2
$\begingroup$

If one looks at Account Settings > End-to-End-Encryption > Add Key and creates a new key then gets the option EC. But you can't choose the bit length nor does one know which curve is used.

Anyone who knows that? Anyone who knows how secure that is?

Improve this question
$\endgroup$
2
  • $\begingroup$ datatracker.ietf.org/doc/html/rfc6637 $\endgroup$
    – kelalaka
    Dec 17, 2021 at 20:54
  • 1
    $\begingroup$ @kelalaka thanks, but I already knew that document ... the problem is: I generated it with Thunderbird (which uses OpenPGP). While there is a variety of EC for OpenPGP, there is only the choice "EC" in Thunderbird. They do not mention bitlength or which curve they use. Which is weird, since OpenPGP provides different curves. $\endgroup$
    – Fukai Mizumi
    Dec 17, 2021 at 23:21

1 Answer 1

Reset to default
2
$\begingroup$

I've downloaded the source code from the developer site. All the encrypted related parts under the libgcrypt

Libgcrypt is a cryptography library developed as a separated module of GnuPG. It can also be used independently of GnuPG, but depends on its error-reporting library Libgpg-error.

And those are listed as the supported curve in the library

  • NIST (P-256, P-384, P-521)
  • SECG (secp256k1)
  • ECC Brainpool / RFC 5639 (P256r1, P384r1, P512r1)
  • Bernstein Curve25519
  • Curve448
  • GOST R 34.10-2012 (RFC 7091)
  • SM2

Which one is used? Search files excluding the libgcrypt directory since it is the exact copy of it.

A grep grep -r curve tools/ build/ calendar/ chat/ mail mailnews/ python/ suite/ taskcluster/ tools/

in the command line displays only

curve25519

This curve is used for public-key signature system as Ed2559.

Curve25519 is also known as the Bernstein curve. It is one of the most secure curves for the listed criteria as in the safecurves.cr.yp.to/.

If you want to read more about how secure the curve is, see our canonical answer;

And note that if ever Shor's period finding algorithm is built with enough qbits the curve will be no more secure.

Also;

Edwards25519 aims for a ‘128-bit security level

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Thank you very much! $\endgroup$
    – Fukai Mizumi
    Dec 18, 2021 at 16:26

Your Answer

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

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