# Crypto: Chunk Norris
## Task Description
Chunk Norris is black belt in fast random number generation.
## First Look
We are given the following Python file, as well as its output
```#!/usr/bin/python3 -u
import randomfrom Crypto.Util.number import *import gmpy2
a = 0xe64a5f84e2762be5chunk_size = 64
def gen_prime(bits):  s = random.getrandbits(chunk_size)
  while True:    s |= 0xc000000000000001    p = 0    for _ in range(bits // chunk_size):      p = (p << chunk_size) + s      s = a * s % 2**chunk_size    if gmpy2.is_prime(p):      return p
n = gen_prime(1024) * gen_prime(1024)e = 65537flag = open("flag.txt", "rb").read()print('n =', hex(n))print('e =', hex(e))print('c =', hex(pow(bytes_to_long(flag), e, n)))```
It is clear from the last 3 lines that this is an RSA encryption scheme we need to break, and in order to do that, we need the prime factors `p` and `q` such that `n = pq`. This is only possible because of the weak prime generation code used in the `gen_prime` function
## Program Analysis
We see that the `gen_prime` function initializes the variable `s` at a random value, and then modifies it in the creation of a number, as well as performing a bitwise `OR` operation. We treat this as effectively randomizing `s` before each attempt to create a prime. From here on, when we refer to `s`, we mean the actual value of `s` that was set at the start of the iteration of the loop when the prime number was generated that is returned.
We also observe that the form of the prime generated is as follows, where each term is calculated modulo `2^64`:
```p = s << 960 + (s * a) << 896 + (s * a^2) << 832 + ... + (s * a^14) << 64 + (s * a^15)```The first observation that we make from this, is that the value of `p` is very close to `s * 2^960`, and in fact, given a certain `p`, we can caculate `s = p // 2^960`. This also means that we are able to use `n = pq` to get an approximation for `s1 * s2`, the product of the two `s` values used for the primes, as `n // 2^1920`. In fact, looking at `s1 * s2` as a 128-bit number, the most significant 64 bits will be at most `1` away from the real value. In order to obtain `s1 * s2`, however, we still need the lower 64 bits.
The second observation we make is that when multiplying two primes of this form, the least significant 64 bits of the product will be equal to `(s1 * a^15) * (s2 * a^15)` modulo `2^64`. This allows us to calculate the least significant 64 bits of `s1 * s2` as being equal to `(n // 2^64) * inverse(a^30, 2^64)`.
## Solution
Combining the above two parts, we get 3 (Or maybe 2? I am not very good at math) possibilities for `s1 * s2`. We can factor these on a site such as [factordb](https://factordb.com) to obtain the prime factors, and simply iterate over all factor pairs for `s1` and `s2`, and check if the primes generated with these two satisfy `pq = n`. Once we found these two primes, reversing the RSA encyption is trivial with `d = inverse(e, (p-1)*(q-1))` and `pt = pow(ct, d, n)`. Our solution script is as follows:
```from Crypto.Util.number import *from z3 import *from isqrt import *import randomfrom functools import reduce
n = 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 = 0x10001ct = 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
a = 0xe64a5f84e2762be5
def get_prime(s):	p = 0	for _ in range(1024 // 64):		p = (p << 64) + s		s = a * s % 2**64	return p
s1s2l = ((n%(2**64))*inverse(a**30, 2**64))%2**64s1s2m = n//(2**(960*2))
print(((s1s2m) // 2**64 - 1) *2**64 + s1s2l)print(((s1s2m) // 2**64 + 0) *2**64 + s1s2l)print(((s1s2m) // 2**64 + 1) *2**64 + s1s2l)
# Use factordb.com to find prime factorization
def find_primes():    arr = [3, 5, 41, 43, 509, 787, 31601, 258737, 28110221, 93627982031]    #arr = [11, 61, 443, 21751, 1933727, 53523187, 340661278587863]    #arr = [79, 30577, 12153143, 7765238529536474698954633]
    for i in range(2**len(arr)):        mask = [(i>>j) & 1 for j in range(len(arr))]        s1 = reduce(lambda x,y: x*y, [arr[j] for j in range(len(arr)) if mask[j]], 1)        s2 = reduce(lambda x,y: x*y, [arr[j] for j in range(len(arr)) if not mask[j]], 1)        if (p := get_prime(s1))*(q := get_prime(s2)) == n:            return (p, q)    return (0, 0)
p, q = find_primes()
d = inverse(e, (p-1)*(q-1))pt = pow(ct, d, n)print(long_to_bytes(pt))```
And neatly outputs the flag on the first try, `CTF{__donald_knuths_lcg_would_be_better_well_i_dont_think_s0__}`