As I noted in this post, RSA encryption is often carried out reusing exponents. Sometimes the exponent is exponent 3, which is subject to an attack we’ll describe below . (The most common exponent is 65537.)
Suppose the same message m is sent to three recipients and all three use exponent e = 3. Each recipient has a different modulus Ni, and each will receive a different encrypted message
ci = m³ mod Ni.
Someone with access to c1, c2, and c3 can recover the message m as follows. We can assume each modulus Ni is relatively prime to the others, otherwise we can recover the private keys using the method described here. Since the moduli are relatively prime, we can solve the three equations for m³ using the Chinese Remainder Theorem. There is a unique x < N1 N2 N3 such that
x = c1 mod N1
x = c2 mod N2
x = c3 mod N3
and m is simply the cube root of x. What makes this possible is knowing m is a positive integer less than each of the Ns, and that x < N1 N2 N3. It follows that we can simply take the cube root in the integers and not the cube root in modular arithmetic.
This is an attack on “textbook” RSA because the weakness in this post could be avoiding by real-world precautions such as adding random padding to each message so that no two recipients are sent the exact same message.
By the way, a similar trick works even if you only have access to one encrypted message. Suppose you’re using a 2048-bit modulus N and exchanging a 256-bit key. If you message m is simply the key without padding, then m³ is less than N, and so you can simply take the cube root of the encrypted message in the integers.
Here we’ll work out a specific example using realistic RSA moduli.
from secrets import randbits, randbelow from sympy import nextprime from sympy.ntheory.modular import crt def modulus(): p = nextprime(randbits(2048)) q = nextprime(randbits(2048)) return p*q N = [modulus() for _ in range(3)] m = randbelow(min(N)) c = [pow(m, 3, N[i]) for i in range(3)] x = crt(N, c) assert(cbrt(x) == m) # integer cube root
crt is the Chinese Remainder Theorem. It returns a pair of numbers, the first being the solution we’re after, hence the
 after the call.
The script takes a few seconds to run. Nearly all the time goes to finding the 2048-bit (617-digit) primes that go into the moduli. Encrypting and decrypting m takes less than a second.
 I don’t know who first discovered this line of attack, but you can find it written up here. At least in the first edition; the link is to the 2nd edition which I don’t have.
2 thoughts on “An attack on RSA with exponent 3”
Is it an attack on e=65537 RSA if you recover the identical message encrypted to 65,537 unique moduli? It seems like it.
+ from sympy import cbrt