RE: Encryption question

From: Prasad S. Athawale (athawale_at_cse.Buffalo.EDU)
Date: 02/26/04

  • Next message: Hollis Johnson: "Re: Encryption question" 
    To: "'Jordan, Jason D. \"Dallas\"'" <Jason.Jordan@honeywell-tsi.com>, "'Preston, Tony'" <Tony.Preston@acs-inc.com>, <security-basics@securityfocus.com>
Date: Wed, 25 Feb 2004 23:45:19 -0500

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    Hi!

     

    The public/private key pair is generated together at the same time -
    and one cannot atleast it's termed 'mathematically difficult'
    determine the other key when one key is known - which is the essence
    of public-key cryptography.

     

    So when you want to generate another private key - to be used with
    Alice's public it effectively has to Alice's private key that you
    somehow generate - or else the decryption just wouldn't work. And
    this like I said earlier is mathematically difficult.

     

    As regards the man-in-the middle attack - this can be done like what
    Jason said - about suppressing the knowledge of Alice's 'true' public
    key and providing Bob with your own fake 'Alice's public key. This
    fake key would be the public key from a public-private key pair that
    you generated. So whenever Bob encrypts something with the public key
    of Alice in his possession, it effectively can be decrypted by you
    (if you can intercept the message of course), re-encrypted with
    Alice's true public key - which you possess and forwarded on to
    Alice. Alice would in turn be able to read the message because it is
    encrypted with her public key.

    Here's the relatively easy to understand math behind RSA public key
    encryption.

    1. Find P and Q, two large (e.g., 1024-bit) prime numbers.
    2. Choose E such that E is greater than 1, E is less than PQ, and E
    and (P-1)(Q-1) are relatively prime, which means they have no prime
    factors in common. E does not have to be prime, but it must be odd.
    (P-1)(Q-1) can't be prime because it's an even number.
    3. Compute D such that (DE - 1) is evenly divisible by (P-1)(Q-1).
    Mathematicians write this as DE = 1 (mod (P-1)(Q-1)), and they call D
    the multiplicative inverse of E. This is easy to do -- simply find an
    integer X which causes D = (X(P-1)(Q-1) + 1)/E to be an integer, then
    use that value of D.
    4. The encryption function is C = (T^E) mod PQ, where C is the
    ciphertext (a positive integer), T is the plaintext (a positive
    integer), and ^ indicates exponentiation. The message being
    encrypted, T, must be less than the modulus, PQ.
    5. The decryption function is T = (C^D) mod PQ, where C is the
    ciphertext (a positive integer), T is the plaintext (a positive
    integer), and ^ indicates exponentiation.

    Your public key is the pair (PQ, E). Your private key is the number D
    (reveal it to no one). The product PQ is the modulus (often called N
    in the literature). E is the public exponent. D is the secret
    exponent.

    The difficult part here is determining D=E^-1(mod(P-1)(Q-1)), when P
    & Q are unknown - you just know their product.

    Also since someone mentioned about the primality of a number - it's a
    known difficult problem in mathematics a solution to which has been
    recently proposed.

    http://mathworld.wolfram.com/news/2002-08-07/primetest/

    And remember PKE is based on the premise - that everyone can keep
    their private key secret, and the authentic public key's are readily
    and reliably available to everyone from an authentic server.

    Hope this helped,

    Prasad

    - -----Original Message-----
    From: Jordan, Jason D. "Dallas"
    [mailto:Jason.Jordan@honeywell-tsi.com]
    Sent: Wednesday, February 25, 2004 12:45 PM
    To: 'Preston, Tony'; 'security-basics@securityfocus.com'
    Subject: RE: Encryption question

     

    Tony,

        I believe, in order to spoof a digital signature of Alice, you
    would need to get her private key....which she should have securely
    stored somewhere. A hash of the message is done and then encrypted
    with Alices private key. The only other key that

    can decrypt it is the public key generated with her original key
    pair. You could substitute Alice's public key with your public key
    so when Bob used that public key to encrypt the message meant for
    Alice, you could intercept it and read the message.

    Then you could re-encrypt it with Alice's real public key and send it
    on to her. Kinda like a man in the middle deal. I think this is how
    it works, I could be wrong. Does that help any?

     

     

    Dallas Jordan MCSE, CCNA, Security+

    Electronics Technician II

    Honeywell Technology Solutions

    1010 Bankton Drive

    Hanahan, SC 29406

    843-744-1221 Ext 11

     

     -----Original Message-----

    From: Preston, Tony [mailto:Tony.Preston@acs-inc.com]

    Sent: Tuesday, February 24, 2004 1:01 PM

    To: security-basics@securityfocus.com

    Subject: Encryption question

     

     

     

    Tony Preston

    Systems Engineer, AS&T Inc.

    Division of L3 Corporation

    (609) 485-0205 x 181

     

    I have what is a rather basic question... I probably am missing
    something

    so I thought I would ask here.

     

    Alice and Bob both have a public and private key.

     

    Alice encrypts her email to Bob using his public key. Sends the
    email and

    Bob decrypts it using his keys..

     

    Since both Bob and Alice's public keys are known, Why can't I take
    Alice's

    public key and create a key pair using any other private key. Now, I
    fake

    an electronic signature from Alice using the pair I created and send
    a bogus

    encrypted message to Bob with my "fake" Alice signature. Bob checks
    the

    signature by using the public key and it is valid. Bob assumes the
    message

    is from Alice...

     

    What prevents me from spoofing someone's electronic signature this
    way?

     

     

     

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