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- Newsgroups: sci.math
- Path: sparky!uunet!cs.utexas.edu!hellgate.utah.edu!math.utah.edu!math-2-1.math.utah.edu!user
- From: gross@math.utah.edu (Fletcher Gross)
- Subject: Re: n doesnt divide 2^n-1
- Sender: news@math.utah.edu
- Date: Thu, 3 Sep 1992 21:48:07 GMT
- Organization: Math Dept, University of Utah
- References: <1992Aug28.194350.19717@cs.rose-hulman.edu> <1992Aug28.203144.13773@super.org> <Sep.3.16.29.43.1992.1193@yoko.rutgers.edu>
- Message-ID: <gross-030992154434@math-2-1.math.utah.edu>
- Followup-To: sci.math
- Lines: 30
-
- In article <Sep.3.16.29.43.1992.1193@yoko.rutgers.edu>,
- gore@yoko.rutgers.edu (Bittu) wrote:
- >
- > From kedlaya@b124.super.org (Kiran Sridhara Kedlaya) Fri Aug 28 16:31:44 1992
- >
- > > Sure. Let n be the smallest integer > 1 such that n divides 2^n - 1.
- > > Let d be the smallest integer such that n divides 2^d - 1 (i.e. the order
- > > of 2, with respect to n). Then d divides n (otherwise n = rd + s, where
- > > 0 < s < d, and since 2^n and 2^d are congruent to 1 mod n, so is
- > > 2^(n - rd) = 2^s, whence n divides 2^s - 1, which is impossible
- > > by the choice of d). Also 2^d - 1 divides 2^n - 1, and d > 1 (otherwise
- > > n divides 2^1 - 1 = 1). So d is an integer > 1 but smaller than n, such
- > > that d divides 2^d - 1. Contradiction.
- >
- > But what if d = n? I don't see the contradiction in that case. Am I
- > missing something? How does one show that d < n? There is probably a
- > theorem that says that the order of 2 w.r.t an odd integer n is always
- > less than n.
- >
- > -Bittu
-
- Since n divides 2^n - 1, n and 2 are relatively prime. If m is the Euler
- phi function of n, then n must divide 2^m - 1. Hence d <= m. Since m is the
- number of positive integers less than n and relatively prime to n, m <= n -
- 1. It follows that d < n.
-
- Fletcher Gross | (801) 581-6567
- Department of Mathematics | gross@math.utah.edu
- University of Utah |
- Salt Lake City, Utah 84112 |
-