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11-05-2003, 06:23 AM
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#1 |
RSA cryptography
Hi, all!
I have a question regarding RSA algorithm. Suppose two users A and B have the RSA public keys (ea, n) and (eb, n). Further assume that ea and eb are relatively prime. C sends a message m to both A and B encrypting it with their respective public keys. If E gets hold of these messages, is it possible for E to find m? If so, how? Thank you very much for your help! Dawnbell |
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11-05-2003, 08:05 PM
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#2 |
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Posts: n/a
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Re: RSA cryptography
"Dawnbell" <> wrote in message news:_I0qb.2658$iE.945@okepread01... > Hi, all! > > I have a question regarding RSA algorithm. Suppose two users A and B have > the RSA public keys (ea, n) and (eb, n). Further assume that ea and eb are > relatively prime. C sends a message m to both A and B encrypting it with > their respective public keys. If E gets hold of these messages, is it > possible for E to find m? If so, how? > > Thank you very much for your help! > Homework due soon? Dave Thornburgh |
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11-06-2003, 12:14 AM
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#3 |
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Posts: n/a
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Re: RSA cryptography
On Wed, 5 Nov 2003 00:23:42 -0600, "Dawnbell" <> wrote:
>Further assume that ea and eb are relatively prime. I know a woman who is slightly pregnant. -- Jim Watt http://www.gibnet.com Jim Watt |
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11-06-2003, 12:52 AM
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#4 |
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Posts: n/a
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Re: RSA cryptography
Not sure what "n" is, key length?
E can only see m if he can get the private key corresponding to either A or B. Hopefully he can't get the private keys.... J -- Check my web site for tips on insuring safe computing in wired and wireless homenetworking environments! www.pccitizen.com "Dawnbell" <> wrote in message news:_I0qb.2658$iE.945@okepread01... > Hi, all! > > I have a question regarding RSA algorithm. Suppose two users A and B have > the RSA public keys (ea, n) and (eb, n). Further assume that ea and eb are > relatively prime. C sends a message m to both A and B encrypting it with > their respective public keys. If E gets hold of these messages, is it > possible for E to find m? If so, how? > > Thank you very much for your help! > > John D Loop |
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