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Re: Quicker than a Fermat PRP for some forms
Taking the remainder away from the quotient and letting the base for the remainder exponentiation be a*b, should speed things up a little... ?
paulunderwooduk 
Re: Quicker than a Fermat PRP for some forms
I may have underestimated the timesaving.
paulunderwooduk 
Quicker than a Fermat PRP for some forms
Using an example, let N=2^n3. Let N=n*q+r where r < n/2. 2 == 2^N (mod N) [Fermat] So 2^(n*q+r) == (2^n)^q*2^r == 3^q*2^r == 2 (mod N) If N is 1 million
paulunderwooduk
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PrimeNumbers is a email discussion list for those
who enjoy prime numbers. Some look for patterns,
some seek to find new records, others try to understand
the distribution. We try to limit our discussion to
prime numbers, primality testing, and very closely
related subjects.
who enjoy prime numbers. Some look for patterns,
some seek to find new records, others try to understand
the distribution. We try to limit our discussion to
prime numbers, primality testing, and very closely
related subjects.
If you enjoy prime numbers there are similar lists you
might enjoy, such as PrimeForm
focusing on primality proving programs and
GIMPS'
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If you think you have a Goldbach proof, then please note that there are groups which deal specifically with such matters, perhaps try those first?
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 1114
 Number Theory
 Dec 27, 2000
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