New PDF release: A Course in Cryptography

By Raphael Pass, Abhi Shelat

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Since inputs are given in binary notation, this requires our procedure to run in time poly(log( a), log( x ), log( N )). 1. 5 computes a x mod N in time O(log( x ) log2 ( N )). i Proof. Rewrite a x mod N as ∏i xi a2 mod N. Since multiplying and squaring modulo N take time log2 ( N ), each iteration of the loop requires O(log2 ( N )) time. Because each iteration divides x by two, the loop runs at most log x times which establishes a running time of O(log( x ) log2 ( N )). 46 chapter 2. computational hardness Later, after we have introduced Euler’s theorem, we present a similar algorithm for modular exponentiation which removes the restriction that x < N.

Consider the set A = { ax | x ∈ Z∗N }. Since Z∗N is a group, every element ax of A must also be in Z∗N and so it follows that A ⊆ Z∗N . Now suppose that | A| < |Z∗N |. By the pidgeonhole principle, this implies that there exist two group element i, j ∈ Z∗N such that i = j but ai = aj. Since a ∈ Z∗N , there exists an inverse a−1 such that aa−1 = 1. Multiplying on both 48 chapter 2. computational hardness sides we have a−1 ai = a−1 aj =⇒ i = j which is a contradiction. Thus, | A| = |Z∗N | which implies that A = Z∗N .

6. Basic Computational Number Theory 47 Closure Suppose, for the sake of reaching a contradiction, that there exist two elements a, b ∈ Z∗N such that ab ∈ Z∗N . This implies that gcd( a, N ) = 1, gcd(b, N ) = 1, but that gcd( ab, N ) = d > 1. The latter condition implies that d has a non-trivial prime factor that divides both ab and N. Thus, the prime factor must also divide either a or b (verify as an exercise), which contradicts the assumption that gcd( a, N ) = 1 and gcd(b, N ) = 1. Inverse Consider an element a ∈ Z∗N .

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A Course in Cryptography by Raphael Pass, Abhi Shelat

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