HTB CyberApocalypse Primary Knowledge Writeup

Given:

import math
from Crypto.Util.number import getPrime, bytes_to_long
from secret import FLAG

m = bytes_to_long(FLAG)

n = math.prod([getPrime(1024) for _ in range(2**0)])
e = 0x10001
c = pow(m, e, n)

with open('output.txt', 'w') as f:
    f.write(f'{n = }\n')
    f.write(f'{e = }\n')
    f.write(f'{c = }\n')

n = math.prod([getPrime(1024) for _ in range(2**0)])

We focus particularly on the 2**0 which is equal to 1 so n is not a product of 2 primes but just one prime

and we know phi (euler totient) for a prime number is φ(n) = (n − 1)

solve script:

from Crypto.Util.number import long_to_bytes, inverse

n = 144595784022187052238125262458232959109987136704231245881870735843030914418780422519197073054193003090872912033596512666042758783502695953159051463566278382720140120749528617388336646147072604310690631290350467553484062369903150007357049541933018919332888376075574412714397536728967816658337874664379646535347
e = 65537
c = 15114190905253542247495696649766224943647565245575793033722173362381895081574269185793855569028304967185492350704248662115269163914175084627211079781200695659317523835901228170250632843476020488370822347715086086989906717932813405479321939826364601353394090531331666739056025477042690259429336665430591623215

phi = n - 1

d = inverse(e, phi)
m = pow(c, d, n)
print(long_to_bytes(m))

flag:

HTB{0h_d4mn_4ny7h1ng_r41s3d_t0_0_1s_1!!!}