RSA

RSA, first described in 1977, is the most famous public-key cryptosystem. It has two main use-cases:

  • Public key encryption enables a user, Alice, to distribute a public key and others can use that public key to encrypt messages to her. Alice can then use her private key to decrypt the messages.
  • Digital signatures enable Alice to use her private key to "sign" a message. Anyone can use Alice's public key to verify that the signature was created with her corresponding private key, and that the message hasn't been tampered with.

Although RSA's security is based on the difficulty of factoring large composite numbers, in recent years the cryptosystem has received criticism for how easy it is to implement incorrectly. Major flaws have been found in common deployments, the most notorious of these being the ROCA vulnerability which led to Estonia suspending 760,000 national ID cards.

These challenges introduce you to the many footguns of RSA, and soon see you performing attacks which have caused millions of dollars of damage in the real world.

Starter

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  • Modular Exponentiation
    10 pts · 11841 Solves
    All operations in RSA involve modular exponentiation.

    Modular exponentiation is an operation that is used extensively in cryptography and is normally written like: $2^{10} \mod 17$

    You can think of this as raising some number to a certain power ($2^{10} = 1024$), and then taking the remainder of the division by some other number ($1024 \mod 17 = 4$). In Python there's a built-in operator for performing this operation: pow(base, exponent, modulus).

    In RSA, modular exponentiation, together with the problem of prime factorisation, helps us to build a "trapdoor function". This is a function that is easy to compute in one direction, but hard to do in reverse unless you have the right information. It allows us to encrypt a message, and only the person with the key can perform the inverse operation to decrypt it.

    To grab the flag, find the solution to $101^{17} \mod 22663$

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  • Public Keys
    15 pts · 11268 Solves · 14 Solutions
    RSA encryption is modular exponentiation of a message with an exponent $e$ and a modulus $N$ which is normally a product of two primes: $N = p \cdot q$.

    Together, the exponent and modulus form an RSA "public key" $(N, e)$. The most common value for $e$ is 0x10001 or $65537$.

    "Encrypt" the number $12$ using the exponent $e = 65537$ and the primes $p = 17$ and $q = 23$. What number do you get as the ciphertext?

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  • Euler's Totient
    20 pts · 10646 Solves · 15 Solutions
    RSA relies on the difficulty of the factorisation of the modulus N. If the prime factors can be deduced, then we can calculate the Euler totient of N and thus decrypt the ciphertext.

    Given $N = p \cdot q$ and two primes:

    p = 857504083339712752489993810777
    q = 1029224947942998075080348647219

    What is Euler's totient $\phi(N)$?

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  • Private Keys
    20 pts · 10046 Solves · 22 Solutions
    The private key $d$ is used to decrypt ciphertexts created with the corresponding public key (it's also used to "sign" a message but we'll get to that later).

    The private key is the secret piece of information, or "trapdoor", which allows us to quickly invert the encryption function. If RSA is implemented well, if you do not have the private key the fastest way to decrypt the ciphertext is to factorise the modulus which is very hard to do for large integers.

    In RSA, the private key is the modular multiplicative inverse of the exponent $e$ modulo $\phi(N)$, Euler's totient of $N$.

    Given the two primes:

    p = 857504083339712752489993810777
    q = 1029224947942998075080348647219

    and the exponent $e = 65537$, what is the private key $d \equiv e^{-1} \mod \phi(N)$?

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  • RSA Decryption
    20 pts · 9693 Solves · 17 Solutions
    I've encrypted a secret number for your eyes only using your public key parameters:

    N = 882564595536224140639625987659416029426239230804614613279163
    e = 65537

    Use the private key that you found for these parameters in the previous challenge to decrypt this ciphertext:

    c = 77578995801157823671636298847186723593814843845525223303932

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  • RSA Signatures
    25 pts · 7405 Solves · 24 Solutions
    How can you ensure that the person receiving your message knows that you wrote it?

    You've been asked out on a date, and you want to send a message telling them that you'd love to go, however a jealous lover isn't so happy about this.

    When you send your message saying yes, your jealous lover intercepts the message and corrupts it so it now says no!

    We can protect against these attacks by cryptographically signing the message.

    Imagine you write a message $m$. You encrypt this message with your friend's public key: $c = m^{e_{0}} \mod N_{0}$.

    To sign this message, you calculate the hash of the message: $H(m)$ and "encrypt" this with your private key: $S = H(m)^{d_{1}} \mod N_{1}$.

    In real cryptosystems, it's best practice to use separate keys for encrypting and signing messages.

    Your friend can decrypt the message using their private key: $m = c^{d_{0}} \mod N_{0}$. Using your public key they calculate $s = S^{e_{1}} \mod N_{1}$.

    Now by computing $H(m)$ and comparing it to $s$: assert H(m) == s, they can ensure that the message you sent them, is the message that they received! As long as your private key is safe, no one else could have signed this message!

    Sign the flag crypto{Immut4ble_m3ssag1ng} using your private key and the SHA256 hash function.

    The output of the hash function needs to be converted into a number that can be used with RSA math. Remember the helpful bytes_to_long() function that can be imported from Crypto.Util.number.

    Challenge files:
      - private.key

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Primes Part 1

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  • Factoring
    15 pts · 7531 Solves · 13 Solutions
    So far we've been using the product of small primes for the modulus, but small primes aren't much good for RSA as they can be factorised using modern methods.

    What is a "small prime"? There was an RSA Factoring Challenge with cash prizes given to teams who could factorise RSA moduli. This gave insight to the public into how long various key sizes would remain safe. Computers get faster, algorithms get better, so in cryptography it's always prudent to err on the side of caution.

    These days, using primes that are at least 1024 bits long is recommended—multiplying two such 1024 primes gives you a modulus that is 2048 bits large. RSA with a 2048-bit modulus is called RSA-2048.

    Some say that to really remain future-proof you should use RSA-4096 or even RSA-8192. However, there is a tradeoff here; it takes longer to generate large prime numbers, plus modular exponentiations are predictably slower with a large modulus.

    Factorise the 150-bit number 510143758735509025530880200653196460532653147 into its two constituent primes. Give the smaller one as your answer.

    Resources:
      - How big an RSA key is considered secure today?
      - primefac-fork

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  • Inferius Prime
    30 pts · 5457 Solves · 12 Solutions
    Here is my super-strong RSA implementation, because it's 1600 bits strong it should be unbreakable... at least I think so!

    Challenge files:
      - inferius.py
      - output.txt

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  • Monoprime
    30 pts · 6758 Solves · 24 Solutions
    Why is everyone so obsessed with multiplying two primes for RSA. Why not just use one?

    Challenge files:
      - output.txt

    Resources:
      - Why do we need in RSA the modulus to be product of 2 primes?

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  • Square Eyes
    35 pts · 4947 Solves · 17 Solutions
    It was taking forever to get a 2048 bit prime, so I just generated one and used it twice.

    If you're stuck, look again at the formula for Euler's totient.

    Challenge files:
      - output.txt

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  • Manyprime
    40 pts · 5954 Solves · 28 Solutions
    Using one prime factor was definitely a bad idea so I'll try using over 30 instead.

    If it's taking forever to factorise, read up on factorisation algorithms and make sure you're using one that's optimised for this scenario.

    Challenge files:
      - output.txt

    Resources:
      - The Elliptic Curve Factorization Method

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Public exponent

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Primes Part 2

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  • Infinite Descent
    50 pts · 2693 Solves · 17 Solutions
    Finding large primes is slow, so I've devised an optimisation.

    Challenge files:
      - descent.py
      - output.txt

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  • Marin's Secrets
    50 pts · 2644 Solves · 18 Solutions
    I've found a super fast way to generate primes from my secret list.

    Challenge files:
      - marin.py
      - output.txt

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  • Fast Primes
    75 pts · 1765 Solves · 6 Solutions
    I need to produce millions of RSA keys quickly and the standard way just doesn't cut it. Here's yet another fast way to generate primes which has actually resisted years of review.

    Challenge files:
      - fast_primes.py
      - key.pem
      - ciphertext.txt

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  • Ron was Wrong, Whit is Right
    90 pts · 1660 Solves · 16 Solutions
    Here's a bunch of RSA public keys I gathered from people on the net together with messages that they sent.

    As excerpt.py shows, everyone was using PKCS#1 OAEP to encrypt their own messages. It shouldn't be possible to decrypt them, but perhaps there are issues with some of the keys?

    Challenge files:
      - excerpt.py
      - keys_and_messages.zip

    Resources:
      - The recent difficulties with RSA

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  • RSA Backdoor Viability
    175 pts · 1240 Solves · 7 Solutions
    It seems like my method to generate fast primes was not completely secure. I came up with a new approach to improve security, including a factorization backdoor in case I ever lose my private key. You'll definitely need some complex techniques to break this!

    You may need to tweak the recursion limit (sys.setrecursionlimit(n) in Python) in your programming language to get your solution working.

    Challenge files:
      - complex_primes.py
      - output.txt


    Challenge contributed by joachim

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Padding

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Signatures Part 1

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  • Signing Server
    60 pts · 1710 Solves · 12 Solutions
    My boss has so many emails he's set up a server to just sign everything automatically. He also stores encrypted messages to himself for easy access. I wonder what he's been saying.

    Connect at socket.cryptohack.org 13374

    Challenge files:
      - 13374.py

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  • Let's Decrypt
    80 pts · 1123 Solves · 5 Solutions
    If you can prove you own CryptoHack.org, then you get access to one of our secrets.

    Connect at socket.cryptohack.org 13391

    Challenge files:
      - 13391.py

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  • Blinding Light
    120 pts · 1210 Solves · 14 Solutions
    Here's my token signing and verification server. I'm not sure it's doing signing properly, but I've implemented some safeguards to ensure it won't hand out admin tokens to just anyone.

    Connect at socket.cryptohack.org 13376

    Challenge files:
      - 13376.py

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Signatures Part 2

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