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Old 2021-09-09, 23:05   #1
bhelmes
 
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Mar 2016

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Default speed up by a linear substitution of a quadratic polynomial ?

A peaceful and pleasant day for you,

Let f(n)=2n²-1 and the linear substitution n=Mp*k+1, Mp is the exponent of the coresponding Mersenne number

I know that there is a factor g | f(n0) if 2^Mp-1 is not prime.
I think the quadratic polynomial after the linear substitution contains also the same g | f(k0)

Is it right that the linear substitution speed up the search for g by successiv increasing the k with one ?

P.S. @Dr Sardonicus, I appreciate your answers, clear words with logical constructions, thanks for that.
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Old 2021-09-15, 23:15   #2
bhelmes
 
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A peaceful and pleasant night for you,

I have successfully implemented the following algorithm:

1. precalculate primes and n with help of f(n)=2n²-1
2. make a linear substitution with n=2*Mp*k+1 so that g(k)=8*Mp*k(Mp*k+1)+1
and calculate for the precalculated primes p the k=(n-1)*(Mp⁻¹)*(2⁻¹) mod p
3. As g(k)=1 mod Mp, I checked if the product of the sieving factors of g(k)=1 mod Mp
4. If the last condition is true, I calculate g(k) and divide by the sieving factors
and finally I checked if the remaining cofactor is a divisor of 2^Mp-1

The program seems to be right but a little bit slow ...

What a wounderful piece of work (600 lines in c), I did not parallel it.



Last fiddled with by bhelmes on 2021-09-15 at 23:16
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Old 2021-09-23, 00:35   #3
bhelmes
 
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A peaceful night for you,

The last program was too slow. I am thinking of using the chinese remainder theorem in order to search for a factor of Mp.

I would suggest a linear substitution for f(n)=2n²-1 with n=Mp*k+1, using some precalculated primes and
sort them in different modulo classes (mod Mp).

As the function value is always f(k)=1 mod Mp and the factor g of Mp is also g=1 mod Mp
I am looking for precalculated primes p=1 mod Mp, so that f(k)=p*g and
in the second run for two precalculated primes p*q=1 mod Mp, so that f(k)=p*q*g

Is this method practical and an improvement, or do I replace a slow version by a worse implementation ?

I am not the fastest guy in programming and if someone could give me a hint, would be nice.


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