Solving k^m = q mod N
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From: Rotwang on 8 Jun 2010 21:26 JSH wrote: > > [...] > > For insight pick up any number theory journal and note how much in it > is NOT practical alongside methods considered more practical. By your > criteria the mathematicians who include such results are idiots for so > doing. Are you able to provide specific references to some of these papers you reckon are to be found in any number theory journal? Or are you one of those posters who "love making claims that they don't feel a need to back up"?
From: Tim Little on 8 Jun 2010 22:23 On 2010-06-08, MichaelW <msjmb(a)tpg.com.au> wrote: > To be fair the time for the MRSA to solve the equation is of the same > order as sequentially checking 1,2,3... allowing that there is a > theoretical possibility of taking infinite time. I suppose that digits of pi are easy enough to compute that this is close enough to the truth, yes. - Tim
From: JSH on 8 Jun 2010 22:34 On Jun 8, 6:26 pm, Rotwang <sg...(a)hotmail.co.uk> wrote: > JSH wrote: > > > [...] > > > For insight pick up any number theory journal and note how much in it > > is NOT practical alongside methods considered more practical. By your > > criteria the mathematicians who include such results are idiots for so > > doing. > > Are you able to provide specific references to some of these papers you > reckon are to be found in any number theory journal? Or are you one of > those posters who "love making claims that they don't feel a need to > back up"? I meant number theory textbook. Pick up any of your choice and look at the first few pages. Usually fairly trivial results are established first. My point is that easy does not equate with worthless. Foundation level results can be very easy, but their generality makes them still important. James Harris
From: Tim Little on 8 Jun 2010 22:35 On 2010-06-08, JSH <jstevh(a)gmail.com> wrote: > Talk about arrogant, readers should note that "MichaelW" is equating > whatever he thought up at the moment against a general method for > solving for k, when k^m = q mod N Why arrogant? What he thought up *is* a general method for solving for k. > Don't expect to be cheered. But don't lose heart when jeered. Just > look at my example. Your example is a good reason *for* losing heart when jeered. If one doesn't, one might end up doggedly pursuing obviously worthless approaches for months at a time, like you. Much of what you call "jeering" (and ignore) actually contains worthwhile criticism that could help you. Usually you end up acknowledging that a given approach is worthless after a couple of months of ignoring people who are directly showing you why it is worthless, and then rant about how nobody ever pointed it out. As you will again. Welcome to the beginning of your next cycle. - Tim
From: JSH on 8 Jun 2010 22:44
On Jun 8, 6:04 pm, MichaelW <ms...(a)tpg.com.au> wrote: > On Jun 9, 10:22 am, JSH <jst...(a)gmail.com> wrote: > > > > > Trivially you find for instance that if k = floor(N/2), this approach > > will quickly find solutions with very small values of a_1 and a_2, > > typically, a_1=a_2 = 1, or a_1 = 1, a_2 = 2. > > > It has other somewhat odd behavior though. > > > For instance if you have q a perfect square such that q = k^2 < N, but > > k^2 is approximately equal to N, that is a large square near N is q, > > then this approach find solutions as well with small a's, doing so > > quickly (may give a very fast square root algorithm in fact). > > James, > > Long post showing you don't understand what you are doing. To pick out > an example if a_1 = a_2 = 1 then the solution is obviously f_1= f_2 = > k. Your algorithm becomes searching q, q+N, q+2N, q+3N... until you > hit k^2. You're wrong. > The stuff about square root algorithms does not have a shred of > evidence to support it. You're wrong again. > I asked for evidence and I got an ego trip. Provide evidence. I've noted before that I have a basic test Java program zipped up in QuadRes.zip in the Files section of mymathgroup: http://groups.google.com/group/mymathgroup/files?hl=en I've been playing with the program for a while and watching some of the behavior, where yeah, if you put in large squares, less than N-- especially near N itself, it solves them freaking fast. And again, that's just m=2. The result covers m, a Natural number. No sane person thinks they can figure out infinity by glancing at some general equations--or by dismissing the source as a supposed crackpot. You're a social animal. You know what people tell you is important. Otherwise, you're lost. I guess Usenet attracts people like you, but it still seems sad. That mathematical beauty to you is just a phrase, as all you know--is what other people tell you. James Harris |