JSH: But is quadratic residue idea new?
JSH
this result new? Now I'm in pondering mode as I consider something so
trivially simple that it seems weird if it's NOT previously known.
Because of the way congruences work, it's possible to do something
interesting with quadratic residues, where I've simplified from the
heavy machinery of the full idea, to make things understandable, and
now it looks like I'm grabbing certain equations from a hat, but at
least it should be easily understood.
Imagine you have k^2 = q mod p, where p is an odd prime.
And now take the equations:
f_1 = k mod p and f_2 = 2k mod p
And let T = f_1*f_2, so T = 2k^2 mod p, and I can see k^2 in there, so
k^2 = 2^{-1}*T mod p, which means that
q = 2^{-1}*T mod p, and T = 2q mod p. Easy.
Now
z^2 - y^2 = T, then (z-y)(z+y) = T, so I have a factorization, and in
fact I have simply enough:
z = (f_1 + f_2)/2, so I can substitute with f_1 and f_2 above, and get
f_1 + f_2 = 3k mod p, so
z = 3k/2 mod p
and I have the implication that the factorization of T is connected
with the value of k, and I have finally:
k = 3^{-1}(f_1 + f_2) mod p, and T = 2q mod p
when k^2 = q mod p.
So this simple approach has connected finding k, when k^2 = q mod p,
with factoring T, where T = 2q mod p.
I'm guessing that the probability of it working is roughly 1/k.
But is this idea new? And what might be serious problems with it?
Can it work well?
James Harris
Mark Murray
> Finding a result is one thing but often I wonder afterwards, why is
> this result new? Now I'm in pondering mode as I consider something so
> trivially simple that it seems weird if it's NOT previously known.
Very often, it turns out that your result is NOT new. You usually ignore
this though, and uselessly puzzle on for ages wondering why no-one
else has found what you have.
M
Calvin Ramsey
Which is a typical, and well-known, symptom of the Narcissistic
Personality Disorder.
One should always keep that in mind while corresponding with JSH.
--C. Ramsey
Enrico
=====================================================
Quadratic residue - Wikipedia, the free encyclopedia... concept from
the branch of number theory known as modular arithmetic, ..... Say
there were an efficient algorithm for finding square roots modulo
a ... The fact that finding a square root of a number modulo a large
composite n is ...
Enrico
JSH
> Mark Murray wrote:
> > JSH wrote:
> >> Finding a result is one thing but often I wonder afterwards, why is
> >> this result new? Now I'm in pondering mode as I consider something so
> >> trivially simple that it seems weird if it's NOT previously known.
>
> > Very often, it turns out that your result is NOT new. You usually ignore
> > this though, and uselessly puzzle on for ages wondering why no-one
> > else has found what you have.
>
> > M
>
> Which is a typical, and well-known, symptom of the Narcissistic
> Personality Disorder.
Give it a rest, accusing me of NPD is something I started years ago.
Thought it might be simpler to toss out that diagnosis than deal with
people using schizophrenia all the time. So I made a post considering
the question of whether or not I had NPD.
As for the math, the deflection would indicate that no, this approach
is not commonly known.
> One should always keep that in mind while corresponding with JSH.
>
> --C. Ramsey
NPD doesn't have the bite of schizophrenia as a negative insult, so in
a way the deflection was a success.
(Yes, people call other people crazy on newsgroups in order to insult
them. That part is easy.)
But for some reason people seem to be more comfortable accusing mental
illness when it's NPD versus schizophrenia, so the accusation is more
common.
So I didn't solve the problem. I merely reduced the bite of the
assertion of mental illness, while dealing with greater frequency of
the insult.
NPD isn't as nasty as schizophrenia. But it is intriguing that the
schizophrenia diagnosis went away completely when I put out the NPD as
a possible instead. People shied away from the harsher negative of
using schizophrenia. Possibly some sense of conscience?
Posters opted for the lesser illness, which is of some interest. Why
the complete replacement?
James Harris
Tom St Denis
Feed the poor and hungry, not the trolls.
Tom
JSH
And none of that seems to address the question of, is this approach
new?
To me it's a simple issue but I guess I can see where for others it's
a competitive question.
But that's where "pure math" as a concept runs into the reality of
human competition.
It is of interest that math people have ignored the simple possibility
that in competitive human interactions with a "pure math" result that
people might simply consider what benefit a discoverer might have from
recognition, and refuse to give it!
Practical mathematics makes that harder.
So "pure math" has built-in a need for social approval. Which selects
for people who treasure group acceptance above all else, including,
ultimately, above mathematical truth itself.
Fascinating. Then social scientists should expect to find a typical
type among modern mathematicians--selected by a particular way of
doing business.
James Harris
rossum
wrote:
>Feed the poor and hungry, not the trolls.
>
>Tom
See http://freerice.com/index.php
rossum
Mark Murray
> And none of that seems to address the question of, is this approach
> new?
Very often, your methods are so poorly written up that it takes ages
and/or multiple iterations to actually understand them well enough
to make a decision.
Most of the time, actually doing this is a waste of time.
M
Mark Murray
> Give it a rest, accusing me of NPD is something I started years ago.
> Thought it might be simpler to toss out that diagnosis than deal with
> people using schizophrenia all the time. So I made a post considering
> the question of whether or not I had NPD.
When did you routinely start accusing people of lying?
M
MichaelW
James,
Do I have this correct:
Say we have a value q such that k^2 = q mod p where p is prime and k is
unknown.
Set T=2q mod p. Factorise T into f_1 and f_2. We can determine k by
k = 3^-1 * (f_1 + f_2) mod p
For example k^2=4 mod 17. Set T = 8 mod 17 = 8. Select f_1 =2 and f_2 =
4. Given 3^-1 mod 17 = 6 we have
k = 6 * (2+4) mod 17 = 36 mod 17 = 2; correct!
Also you are asking two questions: (1) the probability of correctly
determining k is 1/k. Does this include the variant possibilities of
factoring T? (In our example f_1=1 and f_2=8 would fail).
Also (2) is there anything else in the literature that uses this approach?
Have I got the algorithms and questions correct?
Regards, Michael W.
samvaknin
http://health.groups.yahoo.com/group/narcissisticabuse/message/4948
The Narcissist as Know-it-all
http://health.groups.yahoo.com/group/narcissisticabuse/message/4945
Grandiosity, Fantasies, and Narcissism
http://health.groups.yahoo.com/group/narcissisticabuse/message/4923
Jens Stuckelberger
> Finding a result is one thing
Which you have never experienced. You will, when you realize what
a complete moron you are.
JSH
Yes. If k^2 = q mod p, and you find T, such that T = 2q mod p, and it
has factors f_1 and f_2 such that f_1 = k mod p and f_2 = 2k mod p,
then k = 3^{-1}(f_1 + f_2) mod p, which follows just from adding f_1
and f_2 together.
> For example k^2=4 mod 17. Set T = 8 mod 17 = 8. Select f_1 =2 and f_2 =
> 4. Given 3^-1 mod 17 = 6 we have
>
> k = 6 * (2+4) mod 17 = 36 mod 17 = 2; correct!
Yes. I think it's kind of fun even though it follows from such
trivially easy algebra.
> Also you are asking two questions: (1) the probability of correctly
> determining k is 1/k. Does this include the variant possibilities of
> factoring T? (In our example f_1=1 and f_2=8 would fail).
My *guess* is that the probability is roughly 1/k which I mused after
pondering the issue for a little bit.
It seems that the value of k is most important for determining how
many T's and factorizations of T, need to be checked before you find a
solution. For instance the trivial case k=q=1, always returns in a
single iteration. The case k=2, should also return extremely rapidly.
But as k increases in size, my hypothesis is that it will take longer
and longer to find it.
> Also (2) is there anything else in the literature that uses this approach?
Yes. That question is actually what this thread is for, as you can
see in its subject.
> Have I got the algorithms and questions correct?
>
> Regards, Michael W.
Sure, yeah, fine. Two main questions then:
Is this idea new? And what is the probability of success with it?
James Harris
JSH
Ok, I'm curious. What exactly is the point of you posting these
links? And let's say I accept your diagnosis and decide I have
narcissistic personality disorder, what then? What behaviors would I
show to prove to you that I've accepted your diagnosis and have made a
real change in my life?
Can you list some example of what I might then do or NOT do which
would satisfy you?
James Harris
MichaelW
James,
Answering backwards, the probability seems to depend more upon the nature
of q. If q is a prime of the form 3*r+1 then the probability is zero.
Other forms of q are giving similar results.
If q is a perfect square (the case you refer to when k < sqrt(p) ) then
it works every time as long as you select f_1 = k. I would consider this
to be a trivial case.
Since the algorithm does not really work I would have to say that the
idea may or may not be new since no one would bother publishing it.
Regards, Michael W.
P.S. When I say the probability is zero I am not being entirely precise.
If the analysis goes any further I will go into details.
JSH
Let's see. q=13, p=17. T = 26 mod 17 = 9 mod 17
So try T= 26, and go for the trivial case first, so f_1 = 26 and f_2 =
1 are up first:
3^{-1} = 6 mod 17, and 6(26 + 1) = 9 mod 17.
9^2 = 13 mod 17
So your assertion is refuted by counterexample.
> Other forms of q are giving similar results.
It seems you are refusing to accept that given f_1 = k mod p, and f_2
= 2k mod p, it MUST be true that if
T = f_1*f_2 = 2k^2 mod p, that if k^2 = q mod p, you have T = 2q mod
p, and necessarily:
3^{-1}(f_1 + f_2) = k mod p.
> If q is a perfect square (the case you refer to when k < sqrt(p) ) then
> it works every time as long as you select f_1 = k. I would consider this
> to be a trivial case.
>
> Since the algorithm does not really work I would have to say that the
> idea may or may not be new since no one would bother publishing it.
Your opinion is noted.
> Regards, Michael W.
>
> P.S. When I say the probability is zero I am not being entirely precise.
> If the analysis goes any further I will go into details.
You are precise--precisely wrong.
James Harris
Mark Murray
> Ok, I'm curious. What exactly is the point of you posting these
> links? And let's say I accept your diagnosis and decide I have
> narcissistic personality disorder, what then? What behaviors would I
> show to prove to you that I've accepted your diagnosis and have made a
> real change in my life?
1) Check your work. Do not fuel the perception that you think all your
work is somehow automatically superior. Make a serious effort of
checking your detractors' reports of errors. Finding errors in
their work is part of the process. Do this /in lieu of/
accusations of lying. Adopt a far more conciliatory tone when
making announcements.
2) Cease abruptly your habit of making grandiose claims of congressional
hearings, internet collapse and other apocalyptic musings. If these
things happen, they happen; there is no use talking about it.
3) If you have a hard time understanding your own work, then
consider how hard other folks might find it. Use this to
dramatically improve your writing. Your anger at folks'
"deliberate" misunderstanding is better directed at this improvement.
Hope this helps.
M
MichaelW
>
> - Show quoted text -
The last time you did the "precisely wrong" phrase you abandoned the
discussion after I was able to answer all points. This is what is
going to happen this time as well.
If q is a prime of the form 3*r+1 then 2q will produce only two
possible (f_1,f_2) pairs. These are (1,2q) and (2,q). Substituting q=3r
+1 into f_1+f_2 we get one of two possibilities:
k=r+1
k=2r+1
For example if q=13 then r=4 and k=5 or k=9 (which is the case you
used in your example). Note that this value of k is what is generated
irregardless of the value of p.
Now we have k^2 = q mod p so substituting we find
r^2-r = 0 mod p
or
4*r^r+r = 0 mod p
Again using the example of r=4 we require that 12 = 0 mod p or 68 = 0
mod p. The first will not provide a suitable p but the second will
work for p=17 (as you note). Note that there are more solutions if the
requirement that p be a prime is dropped.
So if q=13 there is only one solution generated by the algorithm that
is correct; p=17 and k=9. So for all the infinite solutions where k^2
= 13 mod p (e.g (k,p)=(12,131), (18,311), (24,563) etc) only one
"works". Strictly speaking if the chance of getting a correct solution
is one over infinity then the probability is zero.
I was aware of your counter example; this is why I made the qualifying
postscript. My point would be that your statement in the original post
"I'm guessing that the probability of it working is roughly 1/k" is
not precise. In particular 1/k per *what*? I have taken one possible
interpretation and on that interpretation the probability is indeed
zero. If you meant something else then please clarify. I note and echo
Mark Murray's request on this thread that you improve your writing.
Regards, Michael W.
Mark Murray
> Can you list some example of what I might then do or NOT do which
> would satisfy you?
Stop advertising.
As you already note, your maths already has coverage. The
self-promotion helps fuel the NPD idea.
Maths is either correct or its not. No amount of advertising will
change that.
M
JSH
You make absolute positions, get refuted mathematically, and then
simply make errors in reply.
Why should I bother with the tedium?
> If q is a prime of the form 3*r+1 then 2q will produce only two
> possible (f_1,f_2) pairs. These are (1,2q) and (2,q). Substituting q=3r
<deleted>
> So if q=13 there is only one solution generated by the algorithm that
> is correct; p=17 and k=9. So for all the infinite solutions where k^2
> = 13 mod p (e.g (k,p)=(12,131), (18,311), (24,563) etc) only one
> "works". Strictly speaking if the chance of getting a correct solution
> is one over infinity then the probability is zero.
There are an infinity of values for T that will work.
Whenever f_1 = 9 mod 17, and f_2 = 1 mod 17, the trivial factorization
WILL work.
There are an infinity of cases.
> I was aware of your counter example; this is why I made the qualifying
> postscript. My point would be that your statement in the original post
Which was precisely wrong as I noted.
> "I'm guessing that the probability of it working is roughly 1/k" is
> not precise. In particular 1/k per *what*? I have taken one possible
> interpretation and on that interpretation the probability is indeed
> zero. If you meant something else then please clarify. I note and echo
> Mark Murray's request on this thread that you improve your writing.
>
> Regards, Michael W.
There are an infinity of solutions where you claim 1. And you seem to
be ignoring basic realities of congruence relationships.
But are insulting and condescending in your reply nonetheless.
Fascinating.
James Harris
Jens Stuckelberger
> Hope this helps.
It won't. The guy is a complete moron. He's learned nothing, even
after well over a decade of posting junk, and being repeatedly told that
it is junk. He's good fun though - like using a punching bag to release
stress.
MichaelW
>
> You make absolute positions, get refuted mathematically, and then
> simply make errors in reply.
>
> Why should I bother with the tedium?
Why do you bother posting here at all? Unilaterally claiming to be
right convinces no-one. Ask others here what they think.
>
>
> > So if q=13 there is only one solution generated by the algorithm that
> > is correct; p=17 and k=9. So for all the infinite solutions where k^2
> > = 13 mod p (e.g (k,p)=(12,131), (18,311), (24,563) etc) only one
> > "works". Strictly speaking if the chance of getting a correct solution
> > is one over infinity then the probability is zero.
>
> There are an infinity of values for T that will work.
>
> Whenever f_1 = 9 mod 17, and f_2 = 1 mod 17, the trivial factorization
> WILL work.
>
> There are an infinity of cases.
>
Which comes back to the question you keep not answering. What does a
probability of 1/k mean?
Now by my reading of your original post I understood you to be saying
T=2q mod p and take the lowest possibly value for T. This is what my
and your examples did. I do not see why you would bother with other
values of T and by extension f_1 and f_2 since they will all generate
exactly the same k. However it is your algorithm.
> > I was aware of your counter example; this is why I made the qualifying
> > postscript. My point would be that your statement in the original post
>
> Which was precisely wrong as I noted.
>
> > "I'm guessing that the probability of it working is roughly 1/k" is
> > not precise. In particular 1/k per *what*? I have taken one possible
> > interpretation and on that interpretation the probability is indeed
> > zero. If you meant something else then please clarify. I note and echo
> > Mark Murray's request on this thread that you improve your writing.
>
> > Regards, Michael W.
>
> There are an infinity of solutions where you claim 1. And you seem to
> be ignoring basic realities of congruence relationships.
It is because of the basic realities of congruence relationships that
I took the simplest T.
>
> But are insulting and condescending in your reply nonetheless.
>
> Fascinating.
>
> James Harris
And you are obscuring the main point. So how about I rephrase things
for you.
<point>
For any prime q of the form 3r+1 there are only two values of k
generated, namely r+1 and 2r+1. These values are not correct except in
the case of a finite (and sometimes zero) number of values of p.
</point>
Samplers: q=19, never works. q=31, p=41 only solution that matches.
Is this an acceptable statement?
Regards, Michael W.
Mark Murray
> On Nov 23, 10:45 am, JSH <jst...@gmail.com> wrote:
>> You make absolute positions, get refuted mathematically, and then
>> simply make errors in reply.
>>
>> Why should I bother with the tedium?
>
> Why do you bother posting here at all? Unilaterally claiming to be
> right convinces no-one. Ask others here what they think.
I think James is being a poor listener and an arrogant prat.
He's also wrongly accusing people of exactly the faults HE
routinely commits.
M
Noob
> You are precise--precisely wrong.
How often do you plan on using that line?
cf. Message-ID:
<e482a0d8-35f3-4f2d...@z4g2000prh.googlegroups.com>
> Okay, I need to be more precise. The correct values of j are 2*m as
You already were precise--precisely wrong.
Have you already cleaned up your tracks in Google's archive?
Noob
> I think James is being a poor listener and an arrogant prat.
>
> He's also wrongly accusing people of exactly the faults HE
> routinely commits.
Psychological projection, or "Freudian projection", a defense mechanism
in which one attributes to others, one's own unacceptable or unwanted
thoughts or emotions.
MichaelW
Had a think overnight and reread your posts. I see the mistake now. I
have been assuming that you mean that T=2q only. What you meant was
that T=2q mod p so you expect to check the factoring of each possible
T.
When I originally replied I used my assumption in my example and you
seemed to confirm it so our conversation went at (rather grumpy) cross
purposes from there.
Hope that helps.
Regards, Michael W.
Lits O'Hate
> Ok, I'm curious. What exactly is the point of you posting these
> links? And let's say I accept your diagnosis and decide I have
> narcissistic personality disorder, what then? What behaviors would I
> show to prove to you that I've accepted your diagnosis and have made a
> real change in my life?
You would see a mental health professional and honestly describe
your situation. Be sure to mention collapsing the world economy,
space aliens, and information from the future.
> Can you list some example of what I might then do or NOT do which
> would satisfy you?
You might stop referring to yourself using terms like "a person who
the world will later see as one of its major discoverers."
You might stop claiming that the people you don't like will be
forced to testify before Senate committees.
You might stop contacting the government whenever you think you
found something awesome.
You might stop claiming things like "I know without a doubt that
an entire planet is wrong about my ideas."
You might stop making weak excuses for why you don't try to factor
a large number.
You might stop telling people you don't like things like "you're
pig-headed, arrogant, and also kind of dumb, unfortunately, despite
your pretensions to the contrary."
You might stop exhibiting such paranoid behavior that you write,
"for a while I walked around waiting to be killed on the street. I'd
get home and just be glad to still be alive. I'd glance at windows
wondering where a sniper would setup."
Mark Murray
> You might stop referring to yourself using terms like "a person who
> the world will later see as one of its major discoverers."
But he is!! Look! Twitter reveals all!
# examples: 7+13=20, 13=7 mod 6, 13�2 = 7�2 = 1 mod 6; 13+3=16, 13�2 = 3�2 = 9 mod 10. cool little result.
about 2 hours ago from txt
# given p_1+p_2= C, must exist h_N such that: p_1 = p_c mod h_N and p_1�2 = p_2�2 mod h_N
about 2 hours ago from txt
# noting time and date of discovery as 8:38 am pst, 11/24/09: possible proof of goldbach's conjecture.
about 6 hours ago from txt
# noting reversal of my past position that goldbach's conjecture is false. method appears to allow calculating a p_1 and p_2 given C.
about 6 hours ago from txt
# oh, guess i may have a proof of goldbach's conjecture. but i'm not giving it! i'm holding it. easy proof too, if i didn't make a mistake.
about 6 hours ago from txt
# whew! false alarm. no proof.
about 6 hours ago from txt
# will destroy notes? maybe
about 6 hours ago from txt
# yuck, maybe i've proven goldbach's conjecture. and i thought it was false! hopefully i'm looking now at a mirage.
about 6 hours ago from txt
# closing in on goldbach's conjecture?
about 6 hours ago from txt
M
Serge Paccalin
(dans <news:4b0c5ef0$0$2493$db0f...@news.zen.co.uk>, posté
dans sci.crypt,alt.math.undergrad) :
> # oh, guess i may have a proof of goldbach's conjecture. but i'm
> not giving it! i'm holding it. easy proof too, if i didn't make a
> mistake.
The right way to write this was:
“I've found a marvelous proof for this, but it doesn't fit in Twitter's
140-char limit.”
--
___________
_/ _ \_`_`_`_) Serge PACCALIN -- sp ad mailclub.net
\ \_L_) Il faut donc que les hommes commencent
-'(__) par n'être pas fanatiques pour mériter
_/___(_) la tolérance. -- Voltaire, 1763
Mark Murray
> Le Tue, 24 Nov 2009 22:32:17 +0000, Mark Murray a écrit
> (dans <news:4b0c5ef0$0$2493$db0f...@news.zen.co.uk>, posté
> dans sci.crypt,alt.math.undergrad) :
>
>> # oh, guess i may have a proof of goldbach's conjecture. but i'm
>> not giving it! i'm holding it. easy proof too, if i didn't make a
>> mistake.
>
> The right way to write this was:
>
> “I've found a marvelous proof for this, but it doesn't fit in Twitter's
> 140-char limit.”
What he /actually/ did was unwrite it. That is, he never said it the
first place. Honest.
M
Junoexpress
> Lits O'Hate wrote:
> > You might stop referring to yourself using terms like "a person who
> > the world will later see as one of its major discoverers."
>
> But he is!! Look! Twitter reveals all!
>
>
> # noting time and date of discovery as 8:38 am pst, 11/24/09: possible proof of goldbach's conjecture.
>
M