If I'm reading this right, it's nothing new.

The basic idea is that the RSA function is periodic (it does involve a
modulus after all), so applying it to the same message over and over
again should result in a cycle back to the cleartext (if the author of
the presentation indicated is correct). While this is true, there's no
way for the attacker to know when they've cycled back around to the
message text, unless (s)he can control what the plaintext message is.
And if the attacker already knows part of the plaintext, RSA is
already broken ->
http://en.wikipedia.org/wiki/RSA#Attacks_against_plain_RSA .

This is why RSA messages are padded with pseudorandomized data, to
prevent  attackers from determining these kinds of things.

Please correct me if I am wrong.

Billy3
-------------------------------------------------------------
Computer Science Student - Case Western Reserve University
http://stackoverflow.com/users/82320/billy-oneal

On Thu, Mar 3, 2011 at 11:01 AM, rachid baih <[email protected]> wrote:
> A worked example
> The public key is  (n = 5183, e  = 8609).
> The private key is (n = 5183, d = 209).
>
> m=  127  to encrypt
>                                                               c =
> 127^8609 mod 5183
>                                                               c =
> 2324
>
> to decrypt  c = 2324
>
> 2324 ^8609   mod   5183   =  3748
>
>
> 3748 ^8609   mod   5183   =  123
>
>
> 123   ^8609   mod   5183   =   2257
>
>
> 2257^8609   mod   5183   =   3247
>
>
> 3247^8609   mod   5183   =  127
>
>
> 127  ^8609   mod   5183   =   2324
>
>
>       Now we have successfully decrypted c with m = 127
>
> Take any number c ( rsa encrypted message) You can continue encrypting
> process (with rsa function )
> no matter what number c you start with, you will always eventually
> reach
> m  the decrypted message.
> plez visit
> https://docs.google.com/document/d/1sTHB52526LQW3YnU39HdZ49pXIlQKOXGAsm3oIdUELM/edit?hl=en#
>
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