When using ed25519 what is the maximum size of the message/payload data?

From the specs:

The Diffie-Hellman key exchange is then defined as follows: each party generates a random number 1 <= d < N (the private key), computes Q = d G (the public key). The parties exchange their public keys and compute the shared secret as Z = d Q_peer.

So once both parties have computed the shared secret, what is the maximum size of the message data that could be encrypted and therefore decrypted by the "secret"?

  • 1
    $\begingroup$ Note that Ed25519 is only a signature scheme strictly speaking. $\endgroup$
    – SEJPM
    Mar 26, 2017 at 10:22

1 Answer 1


That depends really on which symmetric cipher and cipher mode is used. Basically, the strength provided by the secret is 128 bits, as indicated in the page you linked to.

Usually the secret is then converted into a key using a Key Derivation Function (KDF) or even into multiple keys, which will each have a security level of 128 bits (as long as the KDF is strong enough, but that's commonly the case).

After that it is up to the scheme used. Probably you don't want to go over $2^{128}$ blocks of encryption, but that's kind of nonsense in the first place.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.