83 points by todsacerdoti on June 24, 2022 | hide | past | favorite | 39 comments

 I think I understand the attack at a high level but I’m still not sure on some details.Could someone explain how the math works to provide the oracle? In particular, the exact values the malicious server inserts as the encrypted sid and q^(-1), and why, if the guess for q is lower than the real q the plaintext sid the client sends back is 0, and why it’s not zero if the guess is higher?Am I right in thinking the attacker simply randomly flips a bit in q^(-1)? Why do they need to do this? Do they flip different bits on different guesses? And then how do they encode their guess for q in the encrypted sid? And why, exactly does this provide an oracle? How does the subsequent lattice search need only half as many guesses as binary search? Finally, does this attack require the victim to actually type their password in 512 times?
 I'll take a stab, but I'm keeping it fuzzy, not because I haven't worked the attack out at a low level but because I don't want you to know that I haven't worked the attack out at a low level. Some of this will be wrong.We're in the bizarre circumstance of being able to get our target to use a partially corrupted version of their own secret RSA key to decrypt a message of our choosing (this never happens! Only with Mega!).We're looking for q, one of two factors of our private key. We set m, the plaintext message we get to choose, to the middle of the range of possible q's. Remember, q is smaller than m; we've got a 2048 bit RSA key, and so q is 1024 bits; the client is effectively going to pad it out, and then blindly extract the 2nd though 45th bytes from it.What we're going to do resembles a fault attack (we're sort of simulating one by scrambling qInv). One of two things will happen:* If m < q, the message is going to miss our faulted qInv (it won't wrap the modulus for the q part of the CRT calculation, by definition, which cancels out the qInv part of the calculation) and thus decrypts properly. After padding and extraction, the client will pull 0's out of the decrypted message --- remember, q is much smaller than the whole message space of m, and remember, the client doesn't bother to check the padding! the number of things that went wrong here is bananas! --- and post it back.* If m >= q, our qInv fucks up the decryption; we decrypt (I assume) random gibberish, all the way through the space of m. We extract 43 bytes of the random gibberish, posting back a non-zero value.It's described in detail in III.B.2 of https://mega-awry.io/pdf/mega-malleable-encryption-goes-awry...
 The fact that you also are a little fuzzy on the details gives me some confidence that I do in fact have some imposter syndrome ;) Thanks for your response. I understand it takes some work to parse these papers and I'm lazily just asking questions here instead of reading the paper myself, but I think I'm quite good at asking precise questions to get to the bottom of stuff, so I hope it will help others who stumble across this thread.> We're in the bizarre circumstance of being able to get our target to use a partially corrupted version of their own secret RSA key to decrypt a message of our choosing (this never happens! Only with Mega!).We overwrite the ciphertext of q^{-1} so that is fully corrupted (or at least, because it is ECB mode, a block of it is corrupted), right? By partially corrupted, do you mean "p and q are fine, but q^{-1} is not what it is supposed to be"? Does it matter how we corrupt q^{-1}? Is it just a bit flip in a single block? Do we have to corrupt it the same every time to make a new guess for q?> We set m, the plaintext message we get to choose, to the middle of the range of possible q's. Remember, q is smaller than m;I'm not quite following this part. To clarify, are you saying (roughly, I know RSA has some checks that q is not outrageously small) q lies in the interval [0, 2^1024] and we, on our first guess, set m = (2^1024 - 0)/2 = 2^1023? Then the adversary encrypts this guess m using the public key N from the client. m, when decrypted by the client at this point, could clearly be either larger or smaller than q, right? Our choice for m is literally in the middle. But the client also pads the resulting m, so it will certainly be larger than q? But to clarify, q is only (almost certainly) smaller than m because of the padding that the client adds?> * If m < q, the message is going to miss our faulted qInvI'm confused again, sorry. Is this m without the padding, right? Why, if m < q, does the RSA calculation not wrap around the modulus? I'm a bit confused how it pulls out only 0s, but this is probably related to my confusion at the last paragraph. Is it that it's that only 0s, but rather a small number followed by a ton of padding of 0s?> * If m >= q, our qInv fucks up the decryption; we decrypt (I assume) random gibberishThe whole reason we fault qInv is so this case screws up while the other doesn't, right? And so it doesn't really matter how we fault qInv each time? This case faults because the calculation wraps around the modulus? It would be good to see the actual calculation and why this is which I think is my main struggle right now.
 1. Yes: the RSA key (as encoded by Mega) is the tuple (p, q, d, qInv), and the only thing we corrupt is qInv. It doesn't matter (in this first attack) how we corrupt qInv! We flip any bit in an AES block that corresponds to where qInv would be in the encrypted key, so that when it's decrypted, the 16 bytes of that block are completely randomized. It doesn't matter what value qInv takes, and the value doesn't have to be consistent.2. I think that's right. I keep making the point about q's size because it's part of why, in the successful RSA decryption case, the SID extracted is 0: our guess for q isn't zero, of course, but a giant chunk of the bits of m are going to be padding, because q doesn't come close to filling m.3. This gets into the details of RSA-CRT, where you take m_p (mod p) and m_q (mod q). The m_q value, by definition, doesn't wrap the modulus q; it's less than q. :) I don't have a high-level way of explaining why this cancels out the qInv part of the calculation, though you can see it pretty clearly in the paper; it just works out that way algebraically (in a "ninth grader" sense) from the RSA-CRT equation.And yeah, it's a number, "left padded" by a whole bunch of zeroes. That's what the paper means by "m is hiding in the padding" --- our guess for q is part of the decrypted message, and it isn't zero, but it doesn't land in the part of the message the Mega client tries to scoop the SID out of. If the Mega client had checked the padding, which is something you're absolutely supposed to do (also, you're supposed to use a real padding scheme, not just zeroes), you couldn't get away with this.4. Yep, in this first attack, the whole point of faulting qInv is to create this difference of behavior.And yeah, you should absolutely implement a model of this attack in Python or whatever. You don't need any of the networking stuff Mega is doing, or any of the encoding. It won't be hard! This is a pretty straightforward attack. I'll probably take a whack at it this weekend.LaterI hadn't seen 'pbsd responded to this, but, of course, "what he said".
 Thanks for the responses. I hope it was useful to go through. Yeah, I think the salient points are that the sid is not all 0 but mostly 0 in one case and Fermat's Little Theorem basically make the math work out. I suspect the authors thought of this clever attack because they were already thinking about fault attacks on RSA where perhaps similar concepts come up.I predict with some confidence this will be made into a CTF soon if it hasn't already so we may as well work on the code now ;-)
 Look at how RSA CRT decryption works:`````` m_1 = c^d_p mod p m_2 = c^d_q mod q h = (m_1 - m_2)*q^-1 mod p m = m_2 + h * m_1 `````` If m < q, m_2 = m and there is nothing left to add. But if m >= q, there needs to be something to add, and this something will very likely be wrong and kind of random, because q^-1 mod p is wrong. The distinguisher between the two cases hinges on the upper bits of the decryption result being (non-)0. So for this attack to work all you have to do is change any part of q^-1; it doesn't really matter much what the change is.Now you have a binary search oracle for q, from which you can iteratively obtain the most significant bits of q, one guess at a time (e.g., try m = 2^1024 first, then (2^1024 or 0) + 2^1023, etc). Once you have half of the most significant bits of q, you can use the Coppersmith method [1] to find a small root of f(x) = msb(q) + x modulo a divisor of n and thus recover the rest of q.In reality doing exactly 512 iterations of guessing plus Coppersmith is pretty costly; if possible, guessing a few more bits will make the Coppersmith step much faster.
 Pretty elucidating, thanks.Could you just elaborate on the part:> If m < q, m_2 = m and there is nothing left to add.Why is this? What is d_p? d mod p?
 There is nothing left to add because m_2 is already the correct value:- If m is also < p, then (m_1 - m_2) is obviously 0, and so will be h.- If m >= p, then m must be of the form m_1 + k * p for some k. But we've already established that m = m_2, so m_1 - m_2 = (m - k * p) - m_2 = -k * p = 0 mod p. Thus h = 0.So regardless of the value of qinv, we get m verbatim if it's less than q. d_p and d_q are d mod (p-1) and d mod (q-1) respectively.Here, this Sage script is short enough that it all fits in a comment. I guess 624 bits and recover the remaining 400 with Coppersmith; this is fast enough to run in a few seconds.`````` def oracle(c, d, p, q, qinv): m_1 = power_mod(c, d % (p-1), p) m_2 = power_mod(c, d % (q-1), q) h = (m_1 - m_2)*qinv % p m = m_2 + h * m_1 return (m >> (211*8)) != 0 # check if sid != 0 # Setup p = random_prime(2^1024, lbound=2^1023+2^1022) q = random_prime(2^1024, lbound=2^1023+2^1022) n = p*q e = 65537 d = inverse_mod(e, (p-1)*(q-1)) qinv = inverse_mod(q, p) + randint(0, 2^128) # random error block inserted in the least significant 128 bits msb_q = 0 for bit in reversed(range(400, 1024)): # if m < q, we are within range if not oracle(power_mod(msb_q + 2^bit, e, n), d, p, q, qinv): msb_q += 2^bit # Find a small root of msb_q + x modulo a divisor of n of size ~n^(1/2) P. = Zmod(n)[] f = msb_q + x [lsb_q] = f.small_roots(beta=0.49, epsilon=1/32) # Check we've reached the correct solution assert(q == msb_q + lsb_q)``````
 Right, but my question is, why is it the correct value? There's some math theorem which makes it true?m_2 = c^d_q mod q`````` = m^(e * d_q) mod q = m because?``````
 That's just how RSA works, via Fermat's little theorem: e * d_q mod (q-1) = 1, so m^(e * d_q) mod q = m^1 mod q = m.
 Got it, thank you!
 This all is so deep, and I'm so sorry I cannot contribute to the questions you've got, but I feel like never coming back on account of "I know fuck all about this". Is this what people call 'Imposter syndrome'?
 Try https://www.cryptopals.com/ or https://www.coursera.org/learn/crypto.There's a lot to know, but you can build up to it!One thing you'll learn about are oracle attacks, in which a bad protocol or implementation can leak secret data to the attacker bit-by-bit (sometimes literally bit-by-bit) in various ways. There are lots of those and they're sadly easy to create and a lot of progress in cryptographic engineering has been about finding ways to make sure oracle attacks can't occur in various real-world deployments.
 context: https://twitter.com/kennyog/status/1539352663770509314> We show that MEGA's system does not protect its users against a malicious server and present five distinct attacks, which together allow for a full compromise of the confidentiality of user files. Additionally, the integrity of user data is damaged to the extent that an attacker can insert malicious files of their choice which pass all authenticity checks of the client. We built proof-of-concept versions of all the attacks, showcasing their practicality and exploitability.
 Weird. This is really just meant to accompany a Twitter thread I wrote; I don't think it's an especially good standalone summary of the paper.
 Couldn't MEGA simply change the client-side javascript to just send them the plain text password?
 The client could in theory see the javascript running and detect this.
 This attack breaks non-browser-Javascript clients.
 This is a totally out of context article.
 I think people here are missing that the attacker here is actually LEA and crypto is Mega's legal defense. If you've ever looked at their crypto in a different way, you did it wrong, sorry.
 Proposal: we should drop support for known bad/unsafe cryptographic primitives from popular cryptography libraries moving forward.When the companies using these bad/unsafe primitives inevitably complain, informed consumers will know to avoid their products and will broadcast a warning to others. The companies will be forced to migrate to safer primitives.We've kind of already seen this with things like TLS 1.3 (only AEAD ciphers allowed).
 Companies would just stay on even older versions of those libraries. Maybe with security patches backported, but probably not.
 How about ratifying some law that says, once a company has been warned about their use of unsafe crypto for X amount of time, if they still don't fix things, then they are fair game to black hats (as in, no black hat will be prosecuted for exploiting the unsafe crypto)?I know it might sound crazy.
 Whats next, enforce fire-safety regulations by arsonists?
 So if you were to use a non-web client, and that client stored the keys locally, would that prevent the attack?
 So how many thousands of logins does the client have to do for this to binary search one key?
 It takes about 512 logins, because after you've leaked a certain number of RSA key bits, you can set up a lattice search for the remaining bits. Because of the way the authentication and key escrow system worked, Mega could have run this attack over long periods of time, in parallel for all its users; the attack doesn't have to occur within a bounded period of time the way, say, an ephemeral-static ECDH TLS attack would work.
 Maybe I misunderstand, but wouldn't the attack be visible to the user when the encryption glitches out and they have to reattempt the login or something?
 No! That's another amazing detail here. The attacker is the server, and so it can just pretend you logged in successfully. You won't have fetched your key properly, but the Mega client automatically re-fetches the key when that happens.
 > You won't have fetched your key properly, but the Mega client automatically re-fetches the key when that happens.Is that something it could hypothetically do, or something that the current code does? In the latter case, is there any legitimate reason for doing that? And this would be detectable right, maybe not by a casual observer, but by a security researcher?
 How would you know? The whole point of the cryptosystem is you can't assume what the server code is doing.
 Well you said the client refetches the key, so that sounds like a network request you could observe.
 Sure; you could also look for all-zero SIDs. That relies on you knowing about this attack a priori, and if you know about it you can just not have the vulnerability.
 Yeah, I was just trying to evaluate the risk that>Mega could have run this attack over long periods of time, in parallel for all its users;without anybody catching on. In light of our discussion, I don't think it's too likely.
 No, I don't agree at all. There is literally new science in this paper; the idea that random "security researchers" looking at benign-looking crypto code and monitoring all the connections their clients made (a thing nobody does) would ever have caught this is not plausible.
 im3w1l on June 24, 2022 Actually I have a second question about this point.. Does this mean it would allow the user in even if the password is mistyped?
 No, if you mistype the password then the server can't look up your escrowed RSA key (it's looked up with an "authentication key" derived from PBKDF2).
 Binary searching for an n-bit number based on a high-or-low oracle takes a max of n attempts.
 I'm pretty sure you should just never create encryption keys client side?

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