>> I have a lame question, though. You mention that `a*B = a'*B` holds >> for the base point. But is it also true for any point in the B's >> subgroup? The reason I ask is that I need to have not just regular >> EdDSA signatures, but also DLEQs (discrete log equality proofs) with >> random generator points. > > Yes. Proof: If P is a point in B's subgroup, then P = p*B for some > scalar p. Thus > > a*P = a*p*B = p*a*B = p*a'*B = a'*p*B = a'*P, > > since multiplication of scalars is associative with multiplication of > curve points, and multiplication of scalars is commutative.
Oh, thanks for pointing this out to me! That was a lame question indeed :) _______________________________________________ Curves mailing list Curves@moderncrypto.org https://moderncrypto.org/mailman/listinfo/curves