Mind your Ps and Qs
In RSA, a small e value can be problematic, but what about N? Can you decrypt this?
Download the file.
Use https://www.dcode.fr/rsa-cipher to obtain the values of p and q.
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def long_to_bytes(n):
return n.to_bytes((n.bit_length() + 7) // 8, 'big')
n = 948406957756830799684818171639547165784816468744946013083947881743680617123566349
e = 65537
c = 15341890103764929939105506004034128738090325640037083301857608662849501626260517
p = 1891771437429478964908181306574287207137
q = 501332739776173570344039681219489434626477
phi = (p - 1) * (q - 1)
# modular inverse
d = pow(e, -1, phi)
m = pow(c, d, n)
print(m)
print(long_to_bytes(m)[::-1])
Use this code to decrypt the ciphertext.
Factoring \(n\) into \(p\) and \(q\) lets you calculate \(\varphi(n) = (p-1)(q-1)\) and the private exponent \(d = e^{-1} \bmod \varphi(n)\). The script then recovers the message integer with pow(c, d, n) and converts it to bytes. Its final [::-1] reverses the byte order. See the RSA decryption primitive for the modular decryption operation.
This reveals the flag!
This post is licensed under CC BY 4.0 by the author.