Nonce reuse. Due to rising entropy prices, Roseacrucis has started to reuse AES-GCM nonces. You must perform the Forbidden Attack in order to recover the authentication key and forge arbitrary ciphertext.
Recall that the AES-GCM ciphertext is computed as the XOR of the keystream and the message. One can modify the bits of the ciphertext arbitrarily to effect the same change in the decrypted plaintext.
Where certain bits of the plaintext are already known, the attacker can fully determine the same bits of the forged plaintext. If nonces are reused, the keystream will be identical, allowing us to recover plaintext via crib dragging, which makes this attack particularly effective: \[ c' = c \oplus m \oplus m'. \]
However, we still need to compute a new MAC over the forged ciphertext. Simplifying for a ciphertext \(c\) of two blocks and no additional authenticated data, the GMAC MAC is computed as \[ mac = s + \vert c\vert h + c_1h^2 + c_0h^3, \] where \(s\) is a constant depending on the AES-GCM key and the nonce, and \(h\) is the authentication key depending only on the AES-GCM key.
If we intercept a second ciphertext \(c'\) encrypted under the same key and nonce, we can compute \[ mac + mac' = (s + s') + (len + len')h + (c_1 + c'_1)h^2 + (c_0+c'_0)h^3, \] Since \(s = s'\) and \(x+x=0\) in \(\mathbb{F}_{2^{128}}\), we are left with the polynomial equation \[ 0 = (mac + mac') + (len + len')h + (c_1 + c'_1)h^2 + (c_0+c'_0)h^3 \] where all variables are known other than \(h\). Thus, recovering \(h\) is a matter of finding the roots by factoring the polynomial.
We plug \(h\) back into the first equation to recover \(s\), and we can forge the MAC for arbitary ciphertext under the same nonce. Note that there may be multiple possible roots; in this case, one can check each possibility against the enemy, or perform the attack twice on two pairs of intercepted messages.
One can use SageMath to compute factors of a polynomial:
K = GF(2**128, name='x', modulus=x^128+x^7+x^2+x+1) x = K.gen() S = PolynomialRing(K, 'y') y = S.gen() p = (1)*y^4 + (x^7)*y^3 + (x^9 + x^4 + 1)*y^2 + (x^12 + x^2)*y + (x^10 + x^5) for factor, _ in p.factor(): if factor.degree() == 1: print('Authentication key:', factor - y)
However, the library powering this demonstration implements polynomial factoring over finite fields from scratch, which is an edifying exercise.
We present advice for those who wish to implement polynomial factorization as well:
Readers who wish to implement this attack themselves can try Cryptopals; specifically Set 8 Problem 62.
from aesgcmanalysis import xor, gmac, gcm_encrypt, gcm_decrypt, nonce_reuse_recover_secrets k = b"tlonorbistertius" nonce = b"jorgelborges" m1, aad1 = b"The universe (which others call the Library)", b"" m2, aad2 = b"From any of the hexagons one can see, interminably", b"" c1, mac1 = gcm_encrypt(k, nonce, aad1, m1) c2, mac2 = gcm_encrypt(k, nonce, aad2, m2) # Recover the authentication key and blind from public information possible_secrets = nonce_reuse_recover_secrets(nonce, aad1, aad2, c1, c2, mac1, mac2) # Forge the ciphertext m_forged = b"As was natural, this inordinate hope" c_forged, aad_forged = xor(c1, xor(m1, m_forged)), b"" for h, s in possible_secrets: print("MAC candidate": gmac(h, s, aad_forged, c_forged))