An attack on RSA with exponent 3

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 [1]. (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 N_i, and each will receive a different encrypted message

c_i = m³ mod N_i.

Someone with access to c₁, c₂, and c₃ can recover the message m as follows. We can assume each modulus N_i 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 avoided 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 your 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.

Python example

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)[0]
    
    assert(cbrt(x) == m) # integer cube root

Note that crt is the Chinese Remainder Theorem. It returns a pair of numbers, the first being the solution we’re after, hence the [0] 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.

[1] 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.

An attack on RSA with exponent 3

Python example

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