HackTheBox TurboCipher Challenge
https://app.hackthebox.com/challenges/401
Description
In a parallel universe, all recursive algorithms are strictly forbidden by the "coding laws". For this reason, students must solve a different recursion-based problem each year at the beginning of their CS studies. This year, it is a difficult encryption method that requires a high level of math skills. Can you solve it?
Exploitation
#!/usr/bin/env python3
from pwn import *
import sys
def legendre(x: int, p: int) -> int:
return pow(x, (p - 1) // 2, p)
def tonelli(n: int, p: int) -> int:
assert legendre(n, p) == 1, 'not a square (mod p)'
q = p - 1
s = 0
while q % 2 == 0:
q //= 2
s += 1
if s == 1:
return pow(n, (p + 1) // 4, p)
for z in range(2, p):
if p - 1 == legendre(z, p):
break
c = pow(z, q, p)
r = pow(n, (q + 1) // 2, p)
t = pow(n, q, p)
m = s
t2 = 0
while (t - 1) % p != 0:
t2 = (t * t) % p
for i in range(1, m):
if (t2 - 1) % p == 0:
break
t2 = (t2 * t2) % p
b = pow(c, 1 << (m - i - 1), p)
r = (r * b) % p
c = (b * b) % p
t = (t * c) % p
m = i
return r
def fast_turbonacci(n: int, p: int, b: int, c: int) -> int:
sqrt_delta = tonelli((pow(b, 2, p) + 4 * c), p)
r1 = (b + sqrt_delta) * pow(2, -1, p) % p
r2 = (b - sqrt_delta) * pow(2, -1, p) % p
return (pow(r1, n, p) - pow(r2, n, p)) * pow(r1 - r2, -1, p) % p
def main():
if len(sys.argv) != 2:
print(f"Usage: {sys.argv[0]} <ip:port>")
sys.exit(1)
host, port = sys.argv[1].split(':')
port = int(port)
try:
io = remote(host, port)
log.info("Connected to server. Reading parameters...")
io.recvuntil(b'p = ')
p = int(io.recvline().decode().strip())
log.info(f"p = {p}")
io.recvuntil(b'b = ')
b = int(io.recvline().decode().strip())
log.info(f"b = {b}")
io.recvuntil(b'c = ')
c = int(io.recvline().decode().strip())
log.info(f"c = {c}")
io.recvuntil(b'nonce = ')
nonce = int(io.recvuntil(b' ').decode().strip())
log.info(f"nonce = {nonce}")
log.info("Calculating OTP...")
otp = fast_turbonacci(nonce, p, b, c)
log.info(f"OTP = {otp}")
io.sendlineafter(b'OTP: ', str(otp).encode())
io.sendlineafter(b'> ', b'1')
io.recvuntil(b'ct = ')
flag_enc = int(io.recvline().strip())
log.info(f"Encrypted flag = {flag_enc}")
pt1 = b'a'
io.sendlineafter(b'> ', b'2')
io.sendlineafter(b'pt = ', pt1)
io.recvuntil(b'ct = ')
ct1 = int(io.recvline().strip())
log.info(f"First ciphertext = {ct1}")
pt2 = b'b'
io.sendlineafter(b'> ', b'2')
io.sendlineafter(b'pt = ', pt2)
io.recvuntil(b'ct = ')
ct2 = int(io.recvline().strip())
log.info(f"Second ciphertext = {ct2}")
io.sendlineafter(b'> ', b'3')
io.close()
log.info("Calculating flag...")
m = ct2 - ct1
k = (ord(pt1) - ct1 * pow(m, -1, p)) % p
flag = (k + flag_enc * pow(m, -1, p)) % p
log.success(f"Flag: {bytes.fromhex(hex(flag)[2:]).decode()}")
except Exception as e:
log.error(f"Error: {str(e)}")
if 'io' in locals():
io.close()
if __name__ == '__main__':
main()
Summary
TurboCipher: model the crypto leak, recover the missing secret, and decrypt the flag.