Monday, June 13, 2011

How to break RSA explicitly with OpenSSL keys

Everybody talks about how vulnerable is RSA and the importance of prime numbers involved in the creation of keys, people knows that the security is based on the complexity of factorization, some days before I posted how to break a key and people asked me to make a post of it. 

We are going to break a RSA key of 256 bits, this size is not too big but is not insignificant in less than 5 minutes, this is another reason to do research in other schemes like discrete logarithm over other algebraic platforms, I have been doing speeches about jacobians of algebraic curves and other abelian varieties. I will introduce RSA for those who don't know how it works and then we are going to break it :) 

Key generation: 
 - Two random primes are generated (p,q) 
 - Compute n=pq this number will be the modulus 
 - Compute phi(n)=(p-1)(q-1) (phi(n) is the function that counts the relative primes to n, and is (p-1)(q-1) because p and q are primes) 
 - Choose e such that 1 less than e less than phi(n) and gcd(phi(n),e) = 1 (OpenSSL generally assigns something near to 2^16 like 65537) 
 - Compute d = e^-1 mod phi(n) \

(d,p,q) is the private key 
(e,n) is the public key 

- A is going to receive M so A sends to B (e,n) 
- B computes c=M^e mod n and sends to A 

 - A computes M=c^d mod n and this is the same as M=(M^e)^d mod n which is the same as M (it's easy to prove that the last equation is M) 

The security is that, if someone gets the public key (e,n) we have that d = e^-1 mod phi(n). Further, it is the same as d = e^-1 mod (p-1)(q-1). An attacker needs to factorize n=pq to get e^-1 mod (p-1)(q-1) because this person must know (p-1)(q-1) to compute the inverse of e.   

How to break RSA given a public key generated with OpenSSL 

First we need a test environment to break.   
We generate a random 256 bit RSA private key (this is p,q) and we save it to privada.pem 

openssl genrsa -out privada.pem 256 

We generate the public key with the previous private.pem (this is e,n, we will break this by factoring n) and we save it to pub.pem 

openssl rsa -in privada.pem -pubout -out pub.pem 

We extract the modulus n and the exponent n from the public key 

openssl rsa -in pub.pem -pubin -text -modulus 

This will give us a lot of info, and in the info you will find 

Exponent: 65537 (0x10001) Modulus=9ABAAD5BBE954A26BDBB15568B7AEB3651D0605EA5687329F849DB1F9871865F  

Exponent is "e" and Modulus is "n" 
We convert to decimal the modulus 

echo "ibase=16; 9ABAAD5BBE954A26BDBB15568B7AEB3651D0605EA5687329F849DB1F9871865F"|bc 69986008711415694391421268580269058232048146719704518153244714221529713444447 

Then we factorize with msieve that is available here, this msieve works using the number field sieve, this algorithm is the fastest to factorize (turing algorithm) you will have to compile it and it requieres GMP and other standard stuff. You can also clone it from git, explore the "march" compiler parameter from Makefile if you would like additional optimization for your processor: 

git clone
cd msieve
make; make install

If msieve compiled correctly and installed correctly, to factorize that modulus n, you just need to run it with msieve as follows:  

msieve -v 69986008711415694391421268580269058232048146719704518153244714221529713444447 

It will take 2 minutes in a home computer (this is because we are working with 256 bits, but in my university theres a cluster called Kanbalam, it can break 768 bits in weeks) When it finishes it will show you the factors 
prp39 factor: 258903250452187592132630852021175987089 
prp39 factor: 270317226953240634960995990331891792623 

Now, the last step is to build the private key with the (e,p,q) , and I made a cgi to my site that generates a PEM private, you just need to put the p,q and e in the GET (URL) and it will show you the private key (note that previously we showed that e=65537) 

Just use it like this: 

The output of the website will be the following: 

Llave privada de 258903250452187592132630852021175987089,270317226953240634960995990331891792623 y 65537 por Eduardo Duarte 

If you compare the initial key (privada.pem) with this new key that I will save in "brokenkey.pem" you will see that they are the same and we did not use any information of the private key generated at the begining to get this brokenkey.pem 

md5sum privada.pem llaverota.pem 
9cd4b5a39be5717978b9035ddc5fc887 privada.pem 
9cd4b5a39be5717978b9035ddc5fc887 brokenkey.pem 

Comments: Eduardo Duarte toorandom at gmail d0t com Twitter: @toorandom


Naoy said...

Hi dude!

Thanks for this post, it's interesting as the codes provided on your site.

Keep it up!

Anonymous said...

Thanks for the fantastic post, Beck.

@madrhouse said...

Muy Interesante


Anonymous said...

Wow, On an Ivy Bridge i3 3225, and performance as my govener, I cracked it in 5 seconds :O

Now I'm trying 512 bit ;)

Cecilia said...


Anonymous said...

Nice article. Thank you!

Could you please publish the source code of the script that generates the private key? I would like to see how it works.

Andy said...

Really nice article thanks !
The CGI on your site doesn't work anymore as of now,
just show input and doesn't generate pem.
Can you fix it please or give some info on how to assemble private key by hand?

Keep up the great work !

Unknown said...

amigo, no tienes el script de nuevo?