Re: Order modulo p^n (Number Theory)

On Mon, 15 Sep 2008 15:38:44 -0700 (PDT), "Achava Nakhash,
the Loving Snake" <achava@xxxxxxxxxxx> wrote:

There is a book called Lectures on Number Theory, or something like
that, by Hans Rademacher which does similar calculations when it shows
that 2^n does not have primitive roots for n > 2, and also when
showing that p^n and 2p^n do have primitive roots when p is an odd
prime. Unfortunately, I can't find my copy, so I can't tell you his
trick.

(Still watching TV!)

Kenneth Ireland and Michael Rosen, /A Classical Introduction to
Modern Number Theory/, 2nd ed. (1990), pages 42f. (edited a bit):

Corollary 1. If n >= 2 and p =/= 2, then (1 + ap)^{p^{n-2}} =
1 + ap^{n-1} (mod p^n) for all a in Z.

Corollary 2. If p =/= 2 and p does not divide a, then p^{n-1}
is the order of 1 + ap (mod p^n).

--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril
.