I can explain easily will give diminishing returns as N
> increases in size.
>
> Do you understand why?  It's trivial to explain.
>
> James Harris- Hide quoted text -
>
> - Show quoted text -


===========================================================
> > How do you select the first q without using
> > the factors of N ?
>
> That's trivial. Just pick a quadratic residue.
>

Oops. I've been doing that all along with X^2 mod n.
For some reason I had the idea that you were taking
a "random" number, testing it somehow to see if it's
a quadratic residue, then using it.



>
> Including k=9, with N=35, gives q = 11, as 81 = 11 mod 35? But the
> trivial factorization didn't work, so I used the non-trivial
> factorization.
>
> From what I saw you're only using the trivial factorization, so how
> would you replicate an example using a non-trivial one?

I put in a division check function when I was starting out so I could
make
sure I was getting the same numbers as you did. Putting in 3 as
the divisor gave me your numbers with k = 9, so I could be reasonably
sure my programming duplicated your method.

If I want to use the trivial factorization, I just type in 1 as the
divisor.
Crude and silly, dividing by 1, but simplest to program on the
spreadsheet, which is just a research tool to test out different
ideas.



>
> My take on what you think the argument is, is that you will keep
> shifting q, the quadratic residue, and you're trying to say you always
> can get to one that will work, with the trivial factorization.
>

This is correct, but turns out to be a very poor strategy as the
following example shows.

N = 187
q = 1, 2, 3, ...

Using trivial factorization, the first solution is at q = 43 giving
43 ^ 2 mod 187 = 166
111^2 mod 187 = 166

Using the strategy of trying all the factorizations gets
the first solution at q = 2 using the factorization of 195.

195 = 3 * 65 fails
195 = 5 * 39 fails

195 = 15*13 works, giving

2 ^ 2 mod 187 = 4
134 ^ 2 mod 187 = 4

and the GCD's produce both facors 11 and 17


Why did I keep on with the trivial factorization before?
I was hoping for an easy swindle on the math.
Didn't work, but was worth a try.


Enrico




From: JSH on 8 Nov 2009 13:12
On Nov 8, 9:08 am, Enrico <ungerne...(a)aol.com> wrote:
> On Nov 7, 10:55 pm, JSH <jst...(a)gmail.com> wrote:
>
>
>
>
>
> > On Nov 7, 9:11 pm, Enrico <ungerne...(a)aol.com> wrote:
>
> > > On Nov 7, 9:19 pm, JSH <jst...(a)gmail.com> wrote:
>
> > <deleted>
>
> > > > > > Mathematically that runs into the reality that if a trivial
> > > > > > factorization will work, the very first one you check will work..
>
> > > > > This is not true.
>
> > > > Yes, it is.  if 2q - f_1^2 is NOT a quadratic residue modulo N, then
> > > > you won't get the correct answer for k.
>
> > > > > A trivial factorization will always work on any odd composite N
>
> > > > No.  It will not.
>
> > > > My guess is you're confused on the details.
>
> > > > If you're looking for q, such that q is a quadratic residue modulo N,
> > > > then you find f_1 and f_2 such that f_1*f_2 = 2q + jN, where j is a
> > > > nonzero integer.
>
> > > > With k^2 = q mod N, you will get k, from k = 3^{-1}(f_1 + f_2) iff 2q
> > > > - f_1^2 is a quadratic residue modulo N.
>
> > > > If you look only at f_1*f_2 where you have a trivial factorization,
> > > > then f_1 will not change, so if it doesn't work once, it will NEVER
> > > > work.  (f_1 and f_2 are interchangeable)
>
> > > > > coprime to 3.  *** eventually ***, NEVER on the first one I check.
>
> > > > I have given an example where the trivial factorization DID work on
> > > > the first one I checked.
>
> > > > > How do _you_ select "the very first one you check" ?
>
> > > > I use 2q + N.  j=1.
>
> > > > I GAVE an example.
>
> > > > > I start at x=1 and increment by 1 for each iteration.
> > > > > I find R = X^2 mod N
> > > > > 2 * R + N factors as 1 and 2 * R + N, so
> > > > > f1 + f2 = 1 + 2 * R + N
> > > > > Multiply by 3^-1 mod N
> > > > > (f1 + f2) * [3^-1 mod N]
> > > > > Y = ((f1 + f2) * [3^-1 mod N]) mod N
> > > > > then check Y^2 mod N to see if it equals X ^2 mod N
>
> > > > > If they are equal, then the factors of N will split between
> > > > > (X+Y) mod N and (X-Y) mod N
>
> > > > >                                           Enrico
>
> > > > I've explained the mathematics to you.  It is a mathematical absolute--
> > > > unless I'm wrong--that 2q - f_1^2 MUST be a quadratic residue modulo N
> > > > for you to get the correct, such that k^2 = q mod N.
>
> > > > If that's wrong then my research is in serious trouble!  If a
> > > > counterexample exists it should be easy for you to show one.
>
> > > > Oh, sorry but I have to add that I have seen cases where when faced
> > > > with absolute mathematics, people just kind of, make things up.  It's
> > > > another way you can have mathematical proof and not get acceptance as
> > > > people claim problems that do not exist.
>
> > > > It is ANOTHER reason I went to the factoring problem, as with my other
> > > > research posters clearly understood that making things up could work
> > > > to prevent widespread knowledge of and acceptance of the research!
>
> > > > But it won't here.  Yes, posters can deny the mathematics if they
> > > > wish, but it will not change anything.
>
> > > > It is your choice.  If you all as a group wish to leave this
> > > > mathematics freely floating out there on the hopes no one will notice
> > > > then you are allowed to so act.
>
> > > > You've done so for a year.  Maybe you wish another?
>
> > > > Remember your brain is what controls you.  Reality is so much about
> > > > perception.  You will tell yourselves whatever you need to hear, but
> > > > it does not mean that reality will not hold you accountable!
>
> > > > Fantasy cannot save you.
>
> > > > You are not royalty.  You never were.  You are not inherently better
> > > > than anyone else.
>
> > > > You are not noble blood.  Let go of the fantasy!!!  Let go.
>
> > > > James Harris- Hide quoted text -
>
> > > > - Show quoted text -
>
> > > ================================================
>
> > > >> How do _you_ select "the very first one you check" ?
> > > >I use 2q + N.  j=1.
>
> > > Terrific.
>
> > > How do you select the first q without using
> > > the factors of N ?
>
> > That's trivial.  Just pick a quadratic residue.
>
> > I like to use the first non-perfect square quadratic residue, but
> > check to be sure that it doesn't share prime factors with N.
>
> > Like with my example I used 8^2 = 29 mod 35 because while 7^2 = 14 mod
> > 35, is a non-perfect square 14 is not coprime to 35.
>
> > > (What you call q is what I call X^2 mod N)
>
> > Yeah the quadratic residue.  I got that part.
>
> > > Remember - I've got this whole thing on a speadsheet.
>
> > Ok.
>
> > > Conditional color coding to make the solutions
>
> > Sounds really funky, and the significance escapes me.
>
> > > really obvious. I can type in any N and see the results
> > > instantly. Most importantly - it duplicates your results
> > > on the few examples you provided.
>
> > Including k=9, with N=35, gives q = 11, as 81 = 11 mod 35?  But the
> > trivial factorization didn't work, so I used the non-trivial
> > factorization.
>
> > From what I saw you're only using the trivial factorization, so how
> > would you replicate an example using a non-trivial one?
>
> > > Experimental results trump theory unless the experiment
>
> > But not mathematical proof.
>
> > The concept of mathematical proof seems to be escaping you.
>
> > It's not like just having a theory.  A mathematical proof trumps all
> > else.
>
> > If an argument fails it was not a proof!!!
>
> > > can be shown to be not a test of the theory. You'll want
> > > that second option, which is why I gave you my algorithm.
>
> > >                                                     Enrico
>
> > I like getting notice of errors.  If I yell "mathematical proof" and
> > you can show a counterexample then I don't have a proof.  And in
> > finding out why, I learn something.
>
> > Here it's an easy process and you keep doing weird stuff, like only
> > doing trivial factorizations.
>
> > My take on what you think the argument is, is that you will keep
> > shifting q, the quadratic residue, and you're trying to say you always
> > can get to one that will work, with the trivial factorization.
>
> > So, you're trying to test this with only the trivial factorization,
> > which should only work with diminishing probability as the number of
> > quadratic residues rises, which would confirm a deep internal need you
> > may have for this to work less the larger N is.
>
> > So you gamed it.  You put in your own arbitrary thing--not what I said
> > to do--which I can explain easily will give diminishing returns as N
> > increases in size.
>
> > Do you understand why?  It's trivial to explain.
>
> > James Harris- Hide quoted text -
>
> > - Show quoted text -
>
> ===========================================================
>
> > > How do you select the first q without using
> > > the factors of N ?
>
> > That's trivial.  Just pick a quadratic residue.
>
> Oops. I've been doing that all along with X^2 mod n.
> For some reason I had the idea that you were taking
> a "random" number, testing it somehow to see if it's
> a quadratic residue, then using it.
>
>
>
> > Including k=9, with N=35, gives q = 11, as 81 = 11 mod 35?  But the
> > trivial factorization didn't work, so I used the non-trivial
> > factorization.
>
> > From what I saw you're only using the trivial factorization, so how
> > would you replicate an example using a non-trivial one?
>
> I put in a division check function when I was starting out so I could
> make
> sure I was getting the same numbers as you did. Putting in 3 as
> the divisor gave me your numbers with k = 9, so I could be reasonably
> sure my programming duplicated your method.
>
> If I want to use the trivial factorization, I just type in 1 as the
> divisor.
> Crude and silly, dividing by 1, but simplest to program on the
> spreadsheet, which is just a research tool to test out different
> ideas.
>
>
>
> > My take on what you think the argument is, is that you will keep
> > shifting q, the quadratic residue, and you're trying to say you always
> > can get to one that will work, with the trivial factorization.
>
> This is correct, but turns out to be a very poor strategy as the
> following example shows.
>
> N = 187
> q = 1, 2, 3, ...
>
> Using trivial factorization, the first solution is at q = 43 giving
> 43 ^ 2 mod 187 = 166
> 111^2 mod 187 = 166
>
> Using the strategy of trying all the factorizations gets
> the first solution at q = 2 using the factorization of 195.
>
> 195 = 3 * 65 fails
> 195 = 5 * 39 fails
>
> 195 = 15*13 works, giving
>
> 2 ^ 2 mod 187 = 4
> 134 ^ 2 mod 187 = 4
>
> and the GCD's produce both facors 11 and 17
>
> Why did I keep on with the trivial factorization before?
> I was hoping for an easy swindle on the math.

If all residues are equally represented then the more residues you
have the less likely you are to get a particular residue, so that
approach would drop in success rate as N increases in size--getting
more residues.

> Didn't work, but was worth a try.
>
>                          Enrico

I don't see a problem with experimentation.

But I get really suspicious with an approach where I've explained why
it can't work well when it's thrown back at me, especially when I know
why that particular approach could play into a hidden bias against the
idea by leading to a drop in success with the increase in size of N.

Now being wrong is one thing, having people twist the data to their
ends is another, especially with the factoring problem.

And it's such a NEAT result!!! Who knew! This fundamental yet simple
relation connecting quadratic residues and factoring, and I'm just
using the most simple one.

There are an infinity of these relations that result from surrogate
factoring where I've just picked the first, which is probably the best
one anyway.

Dragging of the feet on this result proves my point that certain
people see themselves as part of a special society in a class war.
They don't care about the truth--they only care about "truths" that
support their class.

They are wannabe royalty living in their own version of a medieval
world with themselves as the nobility, pretending to be something
else.

Real researchers would be excited with the result, worried about the
consequences, and anxious to see how much more they could learn.

Those of you who are the fakes instead are wishing this thing would go
away, or that hopefully it's not true.

And then you are not real researchers. You simply got away with
playing pretend.

And you are parasites on the world.


James Harris
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