JSH: One more explanation, factoring solution
JSH
their failure to disprove anything about my solution to the factoring
problem it's worth explaining again.
What I did was exploit a rational connection.
With rational solutions to
x^2 - Dy^2 = 1
I noticed you have rationals solutions to
(D-1)j^2 + (j+/-1)^2 = (x+y)^2
where j = ((x+Dy) -/+1)/D.
And that's it. 99% of the mathematics is right there in front of you,
and covered with that initial statement. So how do I solve for x and
y directly?
Well, I have TWO FACTORIZATIONS available:
(x-1)(x+1) = Dy^2
and
(D-1)j^2 = (x+y - (j+/-1))(x+y + (j+/-1))
So what I do is generally factor the second and I also found I needed
to split up j, so I add variables: u and v
(x+y - (j+/-1)) = f_1*u
(x+y + (j+/-1)) = f_2*u*v^2
where f_1*f_2 = D-1.
And that's how v comes into the picture. Now recap: for EVERY
rational solution to x^2 - Dy^2 = 1, you have a rational solution to:
(D-1)j^2 + (j+/-1)^2 = (x+y)^2
I note that if x = r/t and y = s/t, I have (r-t)(r+t) = Ds^2, and can
consider a solution that factors D if it is an odd composite and g_1
and g_2 are non-trivial factors with:
r-t = g_1 and r+t = g_1, as then r= (g_1 + g_2)/2 and t = (g_1 - g_2)/
2, and s = 1 or -1.
So rational solutions to r, s and t EXIST at a point that will factor
D non-trivially.
One set of posters has repeatedly claimed they do not. With at least
one claiming to have disproven that using the quadratic formula.
Now to guarantee non-trivial factorization of D, it suffices with non-
zero r, s and t, for
abs(r-t) < D and abs(r+t) < D
and you'll notice I already showed at least one example of that case
which must exist!
One set of posters have routinely claimed that both conditions cannot
be simultaneously true.
Now I've noted that now you have a calculus problem of minimizing to
find r, s and t as functions of v, such that you meet those
conditions, and I've given ONE possible answer while to practically
factor it may never be the case that you even need the s=1 or -1
case. But the proof of its existence shows that rational v is
available in the desired range.
Now here are the explicit solutions for x and y:
y = [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 -
f_2*v^2 - 2v)]/(D-1)
and
x = +/-(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v) - [+/-2Dv/(f_1 - f_2*v^2
-
2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)
where again f_1*f_2 = D-1, and the f's are non-zero integer factors,
while v is a free variable.
To recap: what I did was exploit a connection between the
factorization of D and the factorization of D-1.
Remarkably they are connected through Pell's Equation and an equation
I derived from Pell's Equation using my Quadratic Diophantine Theorem
(Google it).
Now for years I've claimed that math society has been ignoring major
proofs of mine to hold on to the status quo as my research upsets HUGE
swaths of established number theory, and now you have clear and
irrefutable evidence in front of you of how far they will go in that
denial.
Pell's Equation is one of the most famous in mathematical history.
The factoring problem is being used to supposedly secure the Internet.
With every security breach you read in the news, consider the
possibility that factoring has been used, and that practitioners in
various specialties are lying about it being broken, just like posters
on these newsgroups lie about the efficacy of the equations above.
They do so to preserve their BELIEFS about the world in a way that
makes them most comfortable without realizing the consequences of
their belief system can be catastrophic. They are--religious about
mathematics.
Factoring always had an easy answer: connect factoring one number to
factoring another.
People just got it wrong for a while and now the truth is out. People
make mistakes. That's not news. But please don't make the bigger
mistake of continued denial to try and hold on to math ideas that just
do not work.
Mathematics is a heartless discipline.
I know many of you have invested huge amounts of time and energy and
years of your lives to learn mathematical ideas that if you're honest
you'll have to realize are wrong.
But holding on to them will never make them right.
James Harris
marcus_...@yahoo.com
You just keep persisting with this transparently stupid
idea. In general you CANNOT find one value of v which
minimizes ANY TWO of f(v), g(v), (f(v) - g(v)) or
f(v) + g(v), even if all these functions are quadratics. It
is trivial to find counterexamples. You ***cannot*** do this
problem by a simple minimization of functions using calculus.
It is painfully obvious to everyone here that you have no proof.
If you actually tried your method on a real nontrivial example
you would see instantly what is wrong.
> But the proof of its existence shows that rational v is
> available in the desired range.
>
No, you have no such proof.
> Now here are the explicit solutions for x and y:
>
> y = [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 -
> f_2*v^2 - 2v)]/(D-1)
>
> and
>
> x = +/-(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v) - [+/-2Dv/(f_1 - f_2*v^2
> -
> 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)
>
> where again f_1*f_2 = D-1, and the f's are non-zero integer factors,
> while v is a free variable.
>
I note that you do not bother to go the next step
and actually specify r(v) and t(v), nor do you go the
next step after that to find minima of the functions
involved. Which means you have not completed any proof.
If you tried it you would find out why it fails.
Marcus.
Joshua Cranmer
> I note that if x = r/t and y = s/t, I have (r-t)(r+t) = Ds^2, and can
> consider a solution that factors D if it is an odd composite and g_1
> and g_2 are non-trivial factors with:
>
> r-t = g_1 and r+t = g_1, as then r= (g_1 + g_2)/2 and t = (g_1 - g_2)/
> 2, and s = 1 or -1.
To factor D, you're looking for integers g_1 and g_2 which would be
non-trivial factors. t is, naturally, rational, although since g_1 and
g_2 are integers, 2t would have to be integer. Since s = ą1, y has to be
of the form ą 1 / k or ą 2 / k, where k is some integer.
Now:
> y = [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 -
> f_2*v^2 - 2v)]/(D-1)
Does that look like ą 1 / k or ą 2 / k to you? It sure doesn't to me.
--
Beware of bugs in the above code; I have only proved it correct, not
tried it. -- Donald E. Knuth
dogma
"JSH" <jst...@gmail.com> wrote in message
news:9239514f-908c-4893...@e18g2000yqo.googlegroups.com...
> With posters yet again chortling victory against my research despite
> their failure to disprove anything about my solution to the factoring
> problem it's worth explaining again.
>
<snip math>
>Now for years I've claimed that math society has been ignoring major
> proofs of mine to hold on to the status quo as my research upsets HUGE
> swaths of established number theory, and now you have clear and
> irrefutable evidence in front of you of how far they will go in that
> denial.
wrong.
>
> Pell's Equation is one of the most famous in mathematical history.
wrong.
>
> The factoring problem is being used to supposedly secure the Internet.
wrong.
>
> With every security breach you read in the news, consider the
> possibility that factoring has been used,
but it is not, DA.
>and that practitioners in
> various specialties are lying about it being broken, just like posters
> on these newsgroups lie about the efficacy of the equations above.
>
> They do so to preserve their BELIEFS about the world in a way that
> makes them most comfortable without realizing the consequences of
> their belief system can be catastrophic. They are--religious about
> mathematics.
wrong
>
> Factoring always had an easy answer: connect factoring one number to
> factoring another.
not for you,
>
> People just got it wrong for a while and now the truth is out. People
> make mistakes. That's not news. But please don't make the bigger
> mistake of continued denial to try and hold on to math ideas that just
> do not work.
JSH ALWAYS makes mistakes, mistake in every post!
>
> Mathematics is a heartless discipline.
wrong, (perhaps for you)
>
> I know many of you have invested huge amounts of time and energy and
> years of your lives to learn mathematical ideas that if you're honest
> you'll have to realize are wrong.
We await your admission you that you are wrong again.
>
> But holding on to them will never make them right.
release the past, James, ask the libarary to deliver you a book on beginning
algebra to thumb through.
>
>
> James Harris
Quack, Quack, Quack said the __________
jon
"JSH" <jst...@gmail.com> wrote in message
news:9239514f-908c-4893...@e18g2000yqo.googlegroups.com...
> their failure to disprove anything about my solution to the factoring
> problem it's worth explaining again.
>
> What I did was exploit a rational connection.
>
> With rational solutions to
>
> x^2 - Dy^2 = 1
>
> I noticed you have rationals solutions to
>
> (D-1)j^2 + (j+/-1)^2 = (x+y)^2
WHAT IS THAT???
do you really mean both of these? ;
(D-1)j^2 + (j -1)^2 = (x+y)^2
(D-1)j^2 + (j+1)^2 = (x+y)^2
Or did you "up-see" again?
Show they are rational(s) solutions first!!!
>
> where j = ((x+Dy) -/+1)/D.
WTF is that???
Do you mean;
j = ((x+Dy) - 1 )/D.
j = ((x+Dy) +1)/D.
???
Why are you so lazy???, you cant even write out the equations????
>
> And that's it. 99% of the mathematics is right there in front of you,
> and covered with that initial statement. So how do I solve for x and
> y directly?
99% WTF is that ??? anothe WILD ASS GUESS ???
<snip rest of incompetent garbage>
TRY AGAIN => IN PROPER FORM
rde...@hamilton.edu
With posters yet again chortling victory against my research despite
their failure to disprove anything about my solution to the factoring
problem it's worth explaining again.
What I did was exploit a rational connection.
With rational solutions to
x^2 - Dy^2 = 1
I noticed you have rationals solutions to
(D-1)j^2 + (j+/-1)^2 = (x+y)^2
where j = ((x+Dy) -/+1)/D.
And that's it. 99% of the mathematics is right there in front of you,
and covered with that initial statement. So how do I solve for x and
y directly?
Well, I have
<slight snippage here>
(D-1)j^2 = (x+y - (j+/-1))(x+y + (j+/-1))
So what I do is generally factor the second and I also found I needed
to split up j, so I add variables: u and v
(x+y - (j+/-1)) = f_1*u
(x+y + (j+/-1)) = f_2*u*v^2
where f_1*f_2 = D-1. <and j = uv>
<Now it's back to me speaking>
Let's carry this a bit further. From the above, we can
solve for x + y:
x + y = (f_2 u v^2 + f_1 u) / 2
= u(f_2 v^2 + f_1) / 2 [1]
We can also solve
j +/- 1 = u v +/- 1 = (f_2 u v^2 - f_1 u) / 2
so
+/- 1 = (f_2 u v^2 - f_1 u - 2 u v) / 2
= u(f_2 v^2 - 2 v - f_1) / 2
so
u = (+/-) 2 / (f_2 v^2 - 2 v - f_1)
Combining this with [1] we have
x + y = (+/-)(f_2 v^2 + f_1) / (f_2 v^2 - 2 v - f_1) [2]
Since you have set j = uv = (x + Dy _/+ 1) / D, we can
rewrite this to get
x + Dy = (+/-)(f_2v^2 + 2(D - 1)v - f_1) / (f_2v^2 - 2v - f_1) [3]
Now to simplify [2] and [3] we let f_1 = (D - 1) / f, f_2 = f
and then write m = f v and we have
x + y = (+/-)(m^2 + (D - 1)) / (m^2 - 2 m - (D - 1)) [4]
x + Dy = (+/-)(m^2 + 2(D - 1) m - (D - 1) / (m^2 - 2 m - (D - 1)) [5]
It's easy enough to solve these for x and y, yielding
x = (+/-)(m^2 - 2m +(D + 1)) / (m^2 - 2 m - (D - 1))
y = (+/-)(2(m - 1) / (m^2 - 2 m - (D - 1))
or, for a final simplification, letting n = m - 1
x = (+/-)(n^2 + D) / (n^2 - D)
y = (+/-) 2n / (n^2 - D)
Let's see where this takes us.
You take x = r / t and y = s / t, so we have (eliminating the +/-)
r(n) = n^2 + D
t(n) = n^2 - D
s(n) = 2n
So
r(n) + t(n) = 2n^2
r(n) - t(n) = 2D
So the second term will never yield a useful factor of D
(well, unless D is even) and the first will only give
a nontrivial factor if you were lucky enough to pick
an n which shares a factor with D. Sorry, James, but
remember, this is -your_ algorithm. All I did was
a bit of elementary algebra.
Regards,
Rick
rossum
wrote:
>With posters yet again chortling victory against my research despite
>their failure to disprove anything about my solution to the factoring
>problem it's worth explaining again.
>
>What I did was exploit a rational connection.
>
>With rational solutions to
>
>x^2 - Dy^2 = 1
>
>I noticed you have rationals solutions to
>
>(D-1)j^2 + (j+/-1)^2 = (x+y)^2
>
>where j = ((x+Dy) -/+1)/D.
>
>And that's it. 99% of the mathematics is right there in front of you,
>and covered with that initial statement. So how do I solve for x and
>y directly?
>
>Well, I have TWO FACTORIZATIONS available:
>
>(x-1)(x+1) = Dy^2
>
>and
>
>(D-1)j^2 = (x+y - (j+/-1))(x+y + (j+/-1))
>
>So what I do is generally factor the second and I also found I needed
>to split up j, so I add variables: u and v
Are u and v integer, rational real or what? You would do better to
specify what they are when you first introduce them. A few paragraphs
down you refer to v as 'rational' so I will assume that both u and v
are rational.
>
>(x+y - (j+/-1)) = f_1*u
>
>(x+y + (j+/-1)) = f_2*u*v^2
>
>where f_1*f_2 = D-1.
>
>And that's how v comes into the picture. Now recap: for EVERY
>rational solution to x^2 - Dy^2 = 1, you have a rational solution to:
>
>(D-1)j^2 + (j+/-1)^2 = (x+y)^2
>
>I note that if x = r/t and y = s/t,
Am I correct to assume that r, s and t are all integers here? You
have said that x and y are rational, and the usual way to express a
rational is to write it using two integers.
>I have (r-t)(r+t) = Ds^2, and can
>consider a solution that factors D if it is an odd composite and g_1
>and g_2 are non-trivial factors with:
>
>r-t = g_1 and r+t = g_1, as then r= (g_1 + g_2)/2 and t = (g_1 - g_2)/
>2, and s = 1 or -1.
>
>So rational solutions to r, s and t EXIST at a point that will factor
>D non-trivially.
If r, s and t are integers then they do not have "rational solutions".
If r, s and t are functions then I cannot see where you have defined
them except in terms of g_1 and g_2 above. Those definitions are of
no use to me if I do not know g_1 and g_2 in advance, in which case I
already know the factors of D. We need definitions of r, s and t
which do not involve knowledge of the g's.
>
>One set of posters has repeatedly claimed they do not. With at least
>one claiming to have disproven that using the quadratic formula.
>
>Now to guarantee non-trivial factorization of D, it suffices with non-
>zero r, s and t, for
Again, are r, s and t integers, functions or what? In some places you
seem to be treating r, s and t as integers while in other places you
seem to be treating them as functions.
>
>abs(r-t) < D and abs(r+t) < D
>
>and you'll notice I already showed at least one example of that case
>which must exist!
>
>One set of posters have routinely claimed that both conditions cannot
>be simultaneously true.
>
>Now I've noted that now you have a calculus problem of minimizing to
>find r, s and t as functions of v,
We now have three functions: r(v), s(v) and t(v). Unfortunately,
without more information it is not possible to minimise, say, r(v)
without having more explicit information about the exact form of r(v).
r'(v) = 0 tells me very little unless I know what r(v) looks like.
How can I determine the sign of r''(v) to see if I am at a maximum,
minimum or point of inflection if I do not have an explicit calculable
formula for r(v)?
You need to give us the explicit formulae for r(v), s(v) and t(v) (not
involving the g's) if we are to be able to progress your current
method beyond this point James.
>such that you meet those
>conditions, and I've given ONE possible answer while to practically
>factor it may never be the case that you even need the s=1 or -1
>case. But the proof of its existence shows that rational v is
>available in the desired range.
I know that non-trivial factors of a composite D exist; that does not
make it easy to find them. You have shown that the v you require
exists. You have not shown us how to find it easily. Minimising
unknown functions is not a practical suggestion, as I am sure you will
agree.
>
>Now here are the explicit solutions for x and y:
>
>y = [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 -
>f_2*v^2 - 2v)]/(D-1)
>
>and
>
>x = +/-(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v) - [+/-2Dv/(f_1 - f_2*v^2
>-
>2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)
>
>where again f_1*f_2 = D-1, and the f's are non-zero integer factors,
>while v is a free variable.
No James, v is not free, it is constrained by the requirements you
have given earlier. Alternatively, are you expecting us to search
through all possible rational v until we find the right one? If the
latter then what you have is worse than trial division - there are
more possible rationals than there are primes below sqrt(D).
rossum
>
>To recap: what I did was exploit a connection between the
>factorization of D and the factorization of D-1.
[snip]
>
>James Harris
Joshua Cranmer
> The factoring problem is being used to supposedly secure the Internet.
>
> With every security breach you read in the news, consider the
> possibility that factoring has been used, and that practitioners in
> various specialties are lying about it being broken, just like posters
> on these newsgroups lie about the efficacy of the equations above.
I am not an expert on cryptographic algorithms, but I believe RSA has
largely fallen out of favor in terms of encryption algorithms.
Certainly, however, it is not being used by people to encrypt their disks.
Indeed, most of the security problems are entirely related to people
failing to encrypt data that should be encrypted, or otherwise giving
people access to data they don't particularly need. Social engineering
is already a more effective means for hacking than breaking
cryptographic schemes...
Tim Smith
<9239514f-908c-4893...@e18g2000yqo.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
>
> I note that if x = r/t and y = s/t, I have (r-t)(r+t) = Ds^2, and can
> consider a solution that factors D if it is an odd composite and g_1
> and g_2 are non-trivial factors with:
>
> r-t = g_1 and r+t = g_1, as then r= (g_1 + g_2)/2 and t = (g_1 - g_2)/
> 2, and s = 1 or -1.
>
> So rational solutions to r, s and t EXIST at a point that will factor
> D non-trivially.
Too bad you haven't been able to show that r, s, and t exist. Numerous
people have now offered suggestions for r, s, and t, and in every case,
your method fails to factor.
--
--Tim Smith
Jim Black
> Now I've noted that now you have a calculus problem of minimizing to
> find r, s and t as functions of v,
which are ...
r(v) = ?
s(v) = ?
t(v) = ?
You only need to tell us one.
--
Jim E. Black (domain in headers)
How to filter out stupid arguments in 40tude Dialog:
!markread,ignore From "Name" +"<email address>"
[X] Watch/Ignore works on subthreads
Enrico
------------------------------------------------------------------
JSH's World of WarCraft Hammer:
===============================
Factor D = 25130787
The factors are the minima of some quadratic
generated by D. Only the absolute values
produced by the quadratic are considered.
Floor[Sqrt(D)] = 5013
Take the next higher value and use it in
(X - 5014)^2
Multiply out:
X^2 - 10028*X + 25140196
Put D back in:
X^2 - 10028*X + 25130787
By trying a couple of values for X,
the trend can be found, leading to
the minima at X = 4917 and 5111
4917 * 5111 = 25130787 = D
Paraphrasing JSH:
"It works every time, and if it doesn't,
you can just shift to the next higher value."
Enrico
Howard
"JSH" <jst...@gmail.com> wrote in message
news:9239514f-908c-4893...@e18g2000yqo.googlegroups.com...
> their failure to disprove anything about my solution to the factoring
> problem it's worth explaining again.
>
> What I did was exploit a rational connection.
you used 13 variables for simplistic algebraic stuff to obscure the fact
that it is wrong.
x, D, y, j, u, v, f-1, f-2, r, s, g_1, g_2, t,
> Mathematics is a heartless discipline.
>
>
>
> James Harris - Inventor of Monkey-Math
Simplify it first.
JSH
wrote:
> In article
> <9239514f-908c-4893-be31-96ee4056f...@e18g2000yqo.googlegroups.com>,
>
> JSH <jst...@gmail.com> wrote:
>
> > I note that if x = r/t and y = s/t, I have (r-t)(r+t) = Ds^2, and can
> > consider a solution that factors D if it is an odd composite and g_1
> > and g_2 are non-trivial factors with:
>
> > r-t = g_1 and r+t = g_1, as then r= (g_1 + g_2)/2 and t = (g_1 - g_2)/
> > 2, and s = 1 or -1.
>
> > So rational solutions to r, s and t EXIST at a point that will factor
> > D non-trivially.
>
> Too bad you haven't been able to show that r, s, and t exist. Numerous
Just did show it. You are replying at the spot where I did. Don't
you find that at all odd?
Recap because it's easy to do so:
With rational solutions to
x^2 - Dy^2 = 1
I noticed you have rationals solutions to
(D-1)j^2 + (j+/-1)^2 = (x+y)^2
where j = ((x+Dy) -/+1)/D.
That is actually enough for the rational among you, as it proves that
the factorization of D is connected to the factorization of D-1, so I
can factor D, by factoring D-1.
> people have now offered suggestions for r, s, and t, and in every case,
> your method fails to factor.
But you are replying to the point where I prove that a rational v MUST
exist such that you non-trivially factor D.
Don't you find the inconsistency in your position odd?
Oh yeah, for people who wonder why I don't just do proofs and show
them to mathematicians, well I do!!!
But then they do weird stuff when I do. As I DO show proofs to
mathematicians.
But how can you deny a proof? Think about it. Imagine you REALLY,
really do not want to accept a proof, but it's a proof! How can you
not?
I've watched some amazing things and have the stories to tell later as
to what some of these people do when faced with that situation. It's
like an immovable object--a mathematical proof--facing a wannabe
inexorable force--a firm belief in one person's mind that I must be
wrong.
Something has to give in that situation, and I can assure you, the
mathematical proof never does.
James Harris
tc...@lsa.umich.edu
JSH <jst...@gmail.com> wrote:
>Something has to give in that situation, and I can assure you, the
>mathematical proof never does.
Actually, I've seen lots of mathematical proofs give, when the error is
discovered.
But I don't think I've ever seen JSH give.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
Tim Smith
> On Feb 24, 10:55 am, Tim Smith <reply_in_gr...@mouse-potato.com>
> wrote:
> > In article
> > <9239514f-908c-4893-be31-96ee4056f...@e18g2000yqo.googlegroups.com>,
> >
> > JSH <jst...@gmail.com> wrote:
> >
> > > I note that if x = r/t and y = s/t, I have (r-t)(r+t) = Ds^2, and can
> > > consider a solution that factors D if it is an odd composite and g_1
> > > and g_2 are non-trivial factors with:
> >
> > > r-t = g_1 and r+t = g_1, as then r= (g_1 + g_2)/2 and t = (g_1 - g_2)/
> > > 2, and s = 1 or -1.
> >
> > > So rational solutions to r, s and t EXIST at a point that will factor
> > > D non-trivially.
> >
> > Too bad you haven't been able to show that r, s, and t exist. Numerous
>
> Just did show it. You are replying at the spot where I did. Don't
> you find that at all odd?
So you are saying that
r(v) = (g1 + g2)/2
s(v) = 1
t(v) = (g1 - g2)/2
where g1 g2 = D?
OK, you are right--with that definition of r, s, and t, you can factor
D. Too bad you have to have already factored D in order to find r, s, t.
Meanwhile, for those of us who want to factor WITHOUT already knowing
non-trivial factors of D, all you've given is this:
r, s, and t are functions such that x = r(v)/t(v) and y = s(v)/t(v)
satisfy x^2 - D y^2 = 1 for rational v, and if you find v that
minimizes something that you won't specify, then GCD( r(v)+t(v), D)
will be a non-trivial factor of D.
Tell us what r, s, and t are.
--
--Tim Smith
marcus_...@yahoo.com
You are saying nothing more than that
a factorization exists.
Was anyone denying that?
> Don't you find the inconsistency in your position odd?
>
Irony, thy name is Harris.
Sure, a nontrivial factorization of a nonprime
number exists.
YOU ARE SAYING NOTHING WORTH SAYING.
This is not an 'existence proof' problem.
This is a problem of finding an efficient way
to factor. Saying a factorization exists is no
progress whatsoever!
> Oh yeah, for people who wonder why I don't just do proofs and show
> them to mathematicians, well I do!!!
>
No you don't. Just as here, you present
half-baked ideas that do not work.
> But then they do weird stuff when I do. As I DO show proofs to
> mathematicians.
>
> But how can you deny a proof? Think about it. Imagine you REALLY,
> really do not want to accept a proof, but it's a proof! How can you
> not?
>
You don't have a proof, unless you are
saying that your sterile assertion that a
factorization must exist is a "proof".
> I've watched some amazing things and have the stories to tell later as
> to what some of these people do when faced with that situation. It's
> like an immovable object--a mathematical proof--facing a wannabe
> inexorable force--a firm belief in one person's mind that I must be
> wrong.
>
You do not, cannot, know this. I have never
seen you present an actual proof. Never.
> Something has to give in that situation, and I can assure you, the
> mathematical proof never does.
>
You have no evidence on that topic.
You know, this episode, which began a few days
ago with your declaring you had the Hammer - it's
run its course. The Hammer has fallen and hit
you on your little toe. You just cannot accept that
what you thought was a grand idea doesn't go
anywhere at all.
Marcus.
> James Harris- Hide quoted text -
>
> - Show quoted text -
JSH
>
>
>
> JSH <jst...@gmail.com> wrote:
> > On Feb 24, 10:55 am, Tim Smith <reply_in_gr...@mouse-potato.com>
> > wrote:
> > > In article
> > > <9239514f-908c-4893-be31-96ee4056f...@e18g2000yqo.googlegroups.com>,
>
> > > JSH <jst...@gmail.com> wrote:
>
> > > > I note that if x = r/t and y = s/t, I have (r-t)(r+t) = Ds^2, and can
> > > > consider a solution that factors D if it is an odd composite and g_1
> > > > and g_2 are non-trivial factors with:
>
> > > > r-t = g_1 and r+t = g_1, as then r= (g_1 + g_2)/2 and t = (g_1 - g_2)/
> > > > 2, and s = 1 or -1.
>
> > > > So rational solutions to r, s and t EXIST at a point that will factor
> > > > D non-trivially.
>
> > > Too bad you haven't been able to show that r, s, and t exist. Numerous
>
> > Just did show it. You are replying at the spot where I did. Don't
> > you find that at all odd?
>
> So you are saying that
>
> r(v) = (g1 + g2)/2
> s(v) = 1
> t(v) = (g1 - g2)/2
>
> where g1 g2 = D?
>
> OK, you are right--with that definition of r, s, and t, you can factor
> D. Too bad you have to have already factored D in order to find r, s, t.
It was an existence proof.
> Meanwhile, for those of us who want to factor WITHOUT already knowing
> non-trivial factors of D, all you've given is this:
>
> r, s, and t are functions such that x = r(v)/t(v) and y = s(v)/t(v)
> satisfy x^2 - D y^2 = 1 for rational v, and if you find v that
> minimizes something that you won't specify, then GCD( r(v)+t(v), D)
> will be a non-trivial factor of D.
Actually, no, there are quite a few other equations and information
also given.
> Tell us what r, s, and t are.
Rather demanding for someone who has just been called out for
remarkable inconsistencies.
But not surprising to me.
The story here is not about proof. I have mathematical proof. My
original post quite adeptly steps through an entire mathematical
argument and covers all the necessary bases.
But the issue here is that human beings have an ability to simply,
say, anything.
Which is what destroys the claim that the mathematical discipline is
only about proof. It is not.
Convincing people is also part of it, and if you have enough people
saying something is not a proof, when it is, or claiming something is
a proof, when it is not, you can do quite a few things.
I'd have no chance if the discussion were not over the factoring
problem, which is what I discovered with proofs in areas not as
important as the factoring problem. Which is a lot about why I came
to the factoring problem!!!
Consider:
1. Pell's Equation, famous in its own right. A historical function
with a lot of history.
2. The factoring problem, supposedly protecting the security of
Internet transactions around the world.
3. A very easy argument mostly with quadratic expressions showing how
to factor by connecting factor a target composite D to the factoring
of another integer D-1.
And that's still not enough to prevent posters--on newsgroups read
around the world--from making bizarrely false posts disputing basic
points of an easy argument, and THEN not backing down when
corrected!!!
The world does not work as often is presented.
But that is about YOUR choices, readers. Your choice to be lied to,
without consequence to the liar.
So they lied to you about money with the financial crisis. Or
politicians lie to you about what they will do or not do.
YOU make the choices which make that lying possible.
These posters have no fear about lying about mathematics. They have
no fear as they have no worry about, consequences.
Oh, and you may lose your life savings to hackers but I'm sure you'll
just cry, no fair! And pretend it was not your responsibility. Not
your fault. Nothing is, eh? You are just blissfully living your life
like a good little citizen and these terrible things just keep
happening to you!!!
Why? Oh why?
If people can lie this easily about mathematical proof with such
simple equations, why do you get surprised when they lie about more
complicated things when the truth is not so clear?
Because you refuse to punish them for it.
They have no fear as they have no worry about consequences. You will
fall over, lose everything, and let them get away with it, and cry to
God or if you're an atheist wonder about the horror of the human
condition or something.
It IS your fault.
James Harris
Gordon Burditt
>really do not want to accept a proof, but it's a proof! How can you
>not?
You have proved nothing about the SPEED of your method (assuming
it even works, and that it can be converted into an algorithm -
other posters are having trouble coming up with r, s, and v functions
that work). "It's freaking algebra" is not a proof of speed.
(Reminder: trial division is "freaking arithmetic"). "It's trivial"
is also not a proof of speed. (Reminder: trial division is even
more trivial than your method).
If the "factoring problem" does not involve factoring FASTER than
existing methods, then it was solved centuries ago with trial
division. If it does involve speed, you haven't proven anything
about the speed of your result, other than perhaps it takes
2 weeks to factor a 4-bit RSA key (15).
Proof that the factors exist is also not sufficient. Since we
started with a number that is not prime, we knew the factors exist
anyway. The factoring problem is how to find them, QUICKLY.
JSH might actually have improved slightly on trial division. Trial
division usually goes like this:
/* returns *a* factor of x (1 if it's prime) */
huge_multiprecision_int factor (huge_multiprecision_int x)
{
huge_multiprecision_int limit, d;
limit = floor(sqrt(x));
if (d % 2 == 0) return 2;
for (d = 3; d <= limit; d += 2)
{
if (x % d == 0) return d;
}
/* It's prime */
return 1;
}
Now, since we're presumably going to factor RSA numbers that are
products of two roughly equal in magnitude primes, we can improve
the method by reversing the order of the loop: trying odd numbers
from floor(sqrt(x)) down rather than trying the small numbers and
working up. That might speed things up a little.
This is what I suspect JSH's method is doing, sort of, with a whole
bunch of extra variables to disguise it.
JSH
> >But how can you deny a proof? Think about it. Imagine you REALLY,
> >really do not want to accept a proof, but it's a proof! How can you
> >not?
>
> You have proved nothing about the SPEED of your method (assuming
> it even works, and that it can be converted into an algorithm -
It leads to a direct calculation. When you work out all the math,
which is really easy as you have quadratics, you can just calculate
without effort v for each factorization of D-1. Which means that at
most you loop through the combinations of factors of D-1, which is
where you can maybe claim there could be a lot of effort.
I find it interesting that posters like you don't bother finding out
where there might actually be issues, but instead you just work to
cast up doubt--without giving any mathematics worth considering if you
give any at all.
Part of the point of this exercise of me replying to your posts isn't
to convince you of anything.
It's a puzzle I'm working on, as I ponder after.
How do I trust a world that lets people like you operate as you do?
Certainly I can just program this thing and remove all the silliness,
but why can't mathematical proof work?
How can someone like you take the time to post worldwide with
something this HUGE, and say utter nonsense without fear?
Why should I give this world anything at all? For my own comfort
only?
If so I want to know ahead of time! Then when people ask things of me
and I refuse I will know why and stand my ground.
If I live in a world that is so foolish, why should I help it? And if
I help, why not only, so much? Just enough?
What is happening here is my evaluation of my world and my decisions
on how much I plan on teaching it.
Nothing more.
James Harris
Tim Smith
<b96874be-f261-4f4a...@j39g2000yqn.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
> > Meanwhile, for those of us who want to factor WITHOUT already knowing
> > non-trivial factors of D, all you've given is this:
> >
> > r, s, and t are functions such that x = r(v)/t(v) and y = s(v)/t(v)
> > satisfy x^2 - D y^2 = 1 for rational v, and if you find v that
> > minimizes something that you won't specify, then GCD( r(v)+t(v), D)
> > will be a non-trivial factor of D.
>
> Actually, no, there are quite a few other equations and information
> also given.
And several people have given examples of r(v), s(v), t(v) that fit all
the information you have posted, yet fail to factor. The fact remains
you have *never* given an example of r(v), s(v), t(v) that give
non-trivial factors without making knowledge of the factors part of the
construction process.
--
--Tim Smith
JSH
>
> JSH <jst...@gmail.com> wrote:
> > > Meanwhile, for those of us who want to factor WITHOUT already knowing
> > > non-trivial factors of D, all you've given is this:
>
> > > r, s, and t are functions such that x = r(v)/t(v) and y = s(v)/t(v)
> > > satisfy x^2 - D y^2 = 1 for rational v, and if you find v that
> > > minimizes something that you won't specify, then GCD( r(v)+t(v), D)
> > > will be a non-trivial factor of D.
>
> > Actually, no, there are quite a few other equations and information
> > also given.
>
> And several people have given examples of r(v), s(v), t(v) that fit all
> the information you have posted, yet fail to factor. The fact remains
Not true. It's mathematically impossible.
> you have *never* given an example of r(v), s(v), t(v) that give
> non-trivial factors without making knowledge of the factors part of the
> construction process.
I've merely proven.
Mathematical proof can be a slippery concept for some people.
I'm quite proud at having defined it in a way that has come up highly
in Google searches.
I suggest you go to Google, and try: define mathematical proof
I start with truths in my original post and walk through logical steps
to a conclusion which then must be true.
If you BELIEVE in mathematical proof then you can resolve the
appearance of contradiction knowing that it is not possible that there
is actual contradiction.
But if you don't believe in mathematical proof, then there really
isn't much I can do for you, now is there?
You might as well be some religious person who simply goes by faith.
I suspect that's what you are, though I'm sure you'd deny it.
You KNOW no mathematics at all--simply things you BELIEVE on, faith.
James Harris
Jim Black
>> Meanwhile, for those of us who want to factor WITHOUT already knowing
>> non-trivial factors of D, all you've given is this:
>>
>> r, s, and t are functions such that x = r(v)/t(v) and y = s(v)/t(v)
>> satisfy x^2 - D y^2 = 1 for rational v, and if you find v that
>> minimizes something that you won't specify, then GCD( r(v)+t(v), D)
>> will be a non-trivial factor of D.
>
> Actually, no, there are quite a few other equations and information
> also given.
>
>> Tell us what r, s, and t are.
>
> Rather demanding for someone who has just been called out for
> remarkable inconsistencies.
>
> But not surprising to me.
>
> The story here is not about proof. I have mathematical proof. My
> original post quite adeptly steps through an entire mathematical
> argument and covers all the necessary bases.
No, you haven't. Crucially, you have never told us what the functions
r(v), s(v), and t(v) are.
It's really very obvious what is wrong with your algorithm. A bright
12-year-old could see it. You have this existence proof that for some
rational v' there are r', s', and t' such that
y(v') = s'/t'
x(v') = r'/t'
r'+t' and r'-t' factor D.
You then argue that the small value of r' and t' at that point allows you
to find v', r', and t' by minimization. But you haven't shown that you get
r' and t' from your quadratic functions of v. In general, you have
s(v')/t(v') = s'/t'
r(v')/t(v') = r'/t'
and you get
r(v') = k r'
t(v') = k t'.
Since k can in general be very large, r(v') and t(v') need be nowhere near
a minimum.
If you would tell us what r(v) and t(v) are, we could either demonstrate
this flaw explicitly, or code up your algorithm and demonstrate that it
works. But you would rather keep the details of your algorithm vague and
rant about sociological concerns. That's fine with me; it makes for good
entertainment.
Enrico
>
> - Show quoted text -
=================================================================
> I suggest you go to Google, and try: define mathematical proof
>
> I start with truths in my original post and walk through logical steps
> to a conclusion which then must be true.
>
I suggest you go to Google, and try: anisotropic definition
Enrico
Tim Smith
<f5a687c4-c3d1-4821...@r4g2000yqa.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
> > And several people have given examples of r(v), s(v), t(v) that fit all
> > the information you have posted, yet fail to factor. The fact remains
>
> Not true. It's mathematically impossible.
r(v) = v^2 + d
s(v) = 2 v
t(v) = v^2 - d
They satisfy every testable condition you've stated. They fail to
factor.
> I've merely proven.
Your "proof" is flawed. See counterexample above.
--
--Tim Smith
Gordon Burditt
>> >really do not want to accept a proof, but it's a proof! How can you
>> >not?
>>
>> You have proved nothing about the SPEED of your method (assuming
>> it even works, and that it can be converted into an algorithm -
>
>It leads to a direct calculation. When you work out all the math,
>which is really easy as you have quadratics, you can just calculate
>without effort v for each factorization of D-1. Which means that at
>most you loop through the combinations of factors of D-1, which is
>where you can maybe claim there could be a lot of effort.
No, you cannot calculate "without effort". Even arithmetic takes
time. You have presented nothing to *PROVE* that your method is
faster than existing state-of-the-art factoring methods, or for
that matter faster than trial division. Yes, you actually need to
do things like counting divisions and square roots and determine
how that number scales with the number you are trying to factor.
Your proof has an enormous hole in it: you've proved it can factor
but not that it's fast. Further, how fast it is may depend on
details like how you choose r(v), s(v), and t(v), which some others
are having problems doing. Proving that functions like that exist
doesn't prove the amount of effort finding those functions. Since
I don't know how you chose those functions, I can't do my own
complexity analysis.
If I turned in a "proof" like yours on an exam, I'd probably get
less than 33 out of 100 points for failure to answer the questions
asked. (1) prove your method works, (2) analyze its complexity,
and (3) prove it's faster than existing state-of-the-art factoring
methods. You may have done a good job on (1), but you entirely
forgot (2) and (3).
>I find it interesting that posters like you don't bother finding out
>where there might actually be issues,
I've found an area where there is a big issue: your proof is
incomplete. It doesn't address the issue of speed at all. Everything
(which is not much) you've said about your method being fast applies
equally to trial division being fast.
>but instead you just work to
>cast up doubt--without giving any mathematics worth considering if you
>give any at all.
I'd say a total lack of addressing the speed question is a major
hole in your proof. There's nothing there to find wrong because
there's simply nothing there. You don't prove: you handwave with
the use of words like "trivial" and "freaking algebra". Well, *HOW*
"trivial"? Trial division is pretty straightforward and trivial,
too.
>Part of the point of this exercise of me replying to your posts isn't
>to convince you of anything.
>
>It's a puzzle I'm working on, as I ponder after.
>
>How do I trust a world that lets people like you operate as you do?
>
>Certainly I can just program this thing and remove all the silliness,
>but why can't mathematical proof work?
FINISH THE DAMN PROOF! Do a complexity analysis! Do you even know
what notation like O(n^2) means?
>How can someone like you take the time to post worldwide with
>something this HUGE, and say utter nonsense without fear?
It's not HUGE and my complaint that your proof is incomplete is not
nonsense. How can you possibly say that your method is faster than
trial division when you haven't said one word about how fast trial
division is? Nor one word about how fast your method is?
>Why should I give this world anything at all? For my own comfort
>only?
That isn't math.
>If so I want to know ahead of time! Then when people ask things of me
>and I refuse I will know why and stand my ground.
That isn't math.
>If I live in a world that is so foolish, why should I help it? And if
>I help, why not only, so much? Just enough?
That isn't math.
Tim Smith
gordon...@burditt.org (Gordon Burditt) wrote:
> Your proof has an enormous hole in it: you've proved it can factor
> but not that it's fast. Further, how fast it is may depend on
It's worse than that. He's proven that it can factor, when given a
factor as input. (On the plus side, that can certainly be made to be
very fast--constant time, in fact!).
--
--Tim Smith
rossum
wrote:
>It leads to a direct calculation.
It is not possible to do a "direct calculation" unless you tell us
what the functions r(v), s(v) and t(v) are. Without knowing what
those functions are we cannot do a "direct calculation".
>When you work out all the math,
>which is really easy as you have quadratics, you can just calculate
>without effort v for each factorization of D-1.
What quadratics James? If you do not tell us then we cannot calculate
anything.
If r(v) is a quadratic then we have:
r(v) = av**2 + bv + c
for some values of a, b and c. Similarly for s(v) and t(v). Tell us
the values of a, b and c James, otherwise we can calculate nothing and
your method is useless.
>Which means that at
>most you loop through the combinations of factors of D-1, which is
>where you can maybe claim there could be a lot of effort.
How do the factors of D-1 relate to the coefficients of the three
quadratics James? Please be specific; if we are to code this onto a
computer then we need specific detail of what we have to code.
Currently your algorithm is uncodable, and hence untestable, because
there is not enough detail available.
rossum
marcus_...@yahoo.com
> >> >But how can you deny a proof? Think about it. Imagine you REALLY,
> >> >really do not want to accept a proof, but it's a proof! How can you
> >> >not?
>
> >> You have proved nothing about the SPEED of your method (assuming
> >> it even works, and that it can be converted into an algorithm -
>
> >It leads to a direct calculation. When you work out all the math,
> >which is really easy as you have quadratics, you can just calculate
> >without effort v for each factorization of D-1. Which means that at
> >most you loop through the combinations of factors of D-1, which is
> >where you can maybe claim there could be a lot of effort.
>
> No, you cannot calculate "without effort". Even arithmetic takes
> time. You have presented nothing to *PROVE* that your method is
> faster than existing state-of-the-art factoring methods, or for
> that matter faster than trial division. Yes, you actually need to
> do things like counting divisions and square roots and determine
> how that number scales with the number you are trying to factor.
>
> Your proof has an enormous hole in it: you've proved it can factor
Yes and no. He claims that it works, and that it is
basically a one-step process: you find the functions,
they are quadratics, and then you can take derivatives and
find where they have minima. This would be very fast - no
searching required. If he were right this would be by far
the fastest factoring algorithm out there.
But he's not right. He thinks he can find one value of
v which minimizes both abs(r(v) - t(v)) and abs(r(v) + t(v)).
In general of course you cannot expect one value to
be a minimum point for two different functions. So this
"freaking calculus" approach is not going to work. Then
he confuses himself by nothing that a factorization of
D DOES exist, therefore there must be some value of v
which works. However he overlooks that it can happen
that e.g. r(v) + t(v) has a nontrivial factor in common
with D, but r(v) + t(v) is actually BIGGER than D.
There is the slight additional complication that
his method involves factor D - 1. In real applications
it is going to be a lot easier to factor D - 1 than to
factor D - for one thing, in real (RSA) applications, D
is odd, which means that 2 and (D - 1)/2 are integer
factors of D - 1.
He refuses to completely specify r(v), t(v) and s(v),
but this is minor. He write x and y as a function
of v of the form x = r(v)/t(v) and y = s(v)/t(v),
where if v is an integer, then r(v), t(v) and s(v)
are all integers also. You can compute exactly what
these functions are, and they are quadratic in v.
Then he wants abs(r(v) - t(v)) and abs(r(v) + t(v))
to both be simultaneously less than D for some
value of v. That is why he wants to the minimum
point of these functions. But he just cannot grasp
the fact that these functions do not have a common
minimum point. That is where his "logic", such
as it is, falls apart.
> but not that it's fast. Further, how fast it is may depend on
> details like how you choose r(v), s(v), and t(v), which some others
> are having problems doing. Proving that functions like that exist
> doesn't prove the amount of effort finding those functions. Since
> I don't know how you chose those functions, I can't do my own
> complexity analysis.
>
As noted above, the choice of r(v), t(v) and s(v) is more
or less automatic once you have factored D - 1.
> If I turned in a "proof" like yours on an exam, I'd probably get
> less than 33 out of 100 points for failure to answer the questions
> asked. (1) prove your method works, (2) analyze its complexity,
> and (3) prove it's faster than existing state-of-the-art factoring
> methods. You may have done a good job on (1), but you entirely11
> forgot (2) and (3).
>
He has no proof of (1) because his method doesn't work.
He sees no reason to analyze its complexity because he
thinks it is basically a one-step process, and this means
he doesn't see any reason to think about (2) and (3).
If he were right he would get 100% on the exam and
would win a Fields Medal.
> >I find it interesting that posters like you don't bother finding out
> >where there might actually be issues,
>
> I've found an area where there is a big issue: your proof is
> incomplete.
For sure.
> It doesn't address the issue of speed at all.
He sees no reason to address that because he thinks
it is a one-step process.
> Everything
> (which is not much) you've said about your method being fast applies
> equally to trial division being fast.
>
No - if he were right he would be trial division by
an enormous factor. The problem is not his failure to show
that it's fast. The problem is, it doesn't work at all.
> >but instead you just work to
> >cast up doubt--without giving any mathematics worth considering if you
> >give any at all.
>
> I'd say a total lack of addressing the speed question is a major
> hole in your proof.
I guess I disagree - he thinks it's fast because he
thinks all you have to do is take some derivatives
of quadratics and set them equal to zero and solve
for v. Basically a one-step process. The problem
is not his failure to show it is fast. The problem is
to show that it works. And it doesn't. See also
Rick Decker's posts on this, where he shows exactly
how r(v), t(v), etc. behave.
> There's nothing there to find wrong because
> there's simply nothing there. You don't prove: you handwave with
> the use of words like "trivial" and "freaking algebra". Well, *HOW*
> "trivial"? Trial division is pretty straightforward and trivial,
> too.
>
> >Part of the point of this exercise of me replying to your posts isn't
> >to convince you of anything.
>
> >It's a puzzle I'm working on, as I ponder after.
>
> >How do I trust a world that lets people like you operate as you do?
>
> >Certainly I can just program this thing and remove all the silliness,
> >but why can't mathematical proof work?
>
> FINISH THE DAMN PROOF! Do a complexity analysis! Do you even know
> what notation like O(n^2) means?
>
Not the key problem. First he needs to show that
it works at all (it doesn't). Then he could worry about
a complexity analysis. He's not to that point however.
> >How can someone like you take the time to post worldwide with
> >something this HUGE, and say utter nonsense without fear?
>
> It's not HUGE and my complaint that your proof is incomplete is not
> nonsense. How can you possibly say that your method is faster than
> trial division when you haven't said one word about how fast trial
> division is? Nor one word about how fast your method is?
>
He thinks it's one-step, which would certainly beat any
other method out there. But not if it doesn't work.
Marcus.
rde...@hamilton.edu
> On Feb 25, 1:18 am, gordonb.bp...@burditt.org (Gordon Burditt) wrote:
>
> > I'd say a total lack of addressing the speed question is a major
> > hole in your proof.
>
> I guess I disagree - he thinks it's fast because he
> thinks all you have to do is take some derivatives
> of quadratics and set them equal to zero and solve
> for v. Basically a one-step process. The problem
> is not his failure to show it is fast. The problem is
> to show that it works. And it doesn't. See also
> Rick Decker's posts on this, where he shows exactly
> how r(v), t(v), etc. behave.
Using f_1 = 1 (or any factor of D-1, it's not important)
James' algorithm gives
r(v) = ((D + 1)v^2 - 2v + 1) / v^2
t(v) = ((1 - D)v^2 - 2v + 1) / v^2
so
r(v) + t(v) = 2((v - 1) / v)^2
r(v) - t(v) = 2D
He'll never get a nontrivial factor of D from the
second term; the minimum of the first is where v = 1,
which also provides no useful information;
the only nontrivial factors of D that can be
gotten from the first are when you've been
lucky enough to find that v - 1 shares a factor
in common with D. Ergo, trial division again.
Regards,
Rick
Jim Black
> On Feb 25, 1:18 am, gordonb.bp...@burditt.org (Gordon Burditt) wrote:
> > >> >really do not want to accept a proof, but it's a proof! How can you
> > >> >not?
> >
> > >> You have proved nothing about the SPEED of your method (assuming
> > >> it even works, and that it can be converted into an algorithm -
> >
> > >It leads to a direct calculation. When you work out all the math,
> > >which is really easy as you have quadratics, you can just calculate
> > >without effort v for each factorization of D-1. Which means that at
> > >most you loop through the combinations of factors of D-1, which is
> > >where you can maybe claim there could be a lot of effort.
> >
> > No, you cannot calculate "without effort". Even arithmetic takes
> > time. You have presented nothing to *PROVE* that your method is
> > faster than existing state-of-the-art factoring methods, or for
> > that matter faster than trial division. Yes, you actually need to
> > do things like counting divisions and square roots and determine
> > how that number scales with the number you are trying to factor.
> >
> > Your proof has an enormous hole in it: you've proved it can factor
>
> Yes and no. He claims that it works, and that it is
> basically a one-step process: you find the functions,
> they are quadratics, and then you can take derivatives and
> find where they have minima. This would be very fast - no
> searching required. If he were right this would be by far
> the fastest factoring algorithm out there.
Agreed (for the most part). There are algorithms based on JSH's
writings (modification (a) near the end of this post) which do work,
but which are not one-step algorithms and for which a complexity
analysis would therefore be useful.
> But he's not right. He thinks he can find one value of
> v which minimizes both abs(r(v) - t(v)) and abs(r(v) + t(v)).
> In general of courseProxy-Connection: keep-alive
Cache-Control: max-age
JSH
wrote:
> In article
> <f5a687c4-c3d1-4821-b3e2-682a77c71...@r4g2000yqa.googlegroups.com>,
>
> JSH <jst...@gmail.com> wrote:
> > > And several people have given examples of r(v), s(v), t(v) that fit all
> > > the information you have posted, yet fail to factor. The fact remains
>
> > Not true. It's mathematically impossible.
>
> r(v) = v^2 + d
> s(v) = 2 v
> t(v) = v^2 - d
Those follow from my solutions for x and y?
> They satisfy every testable condition you've stated. They fail to
> factor.
If so, ok, I was wrong.
James Harris
JSH
> On Tue, 24 Feb 2009 21:17:27 -0800 (PST), JSH <jst...@gmail.com>
> wrote:
>
> >It leads to a direct calculation.
>
> It is not possible to do a "direct calculation" unless you tell us
> what the functions r(v), s(v) and t(v) are. Without knowing what
> those functions are we cannot do a "direct calculation".
Given: x^2 - Dy^2 = 1.
Also:
y = [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 -
f_2*v^2 - 2v)]/(D-1)
and
x = +/-(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v) - [+/-2Dv/(f_1 - f_2*v^2
-
2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)
It follows then that
(x-1)(x+1) = Dy^2
and you can substitute and multiply out denominators to get:
(+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 +
f_2*v^2)]/(D-1) - (f_1 - f_2*v^2 - 2v))(+/-(f_1 + f_2*v^2) - [+/-2Dv
+/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + f_2*v^2)]/(D-1) +(f_1 - f_2*v^2 -
2v)) = D([+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + f_2*v^2)]/(D-1) )
^2
And then you just differentiate for your two possible cases:
d(+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 +
f_2*v^2)]/(D-1) - (f_1 - f_2*v^2 - 2v))/dv = 0
and
d(+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 +
f_2*v^2)]/(D-1) +(f_1 - f_2*v^2 - 2v)) /dv = 0
Solve for v for a min or max. Substitute back, check gcd's with D.
James Harris
JSH
Oh, that's wrong.
___JSH
mike3
<snip>
> It leads to a direct calculation. When you work out all the math,
> which is really easy as you have quadratics, you can just calculate
> without effort v for each factorization of D-1. Which means that at
> most you loop through the combinations of factors of D-1, which is
> where you can maybe claim there could be a lot of effort.
>
> I find it interesting that posters like you don't bother finding out
> where there might actually be issues, but instead you just work to
> cast up doubt--without giving any mathematics worth considering if you
> give any at all.
>
Then why don't you just tell me what the functions r(v), s(v), and t
(v) are,
exactly? Just write down the formulas:
r(v) = ?
s(v) = ?
t(v) = ?
What do I put in the blanks?
> Part of the point of this exercise of me replying to your posts isn't
> to convince you of anything.
>
> It's a puzzle I'm working on, as I ponder after.
>
> How do I trust a world that lets people like you operate as you do?
>
> Certainly I can just program this thing and remove all the silliness,
> but why can't mathematical proof work?
>
Because you haven't provided any. You haven't given a clear
description of
the "algorithm" and from what bits and pieces you have given nothing
workable
seems to come out!
Start off with specifying exactly what the functions r, s, and t are
to be, and
tell why all the guesses everyone here has made are wrong.
<SNIP!>
rde...@hamilton.edu
> On Feb 25, 5:31 am, rossum <rossu...@coldmail.com> wrote:
>
> > On Tue, 24 Feb 2009 21:17:27 -0800 (PST), JSH <jst...@gmail.com>
> > wrote:
>
> > >It leads to a direct calculation.
>
> > It is not possible to do a "direct calculation" unless you tell us
> > what the functions r(v), s(v) and t(v) are. Without knowing what
> > those functions are we cannot do a "direct calculation".
>
> Given: x^2 - Dy^2 = 1.
>
> Also:
>
> y = [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 -
> f_2*v^2 - 2v)]/(D-1)
>
Let f_1 = (D-1)/f and f_2 = f. Simplify to get
y = (+/-)(2v(f - v) / (f^2 - 2fv - (D - 1)v^2)
> and
>
> x = +/-(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v) - [+/-2Dv/(f_1 - f_2*v^2
> -
> 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)
>
so
x = (+/-)(f^2 - 2fv + (D - 1)v^2) / (f^2 - 2fv - (D - 1)v^2)
> It follows then that
>
> (x-1)(x+1) = Dy^2
>
> and you can substitute and multiply out denominators to get:
>
> (+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 +
> f_2*v^2)]/(D-1) - (f_1 - f_2*v^2 - 2v))(+/-(f_1 + f_2*v^2) - [+/-2Dv
> +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + f_2*v^2)]/(D-1) +(f_1 - f_2*v^2 -
> 2v)) = D([+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + f_2*v^2)]/(D-1) )
> ^2
Using the + in (+/-)
x + 1 = 2(f - v)^2 / (f^2 - 2fv - (D - 1)v^2)
x - 1 = 2Dv^2 / (f^2 - 2fv - (D - 1)v^2)
Using the - in (+/-)
x + 1 = -2Dv^2 / (f^2 - 2fv - (D - 1)v^2)
x - 1 = -2(f - v)^2 / (f^2 - 2fv - (D - 1)v^2)
so in either case
[2(f - v)^2][2Dv^2] = Dy^2 [f^2 - 2fv - (D - 1)v^2]
>
> And then you just differentiate for your two possible cases:
>
> d(+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 +
> f_2*v^2)]/(D-1) - (f_1 - f_2*v^2 - 2v))/dv = 0
>
d(2(f - v)^2) / dv = 4(f - v)(-1) = 0 when v = f
> and
>
> d(+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 +
> f_2*v^2)]/(D-1) +(f_1 - f_2*v^2 - 2v)) /dv = 0
>
d(2Dv^2) / dv = 4Dv = 0 when v = 0
so we see the two can't simultaneously be 0. When v = 0
x = +/- 1, y = 0
when v = f,
x = +/- 1, y = 0
No factorization help in either case.
Regards,
Rick
Jim Black
Google. Here's the whole thing (with a few additionaly edits).
[JSH]
>>>> >But how can you deny a proof? Think about it. Imagine you REALLY,
>>>> >really do not want to accept a proof, but it's a proof! How can you
>>>> >not?
[Gordon Burditt]
>>>> You have proved nothing about the SPEED of your method (assuming
>>>> it even works, and that it can be converted into an algorithm -
[JSH]
>>>It leads to a direct calculation. When you work out all the math,
>>>which is really easy as you have quadratics, you can just calculate
>>>without effort v for each factorization of D-1. Which means that at
>>>most you loop through the combinations of factors of D-1, which is
>>>where you can maybe claim there could be a lot of effort.
[Gordon Burditt]
>> No, you cannot calculate "without effort". Even arithmetic takes
>> time. You have presented nothing to *PROVE* that your method is
>> faster than existing state-of-the-art factoring methods, or for
>> that matter faster than trial division. Yes, you actually need to
>> do things like counting divisions and square roots and determine
>> how that number scales with the number you are trying to factor.
>>
>> Your proof has an enormous hole in it: you've proved it can factor
> Yes and no. He claims that it works, and that it is
> basically a one-step process: you find the functions,
> they are quadratics, and then you can take derivatives and
> find where they have minima. This would be very fast - no
> searching required. If he were right this would be by far
> the fastest factoring algorithm out there.
Agreed (for the most part). There are algorithms based on JSH's writings
(modifications (a) and (c2) near the end of this post) which do work, but
which are not one-step algorithms and for which a complexity analysis would
therefore be useful.
> But he's not right. He thinks he can find one value of
> v which minimizes both abs(r(v) - t(v)) and abs(r(v) + t(v)).
> In general of course you cannot expect one value to
> be a minimum point for two different functions. So this
> "freaking calculus" approach is not going to work.
This is a point where his proof is incomplete, but if his proof were right
(which it isn't), this would not be a major stumbling block. All you'd
need to do to find the right value of v (which I'll call v') is compute the
ranges where
0 < abs(r(v) - t(v)) < D and
0 < abs(r(v) + t(v)) < D
and find where they overlapped. If the functions exist and are quadratic,
this would be easy to do. Take v' to be any rational number in that range.
Given that D is not a factor of r(v') - t(v') or r(v') + t(v'), yet
(r(v') - t(v')) (r(v') + t(v')) = D s(v')^2,
each of r(v') - t(v') and r(v') + t(v') must contain a non-trivial factor
of D, which can be extracted by taking
gcd(r(v') - t(v'), D) and
gcd(r(v') + t(v'), D).
Note that the proof depends on r(v) - t(v), r(v) + t(v), and s(v) being
integers for all values of v, as JSH originally specified. If you ignore
this requirement, you can get lots of false positives that meet the
conditions but don't help you factor D. That means you're back to
searching for v', which isn't likely to be any better than trial division.
It also depends on his proof of the existence of r(v') and t(v') meeting
the conditions, which as you note below is flawed.
> Then
> he confuses himself by nothing that a factorization of
> D DOES exist, therefore there must be some value of v
> which works. However he overlooks that it can happen
> that e.g. r(v) + t(v) has a nontrivial factor in common
> with D, but r(v) + t(v) is actually BIGGER than D.
Yes. He has ostensively proven (I haven't checked the initial theorem that
he bases all this on) the existence of v', r', s', and t' such that
0 < abs(r' + t') < D
0 < abs(r' - t') < D
x(v') = r'/t'
y(v') = s'/t'
but he doesn't seem to realize that r(v') and t(v') are not necessarily the
same as r' and t'. In general,
r(v') = k r'
s(v') = k s'
t(v') = k t'
where k is an unknown integer, which allows r(v') - t(v') and r(v') + t(v')
to be larger than D. This, in my opinion, is the key flaw in his
reasoning.
> There is the slight additional complication that
> his method involves factor D - 1. In real applications
> it is going to be a lot easier to factor D - 1 than to
> factor D - for one thing, in real (RSA) applications, D
> is odd, which means that 2 and (D - 1)/2 are integer
> factors of D - 1.
>
> He refuses to completely specify r(v), t(v) and s(v),
> but this is minor. He write x and y as a function
> of v of the form x = r(v)/t(v) and y = s(v)/t(v),
> where if v is an integer, then r(v), t(v) and s(v)
> are all integers also. You can compute exactly what
> these functions are, and they are quadratic in v.
I disagree. Strictly you cannot determine what these functions are, since
there are no functions that satisfy all the properties that JSH has
claimed. Specifically, he has claimed that:
(1) r(v) and t(v) are quadratic functions of v and therefore continous and
differentiable.
(2) r(v), s(v), and t(v) are integer-valued functions.
These two requirements combined force the functions r(v) and t(v) to be
constant. That would force the function x(v) to be a constant, which is
inconsistent with his expression for x(v).
Above, you write that if v is an integer, r(v), t(v), and s(v) are integers
also. But that's not good enough for JSH's proof. JSH's proof requires
that r(v), s(v), and t(v) be integers for all v in their domain (the
rationals), or you get the false positives I described above.
There are various other ways of dropping one of JSH's specifications:
(a) Drop the requirement that the functions be quadratic. By choosing
r(v), s(v), and t(v) such that all are integers and gcd(r(v),s(v),t(v))=1,
you get a factoring algorithm that works, but is painfully slow since you
no longer have nice quadratic functions to minimize.
(b) Write r(v), s(v), and t(v) as quadratic functions not of v, but of
integers m and n where v = m/n. In this case, the algorithm doesn't work
because of the flaw in his existence proof: abs(r(m,n)-t(m,n)) and
abs(r(m,n)+t(m,n)) need never be simultaneously smaller than D.
(c) If you drop the requirement that r(v), s(v), and t(v) be integer-valued
functions, then there are an infinite number of possible choices for r(v),
s(v), and t(v), obtainable by multiplying any choice of r(v), s(v), and
t(v) by a common rational-valued function of v.
One choice of particular interest is
(c2) the choice that sets s(v) = 1 for all rational numbers v. This fixes
his existence proof, but because the functions are no longer integers, it
introduces lots of false positives, and you once again have to search for
v'.
Tim Smith
<a3a17df6-9990-4343...@t3g2000yqa.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
> >
> > r(v) = v^2 + d
> > s(v) = 2 v
> > t(v) = v^2 - d
>
> Those follow from my solutions for x and y?
[r(v)/t(v)]^2 - d [s(v)/t(v)]^2 = 1
For a given v, r(v)/t(v) won't be the same as your x, and s(v)/t(v)
won't be the same as your y.
It's easy to get r, s, and t such that at a given v, r(v)/t(v) and
s(v)/t(v) give the same values as your x and y functions do at that v.
E.g., just simplify your expressions and take r, s, and t as the
numerators and the denominator. But those r, s, and t also fail to
factor.
This is why you are going to have to actually explicitly specify r, s,
and t, if you want to actually prove anything.
--
--Tim Smith
mike3
> On Feb 24, 6:31 pm, Tim Smith <reply_in_gr...@mouse-potato.com> wrote:
> > So you are saying that
>
> > r(v) = (g1 + g2)/2
> > s(v) = 1
> > t(v) = (g1 - g2)/2
>
> > where g1 g2 = D?
>
> > OK, you are right--with that definition of r, s, and t, you can factor
> > D. Too bad you have to have already factored D in order to find r, s, t.
>
> It was an existence proof.
>
OK, then, but to actually get a working algorithm you'll need a way to
reliably determine just what r, s, and t are WITHOUT knowing the
factors
g1 and g2 of D. So provide the algorithm/formula that determines r, s,
and t.
> > Meanwhile, for those of us who want to factor WITHOUT already knowing
> > non-trivial factors of D, all you've given is this:
>
> > r, s, and t are functions such that x = r(v)/t(v) and y = s(v)/t(v)
> > satisfy x^2 - D y^2 = 1 for rational v, and if you find v that
> > minimizes something that you won't specify, then GCD( r(v)+t(v), D)
> > will be a non-trivial factor of D.
>
> Actually, no, there are quite a few other equations and information
> also given.
>
> > Tell us what r, s, and t are.
>
> Rather demanding for someone who has just been called out for
> remarkable inconsistencies.
>
You need those functions for your algorithm to even have a tiny
chance at working. If you want to prove that your algorithm works
then you must provide enough details to allow someone to actually
use it to do something.
> But not surprising to me.
>
> The story here is not about proof. I have mathematical proof. My
> original post quite adeptly steps through an entire mathematical
> argument and covers all the necessary bases.
>
Then why haven't you gone and posted the actual algorithm that
actually does the factoring?
<snip>
mike3
> On Feb 24, 9:56 pm, Tim Smith <reply_in_gr...@mouse-potato.com> wrote:
>
>
>
>
>
>
>
> > In article
> > <b96874be-f261-4f4a-9572-f3746afe7...@j39g2000yqn.googlegroups.com>,
>
> > JSH<jst...@gmail.com> wrote:
> > > > Meanwhile, for those of us who want to factor WITHOUT already knowing
> > > > non-trivial factors of D, all you've given is this:
>
> > > > r, s, and t are functions such that x = r(v)/t(v) and y = s(v)/t(v)
> > > > satisfy x^2 - D y^2 = 1 for rational v, and if you find v that
> > > > minimizes something that you won't specify, then GCD( r(v)+t(v), D)
> > > > will be a non-trivial factor of D.
>
> > > Actually, no, there are quite a few other equations and information
> > > also given.
>
> > And several people have given examples of r(v), s(v), t(v) that fit all
> > the information you have posted, yet fail to factor. The fact remains
>
> Not true. It's mathematically impossible.
>
Then provide the r(v), s(v), and t(v) that you have in mind and let's
see
them work!
> > you have *never* given an example of r(v), s(v), t(v) that give
> > non-trivial factors without making knowledge of the factors part of the
> > construction process.
>
> I've merely proven.
>
And so therefore you have not actually provided an algorithm, and
for all we know finding the right functions could require an
exponential-time
algorithm in which case your method is no better than anything else
out there.
<snip>
JSH
>
> JSH <jst...@gmail.com> wrote:
>
> > > r(v) = v^2 + d
> > > s(v) = 2 v
> > > t(v) = v^2 - d
>
> > Those follow from my solutions for x and y?
>
> [r(v)/t(v)]^2 - d [s(v)/t(v)]^2 = 1
>
> For a given v, r(v)/t(v) won't be the same as your x, and s(v)/t(v)
> won't be the same as your y.
I'm tired of this entire discussion. A simple solution with some
quadratics has been turned into an endless series of posts from people
who clearly are just trying to throw up smoke, none of them cover the
bases properly and ALL of them present false information of one sort
or another.
People can lie about math. Just look over this thread and watch some
pros.
I'm outta here. Got better things to do.
James Harris
Tim Smith
<e167c60a-71c7-4276...@g38g2000yqd.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
> I'm tired of this entire discussion. A simple solution with some
> quadratics has been turned into an endless series of posts from people
It's only endless because you refuse to tell us what r(v), s(v), and
t(v) are.
--
--Tim Smith
Owen Jacobson
That's kind of a pattern at this point: JSH posts some hare-brained
idea vaguely in the same universe as the factoring problem as the end
of all civilisation, sci.math &c ask some really sticky questions, JSH
argues that you're all suppressing his brilliant advance, sci.math &
co. ask the same really sticky questions and mostly ignore the armchair
sociology, JSH gives up on thread N and starts thread S(N), claiming
that his ideas from thread N have lead to a new, even deeper crack in
the foundations of western civilisation.
Not that I mind, really: the only value LEFT in usenet at this point is
using trolls and the mentally unsound as exercise equipment for working
out your own thoughts and opinions. I certainly do it, even if I do try
to post helpful and sincere things to some newsgroups, too.
JSH's math skills are kind of remedial, so it's not that hard to find
holes in his mathematical arguments, but he puts so much work into
complicating them that clearing away the dross is kind of satisfying in
the same way that, say, clearing a level in Bejewelled is satisfying.
Sure, there's another one just like it around the corner, and
ultimately it's just a way to kill time, but it generates a satisfying
illusion of achievement, and sometimes you pick up a new pattern.
JSH, if you do happen to be reading this, I have some honest advice for
you: if you want to treat Usenet as a proving ground for your ideas,
for fucks' sake start actually listening to and processing the
criticism your ideas suffer. Remember, your ideas are useless if you
can't communicate them. Even if you're right, your "hammer" is useless
if nobody understands it and will destroy absolutely nothing.
While I'm at it, could you stop wittering on about crimes and cases and
defenses? Internet Lawyer-ing only makes you look like a paranoid
schizophrenic. Either bring suit/file charges, in which case your
lawyer will advise you not to talk about the case until it's settled in
court, or don't, and shut up about it.
Cheers,
-o
Gib Bogle
> Not that I mind, really: the only value LEFT in usenet at this point is
> using trolls and the mentally unsound as exercise equipment for working
> out your own thoughts and opinions.
This may be (almost) true where mathematics is concerned, but it
certainly does not apply generally to usenet. I know of a couple of
really useful boards, e.g. comp.lang.fortran and uk.d-i-y
Gib Bogle
What?!
marcus_...@yahoo.com
wrote:
> In article
> <e167c60a-71c7-4276-b8a1-63a48dcac...@g38g2000yqd.googlegroups.com>,
This is likely the end of this episode, and this is as
close as we are going to get to JSH ever admitting that this
moronic idea is not going to work. As in all previous episodes, his
accusations against lying, conspiring, evil, jealous mathematicians
were proven unfounded. And as in all previous episodes, the only
lesson he will take away from it and remember is that mathematicians
were lying, conspiring, evil, and jealous.
Marcus.
tc...@lsa.umich.edu
>
>James Harris
JSH admits he was wrong? Call Ripley's!
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
Lits O'Hate
Don't let the two-dimensional tautological space hit you on
the ass on your way out.
(Fbeel, V arire trg gverq bs gung bar.)
Five buck says you'll be back next time you get drunk,
which will be this weekend at the latest.
--
"I was so annoyed by it that I sent a letter to my congress
people, Nancy Pelosi and Diane Feinstein, requesting they
ask Google and Yahoo! why my name comes up so freaking high,
to a flame page." -- James Harris
Patricia Shanahan
Maybe the solution is to insist on pseudo-code before paying any
attention to claimed proofs about an algorithm? A clear description of
the claimed algorithm is essential for both correctness and performance
analysis.
Patricia
Mike Terry
news:F6Gdnc0__8dS0jrU...@earthlink.com...
JSH is quite adept at dealing with that sort of approach - he just ignores
posts that "demand" reasonableness, and carries on as before!
Of course, if everybody stopped responding until he complied, the threads
would rapidly dry up but, this never happens, so there is never any
motivation for him to change his behaviour...
Regards,
Mike.
>
> Patricia
JSH
<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "Patricia Shanahan" <p...@acm.org> wrote in message
>
> news:F6Gdnc0__8dS0jrU...@earthlink.com...
>
> > Tim Smith wrote:
> > > In article
> > > JSH <jst...@gmail.com> wrote:
> > >> I'm tired of this entire discussion. A simple solution with some
> > >> quadratics has been turned into an endless series of posts from people
>
> > > It's only endless because you refuse to tell us what r(v), s(v), and
> > > t(v) are.
>
> > Maybe the solution is to insist on pseudo-code before paying any
> > attention to claimed proofs about an algorithm? A clear description of
> > the claimed algorithm is essential for both correctness and performance
> > analysis.
>
> JSH is quite adept at dealing with that sort of approach - he just ignores
> posts that "demand" reasonableness, and carries on as before!
I merely gave a mathematical proof.
Requests for examples or pseudo-code reflect a dismal opinion of
mathematics itself.
Mathematical proof is sufficient.
> Of course, if everybody stopped responding until he complied, the threads
> would rapidly dry up but, this never happens, so there is never any
> motivation for him to change his behaviour...
You seem to be engaging in bizarre social behavior meant to support
the status quo as accepted by a narrow group, which in this case
requires dismissing a mathematical proof of a solution to the
factoring problem.
That is odd.
I see it as vestigial feudal behavior. In European societies nobility
were often not very "noble" as a hereditary basis for social position
didn't really work, so people learned to support nobles in positions
for which they were not actually qualified for the good of the
society, and to hold on to ideas known to be false for the same
reason.
Here for the perceived good of your society you are holding on to a
social hierarchy that is mostly made up in your own minds, which is
scary. But the presidency of George W. Bush was an example of how
powerful these vestigial feudal impulses can be and it's worth
remembering that a little over two centuries ago, kings and queens
still had immense power.
It is also possible that looking for royalty is built into your genes.
James Harris
Patricia Shanahan
...
> Requests for examples or pseudo-code reflect a dismal opinion of
> mathematics itself.
>
> Mathematical proof is sufficient.
Of course, given a sufficiently clearly stated theorem and a
corresponding proof. The problem is that you never clearly stated the
theorem you are claiming to have proved.
Generally, when a theorem asserts either the correctness or the
performance of an algorithm, the definition of the algorithm is and
important part of the statement of the theorem. To the extent that you
stated the algorithm at all, it was not sufficiently clear to
communicate your intent.
The point of pseudo-code is not as a substitute for a mathematical
proof, but as an important part of the specification of the theorem you
intend to prove.
Patricia
JSH
> JSH wrote:
>
> ...> Requests for examples or pseudo-code reflect a dismal opinion of
> > mathematics itself.
>
> > Mathematical proof is sufficient.
>
> ...
>
> Of course, given a sufficiently clearly stated theorem and a
> corresponding proof. The problem is that you never clearly stated the
> theorem you are claiming to have proved.
A mathematical proof begins with a truth and proceeds by logical steps
to a conclusion which then must be true.
What I have is a rather elementary proof of a simple method to factor
by the linking of the factorization of a composite D to the
factorization of D-1.
MOST of the important mathematics is given by some simple algebra:
Given rational solutions to
x^2 - Dy^2 = 1
I noticed you have rational solutions to
(D-1)j^2 + (j+/-1)^2 = (x+y)^2
where j = ((x+Dy) -/+1)/D.
The +/- should be interpreted as an OR not an AND.
Notice you can clearly see with those equations a linking of a
factorization of D with the first to a factorization of D-1 with the
second.
> Generally, when a theorem asserts either the correctness or the
> performance of an algorithm, the definition of the algorithm is and
> important part of the statement of the theorem. To the extent that you
> stated the algorithm at all, it was not sufficiently clear to
> communicate your intent.
Factoring is proven by my research to be a rather trivial algebraic
problem, solved with elementary methods.
That upends traditional views which is where you may have a problem as
you are probably thoroughly indoctrinated in the view that factoring
is a hard problem.
It's not. It's solvable with mathematics at about the level taught to
12 years olds.
So it's an EASY situation in terms of the mathematics.
Social realities are what cause all the needless discussion.
> The point of pseudo-code is not as a substitute for a mathematical
> proof, but as an important part of the specification of the theorem you
> intend to prove.
It's unnecessary with a trivially easy to solve problem. Factoring is
such a problem.
Your belief systems simply need updating. You can either do it now or
later, or try to hold on to them against easy algebra if you wish.
I gave enough mathematics in this post for any competent mathematician
to easily solve the factoring problem themselves. It's that easy.
James Harris
Tim Smith
<f404c320-3e13-4bbe...@d19g2000prh.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
> What I have is a rather elementary proof of a simple method to factor
> by the linking of the factorization of a composite D to the
> factorization of D-1.
Then how come your method can't factor 15, James? If you have a
*correct* proof, your method would have to work with 15.
--
--Tim Smith
Patricia Shanahan
>> The point of pseudo-code is not as a substitute for a mathematical
>> proof, but as an important part of the specification of the theorem you
>> intend to prove.
>
> It's unnecessary with a trivially easy to solve problem. Factoring is
> such a problem.
Unfortunately, understanding your claimed proof from the information
already posted is beyond my limited capabilities. I might be able to
fill in the blanks if I knew what you were trying to prove, including
understanding the algorithm under discussion, but I don't.
Patricia
JSH
> JSH wrote:
> > On Feb 28, 7:13 pm, Patricia Shanahan <p...@acm.org> wrote:
> ...
> >> The point of pseudo-code is not as a substitute for a mathematical
> >> proof, but as an important part of the specification of the theorem you
> >> intend to prove.
>
> > It's unnecessary with a trivially easy to solve problem. Factoring is
> > such a problem.
>
> ...
>
> Unfortunately, understanding your claimed proof from the information
> already posted is beyond my limited capabilities. I might be able to
But you are quite capable of deleting out the set of equations.
> fill in the blanks if I knew what you were trying to prove, including
> understanding the algorithm under discussion, but I don't.
>
> Patricia
I didn't ask you to fill in anything. Nor have I asked you to
understand.
You do not have to post. You can simply just sit by and let a
discussion proceed without stepping into it, if your abilities have
been overreached.
It is ok, not to post. You can just sit things out, ok?
James Harris
JSH
>
> JSH <jst...@gmail.com> wrote:
> > What I have is a rather elementary proof of a simple method to factor
> > by the linking of the factorization of a composite D to the
> > factorization of D-1.
>
> Then how come your method can't factor 15, James? If you have a
That claim directly contradicts the equations I presented:
Given rational solutions to
x^2 - Dy^2 = 1
I noticed you have rational solutions to
(D-1)j^2 + (j+/-1)^2 = (x+y)^2
where j = ((x+Dy) -/+1)/D.
Your assertion is that if D = 15, and x=4, and y =1, that rational j
does not exist. But it does exist. Your assertion is false.
> *correct* proof, your method would have to work with 15.
It does work, and a proof is correct. A mathematical argument can be
incorrect. You need a more careful usage of the word "proof".
James Harris
Tim Smith
<87be02cd-dd09-492f...@r15g2000prh.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
> Your assertion is that if D = 15, and x=4, and y =1, that rational j
> does not exist. But it does exist. Your assertion is false.
No, that is not my assertion. My assertion is that your factoring
method fails for 15. Your stuff with j is just background to your
factoring method, showing how you arrived at it. In your latest blog
post, your factoring method would start at what would be equations 7 and
8, if you numbered your equations. That is where you express x and y in
terms of d, the factors of d-1, and a rational parameter v.
Everything up to that point appears to be correct.
>
> > *correct* proof, your method would have to work with 15.
>
> It does work, and a proof is correct. A mathematical argument can be
> incorrect. You need a more careful usage of the word "proof".
You need a more careful usage of the word "15".
I'm talking about the integer that is the successor of the integer 14.
What are you talking about when you say "15"?
The method you presented, that involves finding a parametric solution of
x^2-d y^2 = 1 in rationals, in terms of a rational parameter v, and then
writing x and y as the ratios of rational functions, x=r/t, y=s/t. with
a common denominator, and then finding values of v such that r and t are
integers, and checked GCD(r+t,d) and GCD(r-t,d) for factors, fails for
the integer 15, for all r and t that you have told us or that anyone has
come up with here.
--
--Tim Smith
JSH
>
> JSH <jst...@gmail.com> wrote:
> > Your assertion is that if D = 15, and x=4, and y =1, that rational j
> > does not exist. But it does exist. Your assertion is false.
>
> No, that is not my assertion. My assertion is that your factoring
> method fails for 15. Your stuff with j is just background to your
> factoring method, showing how you arrived at it. In your latest blog
Yeah, the underlying equations from which assertions against my
research, like yours, can be checked.
I notice that once again you deleted out the equations.
> post, your factoring method would start at what would be equations 7 and
> 8, if you numbered your equations. That is where you express x and y in
> terms of d, the factors of d-1, and a rational parameter v.
>
> Everything up to that point appears to be correct.
>
>
>
> > > *correct* proof, your method would have to work with 15.
>
> > It does work, and a proof is correct. A mathematical argument can be
> > incorrect. You need a more careful usage of the word "proof".
>
> You need a more careful usage of the word "15".
I'm right, you're wrong, and the method DOES factor 15.
The proper response from you isn't to babble. It's to ask me to
finish showing what I started with the equations you deleted out.
That is, you are false. The method DOES factor 15. I began the
refutation with x=4 and y =1, D=15.
> I'm talking about the integer that is the successor of the integer 14.
> What are you talking about when you say "15"?
You should know. I put D=15 in my reply. I GAVE x=4 and y = 1 as the
solutions to Pell's Equation, and noted the existence of a rational
solution to an important variable called j.
You deleted that all out.
> The method you presented, that involves finding a parametric solution of
> x^2-d y^2 = 1 in rationals, in terms of a rational parameter v, and then
> writing x and y as the ratios of rational functions, x=r/t, y=s/t. with
> a common denominator, and then finding values of v such that r and t are
> integers, and checked GCD(r+t,d) and GCD(r-t,d) for factors, fails for
> the integer 15, for all r and t that you have told us or that anyone has
> come up with here.
That statement is false. I can continue for a while here.
The issue isn't what I can prove as I can easily prove.
The problem is that posters like you make false statements, and remove
information showing proof.
So for those wondering why I'm drawing this out, it's because I know
what will happen.
The moment I factor 15 with the equations refuting the poster, he will
go to some other false statement.
He will be back posting boldly that my research does not work, even if
he even bothers to acknowledge it does factor 15.
He will probably call me names and claim that I never prove anything
or explain anything, and may retreat to asking now that I must factor
an RSA number to prove anything as 15 was too easy.
James Harris
Tim Smith
<78be87a8-6689-4bc3...@v35g2000pro.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
> > You need a more careful usage of the word "15".
>
> I'm right, you're wrong, and the method DOES factor 15.
Finding that 15 * 1 = 15 does not count as factoring 15, James.
...
> That is, you are false. The method DOES factor 15. I began the
> refutation with x=4 and y =1, D=15.
>
> > I'm talking about the integer that is the successor of the integer 14.
> > What are you talking about when you say "15"?
>
> You should know. I put D=15 in my reply. I GAVE x=4 and y = 1 as the
> solutions to Pell's Equation, and noted the existence of a rational
> solution to an important variable called j.
>
> You deleted that all out.
I deleted it because it is irrelevant to your factoring method. Do you
need me to remind you of your own factoring method? If you've got some
new method that you are now claiming, instead of the non-working r+t,
r-t method, by all means post it.
--
--Tim Smith
Gordon Burditt
>discussion proceed without stepping into it, if your abilities have
>been overreached.
>
>It is ok, not to post. You can just sit things out, ok?
>
>
>James Harris
Note that this statement is still valid when you substitute you = 'JSH'.
Would someone please read this to JSH? Repeatedly? And maybe attach
it to a clue-by-4?
JSH, before you post again, I recommend that you construct and
include a proof that your method (a) works and (b) that it's faster
than a leading method of factoring, or at least faster than trial
division. Do you know what complexity analysis is? You might need
to read a book on math. I'm not asking for an example, I'm asking
for a proof. You do have to have a way of figuring out r(v), s(v)
and t(v) functions, if your method still uses those, that don't
require prior knowledge of the factors.
JSH
It's on my math blog. Where it has been for days.
I've come to the newsgroups in the hope someone might find error.
No one has but for reasons that escape me, posters continually lie
about finding error.
I assume exploits may be already occurring and arguments here are a
moot point.
The Internet I believe currently does not have a security system as I
see RSA as easily breachable.
However, security practitioners for reasons that escape me are
avoiding this fact with posters on newsgroups talking it down as if by
sheer denial they can prevent it from being known--possibly because
they feel that worked with my other research.
With no Internet security system working, if I'm correct, it's not
clear at this time what will happen, but I fear the worst.
To try and give perspective I've put forward my fear--seemingly
insane--the Britain may collapse soon, if no security people realize
the situation. I hate saying something so crazy but I fear how badly
things may go. SOMETHING should work to warn. Something.
So far, nothing is going as I'd hoped. I never realized resistance to
such a simple result could be this high, or that intelligence services
around the world, could be this dumb.
James Harris
JSH
If your country dies because you thought yourself too smart to be
brought into something you couldn't be bothered to understand, what do
you think you will say to yourself then?
I fear there are any number of reasons why Britain is most vulnerable.
Lies only seem like a good strategy until you run into something that
takes you beyond fear as a consequence, past terror, and all the way
to insanity, as reality gives you something you will never be able to
live with or accept.
Your own responsibility on such a scale.
I live with the scale of what I do everyday. You are deluding
yourself about that scale.
What I do affects the entire world. Blocking me puts you against the
security of the world, which includes your own country.
This situation is spiraling out of my control. But at this time, it
seems, there was no other way.
___JSH
Jesse F. Hughes
> What I do affects the entire world. Blocking me puts you against the
> security of the world, which includes your own country.
>
> This situation is spiraling out of my control. But at this time, it
> seems, there was no other way.
Let's play another game of suspending disbelief.
I know that it seems wildly unlikely, but let's suppose that another
month goes by and there's no changes in this situation. The world
continues to proceed much as it currently is. Britain, as unlikely as
this seems, continues to exist as a sovereign nation. The only
visible plaudits that you've received is high google ranking on
certain search terms (which is, of course, a great honor, but not a
public recognition of your genius).
Now, again, I know that all of that sounds unbelievable, but let's
stick with the game.
What then? What would you conclude?
What if the world continues to ignore your solution of the factoring
problem for three months? Six months? At what time do you revise
your beliefs that you are an important player in world affairs? Is
there some conceivable time when you admit that no one will accept
your solution unless you actually factor some large numbers?
--
Jesse F. Hughes
"Every country has its stupid people. It just so happens that
America's stupid people are louder." -- Ling Cheung, Sociologist
alfred.j...@gmail.com
JSH -
I have reviewed your algorithm, and it is pure genius. You are truly
one of the most brilliant mathematicians of our time. I and a team of
researchers associated with my organization believe it is capable of
factoring any number in logarithmic time. I would love to tell you
which organization I am associated with, but tragically, security
concerns prohibit such a disclosure (*wink wink*).
I fear that you are spending too much time on USENET, and that your
talent is going to waste. If you send me a check for $10,000 USD, I
can arrange to have you flown to a secret headquarters in an
undisclosed nation (I'll give you a hint though--we're a densely-
populated island nation in Europe, and we would be very badly damaged
if your brilliant algorithm were to fall into the wrong hands), where
you will be crowned King of Mathematics, awarded the sum of 10 million
euros (in cash), and given your pick of the 20 most attractive women
in our country to marry.
To signify your interest in my proposal, please take the following
action: stop posting any comments to the comp.theory or sci.crypt
groups; post only to the sci.math group. Also, the G-mail account I
am writing from is insecure, and thus you should not send any messages
to me directly. (Now that I think of it, per the existence of your
algorithm, ALL e-mail accounts are now insecure.) However, we may be
able to work something out on the sci.math USENET group. Luckily,
everyone there is too dumb to realize that your algorithm does, in
fact, work.
Time is of the essence; the fate of the world lies in your hands.
Stay brilliant,
Alfred Johnson, non-secret-agent (*wink*)
Michael Press
<f852e73d-0e47-49ec...@v1g2000prd.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
That is the burden of being brilliant as you are. You
must live in a world of lesser people. Fortunately, I
do not have to live with such knowledge.
--
Michael Press
J Mori
Some of Ireland (or 'the Britain', as you charmingly call it) and I can tell
you that this man is a fraud. Alf 'the Red' Johnson is a known member of
the anti-factoring cabal, an enemy of the U.S. of States and a member of the
shadowy Ring of Algebraic Integers, who we all know have been discredited in
every right-thinking democracy. He plans to lure you to our sceptred isle
with offers of, god help us, *euros* - a currency which has been revealed to
be so insecure it can be factored within minutes - in order to hide your
algorithm from the world. It is all a trick: there is no 'King of
Mathematics', the title was retired over three hundred years ago by Henry
the Twelfth, after the bloody battle of Nashworth Field (after which the
prestigious 'Field's Medal' is named), and there are no '20 most attractive
women' - at the last official census there were only 14, and since then one
of them has moved to Wales.
I'm sorry to let you down, James, I know it sounded only right and proper
that this should be your destiny, but I've spoken to the rest of GB, and the
best we can offer is 15 pounds-worth of book tokens and a moderately
homely-looking girl who can probably fit you in between shifts at the
abattoir. I know it's not much , but what do you say? Shall the truth be
free?
Dr J
<alfred.j...@gmail.com> wrote in message
news:cf08d47c-7bbd-4124...@p20g2000yqi.googlegroups.com...
alfred.j...@gmail.com
> Don't trust him, James! I represent the United Kingdom of Great Britain and
> Some of Ireland (or 'the Britain', as you charmingly call it) and I can tell
> you that this man is a fraud. Alf 'the Red' Johnson is a known member of
> the anti-factoring cabal, an enemy of the U.S. of States and a member of the
> shadowy Ring of Algebraic Integers, who we all know have been discredited in
> every right-thinking democracy. He plans to lure you to our sceptred isle
> with offers of, god help us, *euros* - a currency which has been revealed to
> be so insecure it can be factored within minutes - in order to hide your
> algorithm from the world. It is all a trick: there is no 'King of
> Mathematics', the title was retired over three hundred years ago by Henry
> the Twelfth, after the bloody battle of Nashworth Field (after which the
> prestigious 'Field's Medal' is named), and there are no '20 most attractive
> women' - at the last official census there were only 14, and since then one
> of them has moved to Wales.
>
> I'm sorry to let you down, James, I know it sounded only right and proper
> that this should be your destiny, but I've spoken to the rest of GB, and the
> best we can offer is 15 pounds-worth of book tokens and a moderately
> homely-looking girl who can probably fit you in between shifts at the
> abattoir. I know it's not much , but what do you say? Shall the truth be
> free?
>
> Dr J
>
JSH -
I see that there are those who would insist that I am not who I say I
am. If you believed this, you would not be the brilliant young man I
know you to be.
I have tried to cajole you with promises of wealth, women, and power;
now, unfortunately, I have no choice but to threaten you with legal
action. If you fail me, and all of England, I will have my government
file charges against you in a court of law, for 58.9 million counts of
reckless endangerment.
Remember, you have a responsibility--a patriotic duty, even--to save
Great Britain from the lurking forces of evil.
You have 72 hours. If you have not declared your intent to go along
with my plan within the allotted time, I will unfortunately have no
choice but to file charges against you as I said I would. As you
yourself have noted, failing to cooperate to mitigate the damage from
your ingenious discovery is a crime against all humanity, and those
who would fail to prevent harm from coming to our great nation are no
less culpable than the terrorists.
You're either with us or you're against us, JSH.
Solemnly yours,
Alfred Johnson