20210317, 11:38  #12  
Feb 2017
Nowhere
2·29·103 Posts 
Quote:
Quote:
I also don't know what "ERG" stands for, but it appears to indicate a factor of the number under consideration. And unfortunately, the OP didn't indicate what polynomial and what prime he used to construct M1 for either 2^67  1 ("Mp67") or 2^101  1 ("Mp101"). 

20210317, 11:55  #13 
"Oliver"
Sep 2017
Porta Westfalica, DE
2·571 Posts 

20210317, 13:34  #14  
Feb 2017
Nowhere
1011101010110_{2} Posts 
Quote:
But there's something else I don't understand. The OP's PDF starts with a polynomial, and uses a linear substitution to produce values divisible by a given prime p, then divides out a factor of p. Fair enough. But for the polynomial 2*x^2  1, the matrix M1 does not in any way represent the transformed polynomial. For example, with 2*x^2  1 and p = 31, substituting 31*x + 4 for x gives 2*31^2 *x^2 + 2*2*31*4*x + 2*4^2  1. Dividing by 31 gives 2*31*x^2 + 2*2*4*x + 1. Now "homogenizing" these polynomials gives the binary quadratic forms 2*x^2  y^2 and 62*x^2 + 16*x*y + y^2, which both have discriminant 8, and have the usual matrix representations M = [2,0;0,1] and [62,8;8,1] respectively; matrix multiplication gives [x,y]*[2,0;0,1]*[x;y] = x^2  2*y^2 and [x,y]*[62,8;8,1]*[x;y] = 62*x^2 + 16*x*y + y^2. However, the OP uses the matrix M1 = [31,4;4,1] obtained by dividing the lead coefficient of 2*31*x^2 + 2*2*4*x + 1 by 2 and the coefficient of x by 4. M1 represents the binary quadratic form 31*x^2 + 8*x*y + y^2 which has discriminant 15, and corresponds to the 1variable polynomial 31*x^2 + 8*x + 1, which bears no obvious relation to the polynomial 2*x^2  1. Last fiddled with by Dr Sardonicus on 20210317 at 13:38 Reason: xinfig posty 

20210317, 21:13  #15 
Mar 2016
2×5×41 Posts 
Oh, I think I have made an error in the programming part, sorry for that.
Covid19 is not good for my health and takes too long. I think I will make a small holiday time. Computer is working in the background. Have a pleasant time Bernhard 
20210318, 14:37  #16 
Feb 2017
Nowhere
2×29×103 Posts 
The fact that your procedure works as intended with the "wrong" input is suggestive. It suggests that it has nothing to do with the matrix "representing" your polynomial.
But the matrix you constructed shares an important matrix property with the matrix you intended. It's a special kind of square matrix whose known properties make things clear. 
20211112, 16:36  #17  
Mar 2016
2·5·41 Posts 
A peaceful day for you,
Quote:
I make a linear substitution n=p+no (where p=2(p+n0)²1) for the polynomial f(n)=2n²1 and divide it by p and transform it to a matrix (2p 2n0) (2n0 1) which I use for exponentation with primes < 10^8 This seems to be like two congruences with f(n)=1 mod p and f(n)=0 mod f where f is a factor of the factored number The solution is depending of a+b where a is depending of the first congruence and b of the second congruence. Therefore the solution is not a p1 or p+1 test, but "flexible" cause of the choosing p. This is a huge progress, or ? I add the factor of M137 with runtime (computer is a AMD FX(tm)8300 OctaCore Processor): Mp : 137 factor : 32032215596496435569<20> Matrix : 14, 4 4, 1 Prime exponent = 27978309 elapsed time: 1106.18s Mp : 137 factor : 32032215596496435569<20> Matrix : 194, 14 14, 1 Prime exponent = 27978707 elapsed time: 1122.35s Mp : 137 factor : 32032215596496435569<20> Matrix : 254, 16 16, 1 Prime exponent = 27979159 elapsed time: 1150.24s By the way, the second stage (B2) is still missing in my program. Any suggestion how to perform the second stage (similar to eliptic curves) and how to balance B1 and B2 ? Last fiddled with by bhelmes on 20211112 at 16:57 

20211221, 00:57  #18  
Mar 2016
2×5×41 Posts 
Quote:
Is it a p1 test or is it "flexible" and depends on the choosen matrix. If someone has enough Algebra knowledge it would be nice to get a short answer. 

20211221, 16:40  #19 
17203_{8} Posts 
You may be setting yourself up for failure and "less than" benevolent comments.
As a simple and cursory check look at the "Implementation" section of https://en.wikipedia.org/wiki/Pollar...92_1_algorithm Ask yourself, where are you situated in relation to "anyone else" who has looked at what you're looking at seriously. You will be outclassed in resources such as equipment, theory, time and a few other things but only you think the way you do so if you believe you have a good idea then work it to perfection and try not to present prototypes. There are always consequences so try to anticipate those as well. Do your homework and study relative to your abilities. An easy check would be to look at arXiv papers on any subject that interests you..or better yet..look at Quanta articles. I have a couple of recent books ..The Wall of Fire and Prime Number Conspiracy which I found to be engaging and ties together a number of peer reviewed papers I've read. I'll add one more, Nahin's "Dr. Euler's Fabulous Formula" and read the top paragraph on p.221 for instance. Whatever you present .. and I try to do the same..is to make it flawless regardless of the simplicity/complexity. My communication skills are generally poor and whatever spoor (pun and double entendre) I leave is targeted toward a particular audience keeping in mind all the eyes and bots that review this forum. A word of encouragement, you are in the right ballpark from my perspective but you really need to latch your imagination onto mathematical facts and proofs, try to do some proofs on your own as a self test (ie. Laplace Transform). As a final thought, watch the initial season (or more) of Ted Lasso..cheers. Last fiddled with by jwaltos on 20211221 at 17:05 Reason: corrections 
20220126, 02:12  #20  
Mar 2016
2×5×41 Posts 
Quote:
The special linear group, SL(n, F), is the group of all matrices with determinant 1. When F is R or C, SL(n, F) is a Lie subgroup of GL(n, F) of dimension n^2 − 1. https://en.wikipedia.org/wiki/Genera...l_linear_group 

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