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- Xref: sparky sci.crypt:5727 alt.security.pgp:201
- Newsgroups: sci.crypt,alt.security.pgp
- Path: sparky!uunet!telebit!phr
- From: phr@telebit.com (Paul Rubin)
- Subject: Re: RSA Question (was Re: PKP/RSA comments on PGP legality)
- In-Reply-To: crespin@nadir.uucp's message of Tue, 15 Dec 92 21:50:21 -0800
- Message-ID: <PHR.92Dec16022546@napa.telebit.com>
- Sender: news@telebit.com
- Nntp-Posting-Host: napa.telebit.com
- Organization: Telebit Corporation; Sunnyvale, CA, USA
- References: <1992Dec14.190615.13954@macc.wisc.edu> <dOsLrAnGBh107h@nadir.uucp>
- Date: 16 Dec 92 02:25:46
- Lines: 28
-
- > For the record I am going to publish a number I
- > want to keep secret. I am going to raise it to the power of
- > three modulo 91 which is the product of two primes. The
- > result is 34. My wife knows the factors of 91 and will be able
- > to decode this message.
-
- Please excuse my ignorance, but, how can "your wife" decode the message
- knowing 34, 91, 7, and 13?
-
- i.e. I have a number
- (X^3) moulo 77 (product of two primes) = 48
- what is my number? and how did you get the answer?
-
-
- Generally, given x^3 mod N = y, knowing the factors N = pq,
- recover x by the following:
- 1. Compute phi(N) = (p-1)(q-1). For number theorists, this is
- the number of elements of Z^*_N, the (multiplicative mod N) group
- of integers relatively prime to N.
- 2. Use the Extended Euclidean Algorithm (see Knuth vol. 2, etc.)
- to compute t, so that 3t == 1, mod phi(N). This means that
- x^(3t) mod N will be equal to x.
- This is because x^(phi(N)) mod N == 1, from Lagrange's theorem.
- 3. Now to decrypt y, simply compute x = y^t mod N.
-
- In the case of N=77 and N=91, phi(N) is a multiple of 3 so no
- suitable t exists. You have to have picked your original primes
- p and q to make this not happen.
-