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Math Help - show encryption and decryption are identical

  1. #1
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    Angry show encryption and decryption are identical

    show that the encryption procedure C=P^e(mod n) and decryption procedure P=C^d(mod n) in RSA cryptosystem are identical when encryption is done using modular eponentiation with modulus n=35 and enciphering key e=5
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  2. #2
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    They will be identical as your encryption and decryption key are identical for the following reasons;

    If e=5 and mod m= 35, then in order to get your decryption key you need to find

    \frac{1}{5} mod\ <br />
\phi(35)

    Since this key will be the same as the encryption key, encrypting and decrypting will use the same exponent, not very secure.
    Last edited by jtsab; April 29th 2008 at 05:08 AM.
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  3. #3
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    Quote Originally Posted by mandy123 View Post
    show that the encryption procedure C=P^e(mod n) and decryption procedure P=C^d(mod n) in RSA cryptosystem are identical when encryption is done using modular eponentiation with modulus n=35 and enciphering key e=5
    Just remember that d and e should satisfy de = 1\text{ mod }\phi(35)
    And \phi(35) = (7-1)(5-1) = 24. Now you will have to solve for
    5d = 1\text{ mod }24. Clearly d=5

    Thus d=e=5 and hence encryption and decryption keys are identical
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  4. #4
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    Hello,

    Quote Originally Posted by jtsab View Post

    \frac{1}{5} mod\ <br />
\phi(35)
    Is this writing correct ? Oo
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  5. #5
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    Quote Originally Posted by Moo View Post
    Hello,
    Is this writing correct ? Oo
    There is no harm, if he means 5^{-1} \mod \phi(35), right?
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  6. #6
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    Yep, but the writing still looks weird for me
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  7. #7
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    Quote Originally Posted by Isomorphism View Post
    There is no harm, if he means 5^{-1} \mod \phi(35), right?
    That's what I was implying, since \frac {1}{5} =\ 5^{-1} \mod \phi (35)

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