千米小说网

Chapter 4 Cryptography the secret life of primes（第2页）

Andthisworksquitegenerally。Iofindtheremainderwhensomepowerabisdividedbynsay，weaketheremainderrwhenaisdividedbyraaindersaswetakesuccessivepowersofr。iththeremainderr，whichwillbeaheraoissaythatwearewmodulaiplesofnthatmayarise，astheyleavearemainderof0whendividedbyn，andsotributetothevalueofthefinalremainderr。

AllthisservestogiveanindithatthebyBobtoAlice，memodulon，deedbecalculatedwithouttoomucheffortonbehalfofBob’sputer。Allthesame，thenumbersinvolvedareiybig，somoreexplanatiooshowthattheydled。Thelargepowersinvolvedinebedealtwithinstagesbyaproasfastexpoion。Withoutgoihemethodinvolvessuccessivesquaringandmultiplyioarriveatmemodulonwiththebinaryfuidihmthroughtoquidtherequiredremaiivelyfewsteps。

EuclidshowsAlicehowtofindherdegnumber

&erddusioolthatisover2，300yearsold，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:

andsothehcfof558and396is18。

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。

48=4×11+4

11=2×4+3

4=1×3+1ingthatthehdeisindeed1。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。

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）。

Thisistherequiredfactorizationofbthatfeaturestheprimepasafactor。

In，theheRSAengmakesthesystemsound，althoughvariousprotocolsthathavenotbeenexplaiberespeuardtheiyofthesystem。Thereareissuesofauthentifi（whatifEvetactsAlidingtobeBob?），ion（whatifBobpretendsthatitwasEvewhoseoAlididentityfraud（whatifAliceabusestialidentifittoherbyBobaoimpersonatehimonliherweakhesystembeexposedwheableorrepeatedmessagesproliferate。However，thesedifficultiesmaypotentiallyariseinanypublickey。Theyeandinthemaiotheunderlyieiquesthatensurehighqualityandrobusten。

ThischapterhasdemonstratedamajorappliehetheoryofdivisibilityandremaimathematicsofEudthe18th-turytributionofEulerallowsthistobeexplaionlyinbroadpriiail。

&partofourbookclosester5whiespecialclassesofintegersassociatedwiththeeiourallyroupings。