RFC 5869 describes HMAC-based Extract-and-Expand Key Derivation Function (HKDF). In section 4, entitled "Applications of HKDF", it states that one of the intended uses is:

derivation of symmetric keys from a hybrid public-key encryption scheme

However, it doesn't expand on the meaning of this. Can anyone explain how HKDF can be used in this manner?


2 Answers 2


Many key exchanges output secrets that are either not uniform or are large but splitting them is unsafe because subsets aren't uniform either.

Diffie-Hellman over 4096-bit groups or over 256-bit elliptic curves will not be uniform.

$$k = \operatorname{kdf}(\operatorname{dh}(a, b))$$

RSA-KEM (RFC 5990) where $r$ is a random value $\{0,1,2,3,\dots,N-1\}$ where $N = pq$ and the ciphertext is also in this range. Note the following is slightly modified to include the ciphertext, this eliminates any ciphertext malleability without having to analyse the malleability of each primitive you may pair it with.

$$k = \operatorname{kdf}(\operatorname{encrypt_{pub}}(r) \| r)$$

Critically, when using this key you must include an authenticator, not only to authenticate the message but to authenticate the key exchange itself. Otherwise you've derived a random key and cannot tell if either the key or the message is corrupt.

  • $\begingroup$ I wasn't aware of RSA-KEM-DEM (more familiar with AES-RSA), so more reading for me! Could another example be using HKDF to derive keys from an RSA private key? $\endgroup$
    – Cocowalla
    May 6, 2018 at 17:16
  • $\begingroup$ RSA-KEM-DEM is neat because you don't have to touch OAEP or other paddings as it is already all random and any corruptions to the ciphertext will just randomize it further. OTOH it is implicitly authenticated like Diffie-Hellman so depending on the protocol you may want to include a MAC to verify the key and/or bind the key to some message/payload. $\endgroup$
    – cypherfox
    May 7, 2018 at 2:36
  • $\begingroup$ @Cocowalla But only the (say) server knows the RSA private key? The client cannot derive the same keys. If they cannot, then it sounds more like a CSPRNG is needed. Yes, you could seed a CSPRNG from another static secret like your private key. As always, be careful and recall that RSA public keys aren't uniform, but they can be used as the seed as that only needs to be unique - not uniform. $\endgroup$
    – cypherfox
    May 7, 2018 at 2:38
  • $\begingroup$ By implicitly authenticated, what I mean is that I know if I encrypt $r$ with your public key and you use that key that we're good. But you don't know if the ciphertext was corrupt, which would derive a random key. If I include a MAC, then you can know that we're synchronised without another round trip. $\endgroup$
    – cypherfox
    May 7, 2018 at 2:44
  • 1
    $\begingroup$ Yes. The HKDF should be good so long that the input is unique and secret. Your RSA private key is unique and secret. $\endgroup$
    – cypherfox
    May 7, 2018 at 8:12

Fix an RSA modulus $n$. Here's a way you can send a message $m$ so that only someone who knows the secret factorization of $n$ can read it.

  1. Pick an integer $1 < x < n$ uniformly at random.
  2. Compute $y = x^3 \bmod n$.
  3. Compute $k = \operatorname{HKDF-Extract}(\underline x)$, where $\underline x$ is the little-endian bit encoding of the integer $x$. Note that $\underline x$ itself is not a uniform random bit string: if interpreted as an integer, it is always below $n$. Thus it is unfit to satisfy the security contracts requiring a uniform random key.
  4. Compute $k_\mathrm{enc} = \operatorname{HKDF-Expand}(k, 512, \text{‘Cocowalla's app: encryption key’})$. $k_\mathrm{enc}$ is conjectured to be indistinguishable from a uniform random bit string to an adversary who doesn't know $x$. As such it is fit to satisfy security contracts requiring a uniform random key.
  5. Compute $k_\mathrm{mac} = \operatorname{HKDF-Expand}(k, 256, \text{‘Cocowalla's app: authentication key’})$.
  6. Compute $c = m \oplus (\operatorname{Threefish512}_{\,k_\mathrm{enc}}(0) \mathbin\Vert \operatorname{Threefish512}_{\,k_\mathrm{enc}}(1) \mathbin\Vert \cdots)$, the encryption of $m$ with a one-time pad generated by Threefish-512 in CTR mode.*
  7. Compute $t = \operatorname{Poly1305}_{k_\mathrm{mac}}(c)$, the authentication tag for $c$.
  8. Transmit $(y, c, t)$.

The recipient who knows the factors $p$ and $q$ of $n$ can solve $3d \equiv 1 \pmod{(p - 1)(q - 1)}$ and recover $x = y^d \bmod n$, since $y^d \equiv (x^3)^d \equiv x^{3d} \equiv x^1 \equiv x \pmod n$; then they can derive $k$, etc., as before.

In archaic crypto literature, this is called ‘hybrid’ because we don't use the RSA trapdoor permutation $x \mapsto x^3 \bmod n$ to encrypt the message; rather we use it as a public-key KEM to conceal a secret element $x$ of $\mathbb Z/n\mathbb Z$, and derive from $x$ keys for a symmetric-key cryptosystem—specifically, a DEM, a data encapsulation mechanism. Thus it is a ‘hybrid’ of public-key and symmetric-key cryptography.

* This is not actually how I would recommend doing it: I would rather recommend NaCl crypto_secretbox_xsalsa20poly1305 with a single key. But this illustrated using HKDF-Expand to generate multiple keys of different sizes with different info strings.

The designation of ‘hybrid’ today is archaic because modern cryptographers recognize it was foolish all along to shoehorn structured messages like ‘Help! I'm trapped in an RSA ring.’ or ‘The Magic Words are Squeamish Ossifrage’ into elements of $\mathbb Z/n\mathbb Z$. In fact, for hysterical raisins, we still do approximately this: pick $k \in \{0,1\}^{256}$ uniformly at random, and then shoehorn it with an array of hash functions into an approximately uniform random element of $\mathbb Z/n\mathbb Z$, with a byzantine construction called OAEP. For hysterical raisins most protocols deployed in practice use OAEP, or its even worse predecessors. The much simpler scheme above is called RSA-KEM.

  • 1
    $\begingroup$ Let us discuss in chat. $\endgroup$
    – cypherfox
    May 7, 2018 at 3:26
  • $\begingroup$ Another question: you show using HKDF to derive multiple keys by varying the info parameter - from the RFC and original paper, this is exactly how I've understood it to be used too, but multiple answers on this site, such as this one, say that HKDF should only be used to derive a single key - what's your take on that? $\endgroup$
    – Cocowalla
    May 8, 2018 at 21:11
  • $\begingroup$ @Cocowalla It doesn't make sense. $\operatorname{HKDF}_k(\mathit{salt},\mathit{info})$ is roughly $\operatorname{HMAC}_h(\mathit{info})$ (HKDF-Expand) for $h=\operatorname{HMAC}_{\mathit{salt}}(k)$ (HKDF-Extract). If $h=\operatorname{HMAC}_{\mathit{salt}}(k)$ is distinguishable from a uniform random bit string, you have a problem to begin with. If $\operatorname{HMAC}_h(\mathit{info})$ is not a PRF keyed by $h$, then HMAC is amazingly broken. The outer HMAC in HKDF-Expand serves as exactly the PRF that D.W. is suggesting you use to expand one uniform random key into many uniform random keys. $\endgroup$ May 8, 2018 at 21:29
  • $\begingroup$ @Cocowalla Even a strict interpretation of what D.W. said doesn't stand to scrutiny—namely, that the standard security reduction for HKDF security doesn't cover the use case of HKDF-Expand on the same initial keying material for many info values. It doesn't stand to scrutiny because the theorem does cover that case, and is parametrized by the number of distinct info values. $\endgroup$ May 8, 2018 at 21:44
  • $\begingroup$ Thanks for the clear explanation that confirms my understanding is correct! $\endgroup$
    – Cocowalla
    May 9, 2018 at 6:34

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