General factoring result for k^m = q mod N
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From: JSH on 15 Jun 2010 21:28 On Jun 15, 10:33 am, Mark Murray <w.h.o...(a)example.com> wrote: > On 15/06/2010 15:14, JSH wrote: > > >> I cited your solution, which was a Java program, which used brute force. > > > Why would that be significant? Have you actually *looked* at the > > equations in the post that starts this thread? > > I have looked at the equations. I posted some detail about them too. > > As for significance, the result you cite is (as you admit) trivial. No, the *derivation* is trivial. > Solutions, however, are not efficient, and you are (yet again) Stated without proof. > trumpeting the downfall of civilisation without the faintest > notion that this method won't make a jot of difference. There is no known general method for solving for k, when k^m = q mod N, with m a Natural number besides mine. That is still true. Search the world, none other exists. It's a fundamental result in modular arithmetic. Modular arithmetic is key in a lot of encryption techniques. > > Or are you just hounding me in replies without thinking? > > Nope. I'm pointing out your lack of thinking. Yet, the result I've shown is trivial to derive, but you claim I said it is itself trivial. You've claimed it's equivalent to brute force but given no mathematical proof to back up that claim. Count in false assertion after false assertion you've simply continued to reply--making more false assertions. Doesn't seem like a thoughtful response to me! And why bother? If I'm wrong, I'm wrong, so why make false statements? Why bother? James Harris
From: Mark Murray on 16 Jun 2010 03:19 On 16/06/2010 02:28, JSH wrote: >>> Why would that be significant? Have you actually *looked* at the >>> equations in the post that starts this thread? >> >> I have looked at the equations. I posted some detail about them too. >> >> As for significance, the result you cite is (as you admit) trivial. > > No, the *derivation* is trivial. Same difference. >> Solutions, however, are not efficient, and you are (yet again) > > Stated without proof. Your claim is that solving this requires factoring. Factoring is inefficient. Therefore you are claming a solution that is inefficient. Q.E.D. >> trumpeting the downfall of civilisation without the faintest >> notion that this method won't make a jot of difference. > > There is no known general method for solving for k, when k^m = q mod > N, with m a Natural number besides mine. > > That is still true. > > Search the world, none other exists. A plethora of methods exist, including your brute-force search. Your inability to find solutions by web-search is meaningless. > It's a fundamental result in modular arithmetic. Modular arithmetic > is key in a lot of encryption techniques. Correct, which is why solutions need to be efficient. Your factoring method is not. >>> Or are you just hounding me in replies without thinking? >> >> Nope. I'm pointing out your lack of thinking. > > Yet, the result I've shown is trivial to derive, but you claim I said > it is itself trivial. The word you used is "simple". So call me a liar for not using the exact word you did. > You've claimed it's equivalent to brute force but given no > mathematical proof to back up that claim. No. I said that the only solution you have presented was subset of the original problem (m=4, f_3, f4, a3, a4 all unity), in which you solved by brute force search. > Count in false assertion after false assertion you've simply continued > to reply--making more false assertions. My assertions haven't been NEARLY as false as you'd like them to be. > Doesn't seem like a thoughtful response to me! > > And why bother? If I'm wrong, I'm wrong, so why make false > statements? > > Why bother? If you're wrong, why bother to post? Until I pointed out the Chinese remainder theorem and modular exponentiation, you hadn't a clue they existed, yet you continue to maintain your grandiose claims about how general, unique and important "your" discovery is. What hasn't become aparrent in your writing is a realisation of how much you didn't know to start with. Yet, you hang on, with a white- knuckled grim determination to a trivial, inefficient result, while fanatising about fame and celebrityhood. "I am a discoverer". HAH. M -- Mark "No Nickname" Murray Notable nebbish, extreme generalist.
From: Mark Murray on 16 Jun 2010 14:01
On 13/06/2010 15:36, JSH wrote: > On Jun 13, 3:21 am, Mark Murray<w.h.o...(a)example.com> wrote: >> On 13/06/2010 03:27, JSH wrote: >> > Yet I'm the one who found it, over 200 years since Gauss introduced >> > "mod" in 1801. >> >> 1) Chinese remainder theorem. >> >> 2) Modular exponentiation. > > Interesting, chased the link to Wikipedia for modular exponentiation > and that got me to wondering my result could be used to find e. But your extensive "research" never got you to 3) Discrete logarithm. http://en.wikipedia.org/wiki/Discrete_logarithm > Given c = b^e mod m, where c, b and m are known, yeah, it seems to me > that is should, potentially, maybe be possible using my result to > figure out e. But maybe not. I decided to stop thinking on it after > a point. Kind of overwhelming. So the rest may not be valid, but I > have to toss it out there anyway for national security reasons, as the > "unknown" is not good. It's bad. > > Cool. Well guess that breaks something in encryption. NSA should > start looking for a new method, fast. Not until you tell them how to factor fast enough. M -- Mark "No Nickname" Murray Notable nebbish, extreme generalist. |