Hey! I'm David, the author of the Real-World Cryptography book. I'm a crypto engineer at O(1) Labs on the Mina cryptocurrency, previously I was the security lead for Diem (formerly Libra) at Novi (Facebook), and a security consultant for the Cryptography Services of NCC Group. This is my blog about cryptography and security and other related topics that I find interesting.

more on the next page...

# Hacking Week 2015 : Crypto 4 Write-Up posted June 2015

The Hacking Week ended 2 weeks ago and EISTI got the victory.

I'm the proud creator of the crypto challenge number 4, still available here, which was solved 12 times.

I also wrote a Proof of Solvableness, reading it should teach you about a simple and elegant crypto attack on RSA: the Same Modulus Attack.

(Note that I wrote that back in January)

## Let's start

We are presentend with 4 files:

• alice.pub
• irc.log
• mykey.pem
• secret

Session Start: Thu Feb 05 20:49:04 2015
Session Ident: #mastercsi
[20:49] * Now talking in #mastercsi
[20:49] * Topic is 'http://www.math.u-bordeaux1.fr/CSI/ |||| http://www.youtube.com/watch?v=zuDtACzKGRs   "das boot, ouh, ja" ||| http://www.koreus.com/video/chat-saut-balcon.html ||| http://blog.cryptographyengineering.com/ ||| http://www.youtube.com/watch?v=K1LZ60eMpiw ||| petit chat http://www.youtube.com/watch?v=eu2kVcWKvRo ||| sun : t'as le droit de boire quand même va'
[20:49] * Set by [email protected]:41d0:52:100::65d on Sat Nov 22 00:06:50
[20:49] <asdf> et donc j'ai chopé une vieille clé rsa qu'alice utilisait
[20:49] <qwer> alice la alice? tu te fous de moi ?
[20:49] <asdf> haha non
[20:49] <asdf> mais le truc est corrompu, ça a l'air de marcher pour chiffrer mais la moitié de la clé a disparu
[20:49] <qwer> attend j'ai sa clé publique qui traine quelque part, et même un fichier chiffré avec. me suis toujours demandé ce que c'était...
[20:50] <asdf> je t'ai envoyé le truc, mais ça m'étonnerait que ça soit la même clé non ?
[21:22] * Disconnected
Session Close: Thu Feb 05 21:22:11 2015

so alice.pub seems to be alice public rsa key. secret seems to be the file encrypted under this key and mykey.pem should be the partial key which was found.

Private-Key: (1024 bit)
modulus:
00:c6:c8:35:29:a2:38:8f:14:63:65:c5:f5:fd:4b:
0d:88:89:61:b9:5d:e1:0f:fa:88:53:a3:c2:cb:ed:
4f:35:6a:82:d0:62:75:5e:1e:4f:ed:ae:54:e8:ca:
24:71:fc:8d:13:ac:70:0e:e2:57:20:d4:d9:08:9f:
d6:fb:d4:2f:12:e6:a4:1e:1c:1d:e8:1f:57:8c:32:
13:2a:d0:85:94:e8:51:84:1d:02:39:cd:41:0d:ef:
11:d1:c1:5e:e7:5b:92:f8:6a:04:f7:c6:c7:f3:6b:
90:46:b8:fb:2f:e2:95:65:b1
publicExponent: 3 (0x3)
privateExponent:
00:84:85:78:c6:6c:25:b4:b8:42:43:d9:4e:a8:dc:
b3:b0:5b:96:7b:93:eb:5f:fc:5a:e2:6d:2c:87:f3:
a3:5f:10:e6:7e:0a:a5:a1:d6:c2:1f:9d:1e:21:96:
df:78:f1:ac:8a:ec:4e:3e:be:df:f3:c9:8d:f0:86:
c2:f6:a8:5e:0b:ef:c0:ca:19:c5:e2:49:55:49:fe:
e5:2e:51:3e:7b:e9:f2:22:07:d2:4b:84:7f:bb:0c:
b5:ba:b7:95:c6:90:05:3e:65:2d:11:53:9a:2d:96:
0f:ea:de:cb:9b:17:54:87:00:0f:78:12:ce:ac:f5:
db:83:30:16:06:cc:35:7d:a3
prime1: 245 (0xf5)
prime2: 207 (0xcf)
exponent1: 163 (0xa3)
exponent2: 138 (0x8a)
coefficient: 189 (0xbd)

It looks like prime1, prime2 and some other stuff are pretty short. I guess this is what he meant by "half the key" is messed up.

By the way this is what a RSA PrivateKey should look like:

> RSAPrivateKey ::= SEQUENCE {
version           Version,
modulus           INTEGER,  -- n
publicExponent    INTEGER,  -- e
privateExponent   INTEGER,  -- d
prime1            INTEGER,  -- p
prime2            INTEGER,  -- q
exponent1         INTEGER,  -- d mod (p-1)
exponent2         INTEGER,  -- d mod (q-1)
coefficient       INTEGER,  -- (inverse of q) mod p
otherPrimeInfos   OtherPrimeInfos OPTIONAL
}

So this is what exponent1, exponent2 and coefficient are. Just additional information so that computations are faster thanks to CRT.

Let's ignore that for the moment.

$openssl rsa -pubin -in alice.pub -modulus -noout Modulus=C6C83529A2388F146365C5F5FD4B0D888961B95DE10FFA8853A3C2CBED750E9959BD0FF872C2232F6BAD32624F356A82D062755E1E4FEDAE54E8CA2471FC8D13AC700EE25720D4D9089FD6FBD42F12E6A41E1C1DE81F578C32132AD08594E851841D0239CD410DEF11D1C15EE75B92F86A04F7C6C7F36B9046B8FB2FE29565B1$ openssl rsa -in mykey.pem -modulus -noout
Modulus=C6C83529A2388F146365C5F5FD4B0D888961B95DE10FFA8853A3C2CBED750E9959BD0FF872C2232F6BAD32624F356A82D062755E1E4FEDAE54E8CA2471FC8D13AC700EE25720D4D9089FD6FBD42F12E6A41E1C1DE81F578C32132AD08594E851841D0239CD410DEF11D1C15EE75B92F86A04F7C6C7F36B9046B8FB2FE29565B1

the partial key and alice public key seems to share the same modulus. this is vulnerable. If our public/private exponents are not messed up, this means we can factor the modulus and thus inverse Alice's public key.

Let's retrieve all the info we have and put them in a file:

openssl rsa -pubin -in alice.pub -modulus -noout | sed 's/Modulus=//' | xclip -selection c

Here's the modulus. We know that our public key is 3, let's get the private key in the clipboard as well

openssl asn1parse -in mykey.pem | grep 129 | tail -n1 | awk '{ print $7}' | sed 's/://' | xclip -selection c here I parse mykey.pem with openssl. I select the lines I want with grep. It returns two results, the modulus and the private key. I select only the second line with tail. I select only the 7th column with awk. I remove the : with sed. And now I have a beautiful integer in my clipboard. Okay so let's do a bit of Sage now: # let's write the info we have modulus = int(0xC6C83529A2388F146365C5F5FD4B0D888961B95DE10FFA8853A3C2CBED750E9959BD0FF872C2232F6BAD32624F356A82D062755E1E4FEDAE54E8CA2471FC8D13AC700EE25720D4D9089FD6FBD42F12E6A41E1C1DE81F578C32132AD08594E851841D0239CD410DEF11D1C15EE75B92F86A04F7C6C7F36B9046B8FB2FE29565B1) public = 3 private = int(0x848578C66C25B4B84243D94EA8DCB3B05B967B93EB5FFC5AE26D2C87F3A35F10E67E0AA5A1D6C21F9D1E2196DF78F1AC8AEC4E3EBEDFF3C98DF086C2F6A85E0BEFC0CA19C5E2495549FEE52E513E7BE9F22207D24B847FBB0CB5BAB795C690053E652D11539A2D960FEADECB9B175487000F7812CEACF5DB83301606CC357DA3) # now let's factor the modulus k = (private * public - 1)//2 carre = 1 g = 2 while carre == 1 or carre == modulus - 1: g += 1 carre = power_mod(g, k, modulus) p = gcd(carre - 1, modulus) print(p) This does not work. This should work. Let's re-do the maths: We know that our private and public keys cancel out. This is RSA: private * public = 1 mod phi(N) so we have private * public - 1 = 0 mod phi(N) So for any g in our ring, we should have g^(private * public - 1) = 1 mod N This is how RSA works. Let's write it like that: private * public - 1 = k with k a multiple of phi(n). And we know that phi(n) = (p-1)(q-1) is even. So it could be written like this: k = 2^t * r with r an odd number. Now if we take a random g mod N and we do g^(k/2) it should be the square root of a 1. The Chinese Remainder Theorem tells us that there are 4 square roots mod N: • 1 mod p • -1 mod p • 1 mod q • -1 mod q and two of them should be 1 mod N and -1 mod N. The 2 others should be different from 1 and -1 mod N. That's what I was trying to find in my code. Once we have found this x mod N which is a square root of 1 mod N, we know that it is either x = 1 mod p or x = -1 mod p. If we are in the first case, we shoudl have x - 1 = 0 mod p which translates into x - 1 is a multiple of p. Doing gcd(x - 1, N) should give us p the first prime. If you don't understand it maybe check Dan Boneh's explanation (proof end of page 3) which should be clearer than mine. With p it's easy to get q the other prime. But it doesn't work... Ah! I forgot that g^(k/2) could equal 1 all the time if k/2 were to be a multiple of phi(n). So let's code a loop that divides k by 2 and tries any g^k until it is giving us something else than 1. Then we know how many times we have to divide k by 2 so it's not a multiple of phi(n) anymore. It turns out we just have to do it 3 times. And then it magically works. A bit more of Sage gives us the primes: # p and q our primes p = gcd(carre - 1, modulus) q = modulus // p # now that we have factored N let's find alice decryption key public = 65537 phi = (p - 1) * (q - 1) private = inverse_mod(public, phi) Now that we have Alice's private key there are two ways to decrypt our secret: • recreate a valid rsa key with those values and use openssl rsautl • figure out how openssl rsautl works to do it ourselves Let's do the first one. We'll modify our mykey.pem for this: openssl rsa -in mykey.pem -outform DER -out newkey.bin xxd -p newkey.bin > newkey.hex we get this: 3082012202010002818100c6c83529a2388f146365c5f5fd4b0d888961b9 5de10ffa8853a3c2cbed750e9959bd0ff872c2232f6bad32624f356a82d0 62755e1e4fedae54e8ca2471fc8d13ac700ee25720d4d9089fd6fbd42f12 e6a41e1c1de81f578c32132ad08594e851841d0239cd410def11d1c15ee7 5b92f86a04f7c6c7f36b9046b8fb2fe29565b102010302818100848578c6 6c25b4b84243d94ea8dcb3b05b967b93eb5ffc5ae26d2c87f3a35f10e67e 0aa5a1d6c21f9d1e2196df78f1ac8aec4e3ebedff3c98df086c2f6a85e0b efc0ca19c5e2495549fee52e513e7be9f22207d24b847fbb0cb5bab795c6 90053e652d11539a2d960feadecb9b175487000f7812ceacf5db83301606 cc357da3020200f5020200cf020200a30202008a020200bd This is a DER encoding. One particular encoding from the ASN.1 family. It is a TLV kind of encoding (Type Lenght Value). For example in: 02 8181 00c6c83529a2388f146365c5f5fd4b0d888961b95de10ffa8853 a3c2cbed750e9959bd0ff872c2232f6bad32624f356a82d062755e1e4fed ae54e8ca2471fc8d13ac700ee25720d4d9089fd6fbd42f12e6a41e1c1de8 1f578c32132ad08594e851841d0239cd410def11d1c15ee75b92f86a04f7 c6c7f36b9046b8fb2fe29565b1 first is coded the type 02 (integer), then the length (81 repeated twice because the value block is bigger than 127bits, so we set the first byte to 81 (10000001, the first bit means it is a long way of defining the length, the 7 following bits are the number of byte it will take to define the length, in our case only one and it will be the next one) and the second byte to the actual size), then there is our modulo in hexadecimal. Note that the most significant bit of our value has to be zero if it is a positive integer, that's why we use 41 instead of 40 and lead the payload with 00. So let's take the time and break this apart: 3082 // some header 0122 // the length of everything that follows (in byte) 0201 // integer of size 1 00 028181 // integer of size 0x81 (our modulus) 00c6c83529a2388f146365c5f5fd4b0d888961b95de10ffa8853a3c2cbed750e9959bd0ff872c2232f6bad32624f356a82d062755e1e4fedae54e8ca2471fc8d13ac700ee25720d4d9089fd6fbd42f12e6a41e1c1de81f578c32132ad08594e851841d0239cd410def11d1c15ee75b92f86a04f7c6c7f36b9046b8fb2fe29565b1 0201 // integer of size 1 (our public key) 03 028181 // integer of size 0x81 (our private key) 00848578c66c25b4b84243d94ea8dcb3b05b967b93eb5ffc5ae26d2c87f3a35f10e67e 0aa5a1d6c21f9d1e2196df78f1ac8aec4e3ebedff3c98df086c2f6a85e0b efc0ca19c5e2495549fee52e513e7be9f22207d24b847fbb0cb5bab795c6 90053e652d11539a2d960feadecb9b175487000f7812ceacf5db83301606 cc357da3 0202 // integer of size 2 (prime 1) 00f5 0202 // integer of size 2 (prime 2) 00cf 0202 // integer of size 2 (exponent 1) 00a3 0202 // integer of size 2 (exponent 2) 008a 0202 // integer of size 2 (coefficient) 00bd Now let's remove everything which is after the modulus and let's refill the file with our own values. Let's go back in Sage to calculate them: public = 65537 phi = (p - 1) * (q - 1) private = inverse_mod(public, phi) exponent1 = inverse_mod(private, p - 1) exponent2 = inverse_mod(private, q - 1) coefficient = inverse_mod(q, p) After filling and modifying the header's length accordingly, we obtain a nice hexadecimal file that we can transform back to binary: xxd -r -p new_key.hex | openssl asn1parse -inform DER It works! So now let's decrypt with is shall we? xxd -r -p new_key.hex | openssl rsa -inform DER -outform PEM -out newkey.pem openssl rsautl -decrypt -in secret -inkey newkey.pem We have our secret :)! 3 comments # The Logjam Attack posted May 2015 Since it is now common custom to market a new vulnerability, here is the page: weakdh.org you will notice their lazyness in the non-use of a vulnerability logo. The paper containing most information is here: Imperfect Forward Secrecy: How Diffie-Hellman Fails in Practice, from a impressive amounts of experts (David Adrian, Karthikeyan Bhargavan, Zakir Durumeric, Pierrick Gaudry, Matthew Green, J. Alex Halderman, Nadia Heninger, Drew Springall, Emmanuel Thomé, Luke Valenta, Benjamin VanderSloot, Eric Wustrow, Santiago Zanella-Béguelin, Paul Zimmermann) ## Not an implementation bug, flaw lives in the TLS protocol This is not an implementation bug. This is a direct flaw of the TLS protocol. This is also a Man in The Middle attack. By being in the middle, the attacker can modify the ClientHello packet to force the server to use an Export Ciphersuite, i.e. Export Ephemeral Diffie-Hellman, that uses weak parameters. I already explained what is an "Export" ciphersuite when the FREAK attack happened. The server then generates weak parameters for a public key and sends 4 messages: • ServerHello that specifies the Ciphersuite chosen from the list the Client gave him (if the attacker did things correctly, the server must have chosen an Export ciphersuite) • Certificate which is the server's certificate • ServerKeyExchange which contains the weak parameters and his public key. • ServerHelloDone which signals the end of his transmission. The ServerKeyExchange message is here because an "ephemeral" ciphersuite is used. So the Server and the Client need extra messages to compute an "ephemeral" key together. Using an Export DHE (Ephemeral Diffie-Hellman) or a normal DHE do not change the structure of the ServerKeyExchange message. And that's one of the problem since the server only signs this part with his long term public key. Here you can see the four messages in Wireshark, the signature is computed on the Client.Random, the Server.Random and the ECDH parameters contained in the ServerKeyExchange. Thus, the attacker only has to modify the unsigned part of the ServerHello message to tell the Client his normal ciphersuite has been chosen (and not an Export ciphersuite). Now all the attacker has to do is to crack the private key of either the Client or the Server. Which is easy nowadays because of the low 512bits security of the Export DHE ciphersuite. It can then pass as the server and read any messages the client wants to send to the server (taken from the paper) ## Not an implementation bug, but implementations do help the use of common DHE parameters is making things easier for attackers since they can do a pre-computation phase and use it to quickly crack a private key of a weak DHE parameters during the handshake. This happens, for example when Apache hardcoded a prime for its Export DHE Ciphersuite that is now used in a bunch of servers (taken from the paper) ## Defense from the Server Don't use common DH or DHE parameters! Generate your owns. But even more important, remove the Export Ciphersuites as soon as possible. # Defense from the Client From a client perspective, the only defense is to reject small primes in DHE handshakes. This is the only way of detecting this Man in The Middle attack. You could also remove DHE in your ciphersuite list and try to use the elliptic curve equivalent ECDHE (Elliptic Curve Diffie-Hellman Ephemeral) Another way: if you control both the server and the client, you could modify both ends so that the server signs the ciphersuite he chose, and the client verifies that as well. ## 1024 bits primes? In the 1024-bit case, we estimate that such computations are plausible given nation-state resources, and a close reading of published NSA leaks shows that the agency’s attacks on VPNs are consistent with having achieved such a break. We conclude that moving to stronger key exchange methods should be a priority for the Internet community. Seems like the NSA doesn't even need to downgrade you. So as a server, or as a client, you should refuse primes <= 1024bits ## Where is TLS used? TLS is not only used in https! For example, what about EAP, i.e. wifi authentication? From a quick glance it looks like there are no export ciphersuite. But weak DH and DHE parameters should be checked as well everywhere you make use of Discrete Logarithm crypto comment on this story # Hacking Week 2015 posted May 2015 The Hacking week just started, it's a CTF that happens over a week. You'll find challenges about crypto, network, forensic, reverse and exploit. And also, I have a challenge up there in the crypto challenge ^^ It's in french here: http://hackingweek.fr/challenges/ (click on "voir" next to crypto 4) basically Alice encrypted the secret, you have to find what the secret is. What you have is a key that shares the same modulus as Alice. comment on this story # How to compare password hashes in PHP? posted May 2015 ## The wierdness of == Do you know what happens when you run this code in PHP? <?php var_dump(md5('240610708') == md5('QNKCDZO')); var_dump(md5('aabg7XSs') == md5('aabC9RqS')); var_dump(sha1('aaroZmOk') == sha1('aaK1STfY')); var_dump(sha1('aaO8zKZF') == sha1('aa3OFF9m')); var_dump('0010e2' == '1e3'); var_dump('0x1234Ab' == '1193131'); var_dump('0xABCdef' == ' 0xABCdef'); ?> Check the answer here. That's right, everything is True. This is because == doesn't check for type, if a string looks like an integer it will first try to convert it to an integer first and then compare it. More about PHP == operator here This is weird and you should use === instead. Even better, you can use hash_equals (coupled with crypt) Compares two strings using the same time whether they're equal or not. This function should be used to mitigate timing attacks; for instance, when testing crypt() password hashes. Here's the example from php.net: <?php$expected  = crypt('12345', '$2a$07$usesomesillystringforsalt$');
$correct = crypt('12345', '$2a$07$usesomesillystringforsalt$');$incorrect = crypt('apple',  '$2a$07$usesomesillystringforsalt$');

hash_equals($expected,$correct);
?>

Which will return True.

## But why?

the hashed strings start with 0e, for example both strings are equals in php:

md5('240610708') = 0e462097431906509019562988736854
md5('QNKCDZO')   = 0e830400451993494058024219903391

because php understands them as both being zero to the power something big. So zero.

## Security

Now, if you're comparing unencrypted or unhashed strings and one of them is supposed to be secret, you might have potentialy created the setup for a timing-attack.

Always try to compare hashes instead of the plaintext!

# Previous Links posted April 2015

There is a Link section here that is not very visible, I don't really know how I could show its content on the frontpage here. But here's one way:

## More

And you can find more on the Links section of this blog

comment on this story

# Some research on recovering small RSA private keys posted April 2015

To make it short, I did some research on the Boneh and Durfee bound, made some code and it worked. (The bound that allows you to find private keys if they are lesser than $N^{0.292}$)

I noticed that many times, the lattice was imperfect as many vectors were unhelpful. I figured I could try to remove those and preserve a triangular basis, and I went even further, I removed some helpful vectors when they were annoying. The code is pretty straightforward (compare to the boneh and durfee algorithm here)

So what happens is that I make the lattice smaller, so when I feed it to the lattice reduction algorithm LLL it takes less time, and since the complexity of the whole attack is dominated by LLL, the whole attack takes less time.

It was all just theoric until I had to try the code on the plaid ctf challenge. There I used the normal code and solved it in ~3 minutes. Then I wondered, why not try running the same program but with the research branch?

That’s right, only 10 seconds. Because I removed some unhelpful vectors, I could use the value m=4 and it worked. The original algorithm needed m=5 and needed a lattice of dimension 27 when I successfully found a lattice of dimension 10 that worked out. I guess the same thing happened to the 59 triplets before that and that’s why the program ran way faster. 3 minutes to 10 seconds, I think we can call that a success!

The original code:

comment on this story

# Small RSA private key problem posted April 2015

/!\ this page uses LaTeX, if you do not see this: $\LaTeX$

then refresh the page

## Plaid CTF

The third crypto challenge of the Plaid CTF was a bunch of RSA triplet $N : e : c$ with $N$ the modulus, $e$ the public exponent and $c$ the ciphertext.

The public exponents $e$ are all pretty big, which doesn't mean anything in particular. If you look at RSA's implementation you often see $3$, $17$ or other Fermat primes ($2^m + 1$) because it speeds up calculations. But such small exponents are not forced on you and it's really up to you to decide how big you want your public exponent to be.

But the hint here is that the public exponents are chosen at random. This is not good. When you choose a public exponent you should be careful, it has to be coprime with $\varphi(N)$ so that it is invertible (that's why it is always odd) and its related private exponent $d$ shouldn't be too small.

Maybe one of these public keys are associated to a small private key?

I quickly try my code on a small VM but it takes too much time and I give up.

## Wiener

A few days after the CTF is over, I check some write-ups and I see that it was indeed a small private key problem. The funny thing is that they all used Wiener to solve the challenge.

Since Wiener's algorithm is pretty old, it only solves for private exponents $d < N^{0.25}$. I thought I could give my code a second try but this time using a more powerful machine. I use this implementation of Boneh and Durfee, which is pretty much Wiener's method but with Lattices and it works on higher values of $d$. That means that if the private key was bigger, these folks would not have found the solution. Boneh and Durfee's method allows to find values of private key up to $d < N^{0.292}$!

After running the code (on my new work machine) for 188 seconds (~ 3 minutes) I found the solution :)

Here we can see that a solution was found at the triplet #60, and that it took several time to figure out the correct size of lattice (the values of $m$ and $t$) so that if there was a private exponent $d < N^{0.26}$ a solution could be found.

The lattice basis is shown as a matrix (the ~ represents an unhelpful vector, to try getting rid of them you can use the research branch), and the solution is displayed.

## Boneh and Durfee

Here is the code if you want to try it. What I did is that I started with an hypothesis $delta = 0.26$ which tested for every RSA triplets if there was a private key $d < N^{0.26 }$. It worked, but if it didn't I would have had to re-run the code for $delta = 0.27$, $0.28$, etc...

I setup the problem:

# data is our set of RSA triplets
for index, triplet in enumerate(data):

print "Testing triplet #", index

N = triplet[0]
e = triplet[1]

# Problem put in equation
P.<x,y> = PolynomialRing(ZZ)
A = int((N+1)/2)
pol = 1 + x * (A + y)

I leave the default values and set my hypothesis:

delta = 0.26
X = 2*floor(N^delta)
Y = floor(N^(1/2))

I use strict = true so that if the algorithm will stop if a solution is not sure to be found. Then I increase the values of $m$ and $t$ (which increases the size of our lattice) and try again:

solx = -1
m = 2
while solx == -1:
m += 1
t = int((1-2*delta) * m)  # optimization from Herrmann and May
print "* m: ", m, "and t:", t
solx, soly = boneh_durfee(pol, e, m, t, X, Y)

If no private key lesser than $N^{delta}$ exists, I try the next triplet. However, if a solution is found, I stop everything and display it.

Remember our initial equation:

$e \cdot d = f(x, y)$

And what we found are $x$ and $y$

if solx != 0:
d = int(pol(solx, soly) / e)
print "found the private exponent d!"
print d

m = power_mod(triplet[2], d, N)
hex_string = "%x" % m
import binascii
print "the plaintext:", binascii.unhexlify(hex_string)

break

And that's it!

## More?

If you don't really know about lattices, I bet it was hard to follow. But do not fear! I made a video explaining the basics and a survey of Coppersmith and Boneh & Durfee

Also go here and click on the follow button.

# Same RSA modulus and correlated public exponents posted April 2015

Plaid, The biggest CTF Team, was organizing a Capture The Flag contest last week. There were two crypto challenges that I found interesting, here is the write-up of the second one:

You are given a file with a bunch of triplets:

{N : e : c}

and the hint was that they were all encrypting the same message using RSA. You could also easily see that N was the same modulus everytime.

The trick here is to find two public exponent $e$ which are coprime: $gcd(e_1, e_2) = 1$

This way, with Bézout's identity you can find $u$ and $v$ such that: $u \cdot e_1 + v \cdot e_2 = 1$

So, here's a little sage script to find the right public exponents in the triplets:

for index, triplet in enumerate(truc[:-1]):
for index2, triplet2 in enumerate(truc[index+1:]):
if gcd(triplet[1], triplet2[1]) == 1:
a = index
b = index2
c = xgcd(triplet[1], triplet2[1])
break

Now that have found our $e_1$ and $e_2$ we can do this:

$c_1^{u} * c_2^{v} \pmod{N}$

And hidden underneath this calculus something interesting should happen:

$(m^{e_1})^u * (m^{e_2})^u \pmod{N}$

$= m^{u \cdot e_1 + v \cdot e_2} \pmod{N}$

$= m \pmod{N}$

And since $m < N$ we have our solution :)

Here's the code in Sage:

m = Mod(power_mod(e_1, u, N) * power_mod(e_2, v, N), N)

And after the crypto part, we still have to deal with the presentation part:

hex_string = "%x" % m
import binascii
binascii.unhexlify(hex_string)

Tadaaa!! And thanks @spdevlin for pointing me in the right direction :)

1 comment