fermatandeuler39stheorems-資料下載頁(yè)
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【導(dǎo)讀】Forexample:{2,3,5,7,11,13,17,…}. notveryuseful.n=a*b*c. Or,100=5*5*2*2. 143=11*13. factorsof1,3,7,21.theleastpowers.Forexample,125=53and200=23*52. GCD(125,200)=20*52=25. willbe1.GCD(10,21)=1. itselfand1.PierredeFermat(1601-1665)wasalawyerby. whenn>2.HistoryCont.OneofFermat’sbookscontaineda. moderntechniques.p,then...ap-11(modp).apa(modp).primalitytesting.ton.Asyoucanseefrom(5)and(7),(n)will. *Q)=(P-1)*(Q-1),ifPandQareprime.IfGCD(a,p)=1,anda<p,then. a(p)1(modp).Let’stestthetheorem:. Ifa=5andp=6. Then(6)=(2-1)*(3-1)=2. So,5(6)=25and25=24+1=6*4+1. =>25=1(mod6)OR25%6=1. Italsofollowsthata(p)+1a(modp)sothat. primetoa.
【正文】 lue we will call variable R. So, we select two very large prime numbers U and V and multiply them. ? = (R) = (U1)*(V1). This makes R difficult to factor, since the fewer factors a number has, the longer it takes to find them. ?So What? Cont. ? We also define the variables P and Q. P is an arbitrary number that is relatively prime to (R). Q is the calculated inverse of P in (mod (R)). ? We use P and R to create a public key, and Q and R to create a private key. ? This yields P*Q 1(mod (R) ). ? The result is that too much information is lost in the encryption due to the modulus arithmetic to decipher a privately encrypted RSA message without the use of the public key. Unless the wouldbe decipherer had enough time and processing power to attempt a bruteforce factorization. But, the larger the primes, the longer it takes to factor their product. Information in these slides piled by Christopher Simons ????
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