JSH: Posting realities, residues result
in [Math]
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From: JSH on 11 Jun 2010 10:13 I've been fascinated by the response to my posting of a general result for solving quadratic residues because of the elephant in the room NOT mentioned, which is that it is mod N, whereas mathematicians usually MUST use mod p, where p is a prime. That's important for factoring as if N is a composite then with k^2 = q mod N, you have multiple non-trivial solutions for k, where I mean, not just k, and N - k, whereas if N=p, then there is only one. And in all the discussion that erupted around my postings on this result you may have noticed I noted that the approach tends to prefer large k, though all the details of how that works out are not clear even to me. So imagine N = p_1*p_2, so there are *TWO* values for k, given k^2 = q mod N. If you *picked* k_1 to find q, such that it will tend to be small, and k_2 is more likely to be selected then the approach I discovered could conceivably give you k_2, then you have p_1 or p_2 from (k_2 - k_1) gcd N. So it's a way to deliberately probe for factors of N, picking k_1 in order to try and get k_2, which is a technique not available with any other known method, because no other general way to solve for residues is known! So why would posters argue endlessly with me on other subjects and miss this amazing thing that the approach is mod N and not mod p? Because they hate me. Also they want me to be wrong!!! No matter what as it's personal with them. Mathematics is a sideline to their actual activity. It doesn't rank in importance to what they really want. Human beings are quirky creatures. They can do the damndest things. So if you're a Usenet poster arguing with people on Usenet the LAST THING you wish for a target to be, is actually right! I'll give the result again, and note that it's trivially derived though I won't give the derivation again. I'll also note that the result is basic research so it's not clear how hard it is to get it to work at any level. The devil is in the details. For instance with nuclear weapons it's a far cry from knowing you just have to slap some plutonium together to actually building a working nuclear weapon. Now then, back to national security! It IS quite possible that this information could be of interest to governments, oh, all over the world. Failure to disclose of it for some of you could be seen as a LACK OF LOYALTY in your respective countries. Usenet posters who reply decrying the result could face extraordinary scrutiny in the near future if only from the world press wondering how they could do such a thing. And could find themselves labeled for life. Post in reply now with care. No matter how little you think of Usenet, you can make a decision in this thread which you can't take back, which will end the life you knew, and move you into a Hell on earth that you will not escape until you die. Given an mth residue where m is a natural number, q mod N, to be solved one can find k, where k^m = q mod N, from k = (a_1+a_2+...+a_m)^{-1} (f_1 +...+ f_m) mod N where f_1*...*f_m = T, and T = a_1*...*a_m*q mod N and the a's are free variables as long as they are non-zero and their sum is coprime to N. It's a general result, which may have been known to Gauss and simply didn't get written down, or maybe he did and no one noticed. It's not the sort of thing that had the importance in the past that it MAY have in our modern age of computers and systems based on factoring as a hard problem. It is a general result at the heart of modular arithmetic. No one can really say for sure how big it actually may be as general results have that quality. Kind of like differentiation with the calculus. How big is that? Has humanity determined its limits yet? James Harris
From: José Carlos Santos on 11 Jun 2010 11:34 On 11-06-2010 15:13, JSH wrote: > I've been fascinated by the response to my posting of a general result > for solving quadratic residues You have been fascinated for *decades* now by the way we react to your postings. And I don't think that the reactions have changed that much through the years... > Now then, back to national security! It IS quite possible that this > information could be of interest to governments, oh, all over the > world. Failure to disclose of it for some of you could be seen as a > LACK OF LOYALTY in your respective countries. So, you have already warned the government of *your* country, right?! > Usenet posters who reply decrying the result could face extraordinary > scrutiny in the near future if only from the world press wondering how > they could do such a thing. And could find themselves labeled for > life. Another one among the many pathetic threats that you have been making for decades. Who do you think you will scare? > Post in reply now with care. Why, crackpot? > No matter how little you think of > Usenet, you can make a decision in this thread which you can't take > back, which will end the life you knew, and move you into a Hell on > earth that you will not escape until you die. I'm *so* scared. :-) Best regards, Jose Carlos Santos
From: Jesse F. Hughes on 11 Jun 2010 11:48 JSH <jstevh(a)gmail.com> writes: > Now then, back to national security! It IS quite possible that this > information could be of interest to governments, oh, all over the > world. Failure to disclose of it for some of you could be seen as a > LACK OF LOYALTY in your respective countries. Herc claims that he can answer any question[1] by channeling. If he is right, then this fact is certainly important for national security of every nation on earth. So, if we fail to alert the NSA, say, that Herc claims this power, we might be disloyal to our nations. Is that how it works? Footnotes: [1] Well, any *interesting* question, and a question is only interesting if we cannot prove his answer is false. -- Jesse F. Hughes "[M]oving towards development meetings for new release class viewer 5.0 and since [I]'m the only developer, easy to schedule." --James S. Harris tweets on code development
From: Rick Decker on 11 Jun 2010 12:14 On 6/11/10 10:13 AM, JSH wrote: > I've been fascinated by the response to my posting of a general result > for solving quadratic residues because of the elephant in the room NOT > mentioned, which is that it is mod N, whereas mathematicians usually > MUST use mod p, where p is a prime. Oh, really? On what are you basing this observation, pray tell? > > That's important for factoring as if N is a composite then with k^2 = > q mod N, you have multiple non-trivial solutions for k, where I mean, > not just k, and N - k, whereas if N=p, then there is only one. This has been known for centuries, of course, but thanks for the heads-up. (BTW, it's not true in general.) You might find it helpful to study the Chinese Remainder Theorem. > > And in all the discussion that erupted around my postings on this > result you may have noticed I noted that the approach tends to prefer > large k, though all the details of how that works out are not clear > even to me. > > So imagine N = p_1*p_2, so there are *TWO* values for k, given k^2 = q > mod N. If you *picked* k_1 to find q, such that it will tend to be > small, and k_2 is more likely to be selected then the approach I > discovered could conceivably give you k_2, then you have p_1 or p_2 > from (k_2 - k_1) gcd N. As is often the case with your results, all you have done here is to invent an obfuscated "test some random values" method. > > So it's a way to deliberately probe for factors of N, picking k_1 in > order to try and get k_2, which is a technique not available with any > other known method, because no other general way to solve for residues > is known! This is truly risible, coming from someone who stubbornly refuses to do even the most basic literature search. > > So why would posters argue endlessly with me on other subjects and > miss this amazing thing that the approach is mod N and not mod p? > Because they hate me. Also they want me to be wrong!!! No matter > what as it's personal with them. <snip the rest of the pity-party> Regards, Rick
From: Joshua Cranmer on 11 Jun 2010 17:22
On 06/11/2010 12:14 PM, Rick Decker wrote: > As is often the case with your results, all you have done here > is to invent an obfuscated "test some random values" method. To be fair, most factoring methods boil down to "test for one of these values", but then again, they typically improve on the naive algorithm by drastically narrowing the search space. -- Beware of bugs in the above code; I have only proved it correct, not tried it. -- Donald E. Knuth |