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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Higher residuosity problem ： ウィキペディア英語版 Higher residuosity problem In cryptography, most public key cryptosystems are founded on problems that are believed to be intractable. The higher residuosity problem (also called the n th-residuosity problem) is one such problem. This problem is ''easier'' to solve than integer factorization, so the assumption that this problem is hard to solve is ''stronger'' than the assumption that integer factorization is hard. ==Mathematical Background== If ''n'' is an integer, then the integers modulo ''n'' form a ring. If ''n''=''pq'' where ''p'' and ''q'' are primes, then the Chinese remainder theorem tells us that :$\backslash mathbb/n\backslash mathbb\; \backslash simeq\; \backslash mathbb/p\backslash mathbb\; \backslash times\; \backslash mathbb/q\backslash mathbb$ The group of units of any ring form a group, and the group of units in $\backslash mathbb/n\backslash mathbb$ is traditionally denoted $(\backslash mathbb/n\backslash mathbb)$ ^ *. From the isomorphism above, we have :$(\backslash mathbb/n\backslash mathbb)^$ * \simeq (\mathbb/p\mathbb)^ * \times (\mathbb/q\mathbb)^ * as an isomorphism of ''groups''. Since ''p'' and ''q'' were assumed to be prime, the groups $(\backslash mathbb/p\backslash mathbb)^$ * and $(\backslash mathbb/q\backslash mathbb)^$ * are cyclic of orders ''p''-1 and ''q''-1 respectively. If ''d'' is a divisor of ''p''-1, then the set of ''d''th powers in $(\backslash mathbb/p\backslash mathbb)^$ * form a subgroup of index ''d''. If gcd(''d'',''q''-1) = 1, then ''every'' element in $(\backslash mathbb/q\backslash mathbb)^$ * is a ''d''th power, so the set of ''d''th powers in $(\backslash mathbb/n\backslash mathbb)^$ * is also a subgroup of index ''d''. In general, if gcd(''d'',''q''-1) = ''g'', then there are (''q''-1)/(''g'') ''d''th powers in $(\backslash mathbb/q\backslash mathbb)^$ *, so the set of ''d''th powers in $(\backslash mathbb/n\backslash mathbb)^$ * has index ''dg''. This is most commonly seen when ''d''=2, and we are considering the subgroup of quadratic residues, it is well known that exactly one quarter of the elements in $(\backslash mathbb/n\backslash mathbb)^$ * are quadratic residues (when ''n'' is the product of exactly two primes, as it is here). The important point is that for any divisor ''d'' of ''p''-1 (or ''q''-1) the set of ''d''th powers forms a subgroup of $(\backslash mathbb/n\backslash mathbb)^$ *. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Higher residuosity problem」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.