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Is it possible for an attacker on a Diffie-Hellman key exchange to manipulate both sides in a way so that the secret generated on each side is identical?

Or put differently, would it be possible to detect an attack via MITM if we can detect via a different channel that the secrets of both parties do not match?

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2 Answers 2

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Is it possible for an attacker on a Diffie-Hellman key exchange to manipulate both sides in a way so that the secret generated on each side is identical?

If both parties use a well seeded CSPRNG then this should not be possible to have identical DH private keys.

As for the shared secret, the whole idea of DH - and any key establishment scheme - is that the secrets on both sides match. So if the attacker doesn't do anything the "secret on each side is identical".

Or put differently, would it be possible to detect an attack via MITM if we can detect via a different channel that the secrets on both parties do not match?

We can authenticate the public keys in the scheme so that the keys and the generated keys can be trusted.

We can also make sure that both sides have the correct secret by performing a message authentication code over a known message (such as the communication transcript so far).

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  • $\begingroup$ Thank you for the response. $\endgroup$
    – smatt
    May 16 at 14:57
  • $\begingroup$ You're welcome. If you think this answers your question you can accept it, it's the best way to say thanks (you can also wait a bit to see if any other answers pop up if you want, no need to be hasty). $\endgroup$
    – Maarten Bodewes
    May 16 at 15:06
  • $\begingroup$ If the man-in-the-middle is trying to intercept the communication, he will separately perform DH key exchange with both sides, correct? Is it possible in this scenario for the attacker carefully "choose" his private keys in this interaction that combined with the public keys identical shared secrets are obtained for both sides? $\endgroup$
    – smatt
    May 16 at 15:08
  • $\begingroup$ Ah, yes, good question. I think that you need to take a look at this answer. Basically, you need to perform public key validation. Quite obviously: if you use 0 as private key then the public key $g^x = g^0 = 1$, and you can raise that to any power, but it will stay $1$ (this is the extreme case). One way around this is to use X25519 or X448. $\endgroup$
    – Maarten Bodewes
    May 16 at 15:57
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Can a MITM during Diffie-Hellman key exchange manipulate both sides to generate symmetric secrets?

Let's assume that the MITM attacker can create an exchanged key that is the same on both sides. I.e., the party $A$ gets $g^t$ from the attacker and calculates $g^{at}$ and party $B$ gets $g^u$ from the attacker and calculates $g^{bu}$ such that $g^{at} = g^{bu}$.

If the attacker sends the identity element, then both sides will have the key as the identity element. This is a trivial solution and detectable (Other than the trial solution needs a reduction...)

would it be possible to detect an attack via MITM if we can detect via a different channel that the secrets on both parties do not match?

On the Major systems, protection from MITM attacks is done with certificates. Sometimes we do TOFU ( Trust On the First Usage) like in Signal ( and in the weaker WhatsApp) and later validate the public keys *.

One simple solution is reading the hex values on the phone since it is harder ( not impossible ) to fake real-time speech.


* Have you ever validated one key in the Signal ( or WhatsApp)?

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  • $\begingroup$ This reduction does not work. The mistake is that the attacker does not need to compute $t$ and $u$. They only need to compute $g^t$ and $g^u$. One attack is e.g. sending the neutral element of the group. $\endgroup$
    – Maeher
    May 16 at 20:46
  • $\begingroup$ @Maeher You are right. the trivial solution exists, Can you see the reduction in the non-trivial case? If so, can you write an answer, I'm happy to see and delete this. $\endgroup$
    – kelalaka
    May 16 at 21:20

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