Article View: sci.crypt
Article #2353RSA-PGP help
From: James "wibble" Hawtin
Date: Mon, 23 Nov 1992 07:23
Date: Mon, 23 Nov 1992 07:23
29 lines
1478 bytes
1478 bytes
I am interested in obtaining a standard RSA compatable key, I only have access to PGP
I have managed to get it to display its calculations :-
256bit calculation.
modulus n = 2E0574EB 8331D378 71B184F5 1AE02817
9EB9113C 366AE339 82549484 74125977
exponent e = 00000011
exponent d = 103E2944 102FB40C 645CC583 AF21F008
093EA014 A60689FC 56931557 4A36C7D1
prime p = 63BB6D2D 38765E6E 5930457A C52DD959
prime q = 76218929 D237532A DE839292 5C9E9F4F
inverse u = 6A3CD1F9 36DF2A5E 871C4456 3E785001
However I am not sure have have got the meaning of the numbers correct.....
In standard RSA, 3 things are used general modulus, public key and private key
so does :-
n = the general modulus
d = public key
u = private key
When I use n, d & u in a standard power modulus algorithm it does not work ;-(
What to I have do do to transform this data into the standard RSA numbers? I know PGP takes some short cuts but have a made a mistake in my calculation or am I using the wrong numbers?
James.
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Path: arc1-ca.pugleaf.net!sparky!uunet!pipex!warwick!uknet!news.cs.bham.ac.uk!mother!jwh From: jwh@cs.bham.ac.uk (James "wibble" Hawtin) Newsgroups: sci.crypt Subject: RSA-PGP help Message-ID: <By5r7y.EIL@cs.bham.ac.uk> Date: 23 Nov 92 07:23:58 GMT Sender: news@cs.bham.ac.uk Reply-To: jwh@cs.bham.ac.uk Organization: School of Computer Science, The University of Birmingham, UK Lines: 29 Nntp-Posting-Host: mother