From: rjc@gnu.ai.mit.edu (Ray)
To: cypherpunks@toad.com
Message Hash: 9bb7437cc280ff10288dca1a4aa278da6143c0df346d9162829d03ae7bd1e70f
Message ID: <9404120259.AA11138@geech.gnu.ai.mit.edu>
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UTC Datetime: 1994-04-12 02:59:36 UTC
Raw Date: Mon, 11 Apr 94 19:59:36 PDT
From: rjc@gnu.ai.mit.edu (Ray)
Date: Mon, 11 Apr 94 19:59:36 PDT
To: cypherpunks@toad.com
Subject: Prime Numbers
Message-ID: <9404120259.AA11138@geech.gnu.ai.mit.edu>
MIME-Version: 1.0
Content-Type: text/plain
Eric Hughes:
> It was first claimed that if (2^n-2)/n was an integer, then n was
> prime. That's false.
I thought he said "if p prime, then p|(2^p-2)" which is why
I stated the converse isn't true.
> then:
> > This is fermat's little theorem. What you have written basically
> >says 2^N - 2 = 0 (mod N) or 2^(N-1) = 1 (mod N). Note, the converse
> >doesn't apply. If (2^N-2)/N is an integer, N isn't neccessarily
> >prime. For example, take N=561=(3*11*37)
>
> 561 is the first Carmichael number. If you replace 2 by any other
> number relatively prime to 561, then the congruence still holds. (The
> second Carmichael number is 1729, if I remember right.) It was
Which is why I chose it. Carmichael numbers are pseudoprime in
any valid base so when coming up with a counterexample to the converse
of fermat's little theorem, just memorize a few Carmichael numbers. The key
property of them is if n is a Carmichael number and n=p*q*r, then (p-1),
(q-1), and (r-1) divide (n-1).
I wonder if Carmichael numbers always have some small factors. If true,
PGP's sieve test probably eliminates the very very rare case that
you actually choose one.
-Ray
-- Ray Cromwell | Engineering is the implementation of science; --
-- rjc@gnu.ai.mit.edu | politics is the implementation of faith. --
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1994-04-12 (Mon, 11 Apr 94 19:59:36 PDT) - Prime Numbers - rjc@gnu.ai.mit.edu (Ray)