Description
In RSA, there is a small danger of using small primes. Can you decrypt this? Given: c, n, e.
Setup
Download the values file from the challenge page.
Solution
- Step 1Factor nBecause n is small (only a few hundred bits), it can be factored instantly using factordb.com. Paste the value of n into the search box to retrieve the two prime factors p and q.
Learn more
RSA's security relies entirely on the integer factorization problem: given a large number
n = p * q, it should be computationally infeasible to findpandq. For modern RSA,nis at least 2048 bits (~617 decimal digits), which makes factoring effectively impossible with current technology.However, when
nis small -- say, a few hundred bits or fewer -- factoring is trivial. Tools like factordb.com maintain a database of pre-factored numbers and can factor small integers instantly. For larger numbers they use algorithms like the Quadratic Sieve or General Number Field Sieve (GNFS). The title "Mind Your Ps and Qs" refers to the prime factorspandq-- small primes undermine the entire system.Historical context: In 2012, researchers found that millions of real-world RSA public keys shared prime factors because of weak random number generators used during key generation. By computing GCDs between pairs of public keys, they factored a significant percentage of real keys -- a real-world demonstration of exactly this vulnerability.
- Step 2Compute the private key and decryptWith p and q known, compute phi = (p-1)*(q-1), then the private exponent d = modular_inverse(e, phi). Raise c to the power d mod n to recover the plaintext integer m, then convert it to bytes.python3 << 'EOF' from Crypto.Util.number import inverse, long_to_bytes # Fill in values from the challenge file and factordb c = <c> n = <n> e = <e> p = <p> # from factordb.com q = <q> # from factordb.com phi = (p - 1) * (q - 1) d = inverse(e, phi) m = pow(c, d, n) print(long_to_bytes(m).decode()) EOF
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The RSA decryption process follows directly from the mathematics. Euler's totient function
φ(n) = (p-1)*(q-1)counts how many integers less thannare coprime with it. The private exponentdis defined as the modular inverse ofemoduloφ(n), meaninge*d ≡ 1 (mod φ(n)). This relationship is what makes decryption the inverse of encryption.Modular exponentiation (
pow(c, d, n)) uses Python's built-in three-argument form, which is highly efficient -- it applies the square-and-multiply algorithm internally. Even with a 2048-bit exponent, this completes in milliseconds. Without the three-argument form (computingc**dand then taking mod), the intermediate number would have millions of digits and be completely impractical.The
long_to_bytes()function frompycryptodomeconverts the recovered plaintext integer back to a byte string. RSA operates on integers, so text messages are first converted to integers (using big-endian byte encoding) before encryption.long_to_bytesreverses this step to give the human-readable flag.
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picoCTF{...}
When n is small enough to factor, RSA provides zero security -- always use at least 2048-bit keys.