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 $101^{17} \mod 22663$

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?

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 $\phi(N)$?

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 $d \equiv e^{-1} \mod \phi(N)$?

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

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

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

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

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?

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

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

Smallest exponent should be fastest, right?

Challenge files:
  - salty.py
  - output.txt

My primes should be more than large enough now!

Challenge files:
  - modulus_inutilis.py
  - output.txt

We have a supercomputer at work, so I've made sure my encryption is secure by picking massive numbers!

Challenge files:
  - source.py
  - output.txt

I asked my friends to encrypt our secret flag before sending it to me, but instead of using my key, they've all used their own! Can you help?

Challenge files:
  - source.py
  - output.txt

Resources:
  - RSA: how to factorize N given d

Okay so I got a bit carefree with my last script, but this time I've protected myself while keeping everything really big. Nothing will stop me and my supercomputer now!

Challenge files:
  - source.py
  - output.txt

Resources:
  - Twenty Years of Attacks on the RSA Cryptosystem

Poor Johan has been answering emails all day and many of the students are asking the same question. Can you read his messages?

Challenge files:
  - johan.py
  - output.txt

Resources:
  - Twenty Years of Attacks on the RSA Cryptosystem

Finding large primes is slow, so I've devised an optimisation.

Challenge files:
  - descent.py
  - output.txt

I've found a super fast way to generate primes from my secret list.

Challenge files:
  - marin.py
  - output.txt

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

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

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

Been cooking up my own padding scheme, now my encrypted flag is different everytime!

Connect at socket.cryptohack.org 13386

Challenge files:
  - 13386.py

Can custom padding save one from some of the mistakes we already covered?

Challenge files:
  - pad_encrypt.py
  - output.txt

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

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

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

If you want my flag, you better vote for Pedro! Can you sign your vote to the server as Alice?

Connect at socket.cryptohack.org 13375

Challenge files:
  - 13375.py
  - alice.key

Let's Decrypt was too easy, let's do it again!

Connect at socket.cryptohack.org 13394

Challenge files:
  - 13394.py


Challenge contributed by Robin_Jadoul and Thunderlord