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Aug 20, 2018 at 1:10 comment added poncho @AymanMadkour: well, yes (actually, one could design a multiparty computation protocol between Alice and Bob), however it needn't be the re-encryptor.
Aug 19, 2018 at 19:32 comment added Ayman Madkour But whoever is going to calculate ab-1 must know both a and b, right?
Aug 19, 2018 at 14:38 comment added poncho @AymanMadkour: actually, the method doesn't assume that you know both private keys, it assumes you know the single value $ab^{-1}$. Obviously, if someone had the private key $a$, he could just decrypt the message, and then perhaps reencrypt it with public key $bG$. What $ab^{-1}$ allows is the possibility of doing proxy re-encryption, without the possibility of decrypting.
Aug 19, 2018 at 2:44 comment added Ayman Madkour This approach assumes that I know both private keys, a & b. But what if I didn't? What if I only knew one private key (a) and one public key (bG)? Is there another way to make it work in this case?
Apr 21, 2018 at 10:47 comment added cygnusv Note that this only work in versions of ECIES that don't include an encoding of $rG$ as input to the KDF that generates the symmetric key to encrypt the message.
Apr 20, 2018 at 22:40 vote accept Ayman Madkour
Apr 20, 2018 at 21:14 history answered poncho CC BY-SA 3.0