Talk:Fermat primality test

From Wikipedia, the free encyclopedia

[edit] both true

Both conditions, 1 \le a < p and \operatorname{gcd}(a,p) = 1, are true. if a^{p-1} \equiv 1 \mod p\ ist true, so every (m\cdot p+a)^{p-1} \equiv 1 \mod p for every natural number m\ with 1\le m\. So because every natural a\ with 1 \le a < p is \operatorname{gcd}(a,p) = 1, so is every natural number a\ different from prime m\cdot p a^{p-1} \equiv 1 \mod p\. --Arbol01 00:40, 29 October 2005 (UTC)

If this is in regard to the rollback I did earlier, I did it because they broke the TeX. Anyway, whichever way looks better, go for it. CryptoDerk 00:45, 29 October 2005 (UTC)

[edit] usage section?

I believe that someone should remove the usage section because it is flat-out wrong. PGP is not a program, but rather a system for signing and encrypting information. GPG is a program that implements PGP. I have no idea what GPG uses, but an implementation of PGP, to the best of my knowledge, can use any primality checker it wants. I have not removed the section because I am not completely sure about this, and want to get it right. Does anyone else have any insight into the subject? - Alan 134.173.34.115 08:35, 25 April 2006 (UTC)

Retrieved from "http://en.wikipedia.org../../../f/e/r/Talk%7EFermat_primality_test_9561.html"