0
$\begingroup$

Can anyone dumb it down for me on why you would need to format the first and last byte of a random number to properly generate a curve25519 private key as noted in 'Computing secret keys' section of http://cr.yp.to/ecdh.html#curve25519-paper.

Computing secret keys. Inside your program, to generate a 32-byte Curve25519 secret key, start by generating 32 secret random bytes from a cryptographically safe source: mysecret[0], mysecret1, ..., mysecret[31]. Then do

 mysecret[0] &= 248;
 mysecret[31] &= 127;
 mysecret[31] |= 64;

mysecret[0] &= 248 = 0b1111 1000...setting three lsb's to 0 to ensure multiple of 8? mysecret[31] &= 127 = 0b0111 1111...setting the msb to 0 for? mysecret[31] |= 64 = 0b0100 0000...setting the second msb to 1 for?

I must be missing some important note on this in RFC7748?

$\endgroup$
6
  • 3
    $\begingroup$ Does this answer your question? crypto.stackexchange.com/q/87709/76473 $\endgroup$ May 18, 2021 at 16:30
  • $\begingroup$ RFC's are the standard, you need to look at papers to get the mathematical background, security proofs etc. $\endgroup$
    – Maarten Bodewes
    May 18, 2021 at 16:47
  • $\begingroup$ Yes, thank you that does help. I also see now in section 5 of the RFC where it describes this encoding. Are there any recommended test vectors for curve25519 shared secret calculation? $\endgroup$
    – nikegolf05
    May 18, 2021 at 17:15
  • $\begingroup$ Did you see section 6? $\endgroup$
    – kelalaka
    May 18, 2021 at 19:09
  • $\begingroup$ I did, but was thinking there would be more than just a single vector... $\endgroup$
    – nikegolf05
    May 18, 2021 at 19:12

0

Browse other questions tagged or ask your own question.