- Hi, Group.
Proth theorem extended:
Let Q= k*2^n +1, where 'n'/is a odd natural number/ and k<= 2^n +1
also odd. If for some 'a', a^((Q-1)/4) == +/-1(mod Q), then 'Q'
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 but
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 composite, 'R' will be the product of at least two
primes each of which has minimum value (2^n +1), and it follows
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.
(2^n -1)*2^n +1 = (2^n)*(2^n) - 2^n +1 >= (2^n -1)*(2^n -1)= (2^n)
*(2^n) - 2*(2^n) +1; but the 1's cancel again, so 1 >= 2 is a con-
tradiction, and the smallest product of at least two primes cannot
be derived using factors of (2^n -1).
However, the first result is incompatible with k<= 2^n +1, and so
if, a^((Q-1)/4) == +/-1(mod Q), then 'Q' is prime. *QED
not a revelation, but it works logically... Bill
I firmly believe that if 'n' is an odd prime, and not just an odd
natural number, that... 'a' only has to be '2' to achieve a valid
result. It has to do with 2^k -2 not being super-Poulet with...
2^(k*2^(p-1))-2, but I'm not sure how to prove it.