General factoring result for k^m = q mod N
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From: JSH on 13 Jun 2010 11:28 On Jun 13, 8:12 am, Tom St Denis <t...(a)iahu.ca> wrote: > On Jun 13, 10:36 am, JSH <jst...(a)gmail.com> wrote: > > > 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 > > "not" ... understatement of this century. Hope so!!! > > Cool. Well guess that breaks something in encryption. NSA should > > start looking for a new method, fast. > > I don't get how this follows, and I think that's what makes you > "special." You admit your idea is probably bunk, but then conclude > that it's a valid result afterall with no justification whatsoever. Well, turns out that there is no other known general method--I don't like to call brute force a general method but you can qualify that with "besides brute force"--for finding k, when k^m = q mod N. And my original post GIVES such a method that relies on factoring. So there is this new thing. So there is a new unknown. Given a new foundation level result in modular arithmetic much that is new may now be possible. No rational person would dispute that until the unknown becomes the known. People who post in reply as if they can dispute it; therefore, are not rational. And that is a fair assessment unless they show that the "unknown" is actually a known!!! James Harris
From: Mark Murray on 13 Jun 2010 14:10 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 you completely failed to notice that "your" result is a trivial consequence of the above. > 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. > > Thanks "Mark Murray"! I hadn't realized that route existed. So > you've helped reveal a new potential vulnerability. And on the > appropriate newsgroup! Try reading some books on the subject you claim to be such an expert on. These are so fundamental to the "basic" result you claim, that its embarassing. > Yuck. This result then potentially breaks all known encryption > schemes around modular arithmetic, that involve integer factorization > or discrete logarithms, which is what a foundation level result can > give you. Nope. The Chinese Remainder Theorem and modular exponentiation are both well understood by cryptologists. Pity your "research" only just stumbled upon them. You could have saved an awful lot of trouble. > British and American mathematicians ignore this result--if it does, > still not sure but for national security reasons will leave this > message in here--at their peril as of course people in other countries > may have no reason to acknowledge that they know of this result, > especially to take down the big arrogant boys who claim to be at the > top of the heap. Usual hubris. *Yawn* M -- Mark "No Nickname" Murray Notable nebbish, extreme generalist.
From: JSH on 13 Jun 2010 14:18 On Jun 13, 11:10 am, Mark Murray <w.h.o...(a)example.com> wrote: > 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 you completely failed to notice that "your" result is a trivial > consequence of the above. Um, but no, it's not. Rather than argue with you, assuming your brain may have snapped, try to derive it from the above. > > 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. > > > Thanks "Mark Murray"! I hadn't realized that route existed. So > > you've helped reveal a new potential vulnerability. And on the > > appropriate newsgroup! > > Try reading some books on the subject you claim to be such an expert > on. These are so fundamental to the "basic" result you claim, that > its embarassing. I don't claim to be an expert. I'm not a mathematician. However, there is no other known method besides brute force to in general find k, when k^m = q mod N, where m is a natural number. None. Absolute. No other than mine in all of human history. That is black and white. Absolute. No if's and's or butt's in there. Claiming otherwise is just a matter of GIVING ONE. And you helped make things worse as I realized that those equations may allow you to handle logarithmic whatever, so that handles what the NSA claims it uses. That breaks all main US encryption systems overnight. So it's a national security meltdown. Dude, you really need to get a grip and handle your cognitive dissonance. I don't want you using a mental insanity defense later as I say that doesn't fly. My position is willful disregard of the facts in order to deny your role in helping to create a national security situation worldwide. James Harris
From: Bruce Stephens on 13 Jun 2010 14:25 JSH <jstevh(a)gmail.com> writes: [...] > However, there is no other known method besides brute force to in > general find k, when k^m = q mod N, where m is a natural number. > None. Absolute. No other than mine in all of human history. > > That is black and white. Absolute. No if's and's or butt's in there. > > Claiming otherwise is just a matter of GIVING ONE. > > And you helped make things worse as I realized that those equations > may allow you to handle logarithmic whatever, so that handles what the > NSA claims it uses. > > That breaks all main US encryption systems overnight. So it's a > national security meltdown. Why? Notice that the discrete logarithm problem (as used in DH, for example) is to find m given k, q, N in k^m = q mod N. Finding k is of no importance since it's a public parameter.
From: Mark Murray on 13 Jun 2010 14:38
On 13/06/2010 19:18, JSH wrote: >> ... but you completely failed to notice that "your" result is a trivial >> consequence of the above. > > Um, but no, it's not. Rather than argue with you, assuming your brain > may have snapped, try to derive it from the above. Do your own checking. You are the one making the wild claims. Until an hour or three ago, you weren't even AWARE of these results; now go and do your homework properly this time. I'm not wasting my time on it. >> Try reading some books on the subject you claim to be such an expert >> on. These are so fundamental to the "basic" result you claim, that >> its embarassing. > > I don't claim to be an expert. I'm not a mathematician. Terminology. You claim to be a mathematical discoverer, and you claim that "your" mathematics is the most advanced in the world. The way you use transparent weasel-words like "may" and "could" fools only you. > However, there is no other known method besides brute force to in > general find k, when k^m = q mod N, where m is a natural number. > None. Absolute. No other than mine in all of human history. No. INCLUDING yours. Your method is a brute force search. Only you can't see that. > That is black and white. Absolute. No if's and's or butt's in there. > > Claiming otherwise is just a matter of GIVING ONE. > > And you helped make things worse as I realized that those equations > may allow you to handle logarithmic whatever, so that handles what the > NSA claims it uses. > > That breaks all main US encryption systems overnight. So it's a > national security meltdown. In which case prove it by posting a decrypt. Bet you can't. Bet you won't, citing a weak "protecting national security" excuse (or claims of fear of "the authorities"). Whatever the excuse is, see if you can make it worthy of a T-shirt. Your usual ones are tired and worn out. > Dude, you really need to get a grip and handle your cognitive > dissonance. I don't want you using a mental insanity defense later as > I say that doesn't fly. My position is willful disregard of the facts > in order to deny your role in helping to create a national security > situation worldwide. Can I use a mental insanity defence NOW? (What's "mental insanity", by the way? Same as "insanity"?) Just joking. I'm not insane, I'm definitely outsane. M -- Mark "No Nickname" Murray Notable nebbish, extreme generalist. |