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What is Let's Encrypt?
Basically, it's a way to get a quick x509 certificate for your server without knowing much about what is a x509 certificate:
You have a website. You want people to be able to log in on it from starbucks without the guy sitting at a near table reading the passwords in clear from the packets you're sending everyone around you. So you google a few random keywords like "secure website https" and you end up with a bunch of links and services (which may not be free) and you suddenly have to understand what are certificates, PKI, x509, publickey cryptography, RSA, FQDN, haaa.... Well, worry no more, Let's Encrypt was thought so that you wouldn't have to go through all that trouble and destroy your server on the way.
Oh and yeah. It's for free. But is hasn't been released yet.
How does it work?
You can learn more reading their technical overview, or some of their videos... or read my tl;dr:
 You download their program on your server that has the address www.example.com:
sudo aptget install letsencrypt
 You run it as sudo telling it you want to get a certificate for your domain
letsencrypt example.com
And voila.
Behing the curtains this is what happens:

letsencrypt
will generate a pair of RSA private key/public key and contact the CA with your public key.

The CA will register your public key

The program will then ask the CA to verify your domain.

The CA will answer with a set of challenges. These are some tasks you can complete to prove to the CA you own that domain. One of the common one is to upload a certain file at a certain address of that domain.

The program you installed will then do that for you and poll the CA for a confirmation

The CA will tell you "OK man, all is good".

The program will then generate another long term pair of private key/public key, generate a CSR (Certificate Signing Request) with the new public key and send that CSR to the CA.

The CA will extract the information, create a beautiful x509 certificate, sign it and send it back to you.
letsencrypt
will install the certificate on your server and set certain options (or not) to force https
By the way, the certificate you will get will be a DV certificate, meaning that they only verified that you owned the domain, nothing more. If you want an EV certificate this is what you will have to go through (according to wikipedia):
 Establish the legal identity as well as the operational and physical presence of website owner;
 Establish that the applicant is the domain name owner or has exclusive control over the domain name; and
 Confirm the identity and authority of the individuals acting for the website owner, and that documents pertaining to legal obligations are signed by an authorised officer.
But how does it really work?
So! The letsencrypt
program you run on your server is open sourced here: https://github.com/letsencrypt/letsencryptpreview
. It is called letsencryptpreview
I guess because it isn't done yet. It's written in Python and has to be run in sudo so that it can do most of the work for you. Note that it will install the certificates on your server only if you are using Apache or Nginx. Also, the address of the CA is hardcoded in the program so be sure to use the official letsencrypt
.
The program installed on the CA is also open sourced! So that anyone can publicly review and audit the code. It's called Boulder and it's here: https://github.com/letsencrypt/boulder and written in Go.
Letsencrypt and Boulder both use the protocol ACME
for Automated Certificate Management Environment specified here as a draft: https://letsencrypt.github.io/acmespec/
ACME
ACME is written like a RFC. It actually wants to become an RFC eventually! So if you've read RFCs before you should feel like home.
The whole protocol is happening over TLS. As a result the exchanges are encrypted, you know that you are talking to the CA you want to talk to (eventhough you might not use DNSSEC) and replay attacks should be avoided.
The whole thing is actually a RESTful API. The client, you, can do GET or POST queries to certains URI on the CA webserver.
Registration
The first thing you want to do is register. Registration is actually done by generating a new pair of RSA keys and sending them the public key (along with your info).
ACME specifies that you should use JWS for the transport of data. Json Web Signature. It's basically Json with authentication (so that you can sign your messages). It actually uses a variant of JWS called Jose that doesn't use a normal base64 encoding but that's all you should know for now. If you really want to know more there is an RFC for it.
So here what a request should look like with JWS (your information are sent unencrypted in the payload field (but don't worry, everything is encrypted anyway because the exchange happens over TLS)):
POST /acme/newregistration HTTP/1.1
Host: example.com
{
"payload":"<payload contents>",
"signatures":[
{"protected":"<integrityprotected header 1 contents>",
"header":<nonintegrityprotected header 1 contents>,
"signature":"<signature 1 contents>"}
}
Boulder will check the signature with the public key you passed, verify the information you gave and eventually add you to its database.
Once you are registered, you can perform several actions:
 Update your infos
 Get one domain authorized (well actually as many as you'd like)
The server will authenticate you because you will now send your public key along AND you will sign your requests. This all runs on top of TLS by the way. I know I already said that.
Boulder's guts
Boulder is separated in multiple components, this makes the code clearer and ensure that every piece of code does what it is supposed to do and nothing more.
One of the components is called the WebFrontEnd (WFE) and is the only one accepting queries from the Client. It parses them, verifies them and passes them to the Registration Authority (RA) that combines the other authorities together to produce a response. The response is then passed back to the WFE and to the client over the ACME protocol. Are you following?
TWe'll see what other authorities the RA has access to in the next queries the client can do. But just to talk about the previous query, the new registration query, the RA talks to the Storage Authority that deals with the database (Which is currently SQLlite) to register your new account.
All the components can be run from a single machine (for the exception of the CA that runs on another library), but they can also be run seperately from different machines that will communicate on the same network via AMQP.
New Authorization
Now that you are registered, you have to validate a domain before you can request a certificate.
You make a request to a certain URI.
Well to be exact you make a POST
request to /newauthz
, and the response will be 201
if it works. It will also give you some information about where you can go next to do stuff (/authz
)
Here's the current list of API calls you can do and the relevant answers.
The server will pass the info to the Policy Authority (PA) and ask it if it is willing to accept such a domain. If so, it will then answer with a list of challenges you can complete to prove you own the domain, along with combinations of accepted challenges to complete. For now they only have two challenges and you can complete either one:
If you choose SimpleHTTPS the letsencrypt
client will generate a random value and upload something at the address formed by a random value the CA sent you and the random value you generated.
If you choose DVSNI, the client will create a TLS certificate containing some of the info of the challenge and with the public key associated to his account.
The client then needs to query the CA again and the CA's Validation Authority (VA) will either check that the file has been uploaded or will perform a handshake with the client's server and verify that specific fields of the certificates have been correctly filled. If everything works out the VA will tell the RA that will tell the WFE that will tell you...
After that you are all good, you can now make a Certificate Signing Request :)
New Certificate
The agent will now generate a new pair of private key/public key. And create a CSR with it. So that your long term key used in your certificate is not the same as your let's encrypt account.
a CSR example, containing the identifier and the public key
After reception of it, the WebFrontEnd of Boulder will pass it to the Registration Authority (RA) which will pass it to the Certificate Authority (CA) that will do all the work and will eventually sign it and send it back to the chain.
1: Client newcert> WFE
2: WFE NewCertificate> RA
3: RA IssueCertificate> CA
4: CA > CFSSL
5: CA < CFSSL
6: RA <return CA
7: WFE <return RA
8: Client < WFE
Oh and also. the CA is talking to a CFSSL server which is CloudFlare's PKI Toolkit, a nice go library that you can use for many things and that is used here to act as the CA. CFSSL has recently pushed code to be compatible with the use of HSM which is a hardware device that you HAVE to use to sign keys when you are a CA.
After reception the letsencrypt client will install the fresh certificate along with the chain to the root on your server and voila!
You can now revoke a certificate in the same way, but interestingly you won't need to sign the request with your account key, but with the private key associated to your certificate's public key. So that even if you lose your agent's key you can still revoke the certificate.
Other stuff
There are a bunch of stuff that you will be able to do with the letsencrypt client but that haven't been implemented yet:
 Renew a certificate
 Read the Terms of Service
 Query OCSP requests (see if a certificate has been revoked)
...
Moar
This post is a simplification of the protocol. If you want to know more and don't want to dig in the ACME specs right now you can also take a look at Boulder's flow diagrams.
If you've followed the news you should have seen that Let's Encrypt just generated its root certificate along with several other certificates: https://letsencrypt.org/2015/06/04/isrgcacerts.html
This is because when you are a CA you are suppose to keep the root certificate offline. So you sign a (few) certificate(s) with that root, you lock that root and you use the signed certificate(s) to sign other certificates. This is all very serious because if something goes wrong with the root certificate, you can't revoke anything and well... the internet goes wrong as a result (until vendors start removing these from their list of trusted roots).
So the keys for the certificate have to be generated during a "ceremony" where everything is filmed and everyone must authenticate oneself at least with two different documents, etc... Check Wikipedia's page on Key Ceremony it's "interesting".
Also, I received a note from Seth David Schoen and I thought that was an interesting anecdote to share :)
the name "Boulder" is a joke from the American children's
cartoon Looney Tunes:
https://en.wikipedia.org/wiki/Wile_E._Coyote_and_The_Road_Runner
This is because the protocol that Boulder implements is called ACME,
which was also the name of the company that made the products
unsuccessfully used by the coyote to attempt to catch the roadrunner.
https://en.wikipedia.org/wiki/Acme_Corporation
Two of those products were anvils (the original name of the CA software)
and boulders.
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
the irc.log reads like this:
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.ubordeaux1.fr/CSI/  http://www.youtube.com/watch?v=zuDtACzKGRs "das boot, ouh, ja"  http://www.koreus.com/video/chatsautbalcon.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.
PrivateKey: (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:
75:0e:99:59:bd:0f:f8:72:c2:23:2f:6b:ad:32:62:
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 (p1)
exponent2 INTEGER,  d mod (q1)
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 redo 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) = (p1)(q1)
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 :)!
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.
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 timingattack.
Always try to compare hashes instead of the plaintext!
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:
/!\ 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 writeups 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 rerun 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((12*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.
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 writeup 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 :)
RFC
So, RFC means Request For Comments and they are a bunch of text files that describe different protocols. If you want to understand how SSL, TLS (the new SSL) and x509 certificates (the certificates used for SSL and TLS) all work, for example you want to code your own OpenSSL, then you will have to read the corresponding RFC for TLS: rfc5280 for x509 certificates and rfc5246 for the last version of TLS (1.2).
x509
x509 is the name for certificates which are defined for:
informal internet electronic mail, IPsec, and WWW applications
There used to be a version 1, and then a version 2. But now we use the version 3. Reading the corresponding RFC you will be able to read such structures:
Certificate ::= SEQUENCE {
tbsCertificate TBSCertificate,
signatureAlgorithm AlgorithmIdentifier,
signatureValue BIT STRING }
those are ASN.1 structures. This is actually what a certificate should look like, it's a SEQUENCE of objects.
 The first object contains everything of interest that will be signed, that's why we call it a To Be Signed Certificate
 The second object contains the type of signature the CA used to sign this certificate (ex: sha256)
 The last object is not an object, its just some bits that correspond to the signature of the TBSCertificate after it has been encoded with DER
ASN.1
It looks small, but each object has some depth to it.
The TBSCertificate is the biggest one, containing a bunch of information about the client, the CA, the publickey of the client, etc...
TBSCertificate ::= SEQUENCE {
version [0] EXPLICIT Version DEFAULT v1,
serialNumber CertificateSerialNumber,
signature AlgorithmIdentifier,
issuer Name,
validity Validity,
subject Name,
subjectPublicKeyInfo SubjectPublicKeyInfo,
issuerUniqueID [1] IMPLICIT UniqueIdentifier OPTIONAL,
 If present, version MUST be v2 or v3
subjectUniqueID [2] IMPLICIT UniqueIdentifier OPTIONAL,
 If present, version MUST be v2 or v3
extensions [3] EXPLICIT Extensions OPTIONAL
 If present, version MUST be v3
}
DER
A certificate is of course not sent like this. We use DER to encode this in a binary format.
Every fieldname is ignored, meaning that if we don't know how the certificate was formed, it will be impossible for us to understand what each value means.
Every value is encoded as a TLV triplet: [TAG, LENGTH, VALUE]
For example you can check the GITHUB certificate here
On the right is the hexdump of the DER encoded certificate, on the left is its translation in ASN.1 format.
As you can see, without the RFC near by we don't really know what each value corresponds to. For completeness here's the same certificate parsed by openssl x509
command tool:
How to read the DER encoded certificate
So go back and check the hexdump of the GITHUB certificate, here is the beginning:
30 82 05 E0 30 82 04 C8 A0 03 02 01 02
As we saw in the RFC for x509 certificates, we start with a SEQUENCE.
Certificate ::= SEQUENCE {
Microsoft made a documentation that explains pretty well how each ASN.1 TAG is encoded in DER, here's the page on SEQUENCE
30 82 05 E0
So 30
means SEQUENCE. Since we have a huge sequence (more than 127 bytes) we can't code the length on the one byte that follows:
If it is more than 127 bytes, bit 7 of the Length field is set to 1 and bits 6 through 0 specify the number of additional bytes used to identify the content length.
(in their documentation the least significant bit on the far right is bit zero)
So the following byte 82
, converted in binary: 1000 0010
, tells us that the length of the SEQUENCE will be written in the following 2 bytes 05 E0
(1504 bytes)
We can keep reading:
30 82 04 C8 A0 03 02 01 02
Another Sequence embedded in the first one, the TBSCertificate SEQUENCE
TBSCertificate ::= SEQUENCE {
version [0] EXPLICIT Version DEFAULT v1,
The first value should be the version of the certificate:
A0 03
Now this is a different kind of TAG, there are 4 classes of TAGs in ASN.1: UNIVERSAL, APPICATION, PRIVATE, and contextspecific. Most of what we use are UNIVERSAL tags, they can be understood by any application that knows ASN.1. The A0
is the [0]
(and the following 03
is the length). [0]
is a context specific TAG and is used as an index when you have a series of object. The github certificate is a good example of this, because you can see that the next index used is [3]
the extensions object:
TBSCertificate ::= SEQUENCE {
version [0] EXPLICIT Version DEFAULT v1,
serialNumber CertificateSerialNumber,
signature AlgorithmIdentifier,
issuer Name,
validity Validity,
subject Name,
subjectPublicKeyInfo SubjectPublicKeyInfo,
issuerUniqueID [1] IMPLICIT UniqueIdentifier OPTIONAL,
 If present, version MUST be v2 or v3
subjectUniqueID [2] IMPLICIT UniqueIdentifier OPTIONAL,
 If present, version MUST be v2 or v3
extensions [3] EXPLICIT Extensions OPTIONAL
 If present, version MUST be v3
}
Since those obects are all optionals, skipping some without properly indexing them would have caused trouble parsing the certificate.
Following next is:
02 01 02
Here's how it reads:
_______ tag: integer
 ____ length: 1 byte
  _ value: 2
  
  
v v v
02 01 02
The rest is pretty straight forward except for IOD: Object Identifier.
Object Identifiers
They are basically strings of integers that reads from left to right like a tree.
So in our Github's cert example, we can see the first IOD is 1.2.840.113549.1.1.11
and it is supposed to represent the signature algorithm.
So go to http://www.alvestrand.no/objectid/top.html and click on 1
, and then 1.2
, and then 1.2.840
, etc... until you get down to the latest branch of our tree and you will end up on sha256WithRSAEncryption.
Here's a more detailed explanation on IOD and here's the microsoft doc on how to encode IOD in DER.