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- Newsgroups: sci.math
- Path: sparky!uunet!elroy.jpl.nasa.gov!nntp-server.caltech.edu!stieger
- From: stieger@cco.caltech.edu (Ronald David Stieger)
- Subject: Re: n doesnt divide 2^n-1
- Message-ID: <1992Aug28.223723.18851@cco.caltech.edu>
- Sender: news@cco.caltech.edu
- Nntp-Posting-Host: punisher
- Organization: California Institute of Technology, Pasadena
- References: <1992Aug28.194350.19717@cs.rose-hulman.edu>
- Date: Fri, 28 Aug 1992 22:37:23 GMT
- Lines: 37
-
- goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard) writes:
-
- >If someone could prove the following for me, I would be forever in
- >his debt.
-
- >If n > 1, then n doesnt divide 2^n-1.
-
- >This _is_ a problem in a book, but I don't _DO_ homework anymore, I'm
- >just curious.
-
- OK. I had number theory spring term so let's see what I remember.
-
- (1) a^p = a^q (mod m) iff p = q (mod t)
- where t is the order (?) of m, that is the smallest positive
- number k (< m) s.t. a^k = 1 (mod m)
-
- The proof of this fact is fairly straightforward. p = q+rt so
- a^p = a^(q+rt) = (a^q)*(a^(rt)). a^(rt) = 1 (mod m) so a^p = a^q (mod m)
-
- So... n divides 2^n - 1 implies 2^n = 1 = 2^0 (mod n) which implies by (1)
- n = 0 (mod t) so n divides t, the order of n.
-
- Now, the order of n always divides n-1. I don't remember how to prove this.
- Maybe you know, or someone else can post it, or maybe I'll think of it
- later. You can test it if you like, a^(n-1) = 1 (mod n) for all a and n.
- So t divides n-1, but we also know that n divides t. This
- implies that n divides n-1, or n = 1 (mod n), which is only true if
- n = 1.
-
- So, if n divides 2^n - 1, n = 1, and thus, as you asked,
-
- if n > 1, n does not divide 2^n - 1. QED.
-
- --
- --Ron Stieger
- stieger@cco.caltech.edu
- "Everybody wants a rock to wind a piece of string around."-They Might Be Giants
-