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Old 2009-04-06, 14:45   #1
Bill Bouris
Apr 2009

210 Posts
Default Proth theorem extended

Proth theorem extended:
let Q= k*2^n +1, where n=>3 is a odd natural number & k<= 2^n +1. if for some 'a', a^((Q-1)/4) == +/-1(mod Q), then 'Q' is prime. 'k' doesn't need to be restricted to only 'odd' numbers, either.

if 'm' is from the set of natural numbers, then every odd prime divisor 'q' of a^(2^(m+1))+/-1 implies that q == +/-1(mod a^(m+2)) [concluded from generalized 'Fermat-number' proofs by Proth along with my replacing 'm' with 'm+1'].

now, if 'p' is any prime divisor of 'R', then a^((Q-1)/4) = (a^k)^(2^(n-2)) == +/-1(mod p) implies that p == +/-1 (mod 2^n). thus, if 'R' is compo-site, 'R' will be the product of at least two primes each of which has minimum value of (2^n +1), and it follows that...

k*2^n +1 >= (2^n +1)*(2^n +1) = (2^n)*(2^n) + 2*(2^n) +1; but the
1's cancel, so k*(2^n) >= (2^n)*(2^n) + 2*(2^n) and upon dividing by 2^n... k>= 2^n +2.

moreover, this result is incompatible with our definition, so if k<= 2^n +1 and a^((Q-1)/4) == +/-1(mod Q), then 'Q' is prime.

finally, if either k= 2*m or k= 2*m +1, then 'Q' is still 'odd'; Q= 2*m*2^n +1= m*2^(n+1)+1, or Q= (2m+1)*2^n +1= m*2^(n+1) +2^n+1.
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