On 18/11/14 17:22, Indtiny S wrote: > Hi, > Sorry,, I am bit new to ECC , I Need to just prove the below thing > > Ca.Sa.G) = Sa.Ca.G) . > > * Client *:- private = Ca , public= Ca,G and *Server*:- private=Sa, pub > = Sa.G > > When I read ECC tutorial, its defined that public key = Q (where Q=dG) > > so how to get the CaG and SaG in my case ? and validate the equation ? > > Please guide .. > Ahhh...homework....
See: https://wiki.openssl.org/index.php/Elliptic_Curve_Diffie_Hellman on that page your Ca=dA, CaG=QA, Sa=dB and SaG=QB You need to generate two key pairs using the same curve...one for the client and one for the server. You'll need to use EC_POINT_mul to calculate Ca.Sa.G, and again for Sa.Ca.G (in a real scenario you would use EVP_PKEY_derive or ECDH_compute_key which hides all the "magic" from you...but I guess that's not the point of the proof you've been asked to complete) The rest is left as an exercise for the reader ;-) Matt ______________________________________________________________________ OpenSSL Project http://www.openssl.org Development Mailing List openssl-dev@openssl.org Automated List Manager majord...@openssl.org
