View Single Post
Old 2020-01-28, 08:49   #80
R. Gerbicz
 
R. Gerbicz's Avatar
 
"Robert Gerbicz"
Oct 2005
Hungary

72·31 Posts
Default

Quote:
Originally Posted by ewmayer View Post
I then separately computed 3^(2^(2^m - 1)) mod p and q via (2^m-1) = 1073741823 iterated squarings modulo p and q, separately, and the results match, which anyone can confirm on even quite modest hardware.
Trivial note: it is faster to reduce the exponent mod (p-1) [use Fermat's little theorem].
R. Gerbicz is offline   Reply With Quote