Cryptography

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By Jürgen Müller

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1 . 1 1 1     . . . . . 1 1 . . 1  . 1 . 1 . 1 1 1 1 . 1 1 It turns out that M is given as follows:  1 1  1 1   .   . 1   .   . 1   1 . 1 . has maximal rank rkF2 (M ) = 8, and an F2 -basis of ker(M tr ) 1 . 1 . 1 . 1 . . 1 1 . 1 . 1 1 . . 1 1 . 1 . . 1 . . . 1 1 . 1 . 1 . . 1 1 . . 1 . . . . . 1 . . . . . 1 . . 1 . . . 1 . 1 . . . . . . . . 1 . . 1 . . 1             Each of the 3-subsets {1, 4, 7}, {2, 5, 8}, {3, 6, 9} and {10, 11, 12} of columns is F2 -linearly independent, thus for all ciphertexts y ∈ F32 the matrix of the resulting system of linear equations for the associated plaintext x ∈ F32 has rank 3, and since there is a solution, the affine solution space for x is 0-dimensional, hence x is uniquely determined.

K} and k = max{l, m}. l If l = m, then we may assume l < m, and thus we have (xc )2 = 1 ∈ (Z/pZ)∗ l l and (xc )2 = 1 ∈ (Z/qZ)∗ , implying gcd((xc )2 − 1, n) = p. To compute a prime divisor of n we proceed as follows: choose x ∈ Zn randomly if x = 0 then return fail g := gcd(x, n) ∈ Zn if 1 < g < n then return g for i ∈ [0, . . , s] do i g := gcd((xc )2 − 1, n) ∈ Zn if 1 < g < n then return g return fail An element x ∈ Z/nZ providing a prime divisor of n is called a factorisation witness. We have 2s ≤ ed ≤ n2 , thus s ≤ 2 log2 (n), hence the algorithm runs in polynomial time.

Then we have (Z/pa Z)∗ ∼ = Cpa−1 (p−1) . Proof. Since (Z/pZ)∗ ∼ = Cp−1 there is x ∈ (Z/pa Z)∗ such that x ∈ (Z/pZ)∗ has order p − 1, hence x has order divisible by p − 1, and thus we may assume that x has order p − 1. Assume there is y ∈ (Z/pa Z)∗ having order pa−1 , then xy ∈ (Z/pa Z)∗ has order pa−1 (p − 1).

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