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304 North Cardinal St.
Dorchester Center, MA 02124
Monday to Friday: 7AM - 7PM
Weekend: 10AM - 5PM
As it’s been making the rounds recently, I wanted to try my hand at cracking 256-bit RSA keys.
If you haven’t seen the video yet, Crown Sterling cracked a 256-bit RSA key in front of a live audience in 50 seconds.
I wasn’t sure how impressive this was originally, and I wanted to try it out myself.
For more information about RSA, and the math behind it, you can always check out the Wikipedia article.
Additionally, for another example of the math behind RSA, and cracking it, I recommend the following post.
All of that said, I’m no cryptographer, so this was more an attempt to see how easy it was for me to crack these keys. I’m in no way making any claims about anyone else’s research, or whether something is invalid or fake. This is also a fitting example of verifying claims yourself where possible though!
First, I generated a 256-bit RSA private key using OpenSSL
[email protected]:~/rsa256# openssl genrsa -out private.pem 256 Generating RSA private key, 256 bit long modulus ...+++++++++++++++++++++++++++ ............+++++++++++++++++++++++++++ e is 65537 (0x10001)
Next, I printed the actual private key, as well as the modulus (n)
[email protected]:~/rsa256# openssl rsa -in private.pem -modulus Modulus=B19D0C77A45D2A8FD9B9EEE42BEBE6CE0F0AF88B5FF529982D2D52257412A507 writing RSA key -----BEGIN RSA PRIVATE KEY----- MIGpAgEAAiEAsZ0Md6RdKo/Zue7kK+vmzg8K+Itf9SmYLS1SJXQSpQcCAwEAAQIg OYL194O0W0TLJoaxQXuYd1SIY0QN+97UIyDMjgZAOXkCEQDbKDy09Z3qWI1cn+C+ ls5NAhEAz3jqOoDgWeu/Q/3DmGtyowIQZQolCu0elDelXOndDSGsFQIQO2ErAKWE EJhlfIszoPsXqwIQZ7estf3MM1uuXDAvw6Hvsw== -----END RSA PRIVATE KEY-----
I also generated the public key, as this is what I would be attacking.
[email protected]:~/rsa256# openssl rsa -in private.pem -outform PEM -pubout -out public.pem writing RSA key
As you can see, the exponent and modulus are the same for the public key and the private key.
[email protected]:~/rsa256# openssl rsa -pubin -in public.pem -text -noout Public-Key: (256 bit) Modulus: 00:b1:9d:0c:77:a4:5d:2a:8f:d9:b9:ee:e4:2b:eb: e6:ce:0f:0a:f8:8b:5f:f5:29:98:2d:2d:52:25:74: 12:a5:07 Exponent: 65537 (0x10001)
[email protected]:~/rsa256# openssl rsa -pubin -in public.pem -modulus -noout Modulus=B19D0C77A45D2A8FD9B9EEE42BEBE6CE0F0AF88B5FF529982D2D52257412A507
Finally, I converted the modulus from hex to an integer using ‘bc’, as this is the input I will use for cracking it.
[email protected]:~/rsa256# echo "ibase=16;B19D0C77A45D2A8FD9B9EEE42BEBE6CE0F0AF88B5FF529982D2D52257412A507" | bc 80336855234907714168477675917972994189398342031083238074132216291031\ 761724679
First, to crack the key, I created a 16 CPU DigitalOcean droplet.
I verified the processors by checking cpuinfo after the machine booted.
[email protected]:~/rsa256# cat /proc/cpuinfo | grep "model name" model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz model name : Intel(R) Xeon(R) Platinum 8168 CPU @ 2.70GHz
With the machine up and running, I installed make, cmake, and g++ to compile any tools that I might use.
Next, I found a tool that implemented a Number Field Sieve called msieve.
First, I cloned the GitHub repository.
[email protected]:~/rsa256# git clone https://github.com/radii/msieve Cloning into 'msieve'... remote: Enumerating objects: 1504, done. remote: Total 1504 (delta 0), reused 0 (delta 0), pack-reused 1504 Receiving objects: 100% (1504/1504), 634.68 KiB | 15.87 MiB/s, done. Resolving deltas: 100% (1132/1132), done.
Next, I had to edit the makefile, so that the architecture matched the CPUs that I was using (Skylake).
[email protected]:~/rsa256/msieve# cat Makefile | grep "march" # gcc with basic optimization (-march flag could OPT_FLAGS = -O3 -fomit-frame-pointer -march=skylake -DNDEBUG #OPT_FLAGS = -O3 -fomit-frame-pointer -march=k8 -DNDEBUG $(CC) $(CFLAGS) -march=pentium2 -DBLOCK_KB=64 -DCPU_PENTIUM2 \ $(CC) $(CFLAGS) -march=pentium3 -DBLOCK_KB=64 -DCPU_PENTIUM3 \ $(CC) $(CFLAGS) -march=pentium4 -DBLOCK_KB=64 -DCPU_PENTIUM4 \ $(CC) $(CFLAGS) -march=pentium-m -DBLOCK_KB=32 -DCPU_PENTIUM_M \ $(CC) $(CFLAGS) -march=prescott -DBLOCK_KB=32 -DCPU_CORE \ $(CC) $(CFLAGS) -march=athlon -DBLOCK_KB=64 -DCPU_ATHLON \ $(CC) $(CFLAGS) -march=athlon-xp -DBLOCK_KB=64 -DCPU_ATHLON_XP \ $(CC) $(CFLAGS) -march=k8 -DBLOCK_KB=64 -DCPU_OPTERON \ $(CC) $(CFLAGS) -march=nocona -DBLOCK_KB=64 -DCPU_PENTIUM4 \ $(CC) $(CFLAGS) -march=nocona -DBLOCK_KB=32 -DCPU_CORE \ $(CC) $(CFLAGS) -march=k8 -DBLOCK_KB=64 -DCPU_OPTERON \ [email protected]:~/rsa256/msieve# make x86_64 gcc -D_FILE_OFFSET_BITS=64 -O3 -fomit-frame-pointer -march=skylake -DNDEBUG -Wall -W -I. -Iinclude -Ignfs -Ignfs/poly -Ignfs/poly/stage1 -c -o common/filter/clique.o common/filter/clique.c
After that change, I was able to successfully compile the application and view the help.
[email protected]:~/rsa256/msieve# ./msieve --help Msieve v. 1.46 usage: ./msieve [options] [one_number] numbers starting with '0' are treated as octal, numbers starting with '0x' are treated as hexadecimal options: -s
save intermediate results to instead of the default msieve.dat -l append log information to instead of the default msieve.log -i read one or more integers to factor from (default worktodo.ini) instead of from the command line -m manual mode: enter numbers via standard input -mb hint for number of megabytes of memory for postprocessing (set automatically if unspec- ified or zero) -q quiet: do not generate any log information, only print any factors found -d deadline: if still sieving after minutes, shut down gracefully (default off) -r stop sieving after finding relations -p run at idle priority -v verbose: write log information to screen as well as to logfile -t use at most threads elliptic curve options: -e perform 'deep' ECM, seek factors > 15 digits quadratic sieve options: -c client: only perform sieving number field sieve options: -n use the number field sieve (80+ digits only; performs all NFS tasks in order) -nf read from / write to NFS factor base file instead of the default msieve.fb -np [X,Y] perform only NFS polynomial selection; if specified, only cover leading coefficients in the range from X to Y inclusive -np1 [X,Y] perform stage 1 of NFS polynomial selection; if specified, only cover leading coefficients in the range from X to Y inclusive -np2 perform stage 2 of NFS polynomial selection -ns [X,Y] perform only NFS sieving; if specified, handle sieve lines X to Y inclusive -nc perform only NFS combining (all phases) -nc1 [X,Y] perform only NFS filtering. Filtering will track ideals >= X (determined automatically if 0 or unspecified) and will only use the first Y relations (or all relations, if 0 or unspecified) -nc2 perform only NFS linear algebra -ncr perform only NFS linear algebra, restarting from a previous checkpoint -nc3 [X,Y] perform only NFS square root (compute dependency numbers X through Y, 1<=X<=Y<=64)
With msieve running, I timed it using the q and n flags, as that seemed basic and straightforward.
[email protected]:~/rsa256/msieve# time ./msieve -q -n 80336855234907714168477675917972994189398342031083238074132216291031761724679 80336855234907714168477675917972994189398342031083238074132216291031761724679 prp39: 275778021469467750604832321873164071587 prp39: 291309854232898176366046870573797527117 real 2m44.099s user 2m43.996s sys 0m0.092s
While two minutes and forty-four seconds wasn't bad from a random program, I was hoping that I could do it faster!
I decided to try a few Docker images, to see if any of them could give me a lower time.
First, I installed Docker to my droplet.
Next, I found an image titled rsacrack, which sounded perfect
I pulled down the image to my droplet.
[email protected]:~# docker pull b4den/rsacrack Using default tag: latest latest: Pulling from b4den/rsacrack 5667fdb72017: Pull complete d83811f270d5: Pull complete ee671aafb583: Pull complete 7fc152dfb3a6: Pull complete b83d8b6245c4: Pull complete 36cb8498468a: Pull complete 22aaf58c64c7: Pull complete fb0d231b82e1: Pull complete fc922e878061: Pull complete 84cf9426007f: Pull complete 897258e3faf3: Pull complete 6dc87ced1f13: Pull complete 71db44ed3509: Pull complete b266cfc4f7eb: Pull complete e62159d0a47e: Pull complete 8fc467efa1ec: Pull complete d4b519fd0423: Pull complete da3c0b30b580: Extracting [==================================================>] 3.126kB/3.126kB 365f1efd6c17: Download complete fa0aa39ead30: Download complete 97567ded379a: Download complete ca8390f60942: Download complete 54e964f3a0f6: Download complete 878a41964830: Download complete 85beee49cceb: Download complete
Once I finally figured out how the image worked, I was able to run and time it. Also, it gave me a bit more information about the cracked prime, as well as generating a new private key for me.
(after a few attempts, I was able to time it and run it)
[email protected]:~# time docker run b4den/rsacrack 80336855234907714168477675917972994189398342031083238074132216291031761724679 [*] pubkey.e: 65537 [*] pubkey.n: 80336855234907714168477675917972994189398342031083238074132216291031761724679 [*] Key looks like 256 bits [*] Using cadonfs to compute primes [*] results are: [u'275778021469467750604832321873164071587', u'291309854232898176366046870573797527117', 65537L] [*] Key extraction done. -----BEGIN RSA PRIVATE KEY----- MIGqAgEAAiEAsZ0Md6RdKo/Zue7kK+vmzg8K+Itf9SmYLS1SJXQSpQcCAwEAAQIgOYL194O0W0TL JoaxQXuYd1SIY0QN+97UIyDMjgZAOXkCEQDPeOo6gOBZ679D/cOYa3KjAhEA2yg8tPWd6liNXJ/g vpbOTQIRAMPjbcGJUbxD1IK6JAci8ZcCECHcFgi4g6NUoxoNBUMvszICEG1I5uI3mwxY9rfO3XxE Hcs= -----END RSA PRIVATE KEY----- real 1m13.985s user 0m0.038s sys 0m0.025s
As you can see, it only took one minute and fourteen seconds this time, which was a huge improvement!
Note that the RSA private key is a bit different from the one I generated earlier, even though the modulus is the same.
I also wanted to try cado-nfs, for no particular reason. That said, I was unable to get the boost libraries to properly work on my Ubuntu droplet.
I was able to find a cado-nfs Docker image, so that seemed like the next best choice.
First, I pulled down the image.
[email protected]:~# docker pull cyrilbouvier/cado-nfs.py Using default tag: latest latest: Pulling from cyrilbouvier/cado-nfs.py ... docker.io/cyrilbouvier/cado-nfs.py:latest
When running this image, a lot seemed to be going on, but it eventually printed the same primes as everything else.
[email protected]:~# time docker run cyrilbouvier/cado-nfs.py 80336855234907714168477675917972994189398342031083238074132216291031761724679 Info:root: Using default parameter file /cado-nfs/share/cado-nfs-2.2.1/factor/params.c75 Info:root: No database exists yet Info:root: Created temporary directory /tmp/cado.33dpijhe Info:Database: Opened connection to database /tmp/cado.33dpijhe/c75.db Info:root: Set tasks.threads=16 based on detected physical cpus Info:root: tasks.polyselect.threads = 2 Info:root: tasks.sieve.las.threads = 2 Info:root: slaves.scriptpath is /cado-nfs/bin Info:root: Command line parameters: /cado-nfs/bin/cado-nfs.py 80336855234907714168477675917972994189398342031083238074132216291031761724679 Info:root: If this computation gets interrupted, it can be resumed with /cado-nfs/bin/cado-nfs.py /tmp/cado.33dpijhe/c75.parameters_snapshot.0 ... Info:Linear Algebra: Total cpu/real time for bwc: 16.08/0.000250816 Info:Linear Algebra: Aggregate statistics: Info:Linear Algebra: Krylov: WCT time 2.07 Info:Linear Algebra: Lingen CPU time 4.26, WCT time 0.51 Info:Linear Algebra: Mksol: WCT time 1.62 Info:Quadratic Characters: Total cpu/real time for characters: 0.81/0.234679 Info:Square Root: Total cpu/real time for sqrt: 28.48/2.57451 Info:HTTP server: Shutting down HTTP server Info:Complete Factorization: Total cpu/elapsed time for entire factorization: 324.42/70.4059 Info:root: Cleaning up computation data in /tmp/cado.33dpijhe 275778021469467750604832321873164071587 291309854232898176366046870573797527117 real 1m11.840s user 0m0.040s sys 0m0.095s
As you can see, this brought the time down even further, and in the realm of under one minute!
In the end, I only needed this droplet for around 20 minutes, so it only cost me ~$0.16 to crack my key.
While all these applications had the same output, I wanted to verify that I was actually successful.
First, I used Python to calculate the original modulus from my derived factors.
[email protected]:~/rsa256# python -c 'print (275778021469467750604832321873164071587*291309854232898176366046870573797527117)' 80336855234907714168477675917972994189398342031083238074132216291031761724679
Next, I encrypted a secret message with my public key. Note that the message needs to be shorter than the original key, so I only had 32 bytes.
[email protected]:~/rsa256# cat plaintext.txt Secret message! [email protected]:~/rsa256# openssl rsautl -encrypt -pubin -inkey public.pem -in plaintext.txt -out encrypted.txt [email protected]:~/rsa256# cat encrypted.txt ??=79eUn?B????????Y?x??c???
I saved the output of the rsacrack Docker image as my cracked private key.
[email protected]:~/rsa256# cat cracked.pem -----BEGIN RSA PRIVATE KEY----- MIGqAgEAAiEAsZ0Md6RdKo/Zue7kK+vmzg8K+Itf9SmYLS1SJXQSpQcCAwEAAQIgOYL194O0W0TL JoaxQXuYd1SIY0QN+97UIyDMjgZAOXkCEQDPeOo6gOBZ679D/cOYa3KjAhEA2yg8tPWd6liNXJ/g vpbOTQIRAMPjbcGJUbxD1IK6JAci8ZcCECHcFgi4g6NUoxoNBUMvszICEG1I5uI3mwxY9rfO3XxE Hcs= -----END RSA PRIVATE KEY-----
Using this private key, I was still able to decrypt the message, even though it was different than my original private key!
[email protected]:~/rsa256# openssl rsautl -decrypt -inkey cracked.pem -in encrypted.txt -out decrypted.txt [email protected]:~/rsa256# cat decrypted.txt Secret message!
While I was able to decrypt my secret message using the cracked private key, I was unsure how this key was generated.
In this case, I found a StackOverflow post on how to generate an RSA key using specific input numbers.
First, I created a small Python script (that I'll share below) to create a configuration file for asn1parse.
[email protected]:~/rsa256# python genKey.py asn1=SEQUENCE:rsa_key [rsa_key] version=INTEGER:0 modulus=INTEGER:80336855234907714168477675917972994189398342031083238074132216291031761724679 pubExp=INTEGER:65537 privExp=INTEGER:26013220088499269151307883495663593232543266314163563907943768170179715938681 p=INTEGER:275778021469467750604832321873164071587 q=INTEGER:291309854232898176366046870573797527117 e1=INTEGER:78928976741425555000913069889908578219 e2=INTEGER:134304701857683726732992144809608195093 coeff=INTEGER:145264379791353023195410394043851742667 [email protected]:~/rsa256# python genKey.py > gen.conf
Using this configuration file, I generated another private key using these values.
[email protected]:~/rsa256# openssl asn1parse -genconf gen.conf -out newkey.der 0:d=0 hl=3 l= 169 cons: SEQUENCE 3:d=1 hl=2 l= 1 prim: INTEGER :00 6:d=1 hl=2 l= 33 prim: INTEGER :B19D0C77A45D2A8FD9B9EEE42BEBE6CE0F0AF88B5FF529982D2D52257412A507 41:d=1 hl=2 l= 3 prim: INTEGER :010001 46:d=1 hl=2 l= 32 prim: INTEGER :3982F5F783B45B44CB2686B1417B9877548863440DFBDED42320CC8E06403979 80:d=1 hl=2 l= 17 prim: INTEGER :CF78EA3A80E059EBBF43FDC3986B72A3 99:d=1 hl=2 l= 17 prim: INTEGER :DB283CB4F59DEA588D5C9FE0BE96CE4D 118:d=1 hl=2 l= 16 prim: INTEGER :3B612B00A5841098657C8B33A0FB17AB 136:d=1 hl=2 l= 16 prim: INTEGER :650A250AED1E9437A55CE9DD0D21AC15 154:d=1 hl=2 l= 16 prim: INTEGER :6D48E6E2379B0C58F6B7CEDD7C441DCB [email protected]:~/rsa256# openssl rsa -in newkey.der -inform der -text -check Private-Key: (256 bit) modulus: 00:b1:9d:0c:77:a4:5d:2a:8f:d9:b9:ee:e4:2b:eb: e6:ce:0f:0a:f8:8b:5f:f5:29:98:2d:2d:52:25:74: 12:a5:07 publicExponent: 65537 (0x10001) privateExponent: 39:82:f5:f7:83:b4:5b:44:cb:26:86:b1:41:7b:98: 77:54:88:63:44:0d:fb:de:d4:23:20:cc:8e:06:40: 39:79 prime1: 00:cf:78:ea:3a:80:e0:59:eb:bf:43:fd:c3:98:6b: 72:a3 prime2: 00:db:28:3c:b4:f5:9d:ea:58:8d:5c:9f:e0:be:96: ce:4d exponent1: 3b:61:2b:00:a5:84:10:98:65:7c:8b:33:a0:fb:17: ab exponent2: 65:0a:25:0a:ed:1e:94:37:a5:5c:e9:dd:0d:21:ac: 15 coefficient: 6d:48:e6:e2:37:9b:0c:58:f6:b7:ce:dd:7c:44:1d: cb RSA key ok writing RSA key -----BEGIN RSA PRIVATE KEY----- MIGpAgEAAiEAsZ0Md6RdKo/Zue7kK+vmzg8K+Itf9SmYLS1SJXQSpQcCAwEAAQIg OYL194O0W0TLJoaxQXuYd1SIY0QN+97UIyDMjgZAOXkCEQDPeOo6gOBZ679D/cOY a3KjAhEA2yg8tPWd6liNXJ/gvpbOTQIQO2ErAKWEEJhlfIszoPsXqwIQZQolCu0e lDelXOndDSGsFQIQbUjm4jebDFj2t87dfEQdyw== -----END RSA PRIVATE KEY-----
While this private key is different from either of my two earlier ones, at least I know that I generated it myself.
Using this generated RSA key, I was still able to decrypt my secret message!
[email protected]:~/rsa256# openssl rsautl -decrypt -inkey generated.pem -in encrypted.txt -out decrypted-gen.txt [email protected]:~/rsa256# cat decrypted-gen.txt Secret message!
You can find the code for my genkey.py script below. For more information on RSA cracking, as well as where I got the egcd and modinv methods from, then you should check out this post
#!/usr/bin/python def egcd(a, b): if a == 0: return (b, 0, 1) else: g, y, x = egcd(b % a, a) return (g, x - (b // a) * y, y) def modinv(a, m): gcd, x, y = egcd(a, m) if gcd != 1: return None # modular inverse does not exist else: return x % m def main(): p = 275778021469467750604832321873164071587 # edit with factor #1 q = 291309854232898176366046870573797527117 # edit with factor #2 n = 80336855234907714168477675917972994189398342031083238074132216291031761724679 # edit with modulus e = 65537 # edit if exponent is different phi = (p -1)*(q-1) d = modinv(e,phi) dp = modinv(e,(p-1)) dq = modinv(e,(q-1)) qi = modinv(q,p) print "asn1=SEQUENCE:rsa_key" print "" print "[rsa_key]" print "version=INTEGER:0" print "modulus=INTEGER:" + str(n) print "pubExp=INTEGER:" + str(e) print "privExp=INTEGER:" + str(d) print "p=INTEGER:" + str(p) print "q=INTEGER:" + str(q) print "e1=INTEGER:" + str(dp) print "e2=INTEGER:" + str(dq) print "coeff=INTEGER:" + str(qi) if __name__ == "__main__": main()
As usual, you can find the code and updates in my GitHub repository as well.
This was a fun exercise, and it was much faster than I expected to do the cracking.
I could see a CTF using this as a challenge in the future, so it helps to know how to perform this attack.
Finally, for some more examples of cracking and math, check out Rob's Twitter thread.
Ray Doyle is an avid pentester/security enthusiast/beer connoisseur who has worked in IT for almost 16 years now. From building machines and the software on them, to breaking into them and tearing it all down; he’s done it all. To show for it, he has obtained an OSCE, OSCP, eCPPT, GXPN, eWPT, eWPTX, SLAE, eMAPT, Security+, ICAgile CP, ITIL v3 Foundation, and even a sabermetrics certification!
He currently serves as a Senior Staff Adversarial Engineer for Avalara, and his previous position was a Principal Penetration Testing Consultant for Secureworks.
This page contains links to products that I may receive compensation from at no additional cost to you. View my Affiliate Disclosure page here. As an Amazon Associate, I earn from qualifying purchases.
So, all of our server keys can be cracked ?
Like ssh keys and stuff ?
What should we use it now ?
256-bit RSA is actually rarely used, so you should be fine. For example, when you generate an SSH key, it uses RSA-2048 by default, which is infeasible to crack at this time!
If you really can find the private key, so gind my private key.
This is my public key.
Hi, I have no proof that it is your public key, and I covered how to do this in the post! That said, feel free to give it a try against your own personal keys if you’d like.
This is a fantastic walk-through! Great job and thank you for sharing!
Thanks, glad it helped!
Does this mean Bitcoin wallets are no longer secure?
Not at all! Bitcoin actually uses AES-256-CBC which is far more secure than this.
your work is amazing. Unfortunately the video is not available anymore. How can I see it ?
Eh, they probably pulled it down because it made them look bad haha. Was literally just them on stage making a big deal about cracking 256-bit RSA keys (and slowly).
Thanks alot , you made my day sir
You’re welcome, glad that it helped