天才一秒记住【狂风中文网】地址：https://www.kfzw.net
AllthisservestogiveanindithatthebyBobtoAlice，memodulon，deedbecalculatedwithouttoomucheffortonbehalfofBob'sputer.Allthesame，thenumbersinvolvedareiybig，somoreexplanatiooshowthattheydled.Thelargepowersinvolvedinebedealtwithinstagesbyaproasfastexpoion.Withoutgoihemethodinvolvessuccessivesquaringandmultiplyioarriveatmemodulonwiththebinaryfuidihmthroughtoquidtherequiredremaiivelyfewsteps.
EuclidshowsAlicehowtofindherdegnumber
Aliputerddusioolthatisover2，300yearsold，theEuAlgorithm，whichwillbeexplainedi.Eve'sputercouldofethingifitjustknewwhichequationtosolve.However，sindqareprivatetoAlice，sois（p-1）（q-1）akobegin.
&otheEuAlgorithm，thisbeginsfromtheobservationthatitispossibletofionfabersa〉bbysuccessivesubtra.（Thehcfisalsoknownasthegcd–greatestondivisor.）Wejustr=a-bhasthepropertythatanyonfaytwoofthethreenumbersa，b，andrwillalsobeafactorofthethird.Forexample，ifonfactorofaandb，sothata=db=cb1say，weseethatr=a-b=ca1-cb1=c（a1-b1），givingusafactorizationofrinvolvingthedivisorparticular，thehdbisthesameasthehdr.Sihesenumbersarelessthana，wehesameproblembutappliedtoasmallernumberpair.Repetitionofthisideatheuallyleadtoapairwherethehcfisobvious.（Iwonumbersinhauallybethesame，forifnotwecouldproorestep;theirohenthenumberweseek.）
&ofindthehcfofa=558andb=396，thefirstsubtrawouldgiveusr=558-396=162，soournewpairwouldbe396and162.Since396-162=234，ourthirdpairbees234and162，aihefulllistofnumberpairsis:
（558，396）→（396，162）→（234，162）→（162，72）→（90，72）→→（72，18）→（54，18）→（36，18）→（18，18）
andsothehd396is18.
Itispossibletowritedownthehberpairfromtheprimefactorizationsofthenumbersiion.Inthisexample，558=2×32×31，while396=22×32×11;takingtheonporimeeothefactorizatiohehcfas2×32=18.heless，ferakesmuchlessworktouseEuclid’sAlgorithmasitisgeoperformsubtrasthantofiorizations.
AnotherbonusoftheEuAlgorithmisthatitisalossibletoworkitbadinsodoihehtermsinaltwooseethisinathepreviousexample，itisbesttopressthecalwhenthesamenumberappearsseveraltimesoverinthecourseofthesubtras，representingthisasasiionasfollows:
558=396+162
396=2×162+72
162=2×72+18
72=4×18.
Beginningwiththesedtolastline，weleequatioetheieremaiime.Inthisexample，byusingfirstthepeioheo>
18=162-2×72=162-2×（396-2×162）=5×162-2×396
andfinallyusiequatioethefirstieremainderof162:
=5×（558-396）--2×396=5×558-7×396=18.
Thaterformthisreverseprocedureisimportantforbothpradtheoretis.Inparticular，tofindAlice’sdegnumberd，wewantdtosatisfythethatdeleavesaremainderof1whendividedbyφ（n）.（Forbrevity，weshalldehesinglesymbolk.）Weowseethereasooneandkbeingaepair，asiftheirhighestonfactoris1，wheheEuAlgorithmonthepaireandk，thefihatappearsis，ofcourse，1.Byreversihm，wewilleventuallyexpress1asabinationofeandk;inparticular，wewillfondintegersddsuchthatck+de=1，orinotherwordsde=1-ck，sothatdewillleavearemainderof1whendividedbyk.
ThisrelativelysimpleprocesswillyieldAlice’sdegheinitialvalueofdobtaiheequationmayheraokbutifnot，byaddingasuitablemultipleofk，wewilleventuallyfindtheuniquehatrahasthemagicpropertythatdeleavesaremainderof1wheheuniquenessofdiseasilyproved，butwewoofurtherexplaishowthedegnumberdiscalculatedasweshtotheexamplegivenearlierwherep=5，q=13，sothatn=pq=5×13=65.-1）（q-1）=4×12=48.Alicesetse=11，andsind48aree，thisiswithihegame.TheEuAlgorithmappliedtoφ（n）=k=48ahengives:
48=4×11+4
11=2×4+3
4=1×3+1
ingthatthehdeisindeed1.Reversihm>
1=4-3=4-（11-2×4）=3×4-11=3（48-4×11）-11
=3×48-13×11.
Thisgivesaninitialvalueofd=-13asthesolutiontotherequirementthat11dleavesremainder1upondivisionby48，setapositivevalueofdintherequiredraothisd=48-13=35.
ThereasonwhydworksforAliceisalldowntomodulararithmetidthefactthatdeleavesaremainderof1whendividedbyk=φ（es（me）d=mdemoduloheform1+krforsomeintegerr.Asexplainedbefore，mkleavesaremainderof1whehisisoftenkheorem）andsothesameistrueof（mk）r=mkr.Hencem1+kr=m×mkrleavestheremaindermwhendividedbyailedverifiofthisrequiresalittlealgebra，butthatisens.）Inthisway，AliceretrievesBob’smessage，
AndinpassingitiswelltopointoutthattheEuAlgorithmprovidesthemissinglinkinourproofoftheuniquenessofprimefactorizationasitallowsustoverifytheeupropertythatifaprimepisafactoroftheproductab，sothatab=pcsay，thenpisafactorofatleastoneofaandb.Thereasonforthisisthatifpisnotafactorofathen，sincepisprime，thehdpis1.ByreversingtheEuAlgorithmliedtothepairaahenfindintegersrandssaysuchthatra+sp=1.Thisisenoughtoshoisthenafactorofbfor，sinceab=pc，>
b=b×1=b（ra+sp）=r（ab）+psb=r（pc）+psb=p（rc+sb）.
Thisistherequiredfactorizationofbthatfeaturestheprimepasafa，theheRSAengmakesthesystemsound，althoughvariousprotocolsthathavenotbeenexplaiberespeuardtheiyofthesysteThereareissuesofauthentifi（whatifEvetactsAlidingtobeBob?），ion（whatifBobpretendsthatitwasEvewhoseoAlididentityfraud（whatifAliceabusestialidentifittoherbyBobaoimpersonatehimonliherweakhesystembeexposedwheableorrepeatedmessagesproliferate.However，thesedifficultiesmaypotentiallyariseinanypublickey.Theyeandinthemaiotheunderlyieiquesthatensurehighqualityandrobusten.
ThischapterhasdemonstratedamajorappliehetheoryofdivisibilityandremaimathematicsofEudthe18th-turytributionofEulerallowsthistobeexplaionlyinbroadpriiail.
&partofourbookclosester5whiespecialclassesofintegersassociatedwiththeeiourallyroupings.
本章未完，请点击下一章继续阅读！若浏览器显示没有新章节了，请尝试点击右上角↗️或右下角↘️的菜单，退出阅读模式即可，谢谢！