Will Edgingtom writes:
If I understand P-1 factoring correctly, then using it to a stage one
bound of k to try to factor M(p) will find all possible factors less
than or equal to 2*k*p + 1. I'm assuming that p is less than k (or p
is always used in the powering) and the convention several
[EMAIL PROTECTED] writes:
Am I correct? Or could a factor smaller than 2*k*p + 1 be missed in
some cases?
In the last example a factor 16*97 + 1 could be missed.
Otherwise all factors below 2*k*p + 1 should be found.
One extra squaring will achieve the 2*k*p + 1 bound.
If I understand P-1 factoring correctly, then using it to a stage one
bound of k to try to factor M(p) will find all possible factors less
than or equal to 2*k*p + 1.
Yes, of the form n*p+1 (not 2*n*p+1 :). This is for the simple reason
that every power of a prime =k must divide Q (due to
