🚀 go-pugleaf

RetroBBS NetNews Server

Inspired by RockSolid Light RIP Retro Guy

2 total messages Started by Mr. Lyn R. Kennedy Thu, 17 Dec 1992 22:09
RSA
#2352
Author: Mr. Lyn R. Kennedy
Date: Thu, 17 Dec 1992 22:09
26 lines
1692 bytes
I haven't seen these items mentioned but I'm not a math whiz so maybe
it's hidden in there somewhere.

You can get an encode/decode pair ( e and d ) by taking d*e mod
(p-1)*(q-1) = 1. This implies that the smallest value for d*e
is ((p-1)*(q-1))+1 but it's actually (((p-1)*(q-1))/2)+1. If fact, d*e 
must be some value  (((p-1)*(q-1)/2)*x)+1 where x is any integer >0
 
Also, for ANY value of n ( =p*q), there are at least nine values of m 
(the message) that will encrypt as m. I.e. E(m) = m and D(m) = m. Three 
of these are trivial, m=0, 1, and n-1. The other six allow factoring n. 
For some values of e, there are more cases but I'm not sure how easy it 
would be to get the factors knowing one of those. This brings up the 
question of whether the proceedure should watch for such cases. Only 
while signing a message would the sender be the one to discover key was 
compromised. I know it's not at all likely but it seems like it could 
cause problems.


-------------------------------------------------------------------------
                 
73,            internet     | lrk@k5qwb.lonestar.org
Lyn Kennedy    packet radio | K5QWB @ N5LDD.#NTX.TX.US.NA
               pony express | P.O. Box 5133, Ovilla, TX, USA 75154

-------------- "We have met the enemy and he is us."  Pogo --------------
Re: RSA
#2355
Author: Cetin Kaya Koc
Date: Fri, 18 Dec 1992 02:16
18 lines
1238 bytes
In article <226XVB4w164w@k5qwb.lonestar.org> lrk@k5qwb.lonestar.org (Mr.  
Lyn R. Kennedy) writes:
> Also, for ANY value of n ( =p*q), there are at least nine values of m 
> (the message) that will encrypt as m. I.e. E(m) = m and D(m) = m. Three 
> of these are trivial, m=0, 1, and n-1. The other six allow factoring n. 
> For some values of e, there are more cases but I'm not sure how easy it 

E(m) = m for as many as k messages where
   k = [1+gcd(e-1,p-1)]*[1+gcd(e-1,q-1)].

The smallest value k=[1+2]*[1+2]=9 as pointed out. This is because
   gcd(e-1,p-1) >= 2 
since p and e are odd numbers; p is prime, and gcd(e,(p-1)(q-1))=1 and
thus e is odd.

Thus, you want gcd(e-1,p-1) and gcd(e-1,q-1) as small as possible.

Ref: Blakley & Borosh. Comp. & Math. with App. 5:169-178, 1979.
Thread Navigation

This is a paginated view of messages in the thread with full content displayed inline.

Messages are displayed in chronological order, with the original post highlighted in green.

Use pagination controls to navigate through all messages in large threads.

Back to All Threads