20210317, 11:38  #12  
Feb 2017
Nowhere
11622_{8} 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^{3}·5·17 Posts 

20210317, 13:34  #14  
Feb 2017
Nowhere
5010_{10} 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^{2}×89 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
2·3·5·167 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. 
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