The Obligatory Rsa
This challenge is similar to Really Secure Algorithm, except in this challenge given n are c are big, e is 65537. Still needed to factorize n. Thrown n into factordb.com, it gave the same q and p, which means n=pow(p,2). As p and q are the same, phi=(p-1)*(q-1)=pow(p-1,2)
from Cryptodome.Util import number
n = <very big>
c = <very big>
e = 65537
p = <result from factordb.com>
phi = pow(p-1,2)
d = number.inverse(e, phi)
ans = pow(c, d, p)
print(number.long_to_bytes(ans))Decrypted message is the flag.