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### Quicker than a Fermat PRP for some forms

(3)

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

Feb 5

### Not much response but i still think this is outrageous result

(8)

Oh poo. But not bad for a middle-aged real estate lawyer. I'm working on a follow up but more in the form of a request 4 help on solving problems I'm
Bill Krys

Jan 7

### YA 6 selfridge test

(8)

That counterexample had gcd(Q+1,n)>1. Bouncing back with a wriggle: {tst(n,Q)=gcd(Q+1,n)==1&&kronecker(Q,n)==-1&& kronecker(Q-4,n)==1&&kronecker(Q+4,n)==1&&
paulunderwooduk

Dec 25, 2015

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divineprime

Dec 17, 2015

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