I want to know how I can implement this protocol. I know how Shamir three pass protocol operates without elliptic curve, but I don't know how I can perform it with elliptic curve.

I read about this protocol; if Alice want to send to Bob then:

  1. Alice convert the message in a point $M$ of the elliptic curve. Then she picks a random element $k$ that belongs to the finite field of elliptic curve. Then she encrypt with $k \times M$ to generate $kM$, right?

  2. Bob picks a random element $q$ that belongs to the finite field. Then he encrpyt the point received from alice to generate $q(kM)$, for this he computes $q \times (kM)$.

  3. Alice decrypt the message with the inverse of $k$.

It is possible? The function decrypt would be $k^{-1} \times q(kM)$? Then, Alice send $qM$ to Bob.

  1. Bob decrypt wit the inverse of $q$. He decrypts in the same way that Alice does.

I think that this are the steps for this protocol but it does not work in my SAGE implementation. Can somebody indicate if I missed anything?

  • 1
    $\begingroup$ I could solve the problem. This steps are correct but the problem was in the way to calculate $k^{-1}$. This must be the multiplicative inverse of $k$ over N=# E(Fp ), that is, the order of the elliptic curve. Thanks $\endgroup$ – Cugar19 Jan 13 at 20:03
  • $\begingroup$ Glad you've got it solved. Unfortunately you cannot answer < 15 rep, but feel free to answer yourself if you have gained that amount of reputation. $\endgroup$ – Maarten Bodewes Jan 14 at 11:50

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