27

I am currently experimenting with ed25519 and I noticed that on secret key creation, bit 254 is always set and the lower 3 bits are always cleared. I found that bit 254 is always set to protect against timing attacks in this question: When using Curve25519, why does the private key always have a fixed bit at 2^254?

But why are the lower 3 bits cleared. Obviously it has to do with the formula in the curve25519 paper: The set of secret keys is defined to be $\{\underline{n} : n \in 2^{254} + 8\{0, 1, 2, 3,\ldots, 2^{251}-1\}\}$

It's because of the 8 in there, but why is that 8 there? I suspect it has something to do with theorem 2.1 in the curve25519 paper, but I am not sure, because I do not understand fully what is being proven there.

I am experimenting with ed25519 primitives in some cryptographic routines which need me to add scalars to the secret key (add secret keys). Even if I add two well formed ed25519 secret keys, the result will not always have bit 254 set and the lower 3 bits cleared. Is this a security problem? I understand the risk for bit 254 but not the lower 3.

Community
  • 1
MepAhoo
  • 273
  • 3
  • 4

1 Answers1

30

Clearing the lower 3 bits of the secret key ensures that is it a multiple of 8, which in turn ensures that no information, small as it may be, about the secret key is leaked in the case of an active small-subgroup attack.

The typical simple Diffie-Hellman key exchange works like this:

$$ \text{Alice} \xrightarrow{\hspace{3cm} a G \hspace{3cm}} \text{Bob} \\ \text{Alice} \xleftarrow{\hspace{3cm} b G \hspace{3cm}} \text{Bob} $$

$a$ and $b$ are Alice and Bob's secret keys, and $G$ is the base point, which is $(9,\ldots)$ in curve25519. Alice computes $a\cdot bG$, Bob computes $b \cdot aG$, and the shared secret is derived by passing $abG$ through some key derivation function.

Now, as you have noticed, the order of $G$ is $p_1 = 2^{252} + 27742317777372353535851937790883648493$, whereas the number of points in the curve itself is $8 p_1$. This means that there are a few remaining points that have small order. An active attacker can, e.g., replace Bob's message $bG$ with a point of order $8$ and be able to find $a \bmod 8$ by inspecting following messages. When every valid secret key is $0 \bmod 8$, the attacker gets nothing.

One may argue that this wastes 3 perfectly good key bits to prevent an already ineffective attack. Notice, however, that the actual security of the scheme is tied to the order of the base point, $p_1 \approx 2^{252}$. Clearing those 3 bits (out of 255) does not reduce the keyspace since it still leaves 252 useful bits, giving roughly the same keyspace as the number of points generated by $G$.

Samuel Neves
  • 12,960
  • 46
  • 54