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
I may have underestimated the time-saving.paulunderwooduk
Using an example, let N=2^n-3. 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 millionpaulunderwooduk
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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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- Dec 27, 2000
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