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Breaking RSA if you know m^k = 1 mod n

  1. Feb 5, 2011 #1
    Hello I'm looking for some guidance:
    Last edited: Feb 5, 2011
  2. jcsd
  3. Feb 6, 2011 #2
    You seem to assume that everyone knows what you are talking about--why?

    You certainly have not defined terms, but when I look into this, it is my impression that you have no understanding of modular arithmetic. Why not check that out?
  4. Feb 11, 2011 #3
    Look, RSA is built on the fact in the private key (d,n) and the public key (e,n) d and e are exponential inverses of each other mod n. If x^d or x^e are congruent to 1 then d = phi(n) or e = phi(n) respectively... Take a look at Euler's Theorem which is a generalize form of Fermat's Little Theorem. This is why in RSA there is so much care taken in picking keys.
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