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Old 2021-03-17, 11:38   #12
Dr Sardonicus
 
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Quote:
Originally Posted by bhelmes View Post
Mp67 less than a second
Mp101:

1 sec


M1

A :31
B :4
C :4
D :1



M278557
A :51109652505043588650876826875
B :2347607336006370217211437694510
C :2347607336006370217211437694510
D :191163035652478580518938993307
ERG :7432339208719
Quote:
Originally Posted by LarsNet View Post
Where can you look up the numbers, llke M278557, so i can test as well. ( New here, still learning )
Unfortunately, the OP's notation is very confusing. In the above, M278557 is Mod(M1, 2^101 - 1)^278557 rather than what in standard notation is the Mersenne number M278557.

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").
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Old 2021-03-17, 11:55   #13
kruoli
 
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Quote:
Originally Posted by Dr Sardonicus View Post
I also don't know what "ERG" stands for, but it appears to indicate a factor of the number under consideration.
Most likely the same as "RES", because the German word for "result" is "Ergebnis".
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Old 2021-03-17, 13:34   #14
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Quote:
Originally Posted by kruoli View Post
Quote:
Originally Posted by Dr Sardonicus View Post
I also don't know what "ERG" stands for, but it appears to indicate a factor of the number under consideration.
Most likely the same as "RES", because the German word for "result" is "Ergebnis".
OK, that make sense.

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 1-variable 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 2021-03-17 at 13:38 Reason: xinfig posty
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Old 2021-03-17, 21:13   #15
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Oh, I think I have made an error in the programming part, sorry for that.


Covid-19 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
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Old 2021-03-18, 14:37   #16
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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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