From: Karl Lui Barrus <klbarrus@owlnet.rice.edu>
To: cypherpunks@toad.com
Message Hash: cc40d7e7ee90d57606042a501993bffd6984ca04f0ce393764224f856a20c50b
Message ID: <9310072023.AA06352@flammulated.owlnet.rice.edu>
Reply To: N/A
UTC Datetime: 1993-10-07 20:25:32 UTC
Raw Date: Thu, 7 Oct 93 13:25:32 PDT
From: Karl Lui Barrus <klbarrus@owlnet.rice.edu>
Date: Thu, 7 Oct 93 13:25:32 PDT
To: cypherpunks@toad.com
Subject: Re: Weak RSA keys, 2nd post
Message-ID: <9310072023.AA06352@flammulated.owlnet.rice.edu>
MIME-Version: 1.0
Content-Type: text/plain
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(sorry for posting this twice, but as I sent the previous message, line
noise or something overtook me)
It does appear that there can be multiple decryption keys, but with good
choices of p and q, there will be only two.
Whether or not "ed = 1 mod phi(n)" is a myth or not... depends. Look at
"Cryptography: an Intoduction to Computer Security" by Seberry and
Pieprzyk. On page 92, equation 3.9 gives the relation between encryption
and decryption keys as "ed = 1 mod gamma(n)" where gamma(n) = lcm(p-1,q-1).
Which is what the "F" function described before is.
With proper choices of p and q, the d calulcated by the first forumla and
the d calculated by the second will be the same.
Of course, since you must know how to factor n to calculate the "spare" key,
it isn't clear to me how it will help you decrypt a message. With large
enough properly chosen primes, it won't matter. Solution: pick good primes.
An example with better choices for p and q:
p = 107; q = 167; n = pq = 17869
phi(n) = (p-1)(q-1) = 17596
e = 43
Method 1: d = 43^-1 mod 17596 = 7775
Method 2: lcm(p-1, q-1) = 8798
d = 43^-1 mod 8798 = 7775
Either method: message M = 71
encrypted M' = 71^43 mod 17869 = 10073
decrypt = 10073^7775 mod 17869 = 71
The "spare" key is 7775 + 8798 = 16573, which does indeed work.
But again, I don't see how an attacker can use info about the existence of
a second key.
Karl L. Barrus <klbarrus@owlnet.rice.edu>
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