Browse Groups

• Hi, Group. revised proof... Proth theorem extended: Let Q= k*2^n +1, where n /is a odd natural number/ and k
Message 1 of 1 , Jul 31, 2008
View Source
Hi, Group.

revised proof...

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'
is prime.

Proof:
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
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.

(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.
Your message has been successfully submitted and would be delivered to recipients shortly.
• Changes have not been saved
Press OK to abandon changes or Cancel to continue editing
• Your browser is not supported
Kindly note that Groups does not support 7.0 or earlier versions of Internet Explorer. We recommend upgrading to the latest Internet Explorer, Google Chrome, or Firefox. If you are using IE 9 or later, make sure you turn off Compatibility View.