论文英语翻译

论文英语翻译

ID:38527700

大小:26.37 KB

页数:5页

时间:2019-06-14

论文英语翻译_第1页
论文英语翻译_第2页
论文英语翻译_第3页
论文英语翻译_第4页
论文英语翻译_第5页
资源描述:

《论文英语翻译》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库

1、PublicKeyCryptologyIn1978,RonaldRivest,AdiShamir,andLeonardAdelmanpublished“AMethodforObtainingDigitalSignaturesandPublicKeyCryptosystems.”Inthispaper,theauthorsdescribeamethodofsendingcodedmessagesusingapairofpubliclyavailableintegers.Thismethodiswidelyc

2、alledtheRSApublickeycryptosystem.WebeginwitharesultoncongruencesthatextendsFermat’sLittleTheorem.Theorem1Supposethatpandqaredistinctprimesandkisanyinteger.Then(a)ForanyintegerawithGCD(a,pq)=1,a^k(p-1)(q-1)≡1(modpq)(1)(b)Foranyintegera,a^k(p-1)(q-1)+1≡a(mo

3、dpq)(2)Proof(a)IfGCD(a,pq)=1,thenaisnotdivisiblebyporq;itisrelativelyprimetoboth.ThusbyFermat’sTheorem,wohavea^p-1≡1(modp),andsoa^k(p-1)(q-1)≡1^k(q-1)=1(modp).Similarly,a^k(p-1)(q-1)≡1(modq).Thusthereexistintegersrandswitha^k(p-1)(q-1)=1+rp=1+sq.Itfollows

4、thatrp=sq,andsinceqisnotdivisiblebyp,smustbe,says=pt.Thena^k(p-1)(q-1)=1+pqtanda^k(p-1)(q-1)≡1(modpq).(b)Ifaisrelativelyprimetopq,theresultfollowsfrom(1)bymultiplyingbothsidesbya.Ifnot,thenaisdivisiblebyeitherporqorboth.Ifaisdivisiblebypq,thenbothsidesof(

5、2)arecongruentto0modpqandarethereforecongruenttoeachother.Intheremainingcase,aisdivisiblebyexactlyoneoftheintegersporq,andwithoutlossofgenerality,wemaysupposethatitisp.Thena=b^s,withs≥1andbrelativelyprimetopq.Wenoteforlaterreferencethatbmustsatisfy(2).Sin

6、cepisrelativelyprimetoq,wecanshowasintheproofofpart(a)thatforsomeintegerr,p^k(p-1)(q-1)=1+rq.Multiplyingbypthenshowsthatp^k(p-1)(q-1)≡p(modpq),andtherefore(p^s)^k(p-1)(q-1)=(p^k(p-1)(q-1)+1)^s≡p^s(modpq).Weseethatbothbandp^ssatify(2),andthereforesodoesthe

7、irproducta.DecodingSincesischosentoberelativelyprimeton,s,theremainderclassofsmodn,hasamultiplicativeinversetintheringZn.Thusforsomeintegerrwehavest≡1(modn)orst=1+k(p-1)(q-1)forsomeintegerk.WecanfindtbyusingtheEuclideanalgorithm.Ifwereceivetheintegery=x^s

8、(modm),wecomputey^t(modm)andapplyTheorem1.Sincem=pq,Theorem1(a)guaranteesthaty^t=x^st=x^1+k(p-1)(q-1)≡x(modm).Sincexdoesnotexceedm,wehavey^t(modm)=x,sowehaverecoveredtheoriginalintegerx.Wedothistoallreceivedintegers

当前文档最多预览五页,下载文档查看全文

此文档下载收益归作者所有

当前文档最多预览五页,下载文档查看全文
温馨提示:
1. 部分包含数学公式或PPT动画的文件,查看预览时可能会显示错乱或异常,文件下载后无此问题,请放心下载。
2. 本文档由用户上传,版权归属用户,天天文库负责整理代发布。如果您对本文档版权有争议请及时联系客服。
3. 下载前请仔细阅读文档内容,确认文档内容符合您的需求后进行下载,若出现内容与标题不符可向本站投诉处理。
4. 下载文档时可能由于网络波动等原因无法下载或下载错误,付费完成后未能成功下载的用户请联系客服处理。