--- In

primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...>

wrote:

>

> Hello, Group.

>

> modifying Proth's Theorem...

>

> choose an odd 'n' such that 2^k +1 < n < 2*(2^k +1) for some k --

> notice that this is different from Proth's notion that 'n' must be

> < 2^k +1.

>

> if Z = n*(2^k) +1, then any number relatively prime to, less than,

> and up to the lower limit for 'n' would rather quickly expose 'Z'

as

> composite using the q^(Z-1) != 1 (mod Z) test.

>

> choose n = 15, k = 3 and 9 < 15 < 18 and q = 7 since gcd(3 or 5,

15) !

> = 1; Z = 121. compute 7^120 == 109^24 == 45^6 == 23 (mod 121). I

> could have also chosen q = 2, 4, or 8 to prove it not prime.

Oddly enough, 121 is a psuedoprime base 3 according to Sloane's list.

>

> Conversely, however, a single test might also confirm primality

when

> the value for 'q' is chosen relatively prime to, less than, and up

to

> the lower limit for 'n' as opposed to performing several

classical 'p-

> 1' tests.

>

> The two notable counter examples for Z < 1000 (that are shown prime

> via 2^(p-1) == 1 mod p test but are actually composite) are 341 &

561 and

> both of them can't be produced using the above criteria since 341 =

> 2^2 * 85 +1 (85 is much larger than n < 2*(2^2 +1)) and 561 = 2^4 *

> 35 +1 (35 is just larger than n < 2*(2^4 +1) = 34.

>

> Can someone any counter-example using this modification of Proth's

theorem?

>

> i.e. After 'Z' is produced and 'q' is so carefully chosen... Can Z

be

> composite and still return a value of '1' from a classical 'p-1'

test?

>

> I can't show it... or prove it, but it would seem so... I think the

> classical counter-examples would not fall into this class of primes.

>

> Bill

>

I would name this class of primes... "2nd kind Proth primes"