Hi,
I was looking at the LL-test and thought of the following:
To succeed a LL-test the final value (mod 2^p - 1) has to be zero.
To be able to get this there has to be a number x (between 0 and 2^p - 2)
such that x^2 = 2. This because in the final step of the
LL test the value is squared and substracted by two.
In some cases for example 15 = 2^4 - 1 (this cannot be prime because
four isn't but that's not the point) such a value does not exist:
0^2 = 0, 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 1, 5^2 = 10, 6^2 = 6, 7^2 = 4,
8^2 = 4, 9^2 = 6, 10^2 = 10, 11^2 = 1, 12^2 = 9, 13^2 = 4 and 14^2 = 1.
Of course in this case p = 4 is not prime so we would never do a LL-test
anyway. My question is whether anyone knows if such an x always exists if p
is indeed a prime ? Perhaps this is only true for a subset of all primes,
and this would allow us to focus on this subset only (if it can be
determined more easily) ?
-------------------------------------------------------------------
Benny Van Houdt,
University of Antwerp
Dept. Math. and Computer Science
PATS - Performance Analysis of Telecommunication
Systems Research Group
Universiteitsplein, 1
B-2610 Antwerp
Belgium
email: [EMAIL PROTECTED]
--------------------------------------------------------------------
________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
