Paper 2026/1286

Multiple-of Property for Related-Differential Distinguishers on 5-Round AES

Abstract

A key-independent secret-key distinguisher identifies inherent structural deviations of a block cipher from an ideal random permutation without recovering any information about the secret key. For the Advanced Encryption Standard (AES), various key-independent secret-key distinguishers have been proposed on reduced-round versions. In this paper, we study related-differential distinguishers for 5-round AES that combine a 1-round related differential trail with the 4-round generalized zero-difference property. We prove that the number of valid quartets satisfying the underlying property takes the multiple-of form, N_q = 2^{2-n_z}(2^w)^{n_z} A + 8B, where A, B are non-negative integers, w is the cell size, and n_z is the number of inactive bytes of Delta X_2 (the difference at the second-round input) under the chosen pairing of the 4-plaintext quartet into two pairs. The pairing fixes the bundle size 2^{2-n_z}(2^w)^{n_z}, which sets the variance of the valid-quartet count. We compare the three pairings n_z in {0, 1, 2}. In the chosen-plaintext setting, where a single structure is examined as a whole, a smaller bundle keeps the count closer to its mean, so n_z=0 is the most reliable single-structure distinguisher, n_z=1 is close behind, and n_z=2 almost fails. The pairing n_z=1 is that of Yan et al. Its 2*2^w bundle raises the variance, so it reaches 63% only at 2^{27.2}, rather than at the 2^{27} where one valid quartet is expected. At the same 2^{27.2}, the smaller bundle of n_z=0 reaches 65%. In the adaptively chosen-plaintext setting the bundle does not form across the separate base collisions, so n_z=1, which produces more valid quartets, is the better pairing. All claims are verified experimentally on both small-AES (w=4) and the standard AES (w=8).